{{task}} '''[[wp:Hamming numbers|Hamming numbers]]''' are numbers of the form H = 2ⁱ × 3^j × 5^k where i, j, k ≥ 0

''Hamming numbers'' are also known as ''ugly numbers'' and also ''5-smooth numbers'' (numbers whose prime divisors are less or equal to 5).

;Task: Generate the sequence of Hamming numbers, ''in increasing order''. In particular:

Show the first twenty Hamming numbers.

Show the 1691^st Hamming number (the last one below 2³¹).

Show the one million^th Hamming number (if the language – or a convenient library – supports arbitrary-precision integers).

;Related tasks:

• [https://rosettacode.org/wiki/Humble_numbers humble numbers]

;References:

• Wikipedia entry: [[wp:Hamming numbers|Hamming numbers]] (this link is re-directed to '''Regular number''').
• Wikipedia entry: [[wp:Smooth number|Smooth number]]
• OEIS entry: [[oeis:A051037|A051037 5-smooth or Hamming numbers]]
• [http://dobbscodetalk.com/index.php?option=com_content&task=view&id=913&Itemid=85 Hamming problem] from Dr. Dobb's CodeTalk (dead link as of Sep 2011; parts of the thread [http://drdobbs.com/blogs/architecture-and-design/228700538 here] and [http://www.jsoftware.com/jwiki/Essays/Hamming%20Number here]).

360 Assembly

*        Hamming numbers           12/03/2017
HAM      CSECT
         USING  HAM,R13            base register
         B      72(R15)            skip savearea
         DC     17F'0'             savearea
         STM    R14,R12,12(R13)    save previous context
         ST     R13,4(R15)         link backward
         ST     R15,8(R13)         link forward
         LR     R13,R15            set addressability
         LA     R6,1               ii=1
       DO WHILE=(C,R6,LE,=F'20')   do ii=1 to 20
         BAL    R14,PRTHAM           call prtham
         LA     R6,1(R6)             ii++
       ENDDO    ,                  enddo ii
         LA     R6,1691            ii=1691
         BAL    R14,PRTHAM         call prtham
         L      R13,4(0,R13)       restore previous savearea pointer
         LM     R14,R12,12(R13)    restore previous context
         XR     R15,R15            rc=0
         BR     R14                exit
PRTHAM   EQU    *             ---- prtham
         ST     R14,R14PRT         save return addr
         LR     R1,R6              ii
         XDECO  R1,XDEC            edit
         MVC    PG+2(4),XDEC+8     output ii
         LR     R1,R6              ii
         BAL    R14,HAMMING        call hamming(ii)
         XDECO  R0,XDEC            edit
         MVC    PG+8(10),XDEC+2    output hamming(ii)
         XPRNT  PG,L'PG            print buffer
         L      R14,R14PRT         restore return addr
         BR     R14           ---- return
HAMMING  EQU    *             ---- hamming(ll)
         ST     R14,R14HAM         save return addr
         ST     R1,LL              ll
         MVC    HH,=F'1'           h(1)=1
         SR     R0,R0              0
         ST     R0,I               i=0
         ST     R0,J               j=0
         ST     R0,K               k=0
         MVC    X2,=F'2'           x2=2
         MVC    X3,=F'3'           x3=3
         MVC    X5,=F'5'           x5=5
         LA     R7,1               n=1
         L      R2,LL              ll
         BCTR   R2,0               -1
         ST     R2,LLM1            ll-1
       DO WHILE=(C,R7,LE,LLM1)     do n=1 to ll-1
         L      R4,X2                m=x2
       IF C,R4,GT,X3 THEN            if m>x3 then
         L      R4,X3                  m=x3
       ENDIF    ,                    endif
       IF C,R4,GT,X5 THEN            if m>x5 then
         L      R4,X5                  m=x5
       ENDIF    ,                    endif
         LR     R1,R7                n
         SLA    R1,2                 *4
         ST     R4,HH(R1)            h(n+1)=m
       IF C,R4,EQ,X2 THEN            if m=x2 then
         L      R1,I                   i
         LA     R1,1(R1)               i+1
         ST     R1,I                   i=i+1
         SLA    R1,2                   *4
         L      R2,HH(R1)              h(i+1)
         MH     R2,=H'2'               *2
         ST     R2,X2                  x2=2*h(i+1)
       ENDIF    ,                    endif
       IF C,R4,EQ,X3 THEN            if m=x3 then
         L      R1,J                   j
         LA     R1,1(R1)               j+1
         ST     R1,J                   j=j+1
         SLA    R1,2                   *4
         L      R2,HH(R1)              h(j+1)
         MH     R2,=H'3'               *3
         ST     R2,X3                  x3=3*h(j+1)
       ENDIF    ,                    endif
       IF C,R4,EQ,X5 THEN            if m=x5 then
         L      R1,K                   k
         LA     R1,1(R1)               k+1
         ST     R1,K                   k=k+1
         SLA    R1,2                   *4
         L      R2,HH(R1)              h(k+1)
         MH     R2,=H'5'               *5
         ST     R2,X5                  x5=5*h(k+1)
       ENDIF    ,                    endif
         LA     R7,1(R7)             n++
       ENDDO    ,                  enddo n
         L      R1,LL              ll
         SLA    R1,2               *4
         L      R0,HH-4(R1)        return h(ll)
         L      R14,R14HAM         restore return addr
         BR     R14           ---- return
R14HAM   DS     A                  return addr of hamming
R14PRT   DS     A                  return addr of print
LL       DS     F                  ll
LLM1     DS     F                  ll-1
I        DS     F                  i
J        DS     F                  j
K        DS     F                  k
X2       DS     F                  x2
X3       DS     F                  x3
X5       DS     F                  x5
PG       DC     CL80'H(xxxx)=xxxxxxxxxx'
XDEC     DS     CL12               temp
         LTORG                     positioning literal pool
HH       DS     1691F              array h(1691)
         YREGS
         END    HAM

{{out}}

H(   1)=         1
H(   2)=         2
H(   3)=         3
H(   4)=         4
H(   5)=         5
H(   6)=         6
H(   7)=         8
H(   8)=         9
H(   9)=        10
H(  10)=        12
H(  11)=        15
H(  12)=        16
H(  13)=        18
H(  14)=        20
H(  15)=        24
H(  16)=        25
H(  17)=        27
H(  18)=        30
H(  19)=        32
H(  20)=        36
H(1691)=2125764000

Ada

{{works with|GNAT}} GNAT provides the datatypes Integer, Long_Integer and Long_Long_Integer, which are not large enough to store hamming numbers. In this program, we represent them as the factors for each of the prime numbers 2, 3 and 5, and only convert them to a base-10 numbers for display. We use the gmp library binding part of GNATCOLL, though a simple 'pragma import' would be enough.

This version is very fast (20ms for the million-th hamming number), thanks to a good algorithm. We also do not manipulate large numbers directly (gmp lib), but only the factors of the prime.

It will fail to compute the billion'th number because we use an array of the stack to store all numbers. It is possible to get rid of this array though it will make the code slightly less readable.

with Ada.Numerics.Generic_Elementary_Functions;
with Ada.Text_IO; use Ada.Text_IO;
with GNATCOLL.GMP.Integers;
with GNATCOLL.GMP.Lib;

procedure Hamming is

   type Log_Type is new Long_Long_Float;
   package Funcs is new Ada.Numerics.Generic_Elementary_Functions (Log_Type);

   type Factors_Array is array (Positive range <>) of Positive;

   generic
      Factors : Factors_Array := (2, 3, 5);
      --  The factors for smooth numbers. Hamming numbers are 5-smooth.
   package Smooth_Numbers is
      type Number is private;
      function Compute (Nth : Positive) return Number;
      function Image (N : Number) return String;

   private
      type Exponent_Type is new Natural;
      type Exponents_Array is array (Factors'Range) of Exponent_Type;
      --  Numbers are stored as the exponents of the prime factors.

      type Number is record
         Exponents : Exponents_Array;
         Log       : Log_Type;
         --  The log of the value, used to ease sorting.
      end record;

      function "=" (N1, N2 : Number) return Boolean
        is (for all F in Factors'Range => N1.Exponents (F) = N2.Exponents (F));
   end Smooth_Numbers;

   package body Smooth_Numbers is
      One : constant Number := (Exponents => (others => 0), Log => 0.0);
      Factors_Log : array (Factors'Range) of Log_Type;

      function Image (N : Number) return String is
         use GNATCOLL.GMP.Integers, GNATCOLL.GMP.Lib;
         R, Tmp : Big_Integer;
      begin
         Set (R, "1");
         for F in Factors'Range loop
            Set (Tmp, Factors (F)'Image);
            Raise_To_N (Tmp, GNATCOLL.GMP.Unsigned_Long (N.Exponents (F)));
            Multiply (R, Tmp);
         end loop;
         return Image (R);
      end Image;

      function Compute (Nth : Positive) return Number is
         Candidates : array (Factors'Range) of Number;

         Values     : array (1 .. Nth) of Number;
         --  Will result in Storage_Error for very large values of Nth

         Indices    : array (Factors'Range) of Natural :=
            (others => Values'First);
         Current    : Number;
         Tmp        : Number;
      begin
         for F in Factors'Range loop
            Factors_Log (F) := Funcs.Log (Log_Type (Factors (F)));
            Candidates (F) := One;
            Candidates (F).Exponents (F) := 1;
            Candidates (F).Log := Factors_Log (F);
         end loop;

         Values (1) := One;

         for Count in 2 .. Nth loop
            --  Find next value (the lowest of the candidates)
            Current := Candidates (Factors'First);
            for F in Factors'First + 1 .. Factors'Last loop
               if Candidates (F).Log < Current.Log then
                  Current := Candidates (F);
               end if;
            end loop;

            Values (Count) := Current;

            --  Update the candidates. There might be several candidates with
            --  the same value
            for F in Factors'Range loop
               if Candidates (F) = Current then
                  Indices (F) := Indices (F) + 1;

                  Tmp := Values (Indices (F));
                  Tmp.Exponents (F) := Tmp.Exponents (F) + 1;
                  Tmp.Log := Tmp.Log + Factors_Log (F);

                  Candidates (F) := Tmp;
               end if;
            end loop;
         end loop;

         return Values (Nth);
      end Compute;
   end Smooth_Numbers;

   package Hamming is new Smooth_Numbers ((2, 3, 5));

begin
   for N in 1 .. 20 loop
      Put (" " & Hamming.Image (Hamming.Compute (N)));
   end loop;
   New_Line;

   Put_Line (Hamming.Image (Hamming.Compute (1691)));
   Put_Line (Hamming.Image (Hamming.Compute (1_000_000)));
end Hamming;

{{out}}

 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

ALGOL 68

{{works with|Algol 68 Genie 1.19.0}} Hamming numbers are generated in a trivial iterative way as in the Python version below. This program keeps the series needed to generate the numbers as short as possible using flexible rows; on the downside, it spends considerable time on garbage collection.

PR precision=100 PR

MODE SERIES = FLEX [1 : 0] UNT, # Initially, no elements #
     UNT = LONG LONG INT; # A 100-digit unsigned integer #

PROC hamming number = (INT n) UNT: # The n-th Hamming number #
     CASE n
     IN 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 # First 10 in a table #
     OUT # Additional operators #
         OP MIN = (INT i, j) INT: (i < j | i | j), MIN = (UNT i, j) UNT: (i < j | i | j);
         PRIO MIN = 9;
         OP LAST = (SERIES h) UNT: h[UPB h]; # Last element of a series #
         OP +:= = (REF SERIES s, UNT elem) VOID:
            # Extend a series by one element, only keep the elements you need #
            (INT lwb = (i MIN j) MIN k, upb = UPB s;
             REF SERIES new s = HEAP FLEX [lwb : upb + 1] UNT;
             (new s[lwb : upb] := s[lwb : upb], new s[upb + 1] := elem);
             s := new s
            );
         # Determine the n-th hamming number iteratively #
         SERIES h := 1, # Series, initially one element #
         UNT m2 := 2, m3 := 3, m5 := 5, # Multipliers #
         INT i := 1, j := 1, k := 1; # Counters #
         TO n - 1
         DO h +:= (m2 MIN m3) MIN m5;
            (LAST h = m2 | m2 := 2 * h[i +:= 1]);
            (LAST h = m3 | m3 := 3 * h[j +:= 1]);
            (LAST h = m5 | m5 := 5 * h[k +:= 1])
         OD;
         LAST h
     ESAC;

FOR k TO 20
DO print ((whole (hamming number (k), 0), blank))
OD;
print ((newline, whole (hamming number (1 691), 0)));
print ((newline, whole (hamming number (1 000 000), 0)))

{{out}}

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

ALGOL W

Algol W only has 32 bit integers, so we just show the first 20 Hamming Numbers and Hamming number 1691.

This uses the algorithm from Dr Dobbs (as in the Python version). The Coffee Script solution has some notes on how it works.

begin
    % returns the minimum of a and b                    %
    integer procedure min ( integer value a, b ) ; if a < b then a else b;
    % find and print Hamming Numbers                        %
    % Algol W only supports 32-bit integers so we just find %
    % the 1691 32-bit Hamming Numbers                       %
    integer MAX_HAMMING;
    MAX_HAMMING := 1691;
    begin
        integer array H( 1 :: MAX_HAMMING );
        integer p2, p3, p5, last2, last3, last5;
        H( 1 ) := 1;
        last2  := last3 := last5 := 1;
        p2     := 2;
        p3     := 3;
        p5     := 5;
        for hPos := 2 until MAX_HAMMING do begin
            integer m;
            % the next Hamming number is the lowest of the next multiple of 2, 3, and 5 %
            m := min( min( p2, p3 ), p5 );
            H( hPos ) := m;
            if m = p2 then begin
                last2 := last2 + 1;
                p2    := 2 * H( last2 )
            end if_used_power_of_2 ;
            if m = p3 then begin
                last3 := last3 + 1;
                p3    := 3 * H( last3 )
            end if_used_power_of_3 ;
            if m = p5 then begin
                last5 := last5 + 1;
                p5    := 5 * H( last5 )
            end if_used_power_of_5 ;
        end for_hPos ;
        i_w := 1;
        s_w := 1;
        write( H( 1 ) );
        for i := 2 until 20 do writeon( H( i ) );
        write( H( MAX_HAMMING ) )
    end
end.

{{out}}

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000

ATS

//
// How to compile:
// patscc -DATS_MEMALLOC_LIBC -o hamming hamming.dats
//
#include
"share/atspre_staload.hats"

fun
min3
(
  A: arrayref(int, 3)
) : natLt(3) = i where
{
  var x: int = A[0]
  var i: natLt(3) = 0
  val () = if A[1] < x then (x := A[1]; i := 1)
  val () = if A[2] < x then (x := A[2]; i := 2)
} (* end of [min3] *)

fun
hamming
{n:pos}
(
  n: int(n)
) : int = let
//
var A = @[int](2, 3, 5)
val A = $UNSAFE.cast{arrayref(int, 3)}(addr@A)
var I = @[int](1, 1, 1)
val I = $UNSAFE.cast{arrayref(int, 3)}(addr@I)
val H = arrayref_make_elt<int> (i2sz(succ(n)), 0)
val () = H[0] := 1
//
fun
loop{k:pos}
  (k: int(k)) : void =
(
//
if
k < n
then let
  val i = min3(A)
  val k =
  (
    if A[i] > H[k-1] then (H[k] := A[i]; k+1) else k
  ) : intBtwe(k, k+1)
  val ii = I[i]
  val () = I[i] := ii+1
  val ii = $UNSAFE.cast{natLte(n)}(ii)
  val () = if i = 0 then A[i] := 2*H[ii]
  val () = if i = 1 then A[i] := 3*H[ii]
  val () = if i = 2 then A[i] := 5*H[ii]
in
  loop(k)
end // end of [then]
else () // end of [else]
//
) (* end of [loop] *)
//
in
  loop (1); H[n-1]
end (* end of [hamming] *)

implement
main0 () =
{
val () =
loop(1) where
{
fun
loop
{n:pos}
(
  n: int(n)
) : void =
if
n <= 20
then let
  val () =
  println! ("hamming(",n,") = ", hamming(n))
in
  loop(n+1)
end // end of [then]
// end of [if]
} (* end of [val] *)
val n = 1691
val () = println! ("hamming(",n,") = ", hamming(n))
//
} (* end of [main0] *)

{{out}}

hamming(1) = 1
hamming(2) = 2
hamming(3) = 3
hamming(4) = 4
hamming(5) = 5
hamming(6) = 6
hamming(7) = 8
hamming(8) = 9
hamming(9) = 10
hamming(10) = 12
hamming(11) = 15
hamming(12) = 16
hamming(13) = 18
hamming(14) = 20
hamming(15) = 24
hamming(16) = 25
hamming(17) = 27
hamming(18) = 30
hamming(19) = 32
hamming(20) = 36
hamming(1691) = 2125764000

AutoHotkey

SetBatchLines, -1
Msgbox % hamming(1,20)
Msgbox % hamming(1690)
return

hamming(first,last=0)
{
	if (first < 1)
		ans=ERROR

	if (last = 0)
		last := first

	i:=0, j:=0, k:=0

	num1 := ceil((last * 20)**(1/3))
	num2 := ceil(num1 * ln(2)/ln(3))
	num3 := ceil(num1 * ln(2)/ln(5))

	loop
	{
		H := (2**i) * (3**j) * (5**k)
		if (H > 0)
			ans = %H%`n%ans%
		i++
		if (i > num1)
		{
			i=0
			j++
			if (j > num2)
			{
				j=0
				k++
			}
		}
		if (k > num3)
			break
	}
	Sort ans, N

	Loop, parse, ans, `n, `r
	{
		if (A_index > last)
			break
		if (A_index < first)
			continue
		Output = %Output%`n%A_LoopField%
	}

	return Output
}

AWK

# syntax: gawk -M -f hamming_numbers.awk
BEGIN {
    for (i=1; i<=20; i++) {
      printf("%d ",hamming(i))
    }
    printf("\n1691: %d\n",hamming(1691))
    printf("\n1000000: %d\n",hamming(1000000))
    exit(0)
}
function hamming(limit,  h,i,j,k,n,x2,x3,x5) {
    h[0] = 1
    x2 = 2
    x3 = 3
    x5 = 5
    for (n=1; n<=limit; n++) {
      h[n] = min(x2,min(x3,x5))
      if (h[n] == x2) { x2 = 2 * h[++i] }
      if (h[n] == x3) { x3 = 3 * h[++j] }
      if (h[n] == x5) { x5 = 5 * h[++k] }
    }
    return(h[limit-1])
}
function min(x,y) {
    return((x < y) ? x : y)
}

{{out}}

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
1691: 2125764000
1000000: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

BBC BASIC

      @% = &1010
      FOR h% = 1 TO 20
        PRINT "H("; h% ") = "; FNhamming(h%)
      NEXT
      PRINT "H(1691) = "; FNhamming(1691)
      END

      DEF FNhamming(l%)
      LOCAL i%, j%, k%, n%, m, x2, x3, x5, h%()
      DIM h%(l%) : h%(0) = 1
      x2 = 2 : x3 = 3 : x5 = 5
      FOR n% = 1 TO l%-1
        m = x2
        IF m > x3 m = x3
        IF m > x5 m = x5
        h%(n%) = m
        IF m = x2 i% += 1 : x2 = 2 * h%(i%)
        IF m = x3 j% += 1 : x3 = 3 * h%(j%)
        IF m = x5 k% += 1 : x5 = 5 * h%(k%)
      NEXT
      = h%(l%-1)

{{out}}

H(1) = 1
H(2) = 2
H(3) = 3
H(4) = 4
H(5) = 5
H(6) = 6
H(7) = 8
H(8) = 9
H(9) = 10
H(10) = 12
H(11) = 15
H(12) = 16
H(13) = 18
H(14) = 20
H(15) = 24
H(16) = 25
H(17) = 27
H(18) = 30
H(19) = 32
H(20) = 36
H(1691) = 2125764000

BC

cat hamming_numbers.bc
define min(x,y) {
 if (x < y) {
	return x
 } else {
	return y
 }
}
define hamming(limit) {
 i = 0
 j = 0
 k = 0
 h[0] = 1
 x2 = 2
 x3 = 3
 x5 = 5
 for (n=1; n<=limit; n++) {
  h[n] = min(x2,min(x3,x5))
  if (h[n] == x2) { x2 = 2 * h[++i] }
  if (h[n] == x3) { x3 = 3 * h[++j] }
  if (h[n] == x5) { x5 = 5 * h[++k] }
 }
 return (h[limit-1])
}
for (lab=1; lab<=20; lab++) {
 hamming(lab)
}
hamming(1691)
hamming(1000000)
quit

{{out}}

$ bc hamming_numbers.bc
bc 1.06.95
Copyright 1991-1994, 1997, 1998, 2000, 2004, 2006 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type `warranty'.
1
2
3
4
5
6
8
9
10
12
15
16
18
20
24
25
27
30
32
36
2125764000
51931278044838873608958984375000000000000000000000000000000000000000\
0000000000000000

Bracmat

{{trans|D}}

( ( hamming
  =   x2 x3 x5 n i j k min
    .   tbl$(h,!arg)        { This creates an array. Arrays are always global in Bracmat. }
      & 1:?(0$h)
      & 2:?x2
      & 3:?x3
      & 5:?x5
      & 0:?n:?i:?j:?k
      &   whl
        ' ( !n+1:<!arg:?n
          & !x2:?min
          & (!x3:<!min:?min|)
          & (!x5:<!min:?min|)
          & !min:?(!n$h)               { !n is index into array h }
          & (   !x2:!min
              & 2*!((1+!i:?i)$h):?x2
            |
            )
          & (   !x3:!min
              & 3*!((1+!j:?j)$h):?x3
            |
            )
          & (   !x5:!min
              & 5*!((1+!k:?k)$h):?x5
            |
            )
          )
      & !((!arg+-1)$h) (tbl$(h,0)&)    { We delete the array by setting its size to 0 }
  )
& 0:?I
& whl'(!I+1:~>20:?I&put$(hamming$!I " "))
& out$
& out$(hamming$1691)
& out$(hamming$1000000)
);

{{out}}

1  2  3  4  5  6  8  9  10  12  15  16  18  20  24  25  27  30  32  36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

C

Using a min-heap to keep track of numbers. Does not handle big integers.

#include <stdio.h>
#include <stdlib.h>

typedef unsigned long long ham;

size_t alloc = 0, n = 1;
ham *q = 0;

void qpush(ham h)
{
	int i, j;
	if (alloc <= n) {
		alloc = alloc ? alloc * 2 : 16;
		q = realloc(q, sizeof(ham) * alloc);
	}

	for (i = n++; (j = i/2) && q[j] > h; q[i] = q[j], i = j);
	q[i] = h;
}

ham qpop()
{
	int i, j;
	ham r, t;
	/* outer loop for skipping duplicates */
	for (r = q[1]; n > 1 && r == q[1]; q[i] = t) {
		/* inner loop is the normal down heap routine */
		for (i = 1, t = q[--n]; (j = i * 2) < n;) {
			if (j + 1 < n && q[j] > q[j+1]) j++;
			if (t <= q[j]) break;
			q[i] = q[j], i = j;
		}
	}

	return r;
}

int main()
{
	int i;
	ham h;

	for (qpush(i = 1); i <= 1691; i++) {
		/* takes smallest value, and queue its multiples */
		h = qpop();
		qpush(h * 2);
		qpush(h * 3);
		qpush(h * 5);

		if (i <= 20 || i == 1691)
			printf("%6d: %llu\n", i, h);
	}

	/* free(q); */
	return 0;
}

Alternative

Standard algorithm. Numbers are stored as exponents of factors instead of big integers, while GMP is only used for display. It's much more efficient this way.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <gmp.h>

/* number of factors.  best be mutually prime -- duh. */
#define NK 3
#define MAX_HAM (1 << 24)
#define MAX_POW 1024
int n_hams = 0, idx[NK] = {0}, fac[] = { 2, 3, 5, 7, 11};

/*  k-smooth numbers are stored as their exponents of each factor;
    v is the log of the number, for convenience. */
typedef struct {
	int e[NK];
	double v;
} ham_t, *ham;

ham_t *hams, values[NK] = {{{0}, 0}};
double inc[NK][MAX_POW];

/* most of the time v can be just incremented, but eventually
 * floating point precision will bite us, so better recalculate */
inline
void _setv(ham x) {
	int i;
	for (x->v = 0, i = 0; i < NK; i++)
		x->v += inc[i][x->e[i]];
}

inline
int _eq(ham a, ham b) {
	int i;
	for (i = 0; i < NK && a->e[i] == b->e[i]; i++);

	return i == NK;
}

ham get_ham(int n)
{
	int i, ni;
	ham h;

	n--;
	while (n_hams < n) {
		for (ni = 0, i = 1; i < NK; i++)
			if (values[i].v < values[ni].v)
				ni = i;

		*(h = hams + ++n_hams) = values[ni];

		for (ni = 0; ni < NK; ni++) {
			if (! _eq(values + ni, h)) continue;
			values[ni] = hams[++idx[ni]];
			values[ni].e[ni]++;
			_setv(values + ni);
		}
	}

	return hams + n;
}

void show_ham(ham h)
{
	static mpz_t das_ham, tmp;
	int i;

 	mpz_init_set_ui(das_ham, 1);
	mpz_init_set_ui(tmp, 1);
	for (i = 0; i < NK; i++) {
		mpz_ui_pow_ui(tmp, fac[i], h->e[i]);
		mpz_mul(das_ham, das_ham, tmp);
	}
	gmp_printf("%Zu\n", das_ham);
}

int main()
{
	int i, j;
	hams = malloc(sizeof(ham_t) * MAX_HAM);

	for (i = 0; i < NK; i++) {
		values[i].e[i] = 1;
		inc[i][1] = log(fac[i]);
		_setv(values + i);

		for (j = 2; j < MAX_POW; j++)
			inc[i][j] = j * inc[i][1];
	}

	printf("     1,691: "); show_ham(get_ham(1691));
	printf(" 1,000,000: "); show_ham(get_ham(1e6));
	printf("10,000,000: "); show_ham(get_ham(1e7));
	return 0;
}

{{out}}

     1,691: 2125764000
 1,000,000: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
10,000,000: 16244105063830431823239 ..<a gadzillion digits>.. 000000000000000000000

C#

{{trans|D}}

using System;
using System.Numerics;
using System.Linq;

namespace Hamming {

    class MainClass {

        public static BigInteger Hamming(int n) {
            BigInteger two = 2, three = 3, five = 5;
            var h = new BigInteger[n];
            h[0] = 1;
            BigInteger x2 = 2, x3 = 3, x5 = 5;
            int i = 0, j = 0, k = 0;

            for (int index = 1; index < n; index++) {
                h[index] = BigInteger.Min(x2, BigInteger.Min(x3, x5));
                if (h[index] == x2) x2 = two * h[++i];
                if (h[index] == x3) x3 = three * h[++j];
                if (h[index] == x5) x5 = five * h[++k];
            }
            return h[n - 1];
        }

        public static void Main(string[] args) {
            Console.WriteLine(string.Join(" ", Enumerable.Range(1, 20).ToList().Select(x => Hamming(x))));
            Console.WriteLine(Hamming(1691));
            Console.WriteLine(Hamming(1000000));
        }
    }
}

{{out}}

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Generic version for any set of numbers

The algorithm is similar to the one above.

using System;
using System.Numerics;
using System.Linq;

namespace Hamming {

    class MainClass {

        public static BigInteger[] Hamming(int n, int[] a) {
            var primes = a.Select(x => (BigInteger)x).ToArray();
            var values = a.Select(x => (BigInteger)x).ToArray();
            var indexes = new int[a.Length];
            var results = new BigInteger[n];
            results[0] = 1;
            for (int iter = 1; iter < n; iter++) {
                results[iter] = values[0];
                for (int p = 1; p < primes.Length; p++)
                    if (results[iter] > values[p])
                        results[iter] = values[p];
                for (int p = 0; p < primes.Length; p++)
                    if (results[iter] == values[p])
                        values[p] = primes[p] * results[++indexes[p]];
            }
            return results;
        }

        public static void Main(string[] args) {
            foreach (int[] primes in new int[][] { new int[] {2,3,5}, new int[] {2,3,5,7} }) {
                Console.WriteLine("{0}-Smooth:", primes.Last());
                Console.WriteLine(string.Join(" ", Hamming(20, primes)));
                Console.WriteLine(Hamming(1691, primes).Last());
                Console.WriteLine(Hamming(1000000, primes).Last());
                Console.WriteLine();
            }
        }
    }
}

{{out}}

5-Smooth:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

7-Smooth:
1 2 3 4 5 6 7 8 9 10 12 14 15 16 18 20 21 24 25 27
3317760
4157409948433216829957008507500000000

Fast version

Like some of the other implementations on this page, this version represents each number as a list of exponents which would be applied to each prime number. So the number 60 would be represented as int[3] { 2, 1, 1 } which is interpreted as 2^2 * 3^1 * 5^1.

As often happens, optimizing for speed caused a marked increase in code size and complexity. Clearly the versions I wrote above are easier to read & understand. They were also much quicker to write. But the generic version above runs in 3+ seconds for the 1000000th 5-smooth number whereas this version does it in 0.35 seconds, 8-10 times faster.

I've tried to comment it as best I could, without bloating the code too much.

--Mike Lorenz

using System;
using System.Linq;
using System.Numerics;

namespace HammingFast {

    class MainClass {

        private static int[] _primes = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 };

        public static BigInteger Big(int[] exponents) {
            BigInteger val = 1;
            for (int i = 0; i < exponents.Length; i++)
                for (int e = 0; e < exponents[i]; e++)
                    val = val * _primes[i];
            return val;
        }

        public static int[] Hamming(int n, int nprimes) {
            var hammings  = new int[n, nprimes];                    // array of hamming #s we generate
            var hammlogs  = new double[n];                          // log values for above
            var primelogs = new double[nprimes];                    // pre-calculated prime log values
            var indexes   = new int[nprimes];                       // intermediate hamming values as indexes into hammings
            var listheads = new int[nprimes, nprimes];              // intermediate hamming list heads
            var listlogs  = new double[nprimes];                    // log values of list heads
            for (int p = 0; p < nprimes; p++) {
                listheads[p, p] = 1;                                // init list heads to prime values
                primelogs[p]    = Math.Log(_primes[p]);             // pre-calc prime log values
                listlogs[p]     = Math.Log(_primes[p]);             // init list head log values
            }
            for (int iter = 1; iter < n; iter++) {
                int min = 0;                                        // find index of min item in list heads
                for (int p = 1; p < nprimes; p++)
                    if (listlogs[p] < listlogs[min])
                        min = p;
                hammlogs[iter] = listlogs[min];                     // that's the next hamming number
                for (int i = 0; i < nprimes; i++)
                    hammings[iter, i] = listheads[min, i];
                for (int p = 0; p < nprimes; p++) {                 // update each list head if it matches new value
                    bool equal = true;                              // test each exponent to see if number matches
                    for (int i = 0; i < nprimes; i++) {
                        if (hammings[iter, i] != listheads[p, i]) {
                            equal = false;
                            break;
                        }
                    }
                    if (equal) {                                    // if it matches...
                        int x = ++indexes[p];                       // set index to next hamming number
                        for (int i = 0; i < nprimes; i++)           // copy each hamming exponent
                            listheads[p, i] = hammings[x, i];
                        listheads[p, p] += 1;                       // increment exponent = mult by prime
                        listlogs[p] = hammlogs[x] + primelogs[p];   // add log(prime) to log(value) = mult by prime
                    }
                }
            }

            var result = new int[nprimes];
            for (int i = 0; i < nprimes; i++)
                result[i] = hammings[n - 1, i];
            return result;
        }

        public static void Main(string[] args) {
            foreach (int np in new int[] { 3, 4, 5 }) {
                Console.WriteLine("{0}-Smooth:", _primes[np - 1]);
                Console.WriteLine(string.Join(" ", Enumerable.Range(1, 20).Select(x => Big(Hamming(x, np)))));
                Console.WriteLine(Big(Hamming(1691, np)));
                Console.WriteLine(Big(Hamming(1000000, np)));
                Console.WriteLine();
            }
        }
    }
}

{{out}}

5-Smooth:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

7-Smooth:
1 2 3 4 5 6 7 8 9 10 12 14 15 16 18 20 21 24 25 27
3317760
4157409948433216829957008507500000000

11-Smooth:
1 2 3 4 5 6 7 8 9 10 11 12 14 15 16 18 20 21 22 24
296352
561912530929780078125000

C# Enumerator Version

I wanted to fix the enumerator (old) version, as it wasn't working. It became a bit of an obsession... after a few iterations I came up with the following, which is the fastest C# version on my computer - your mileage may vary. It combines the speed of the Log method; Log(2)+Log(3)=Log(2*3) to help determine which is the next one to use. Then I have added some logic (using the series property) to ensure that exponent sets are never duplicated - which speeds the calculations up a bit.... Adding this trick to the Fast Version will probably result in the fastest version, but I'll leave that to someone else to implement. Finally it's all enumerated through a crazy one-way-linked-list-type-structure that only exists as long as the enumerator and is left up to the garbage collector to remove the bits no longer needed... I hope it's commented enough... follow it if you dare!

using System;
using System.Collections.Generic;
using System.Linq;
using System.Numerics;

namespace HammingTest
{
    class HammingNode
    {
        public double log;
        public int[] exponents;
        public HammingNode next;
        public int series;
    }

    class HammingListEnumerator : IEnumerable<BigInteger>
    {
        private int[] primes;
        private double[] primelogs;
        private HammingNode next;
        private HammingNode[] values;
        private HammingNode[] indexes;

        public HammingListEnumerator(IEnumerable<int> seeds)
        {
            // Ensure our seeds are properly ordered, and generate their log values
            primes = seeds.OrderBy(x => x).ToArray();
            primelogs = primes.Select(x => Math.Log10(x)).ToArray();
            // Start at 1 (log(1)=0, exponents are all 0, series = none)
            next = new HammingNode { log = 0, exponents = new int[primes.Length], series = primes.Length };
            // Set all exponent sequences to the start, and calculate the first value for each exponent
            indexes = new HammingNode[primes.Length];
            values = new HammingNode[primes.Length];
            for(int i = 0; i < primes.Length; ++i)
            {
                indexes[i] = next;
                values[i] = AddExponent(next, i);
            }
        }

        // Make a copy of a node, and increment the specified exponent value
        private HammingNode AddExponent(HammingNode node, int i)
        {
            HammingNode ret = new HammingNode { log = node.log + primelogs[i], exponents = (int[])node.exponents.Clone(), series = i };
            ++ret.exponents[i];
            return ret;
        }

        private void GetNext()
        {
            // Find which exponent value is the lowest
            int min = 0;
            for(int i = 1; i < values.Length; ++i)
                if(values[i].log < values[min].log)
                    min = i;

            // Add it to the end of the 'list', and move to it
            next.next = values[min];
            next = values[min];

            // Find the next node in an allowed sequence (skip those that would be duplicates)
            HammingNode val = indexes[min].next;
            while(val.series < min)
                val = val.next;

            // Keep the current index, and calculate the next value in the series for that exponent
            indexes[min] = val;
            values[min] = AddExponent(val, min);
        }

        // Skip values without having to calculate the BigInteger value from the exponents
        public HammingListEnumerator Skip(int count)
        {
            for(int i = count; i > 0; --i)
                GetNext();

            return this;
        }

        // Calculate the BigInteger value from the exponents
        internal BigInteger ValueOf(HammingNode n)
        {
            BigInteger val = 1;
            for(int i = 0; i < n.exponents.Length; ++i)
                for(int e = 0; e < n.exponents[i]; e++)
                    val = val * primes[i];
            return val;
        }

        public IEnumerator<BigInteger> GetEnumerator()
        {
            while(true)
            {
                yield return ValueOf(next);
                GetNext();
            }
        }

        System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
        {
            return this.GetEnumerator();
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            foreach(int[] primes in new int[][] {
                new int[] { 2, 3, 5 },
                new int[] { 2, 3, 5, 7 },
                new int[] { 2, 3, 5, 7, 9}})
            {
                HammingListEnumerator hammings = new HammingListEnumerator(primes);
                System.Diagnostics.Debug.WriteLine("{0}-Smooth:", primes.Last());
                System.Diagnostics.Debug.WriteLine(String.Join(" ", hammings.Take(20).ToArray()));
                System.Diagnostics.Debug.WriteLine(hammings.Skip(1691 - 20).First());
                System.Diagnostics.Debug.WriteLine(hammings.Skip(1000000 - 1691).First());
                System.Diagnostics.Debug.WriteLine("");
            }
        }
    }
}

{{out}}

5-Smooth:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

7-Smooth:
1 2 3 4 5 6 7 8 9 10 12 14 15 16 18 20 21 24 25 27
3317760
4157409948433216829957008507500000000

11-Smooth:
1 2 3 4 5 6 7 8 9 10 11 12 14 15 16 18 20 21 22 24
296352
561912530929780078125000

'''Alternate Generic Enumerating version'''

YMMV, but unlike the author of the above code, I found the above version to be much slower on my machine than the "Generic version". The following version is actually just a little slower than the Generic version but uses much less memory due to avoiding duplicates and only keeping in memory those "lazy list" streams necessary for calculation from 1/5 of the current range to 1/2 (for Smooth-5 numbers), and not successive values in those ranges but only the values the are the multiples of previous ranges. Like the Haskell code from which it is translated, the head of the streams is not retained so can be garbage collected when no longer necessary and it is recommended that the primes be processed in reverse order so that the least dense streams are processed first for slightly less memory use and operations.

It also shows that one can use somewhat functional programming techniques in C#.

The class implements its own partial version of a lazy list using the Lazy class and uses lambda closures for the recursive use of the successive streams to avoid stack use. It uses Aggregate to implement the Haskell "foldl" function. The code demonstrates that even though C# is primarily imperative in paradigm, with its ability to implement closures using delegates/lambdas, it can express some algorithms such as this mostly functionally.

It isn't nearly as fast as a Haskell, Scala or even Clojure and Scheme (GambitC) versions of this algorithm, being about five times slower is primarily due to its use of many small heap based instances of classes, both for the LazyList's and for closures (implemented using at least one class to hold the captured free variables) and the inefficiency of DotNet's allocation and garbage collection of many small instance objects (although about twice as fast as F#'s implementation, whose closures must require even more small object instances); it seems Haskell and the (Java) JVM are much more efficient at doing these allocations/garbage collections for many small objects. The slower speed to a relatively minor extent is also due to less efficient BigInteger operations: {{trans|Haskell}}

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.Numerics;

namespace Hamming {

  class Hammings : IEnumerable<BigInteger> {
    private class LazyList<T> {
      public T v; public Lazy<LazyList<T>> cont;
      public LazyList(T v, Lazy<LazyList<T>> cont) {
        this.v = v; this.cont = cont;
      }
    }
    private uint[] primes;
    private Hammings() { } // must have an argument!!!
    public Hammings(uint[] prms) { this.primes = prms; }
    private LazyList<BigInteger> merge(LazyList<BigInteger> xs,
                                       LazyList<BigInteger> ys) {
      if (xs == null) return ys; else {
        var x = xs.v; var y = ys.v;
        if (BigInteger.Compare(x, y) < 0) {
          var cont = new Lazy<LazyList<BigInteger>>(() =>
                       merge(xs.cont.Value, ys));
          return new LazyList<BigInteger>(x, cont);
        }
        else {
          var cont = new Lazy<LazyList<BigInteger>>(() =>
                       merge(xs, ys.cont.Value));
          return new LazyList<BigInteger>(y, cont);
        }
      }
    }
    private LazyList<BigInteger> llmult(uint mltplr,
                                        LazyList<BigInteger> ll) {
      return new LazyList<BigInteger>(mltplr * ll.v,
                                      new Lazy<LazyList<BigInteger>>(() =>
                                        llmult(mltplr, ll.cont.Value)));
    }
    public IEnumerator<BigInteger> GetEnumerator() {
      Func<LazyList<BigInteger>,uint,LazyList<BigInteger>> u =
        (acc, p) => { LazyList<BigInteger> r = null;
                      var cont = new Lazy<LazyList<BigInteger>>(() => r);
                      r = new LazyList<BigInteger>(1, cont);
                      r = this.merge(acc, llmult(p, r));
                      return r; };
      yield return 1;
      for (var stt = primes.Aggregate(null, u); ; stt = stt.cont.Value)
        yield return stt.v;
    }
    IEnumerator IEnumerable.GetEnumerator() {
      return this.GetEnumerator();
    }
  }

  class Program {
    static void Main(string[] args) {
      Console.WriteLine("Calculates the Hamming sequence of numbers.\r\n");

      var primes = new uint[] { 5, 3, 2 };
      Console.WriteLine(String.Join(" ", (new Hammings(primes)).Take(20).ToArray()));
      Console.WriteLine((new Hammings(primes)).ElementAt(1691 - 1));

      var n = 1000000;

      var elpsd = -DateTime.Now.Ticks;

      var num = (new Hammings(primes)).ElementAt(n - 1);

      elpsd += DateTime.Now.Ticks;

      Console.WriteLine(num);
      Console.WriteLine("The {0}th hamming number took {1} milliseconds", n, elpsd / 10000);

      Console.Write("\r\nPress any key to exit:");
      Console.ReadKey(true);
      Console.WriteLine();
    }
  }
}

Fast enumerating logarithmic version

The so-called "fast" generic version above isn't really all that fast due to all the extra array accesses required by the generic implementation and that it doesn't avoid duplicates as the above functional code does avoid. It also uses a lot of memory as it has arrays that are the size of the range for which the Hamming numbers are calculated.

The following code eliminates or reduces all of those problems by being non-generic, eliminating duplicate calculations, saving memory by "draining" the growable List's used in blocks as back pointer indexes are used (thus using memory at the same rate as the functional version), thus avoiding excessive allocations/garbage collections; it also is enumerates through the Hamming numbers although that comes at a slight cost in overhead function calls: {{trans|Nim}}

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.Numerics;

class HammingsLogArr : IEnumerable<Tuple<uint, uint, uint>> {
  public static BigInteger trival(Tuple<uint, uint, uint> tpl) {
    BigInteger rslt = 1;
    for (var i = 0; i < tpl.Item1; ++i) rslt *= 2;
    for (var i = 0; i < tpl.Item2; ++i) rslt *= 3;
    for (var i = 0; i < tpl.Item3; ++i) rslt *= 5;
    return rslt;
  }
  private const double lb3 = 1.5849625007211561814537389439478; // Math.Log(3) / Math.Log(2);
  private const double lb5 = 2.3219280948873623478703194294894; // Math.Log(5) / Math.Log(2);
  private struct logrep {
    public uint x2, x3, x5;
    public double lg;
    public logrep(uint x, uint y, uint z, double lg) {
      this.x2 = x; this.x3 = y; this.x5 = z; this.lg = lg;
    }
    public static bool operator <(logrep x, logrep y) {
      return x.lg < y.lg;
    }
    public static bool operator >(logrep x, logrep y) {
      return x.lg > y.lg;
    }
    public logrep mul2() {
      return new logrep(this.x2 + 1, this.x3, this.x5, this.lg + 1.0);
    }
    public logrep mul3() {
      return new logrep(this.x2, this.x3 + 1, this.x5, this.lg + lb3);
    }
    public logrep mul5() {
      return new logrep(this.x2, this.x3, this.x5 + 1, this.lg + lb5);
    }
  }
  public IEnumerator<Tuple<uint, uint, uint>> GetEnumerator() {
    var one = new logrep();
    var m = new List<logrep>(); var h = new List<logrep>();
    var x5 = one.mul5();
    var mrg = one.mul3();
    var x53 = one.mul3().mul3(); // already advanced one step
    var x532 = one.mul2();
    var i = 0; var j = 0;
    yield return Tuple.Create(0u, 0u, 0u); // trivial case for one representation
    while (true) {
      if (i >= h.Capacity >> 1) { h.RemoveRange(0, i); i = 0; } // assume capacity stays the same...
      if (x532 < mrg) { h.Add(x532); x532 = h[i].mul2(); i++; }
      else {
        h.Add(mrg);
        if (j >= m.Capacity) { m.RemoveRange(0, j); j = 0; }
        if (x53 < x5) { mrg = x53; x53 = m[j].mul3(); j++; }
        else { mrg = x5; x5 = x5.mul5(); }
        m.Add(mrg);
      }
      var rslt = h[h.Count - 1];
      yield return Tuple.Create(rslt.x2, rslt.x3, rslt.x5);
    }
  }
  IEnumerator IEnumerable.GetEnumerator() {
    return this.GetEnumerator();
  }
}

class Program {
  static void Main(string[] args) {
    Console.WriteLine(String.Join(" ", (new HammingsLogArr()).Take(20)
                                        .Select(t => HammingsLogArr.trival(t))
                                        .ToArray()));
    Console.WriteLine(HammingsLogArr.trival(NthHamming.findNth(1691)));

    var n = 1000000UL;
    var elpsd = -DateTime.Now.Ticks;

    var rslt = (new HammingsLogArr()).ElementAt((int)n - 1);

    elpsd += DateTime.Now.Ticks;

    Console.WriteLine("2^{0} times 3^{1} times 5^{2}", rslt.Item1, rslt.Item2, rslt.Item3);
    var lgrthm = Math.Log10(2.0) * ((double)rslt.Item1 +
                  ((double)rslt.Item2 * Math.Log(3.0) + (double)rslt.Item3 * Math.Log(5.0)) / Math.Log(2.0));
    var pwr = Math.Floor(lgrthm); var mntsa = Math.Pow(10.0, lgrthm - pwr);
    Console.WriteLine("Approximately:  {0}E+{1}", mntsa, pwr);
    var s = HammingsLogArr.trival(rslt).ToString();
    var lngth = s.Length;
    Console.WriteLine("Decimal digits:  {0}", lngth);
    if (lngth <= 10000) {
      var i = 0;
      for (; i < lngth - 100; i += 100) Console.WriteLine(s.Substring(i, 100));
      Console.WriteLine(s.Substring(i));
    }
    Console.WriteLine("The {0}th hamming number took {1} milliseconds", n, elpsd / 10000);

    Console.Write("\r\nPress any key to exit:");
    Console.ReadKey(true);
    Console.WriteLine();
  }
}

{{output}}

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
2^55 times 3^47 times 5^64
Approximately:  5.19312780448414E+83
Decimal digits:  84
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
The 1000000th hamming number took 55 milliseconds

The above code is about 30 times faster than the functional code due to both eliminating the lambda closures that were the main problem with that code as well as eliminating the BigInteger operations. It has about O(n) empirical performance and can find the billionth Hamming number in about 60 seconds.

===Extremely fast non-enumerating version calculating the error band===

The above code is about as fast as one can go generating sequences; however, if one is willing to forego sequences and just calculate the nth Hamming number (again), then some reading on the relationship between the size of numbers to the sequence numbers is helpful (Wikipedia: regular number). One finds that there is a very distinct relationship and that it quite quickly reduces to quite a small error band proportional to the log of the output value for larger ranges. Thus, the following code just scans for logarithmic representations to insert into a sequence for this top error band and extracts the correct nth representation from that band. It reduces time complexity to O(n^(2/3)) from O(n) for the sequence versions, but even more amazingly, reduces memory requirements to O(n^(1/3)) from O(n^(2/3)) and thus makes it possible to calculate very large values in the sequence on common personal computers. The code is as follows: {{trans|Nim}}

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.Numerics;

static class NthHamming {
  public static BigInteger trival(Tuple<uint, uint, uint> tpl) {
    BigInteger rslt = 1;
    for (var i = 0; i < tpl.Item1; ++i) rslt *= 2;
    for (var i = 0; i < tpl.Item2; ++i) rslt *= 3;
    for (var i = 0; i < tpl.Item3; ++i) rslt *= 5;
    return rslt;
  }

  private struct logrep {
    public uint x2, x3, x5;
    public double lg;
    public logrep(uint x, uint y, uint z, double lg) {
      this.x2 = x; this.x3 = y; this.x5 = z; this.lg = lg;
    }
  }

  private const double lb3 = 1.5849625007211561814537389439478; // Math.Log(3) / Math.Log(2);
  private const double lb5 = 2.3219280948873623478703194294894; // Math.Log(5) / Math.Log(2);
  private const double fctr = 6.0 * lb3 * lb5;
  private const double crctn = 2.4534452978042592646620291867186; // Math.Log(Math.sqrt(30.0)) / Math.Log(2.0)

  public static Tuple<uint, uint, uint> findNth(UInt64 n) {
    if (n < 1) throw new Exception("NthHamming.findNth:  argument must be > 0!");
    if (n < 2) return Tuple.Create(0u, 0u, 0u); // trivial case for argument of one
    var lgest = Math.Pow(fctr * (double)n, 1.0/3.0) - crctn; // from WP formula
    var frctn = (n < 1000000000) ? 0.509 : 0.105;
    var lghi = Math.Pow(fctr * ((double)n + frctn * lgest), 1.0/3.0) - crctn;
    var lglo = 2.0 * lgest - lghi; // upper and lower bound of upper "band"
    var count = 0UL; // need 64 bit precision in case...
    var bnd = new List<logrep>();
    for (uint k = 0, klmt = (uint)(lghi / lb5) + 1; k < klmt; ++k) {
      var p = (double)k * lb5;
      for (uint j = 0, jlmt = (uint)((lghi - p) / lb3) + 1; j < jlmt; ++j) {
        var q = p + (double)j * lb3;
        var ir = lghi - q;
        var lg = q + Math.Floor(ir); // current log2 value (estimated)
        count += (ulong)ir + 1;
        if (lg >= lglo) bnd.Add(new logrep((UInt32)ir, j, k, lg));
      }
    }
    if (n > count) throw new Exception("NthHamming.findNth:  band high estimate is too low!");
    var ndx = (int)(count - n);
    if (ndx >= bnd.Count) throw new Exception("NthHamming.findNth:  band low estimate is too high!");
    bnd.Sort((a, b) => (b.lg < a.lg) ? -1 : 1); // sort in decending order

    var rslt = bnd[ndx];
    return Tuple.Create(rslt.x2, rslt.x3, rslt.x5);
  }
}

class Program {
  static void Main(string[] args) {
    Console.WriteLine(String.Join(" ", Enumerable.Range(1,20).Select(i =>
                                          NthHamming.trival(NthHamming.findNth((ulong)i))).ToArray()));
    Console.WriteLine(NthHamming.trival((new HammingsLogArr()).ElementAt(1691 - 1)));

    var n = 1000000000000UL;
    var elpsd = -DateTime.Now.Ticks;

    var rslt = NthHamming.findNth(n);

    elpsd += DateTime.Now.Ticks;

    Console.WriteLine("2^{0} times 3^{1} times 5^{2}", rslt.Item1, rslt.Item2, rslt.Item3);
    var lgrthm = Math.Log10(2.0) * ((double)rslt.Item1 +
                  ((double)rslt.Item2 * Math.Log(3.0) + (double)rslt.Item3 * Math.Log(5.0)) / Math.Log(2.0));
    var pwr = Math.Floor(lgrthm); var mntsa = Math.Pow(10.0, lgrthm - pwr);
    Console.WriteLine("Approximately:  {0}E+{1}", mntsa, pwr);
    var s = HammingsLogArr.trival(rslt).ToString();
    var lngth = s.Length;
    Console.WriteLine("Decimal digits:  {0}", lngth);
    if (lngth <= 10000) {
      var i = 0;
      for (; i < lngth - 100; i += 100) Console.WriteLine(s.Substring(i, 100));
      Console.WriteLine(s.Substring(i));
    }
    Console.WriteLine("The {0}th hamming number took {1} milliseconds", n, elpsd / 10000);

    Console.Write("\r\nPress any key to exit:");
    Console.ReadKey(true);
    Console.WriteLine();
  }
}

The output is the same as above except that the time is too small to be measured. The billionth number in the sequence can be calculated in just about 10 milliseconds, the trillionth in about one second, the thousand trillionth in about a hundred seconds, and it should be possible to calculate the 10^19th value in less than a day (untested) on common personal computers. The (2^64 - 1)th value (18446744073709551615) cannot be calculated due to a slight overflow problem as it approaches that limit.

C++

C++11 For Each Generator

#include <iostream>
#include <vector>
// Hamming like sequences Generator
//
// Nigel Galloway. August 13th., 2012
//
class Ham {
private:
	std::vector<unsigned int> _H, _hp, _hv, _x;
public:
	bool operator!=(const Ham& other) const {return true;}
	Ham begin() const {return *this;}
        Ham end() const {return *this;}
	unsigned int operator*() const {return _x.back();}
	Ham(const std::vector<unsigned int> &pfs):_H(pfs),_hp(pfs.size(),0),_hv({pfs}),_x({1}){}
	const Ham& operator++() {
	  for (int i=0; i<_H.size(); i++) for (;_hv[i]<=_x.back();_hv[i]=_x[++_hp[i]]*_H[i]);
	  _x.push_back(_hv[0]);
	  for (int i=1; i<_H.size(); i++) if (_hv[i]<_x.back()) _x.back()=_hv[i];
	  return *this;
	}
};

===5-Smooth===

int main() {
  int count = 1;
  for (unsigned int i : Ham({2,3,5})) {
    if (count <= 62) std::cout << i << ' ';
    if (count++ == 1691) {
      std::cout << "\nThe one thousand six hundred and ninety first Hamming Number is " << i << std::endl;
      break;
    }
  }
  return 0;
}

Produces:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 40 45 48 50 54 60 64 72 75 80 81 90 96 100 108 120 125 128 135 144 150 160 162 180 192 200 216 225 240 243 250 256 270 288 300 320 324 360 375 384 400 405
The one thousand six hundred and ninety first Hamming Number is 2125764000

===7-Smooth===

int main() {
  int count = 1;
  for (unsigned int i : Ham({2,3,5,7})) {
    std::cout << i << ' ';
    if (count++ == 64) break;
  }
  std::cout << std::endl;
  return 0;
}

Produces:

1 2 3 4 5 6 7 8 9 10 12 14 15 16 18 20 21 24 25 27 28 30 32 35 36 40 42 45 48 49 50 54 56 60 63 64 70 72 75 80 81 84 90 96 98 100 105 108 112 120 125 126 128 135 140 144 147 150 160 162 168 175 180 189

Avoiding Duplicates with Functional Code

If converted to use multi-precision integers (via GMP, as in this code), the above code is slower than it needs to be due to several reasons: 1) It uses the an adaptation of the original Dijkstra's algorithm, which is somewhat slower due to repeated calculations (2 time 3, 3 times 2, etc.), 2) the generator is written generally to handle any series of multiples, and 3) for finding the nth Hamming number, the code as for (auto hmg : Ham({5, 3, 2}) means that there is a copy of the sometimes very large multi-precision number which can consume more time than the actual computation. The following code isn't particularly fast due to other reasons that will be discussed, but avoids duplication of calculations by a modification of the algorithm; it is written functionally as a LazyList (which could easily also have iteration abilities added, with the a basic LazyList type defined here since there is no library available: {{trans|Haskell}} {{works with|C++11}}

#include <chrono>

#include <iostream>
#include <gmpxx.h>
#include <functional>
#include <memory>

template<class T>
class Lazy {
public:
	T _v;
private:
	std::function<T()> _f;

public:
	explicit Lazy(std::function<T()> thnk)
		: _v(T()), _f(thnk) {};
	T value() { // not thread safe!
		if (this->_f != nullptr) {
			this->_v = this->_f();
			this->_f = nullptr;
		}
		return this->_v;
	}
};

template<class T>
class LazyList {
public:
	T head;
	std::shared_ptr<Lazy<LazyList<T>>> tail;
	LazyList(): head(T()) {} // only used in initializing Lazy...
	LazyList(T head, std::function<LazyList<T>()> thnk)
		: head(head), tail(std::make_shared<Lazy<LazyList<T>>>(thnk)) {}
	// default Copy/Move constructors and assignment operators seem to work well enough
	bool isEmpty() { return this->tail == nullptr; }
};

typedef std::shared_ptr<mpz_class> PBI;
typedef LazyList<PBI> LL;
typedef std::function<LL(LL)> FLL2LL;

LL merge(LL a, LL b) {
	auto ha = a.head; auto hb = b.head;
	if (*ha < *hb) {
		return LL(ha, [=]() { return merge(a.tail->value(), b); });
	} else {
		return LL(hb, [=]() { return merge(a, b.tail->value()); });
	}
}

LL smult(int m, LL s) {
	const auto im = mpz_class(m);
	const auto psmlt =
			std::make_shared<FLL2LL>([](LL ss) { return ss; });
	*psmlt = [=](LL ss) {
		return LL(std::make_shared<mpz_class>(*ss.head * im),
				  [=]() { return (*psmlt)(ss.tail->value()); });
	};
	return (*psmlt)(s); // worker wrapper pattern with recursive closure as worker...
}

LL u(LL s, int n) {
	const auto r = std::make_shared<LL>(LL()); // interior mutable...
	*r = smult(n, LL(std::make_shared<mpz_class>(1), [=]() { return *r; }));
	if (!s.isEmpty()) { *r = merge(s, *r); }
	return *r;
}

LL hammings() {
	auto r = LL();
	for (auto pn : std::vector<int>({5, 3, 2})) {
		r = u(r, pn);
	}
	return LL(std::make_shared<mpz_class>(1), [=]() { return r; });
}

int main() {
	auto hmgs = hammings();
	for (auto i = 0; i < 20; ++i) {
		std::cout << *hmgs.head << " ";
		hmgs = hmgs.tail->value();
	}
	std::cout << "\n";

	hmgs = hammings();
	for (auto i = 1; i < 1691; ++i) hmgs = hmgs.tail->value();
	std::cout << *hmgs.head << "\n";

	auto start = std::chrono::steady_clock::now();
	hmgs = hammings();
	for (auto i = 1; i < 1000000; ++i) hmgs = hmgs.tail->value();
	auto stop = std::chrono::steady_clock::now();

	auto ms = std::chrono::duration_cast<std::chrono::milliseconds>(stop - start);
	std::cout << *hmgs.head << " in " << ms.count() << "milliseconds.\n";
}

{{output}}

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000 in 1079 milliseconds.

Note that the repeat loop to increment to the desired value is written so as not to oopy unnecessary Hamming values, unlike the use of the first Gnerator class.

This shows that one can program functionally in C++, but the performance is many times slower than a language more suitable for functional paradigms such as Haskell or even Kotlin; this is likely due to the cost of memory allocation with (multi-threading-safe) reference counting and that the memory system isn't tuned to many small allocations/de-allocations such as are generally necessary with functional programming. One can easily see how to adapt this algorithm to make it work for the general case by just haveing an argument which is an vector of the required multipliers used in the hammings function.

Avoiding Duplicates with Imperative Code

To show that it is the execution time for the functional LazyList that is taking the time, here is the same algorithm implemented imperatively using vectors, also avoiding duplicate calculations; it is not written as a general function for any set of multipliers as the extra vector addressing takes some extra time; again, it minimizes copying of values: {{trans|Rust}} {{works with|C++11}}

#include <chrono>

#include <iostream>
#include <vector>
#include <gmpxx.h>

class Hammings {
private:
	const mpz_class _two = 2, _three = 3, _five = 5;
	std::vector<mpz_class> _m = {}, _h = {1};
	mpz_class _x5 = 5, _x53 = 9, _mrg = 3, _x532 = 2;
	int _i = 1, _j = 0;
public:
	Hammings() {_m.reserve(65536); _h.reserve(65536); };
	bool operator!=(const Hammings& other) const { return true; }
	Hammings begin() const { return *this; }
	Hammings end() const { return *this; }
	mpz_class operator*() { return _h.back(); }
	const Hammings& operator++() {
		if (_i > _h.capacity() / 2) {
			_h.erase(_h.begin(), _h.begin() + _i);
			_i = 0;
		}
		if (_x532 < _mrg) {
			_h.push_back(_x532);
			_x532 = _h[_i++] * _two;
		} else {
			_h.push_back(_mrg);
			if (_x53 < _x5) {
				_mrg = _x53;
				_x53 = _m[_j++] * _three;
			} else {
				_mrg = _x5;
				_x5 = _x5 * _five;
			}
			if (_j > _m.capacity() / 2) {
				_m.erase(_m.begin(), _m.begin() + _j);
				_j = 0;
			}
			_m.push_back(_mrg);
		}
		return *this;
	}
};

int main() {
	auto cnt = 1;
	for (auto hmg : Hammings()) {
		if (cnt <= 20) std::cout << hmg << " ";
		if (cnt == 20) std::cout << "\n";
		if (cnt++ >= 1691) {
			std::cout << hmg << "\n";
			break;
		}
	}

	auto start = std::chrono::steady_clock::now();
	hmgs = hammings();
	auto&& hmgitr = Hammings();
	for (auto i = 1; i < 1000000; ++i) ++hmgitr;
	auto stop = std::chrono::steady_clock::now();

	auto ms = std::chrono::duration_cast<std::chrono::milliseconds>(stop - start);
	std::cout << *hmgitr << " in " << ms.count() << "milliseconds.\n";
}

{{output}}

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000 in 79 milliseconds.

This code takes about the same amount of time as Haskell for the functional algorithm calculating multi-precision values (also uses GMP; not including the list processing time), but greatly reduces the C++ processing time compared to the functional version due to the use of imperative code and vectors.

Clojure

This version implements Dijkstra's merge solution, so is closely related to the Haskell version.

(defn smerge [xs ys]
  (lazy-seq
    (let [x (first xs),
          y (first ys),
          [z xs* ys*]
          (cond
            (< x y) [x (rest xs) ys]
            (> x y) [y xs (rest ys)]
            :else   [x (rest xs) (rest ys)])]
      (cons z (smerge xs* ys*)))))

(def hamming
  (lazy-seq
    (->> (map #(*' 5 %) hamming)
         (smerge (map #(*' 3 %) hamming))
         (smerge (map #(*' 2 %) hamming))
         (cons 1))))

Note that the above version uses a lot of space and time after calculating a few hundred thousand elements of the sequence. This is no doubt due to not avoiding the generation of duplicates in the sequences as well as its "holding on to the head": it maintains the entire generated sequences in memory.

'''Avoiding duplicates and reducing memory use'''

In order to fix the problems with the above program as to memory use and extra time expended, the following code implements the Haskell idea as a function so that it does not retain the pointers to the streams used so that they can be garbage collected from the beginning as they are consumed. it avoids duplicate number generation by using intermediate streams for each of the multiples and building each on the results of the last; also, it orders the streams from least dense to most so that the intermediate streams retained are as short as possible, with the "s5" stream only from one fifth to a third of the current value, the "s35" stream only between a third and a half of the current output value, and the s235 stream only between a half and the current output - as the sequence is not very dense with increasing range, mot many values need be retained: {{trans|Haskell}}

(defn hamming
  "Computes the unbounded sequence of Hamming 235 numbers."
  []
  (letfn [(merge [xs ys]
            (if (nil? xs) ys
              (let [xv (first xs), yv (first ys)]
                (if (< xv yv) (cons xv (lazy-seq (merge (next xs) ys)))
                              (cons yv (lazy-seq (merge xs (next ys)))))))),
          (smult [m s] ;; equiv to map (* m) s -- faster
            (cons (*' m (first s)) (lazy-seq (smult m (next s))))),
          (u [s n] (let [r (atom nil)]
                      (reset! r (merge s (smult n (cons 1 (lazy-seq @r)))))))]
    (cons 1 (lazy-seq (reduce u nil (list 5 3 2))))))

Much of the time expended for larger ranges (say 10 million or more) is due to the time doing extended precision arithmetic, with also a significant percentage spent in garbage collection. Following is the output from the REPL after compiling the program:

After compiling code in REPL: {{output}}

(take 20 (hamming))
(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36)
(->> (hamming) (drop 1690) (first) (time))
"Elapsed time: 1.105582 msecs"
2125764000
(->> (hamming) (drop 999999) (first) (time))
"Elapsed time: 447.561128 msecs"
519312780448388736089589843750000000000000000000000000000000000000000000000000000000N

So that generated '.class' files in a folder or a generated '.jar' file (possibly standalone, containing the run time library) run at about the same speed as inside the IDE (after compilation), the Leiningen "project.clj" file needs to be modified to contain the following line so as to eliminate JVM options that slow the performance:

  :jvm-opts ^:replace []

CoffeeScript

# Generate hamming numbers in order.  Hamming numbers have the
# property that they don't evenly divide any prime numbers outside
# a given set, such as [2, 3, 5].

generate_hamming_sequence = (primes, max_n) ->
  # We use a lazy algorithm, only ever keeping N candidates
  # in play, one for each of our seed primes.  Let's say
  # primes is [2,3,5].  Our virtual streams are these:
  #
  # hammings:    1,2,3,4,5,6,8,10,12,15,16,18,20,...
  # hammings*2:  2,4,6,9.10,12,16,20,24,30,32,36,40...
  # hammings*3:  3,6,9,12,15,18,24,30,36,45,...
  # hammings*5:  5,10,15,20,25,30,40,50,...
  #
  # After encountering 40 for the last time, our candidates
  # will be
  #   50 = 2 * 25
  #   45 = 3 * 15
  #   50 = 5 * 10
  # Then, after 45
  #   50 = 2 * 25
  #   48 = 3 * 16 <= new
  #   50 = 5 * 10
  hamming_numbers = [1]
  candidates = ([p, p, 1] for p in primes)
  last_number = 1
  while hamming_numbers.length < max_n
    # Get the next candidate Hamming Number tuple.
    i = min_idx(candidates)
    candidate = candidates[i]
    [n, p, seq_idx] = candidate

    # Add to sequence unless it's a duplicate.
    if n > last_number
      hamming_numbers.push n
      last_number = n

    # Replace the candidate with its successor (based on
    # p = 2, 3, or 5).
    #
    # This is the heart of the algorithm.  Let's say, over the
    # primes [2,3,5], we encounter the hamming number 32 based on it being
    # 2 * 16, where 16 is the 12th number in the sequence.
    # We'll be passed in [32, 2, 12] as candidate, and
    # hamming_numbers will be [1,2,3,4,5,6,8,9,10,12,16,18,...]
    # by now.  The next candidate we need to enqueue is
    # [36, 2, 13], where the numbers mean this:
    #
    #    36 - next multiple of 2 of a Hamming number
    #     2 - prime number
    #    13 - 1-based index of 18 in the sequence
    #
    # When we encounter [36, 2, 13], we will then enqueue
    # [40, 2, 14], based on 20 being the 14th hamming number.
    q = hamming_numbers[seq_idx]
    candidates[i] = [p*q, p, seq_idx+1]

  hamming_numbers

min_idx = (arr) ->
  # Don't waste your time reading this--it just returns
  # the index of the smallest tuple in an array, respecting that
  # the tuples may contain integers. (CS compiles to JS, which is
  # kind of stupid about sorting.  There are libraries to work around
  # the limitation, but I wanted this code to be standalone.)
  less_than = (tup1, tup2) ->
    i = 0
    while i < tup2.length
      return true if tup1[i] <= tup2[i]
      return false if tup1[i] > tup2[i]
      i += 1

  min_i = 0
  for i in [1...arr.length]
    if less_than arr[i], arr[min_i]
      min_i = i
  return min_i

primes = [2, 3, 5]
numbers = generate_hamming_sequence(primes, 10000)
console.log numbers[1690]
console.log numbers[9999]

Common Lisp

Maintaining three queues, popping the smallest value every time.

(defun next-hamm (factors seqs)
  (let ((x (apply #'min (map 'list #'first seqs))))
    (loop for s in seqs
	  for f in factors
	  for i from 0
	  with add = t do
	  (if (= x (first s)) (pop s))
	  ;; prevent a value from being added to multiple lists
	  (when add
	    (setf (elt seqs i) (nconc s (list (* x f))))
	    (if (zerop (mod x f)) (setf add nil)))
    finally (return x))))

(loop with factors = '(2 3 5)
      with seqs    = (loop for i in factors collect '(1))
      for n from 1 to 1000001 do
      (let ((x (next-hamm factors seqs)))
	(if (or (< n 21)
		(= n 1691)
		(= n 1000000)) (format t "~d: ~d~%" n x))))

A much faster method:

(defun hamming (n)
  (let ((fac '(2 3 5))
	(idx (make-array 3 :initial-element 0))
	(h (make-array (1+ n)
		       :initial-element 1
		       :element-type 'integer)))
    (loop for i from 1 to n
	  with e with x = '(1 1 1) do
	  (setf e (setf (aref h i) (apply #'min x))
		x (loop for y in x
			for f in fac
			for j from 0
			collect (if (= e y) (* f (aref h (incf (aref idx j)))) y))))
    (aref h n)))

(loop for i from 1 to 20 do
      (format t "~2d: ~d~%" i (hamming i)))

(loop for i in '(1691 1000000) do
      (format t "~d: ~d~%" i (hamming i)))

{{out}}

 1: 1
 2: 2
 3: 3
 4: 4
 5: 5
 6: 6
 7: 8
 8: 9
 9: 10
10: 12
11: 15
12: 16
13: 18
14: 20
15: 24
16: 25
17: 27
18: 30
19: 32
20: 36
1691: 2125764000
1000000: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

D

Basic Version

This version keeps all numbers in memory, computing all the Hamming numbers up to the needed one. Performs constant number of operations per Hamming number produced.

import std.stdio, std.bigint, std.algorithm, std.range, core.memory;

auto hamming(in uint n) pure nothrow /*@safe*/ {
    immutable BigInt two = 2, three = 3, five = 5;
    auto h = new BigInt[n];
    h[0] = 1;
    BigInt x2 = 2, x3 = 3, x5 = 5;
    size_t i, j, k;

    foreach (ref el; h.dropOne) {
        el = min(x2, x3, x5);
        if (el == x2) x2 = two   * h[++i];
        if (el == x3) x3 = three * h[++j];
        if (el == x5) x5 = five  * h[++k];
    }
    return h.back;
}

void main() {
    GC.disable;
    iota(1, 21).map!hamming.writeln;
    1_691.hamming.writeln;
    1_000_000.hamming.writeln;
}

{{out}}

[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Runtime is about 1.6 seconds with LDC2.

Alternative Version 1

This keeps $O(n^\{2/3\})$ numbers in memory, but over-computes a sequence by a factor of about $\backslash Theta(n^\{2/3\})$, calculating extra multiples past that as well. Incurs an extra $O(\backslash log\; n)$ factor of operations per each number produced (reinserting its multiples into a tree). Doesn't stop when the target number is reached, instead continuing until it is no longer needed: {{trans|Java}}

import std.stdio, std.bigint, std.container, std.algorithm, std.range,
       core.memory;

BigInt hamming(in int n)
in {
   assert(n > 0);
} body {
    auto frontier = redBlackTree(2.BigInt, 3.BigInt, 5.BigInt);
    auto lowest = 1.BigInt;
    foreach (immutable _; 1 .. n) {
        lowest = frontier.front;
        frontier.removeFront;
        frontier.insert(lowest * 2);
        frontier.insert(lowest * 3);
        frontier.insert(lowest * 5);
    }
    return lowest;
}

void main() {
    GC.disable;
    writeln("First 20 Hamming numbers: ", iota(1, 21).map!hamming);
    writeln("hamming(1691) = ", 1691.hamming);
    writeln("hamming(1_000_000) = ", 1_000_000.hamming);
}

{{out}}

First 20 Hamming numbers: [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
hamming(1691) = 2125764000
hamming(1_000_000) = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

About 3.2 seconds run time with LDC2.

Alternative Version 2

Does exactly what the first version does, creating an array and filling it with Hamming numbers, keeping the three back pointers into the sequence for next multiples calculations, except that it represents the numbers as their coefficients triples and their logarithm values (for comparisons), thus saving on BigInt calculations. {{trans|C}}

import std.stdio: writefln;
import std.bigint: BigInt;
import std.conv: text;
import std.numeric: gcd;
import std.algorithm: copy, map;
import std.array: array;
import core.stdc.stdlib: calloc;
import std.math: log; // ^^

// Number of factors.
enum NK = 3;

enum MAX_HAM = 10_000_000;
static assert(gcd(NK, MAX_HAM) == 1);

enum int[NK] factors = [2, 3, 5];


/// K-smooth numbers (stored as their exponents of each factor).
struct Hamming {
    double v; // Log of the number, for convenience.
    ushort[NK] e; // Exponents of each factor.

    public static __gshared immutable double[factors.length] inc =
        factors[].map!log.array;

    bool opEquals(in ref Hamming y) const pure nothrow @nogc {
        //return this.e == y.e; // Too much slow.
        foreach (immutable i; 0 .. this.e.length)
            if (this.e[i] != y.e[i])
                return false;
        return true;
    }

    void update() pure nothrow @nogc {
        //this.v = dotProduct(inc, this.e); // Too much slow.
        this.v = 0.0;
        foreach (immutable i; 0 .. this.e.length)
            this.v += inc[i] * this.e[i];
    }

    string toString() const {
        BigInt result = 1;
        foreach (immutable i, immutable f; factors)
            result *= f.BigInt ^^ this.e[i];
        return result.text;
    }
}

// Global variables.
__gshared Hamming[] hams;
__gshared Hamming[NK] values;

nothrow @nogc static this() {
    // Slower than calloc if you don't use all the MAX_HAM items.
    //hams = new Hamming[MAX_HAM];

    auto ptr = cast(Hamming*)calloc(MAX_HAM, Hamming.sizeof);
    static const err = new Error("Not enough memory.");
    if (!ptr)
        throw err;
    hams = ptr[0 .. MAX_HAM];

    foreach (immutable i, ref v; values) {
        v.e[i] = 1;
        v.v = Hamming.inc[i];
    }
}


ref Hamming getHam(in size_t n) nothrow @nogc
in {
    assert(n <= MAX_HAM);
} body {
    // Most of the time v can be just incremented, but eventually
    // floating point precision will bite us, so better recalculate.
    __gshared static size_t[NK] idx;
    __gshared static int n_hams;

    for (; n_hams < n; n_hams++) {
        {
            // Find the index of the minimum v.
            size_t ni = 0;
            foreach (immutable i; 1 .. NK)
                if (values[i].v < values[ni].v)
                    ni = i;

            hams[n_hams] = values[ni];
            hams[n_hams].update;
        }

        foreach (immutable i; 0 .. NK)
            if (values[i] == hams[n_hams]) {
                values[i] = hams[idx[i]];
                idx[i]++;
                values[i].e[i]++;
                values[i].update;
            }
    }

    return hams[n - 2];
}


void main() {
    foreach (immutable n; [1691, 10 ^^ 6, MAX_HAM])
        writefln("%8d: %s", n, n.getHam);
}

The output is similar to the second C version. Runtime is about 0.11 seconds if MAX_HAM = 1_000_000 (as the task requires), and 0.90 seconds if MAX_HAM = 10_000_000.

Alternative Version 3

This version is similar to the precedent, but frees unused values. It's a little slower than the precedent version, but it uses much less RAM, so it allows to compute the result for larger n.

import std.stdio: writefln;
import std.bigint: BigInt;
import std.conv: text;
import std.algorithm: map;
import std.array: array;
import core.stdc.stdlib: malloc, calloc, free;
import std.math: log; // ^^

// Number of factors.
enum NK = 3;

__gshared immutable int[NK] primes = [2, 3, 5];
__gshared immutable double[NK] lnPrimes = primes[].map!log.array;

/// K-smooth numbers (stored as their exponents of each factor).

struct Hamming {
    double ln; // Log of the number.
    ushort[NK] e; // Exponents of each factor.
    Hamming* next;
    size_t n;

    // Recompute the logarithm from the exponents.
    void recalculate() pure nothrow @safe @nogc {
        this.ln = 0.0;
        foreach (immutable i, immutable ei; this.e)
            this.ln += lnPrimes[i] * ei;
    }

    string toString() const {
        BigInt result = 1;
        foreach (immutable i, immutable f; primes)
            result *= f.BigInt ^^ this.e[i];
        return result.text;
    }
}

Hamming getHam(in size_t n) nothrow @nogc
in {
    assert(n && n != size_t.max);
} body {
    static struct Candidate {
        typeof(Hamming.ln) ln;
        typeof(Hamming.e) e;

        void increment(in size_t n) pure nothrow @safe @nogc {
            e[n] += 1;
            ln += lnPrimes[n];
        }

        bool opEquals(T)(in ref T y) const pure nothrow @safe @nogc {
            // return this.e == y.e; // Slow.
            return !((this.e[0] ^ y.e[0]) |
                     (this.e[1] ^ y.e[1]) |
                     (this.e[2] ^ y.e[2]));
        }

        int opCmp(T)(in ref T y) const pure nothrow @safe @nogc {
            return (ln > y.ln) ? 1 : (ln < y.ln ? -1 : 0);
        }
    }

    static struct HammingIterator { // Not a Range.
        Candidate cand;
        Hamming* base;
        size_t primeIdx;

        this(in size_t i, Hamming* b) pure nothrow @safe @nogc {
            primeIdx = i;
            base = b;
            cand.e = base.e;
            cand.ln = base.ln;
            cand.increment(primeIdx);
        }

        void next() pure nothrow @safe @nogc {
            base = base.next;
            cand.e = base.e;
            cand.ln = base.ln;
            cand.increment(primeIdx);
        }
    }

    HammingIterator[NK] its;
    Hamming* head = cast(Hamming*)calloc(Hamming.sizeof, 1);
    Hamming* freeList, cur = head;
    Candidate next;

    foreach (immutable i, ref it; its)
        it = HammingIterator(i, cur);

    for (size_t i = cur.n = 1; i < n; ) {
        auto leastReferenced = size_t.max;
        next.ln = double.max;
        foreach (ref it; its) {
            if (it.cand == *cur)
                it.next;
            if (it.base.n < leastReferenced)
                leastReferenced = it.base.n;
            if (it.cand < next)
                next = it.cand;
        }

        // Collect unferenced numbers.
        while (head.n < leastReferenced) {
            auto tmp = head;
            head = head.next;
            tmp.next = freeList;
            freeList = tmp;
        }

        if (!freeList) {
            cur.next = cast(Hamming*)malloc(Hamming.sizeof);
        } else {
            cur.next = freeList;
            freeList = freeList.next;
        }

        cur = cur.next;
        version (fastmath) {
            cur.ln = next.ln;
            cur.e = next.e;
        } else {
            cur.e = next.e;
            cur.recalculate; // Prevent FP error accumulation.
        }

        cur.n = i++;
        cur.next = null;
    }

    auto result = *cur;
    version (leak) {}
    else {
        while (head) {
            auto tmp = head;
            head = head.next;
            tmp.free;
        }

        while (freeList) {
            auto tmp = freeList;
            freeList = freeList.next;
            tmp.free;
        }
    }

    return result;
}

void main() {
    foreach (immutable n; [1691, 10 ^^ 6, 10_000_000])
        writefln("%8d: %s", n, n.getHam);
}

The output is the same as the second alternative version.

=={{Header|Dart}}==

In order to produce reasonable ranges of Hamming numbers, one needs the BigInt type, but processing of many BigInt's in generating a sequence slows the code; for that reason the following code records the determined values as a combination of an approximation of the log base two value and the triple of the powers of two, three and five, only generating the final output values as BigInt's as required:

import 'dart:math';

final lb2of2 = 1.0;
final lb2of3 = log(3.0) / log(2.0);
final lb2of5 = log(5.0) / log(2.0);

class Trival {
  final double log2;
  final int twos;
  final int threes;
  final int fives;
  Trival mul2() {
    return Trival(this.log2 + lb2of2, this.twos + 1, this.threes, this.fives);
  }
  Trival mul3() {
    return Trival(this.log2 + lb2of3, this.twos, this.threes + 1, this.fives);
  }
  Trival mul5() {
    return Trival(this.log2 + lb2of5, this.twos, this.threes, this.fives + 1);
  }
  @override String toString() {
    return this.log2.toString() + " "
      + this.twos.toString() + " "
      + this.threes.toString() + " "
      + this.fives.toString();
  }
  const Trival(this.log2, this.twos, this.threes, this.fives);
}

Iterable<Trival> makeHammings() sync* {
  var one = Trival(0.0, 0, 0, 0);
  yield(one);
  var s532 = one.mul2();
  var mrg = one.mul3();
  var s53 = one.mul3().mul3(); // equivalent to 9 for advance step
  var s5 = one.mul5();
  var i = -1; var j = -1;
  List<Trival> h = [];
  List<Trival> m = [];
  Trival rslt;
  while (true) {
    if (s532.log2 < mrg.log2) {
      rslt = s532; h.add(s532); ++i; s532 = h[i].mul2();
    } else {
      rslt = mrg; h.add(mrg);
      if (s53.log2 < s5.log2) {
        mrg = s53; m.add(s53); ++j; s53 = m[j].mul3();
      } else {
        mrg = s5; m.add(s5); s5 = s5.mul5();
      }
      if (j > (m.length >> 1)) {m.removeRange(0, j); j = 0; }
    }
    if (i > (h.length >> 1)) {h.removeRange(0, i); i = 0; }
    yield(rslt);
  }
}

BigInt trival2Int(Trival tv) {
  return BigInt.from(2).pow(tv.twos)
           * BigInt.from(3).pow(tv.threes)
           * BigInt.from(5).pow(tv.fives);
}

void main() {
  final numhams = 1000000000000;
  var hamseqstr = "The first 20 Hamming numbers are:  ( ";
  makeHammings().take(20)
      .forEach((h) => hamseqstr += trival2BigInt(h).toString() + " ");
  print(hamseqstr + ")");
  var nthhamseqstr = "The first 20 Hamming numbers are:  ( ";
  for (var i = 1; i <= 20; ++i) {
    nthhamseqstr += trival2BigInt(nthHamming(i)).toString() + " ";
  }
  print(nthhamseqstr + ")");
  final strt = DateTime.now().millisecondsSinceEpoch;
  final answr = makeHammings().skip(999999).first;
  final elpsd = DateTime.now().millisecondsSinceEpoch - strt;
  print("The ${numhams}th Hamming number is:  $answr");
  print("in full as:  ${trival2BigInt(answr)}");
  print("This test took $elpsd milliseconds.");
}

{{Output}}

The first 20 Hamming numbers are:  ( 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 )
The 1000000th Hamming number is:  278.096635606686 55 47 64
in full as:  519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This test took 311 milliseconds.

Due to using a mutable extendable List (Array) and mutation, the above generator is reasonably fast, and as well has the feature that List memory is recovered as it is no longer required, with a considerable saving in both execution speed and memory requirement.

'''Alternate extremely fast version using an "error band"'''

Although not a Hamming sequence generator, the following code uses the known characteristics of the distribution of Hamming numbers to just scan through to find all possibilities in a relatively narrow "error band" which then can be sorted based on the log base two approximation and the nth element determined inside that band; it has a huge advantage that memory requirements drop to O(n^(1/3)) and asymptotic execution complexity drops from O(n) to O(n^(2/3)) for an extremely fast execution speed (thanks to WillNess for the start of this algorithm as referenced in the Haskell section): {{translated from|Haskell}}

import 'dart:math';

final lb2of2 = 1.0;
final lb2of3 = log(3.0) / log(2.0);
final lb2of5 = log(5.0) / log(2.0);

class Trival {
  final double log2;
  final int twos;
  final int threes;
  final int fives;
  Trival mul2() {
    return Trival(this.log2 + lb2of2, this.twos + 1, this.threes, this.fives);
  }
  Trival mul3() {
    return Trival(this.log2 + lb2of3, this.twos, this.threes + 1, this.fives);
  }
  Trival mul5() {
    return Trival(this.log2 + lb2of5, this.twos, this.threes, this.fives + 1);
  }
  @override String toString() {
    return this.log2.toString() + " "
      + this.twos.toString() + " "
      + this.threes.toString() + " "
      + this.fives.toString();
  }
  const Trival(this.log2, this.twos, this.threes, this.fives);
}

BigInt trival2BigInt(Trival tv) {
  return BigInt.from(2).pow(tv.twos)
           * BigInt.from(3).pow(tv.threes)
           * BigInt.from(5).pow(tv.fives);
}

Trival nthHamming(int n) {
  if (n < 1) throw Exception("nthHamming:  argument must be higher than 0!!!");
  if (n < 7) {
    if (n & (n - 1) == 0) {
      final bts = n.bitLength - 1;
      return Trival(bts.toDouble(), bts, 0, 0);
    }
    switch (n) {
      case 3: return Trival(lb2of3, 0, 1, 0);
      case 5: return Trival(lb2of5, 0, 0, 1);
      case 6: return Trival(lb2of2 + lb2of3, 1, 1, 0);
    }
  }
  final fctr = 6.0 * lb2of3 * lb2of5;
  final crctn = log(sqrt(30.0)) / log(2.0);
  final lb2est = pow(fctr * n.toDouble(), 1.0/3.0) - crctn;
  final lb2rng = 2.0/lb2est;
  final lb2hi = lb2est + 1.0/lb2est;
  List<Trival> ebnd = [];
  var cnt = 0;
  for (var k = 0; k < (lb2hi / lb2of5).ceil(); ++k) {
    final lb2p = lb2hi - k * lb2of5;
    for (var j = 0; j < (lb2p / lb2of3).ceil(); ++j) {
      final lb2q = lb2p - j * lb2of3;
      final i = lb2q.floor(); final lb2frac = lb2q - i;
      cnt += i + 1;
      if (lb2frac <= lb2rng) {
        final lb2v = i * lb2of2 + j * lb2of3 + k * lb2of5;
        ebnd.add(Trival(lb2v, i, j, k));
      }
    }
  }
  ebnd.sort((a, b) => b.log2.compareTo(a.log2)); // descending order
  final ndx = cnt - n;
  if (ndx < 0) throw Exception("nthHamming:  not enough triples generated!!!");
  if (ndx >= ebnd.length) throw Exception("nthHamming:  error band is too narrow!!!");
  return ebnd[ndx];
}

void main() {
  final numhams = 1000000;
  var nthhamseqstr = "The first 20 Hamming numbers are:  ( ";
  for (var i = 1; i <= 20; ++i) {
    nthhamseqstr += trival2BigInt(nthHamming(i)).toString() + " ";
  }
  print(nthhamseqstr + ")");
  final strt = DateTime.now().millisecondsSinceEpoch;
  final answr = nthHamming(numhams);
  final elpsd = DateTime.now().millisecondsSinceEpoch - strt0;
  print("The ${numhams}th Hamming number is:  $answr");
  print("in full as:  ${trival2BigInt(answr)}");
  print("This test took $elpsd milliseconds.");
}

The output from the above code is the same as the above version but it is so fast that the time to find the millionth Hamming number is too small to be measured other than the Dart VM JIT time. It can find the billionth prime in a fraction of a second and the trillionth prime in seconds.

'''Increasing the range above 1e13 by using a BigInt log base two representation'''

For arguments higher than about 1e13, the precision of the Double log base two approximations used above is not adequate to do an accurate sort, but the algorithm continues to work (although perhaps slightly slower) by changing the code to use BigInt log base two representations as follows:

import 'dart:math';

final biglb2of2 = BigInt.from(1) << 100; // 100 bit representations...
final biglb2of3 = (BigInt.from(1784509131911002) << 50) + BigInt.from(134114660393120);
final biglb2of5 = (BigInt.from(2614258625728952) << 50) + BigInt.from(773584997695443);

class BigTrival {
  final BigInt log2;
  final int twos;
  final int threes;
  final int fives;
  @override String toString() {
    return this.log2.toString() + " "
      + this.twos.toString() + " "
      + this.threes.toString() + " "
      + this.fives.toString();
  }
  const BigTrival(this.log2, this.twos, this.threes, this.fives);
}

BigInt bigtrival2BigInt(BigTrival tv) {
  return BigInt.from(2).pow(tv.twos)
           * BigInt.from(3).pow(tv.threes)
           * BigInt.from(5).pow(tv.fives);
}

BigTrival nthHamming(int n) {
  if (n < 1) throw Exception("nthHamming:  argument must be higher than 0!!!");
  if (n < 7) {
    if (n & (n - 1) == 0) {
      final bts = n.bitLength - 1;
      return BigTrival(BigInt.from(bts) << 100, bts, 0, 0);
    }
    switch (n) {
      case 3: return BigTrival(biglb2of3, 0, 1, 0);
      case 5: return BigTrival(biglb2of5, 0, 0, 1);
      case 6: return BigTrival(biglb2of2 + biglb2of3, 1, 1, 0);
    }
  }
  final fctr = lb2of3 * lb2of5 * 6;
  final crctn = log(sqrt(30.0)) / log(2.0);
  final lb2est = pow(fctr * n.toDouble(), 1.0/3.0) - crctn;
  final lb2rng = 2.0/lb2est;
  final lb2hi = lb2est + 1.0/lb2est;
  List<BigTrival> ebnd = [];
  var cnt = 0;
  for (var k = 0; k < (lb2hi / lb2of5).ceil(); ++k) {
    final lb2p = lb2hi - k * lb2of5;
    for (var j = 0; j < (lb2p / lb2of3).ceil(); ++j) {
      final lb2q = lb2p - j * lb2of3;
      final i = lb2q.floor(); final lb2frac = lb2q - i;
      cnt += i + 1;
      if (lb2frac <= lb2rng) {
//        final lb2v = i * lb2of2 + j * lb2of3 + k * lb2of5;
//        ebnd.add(Trival(lb2v, i, j, k));
        final lb2v = BigInt.from(i) * biglb2of2
                        + BigInt.from(j) * biglb2of3
                        + BigInt.from(k) * biglb2of5;
        ebnd.add(BigTrival(lb2v, i, j, k));
      }
    }
  }
  ebnd.sort((a, b) => b.log2.compareTo(a.log2)); // descending order
  final ndx = cnt - n;
  if (ndx < 0) throw Exception("nthHamming:  not enough triples generated!!!");
  if (ndx >= ebnd.length) throw Exception("nthHamming:  error band is too narrow!!!");
  return ebnd[ndx];
}

void main() {
  final numhams = 1000000000;
  var nthhamseqstr = "The first 20 Hamming numbers are:  ( ";
  for (var i = 1; i <= 20; ++i) {
    nthhamseqstr += bigtrival2BigInt(nthHamming(i)).toString() + " ";
  }
  print(nthhamseqstr + ")");
  final strt = DateTime.now().millisecondsSinceEpoch;
  final answr = nthHamming(numhams);
  final elpsd = DateTime.now().millisecondsSinceEpoch - strt;
  print("The ${numhams}th Hamming number is:  $answr");
  print("in full as:  ${bigtrival2BigInt(answr)}");
  print("This test took $elpsd milliseconds.");
}

With these changes, the algorithm can find the 1e19'th prime in the order af days depending on the CPU used.

DCL

$ limit = p1
$
$ n = 0
$ h_'n = 1
$ x2 = 2
$ x3 = 3
$ x5 = 5
$ i = 0
$ j = 0
$ k = 0
$
$ n = 1
$ loop:
$  x = x2
$  if x3 .lt. x then $ x = x3
$  if x5 .lt. x then $ x = x5
$  h_'n = x
$  if x2 .eq. h_'n
$  then
$   i = i + 1
$   x2 = 2 * h_'i
$  endif
$  if x3 .eq. h_'n
$  then
$   j = j + 1
$   x3 = 3 * h_'j
$  endif
$  if x5 .eq. h_'n
$  then
$   k = k + 1
$   x5 = 5 * h_'k
$  endif
$  n = n + 1
$  if n .le. limit then $ goto loop
$
$ i = 0
$ loop2:
$  write sys$output h_'i
$  i = i + 1
$  if i .lt. 20 then $ goto loop2
$
$ n = limit - 1
$ write sys$output h_'n

{{out}}

Here's the output;

$ @hamming 1691
1
2
3
4
5
6
8
9
10
12
15
16
18
20
24
25
27
30
32
36
2125764000

Eiffel

note
	description    : "Initial part, in order, of the sequence of Hamming numbers"
	math           : "[
			   Hamming numbers, also known as regular numbers and 5-smooth numbers, are natural integers
			   that have 2, 3 and 5 as their only prime factors.
		         ]"
	computer_arithmetic :
	                 "[
			   This version avoids integer overflow and stops at the last representable number in the sequence.
		         ]"
	output         : "[
    			   Per requirements of the RosettaCode example, execution will produce items of indexes 1 to 20 and 1691.
    			   The algorithm (procedure `hamming') is more general and will produce the first `n' Hamming numbers
    			   for any `n'.
    			  ]"
	source         : "This problem was posed in Edsger W. Dijkstra, A Discipline of Programming, Prentice Hall, 1978"
	date           : "8 August 2012"
	authors        : "Bertrand Meyer", "Emmanuel Stapf"
	revision       : "1.0"
	libraries      : "Relies on SORTED_TWO_WAY_LIST from EiffelBase"
	implementation : "[
			   Using SORTED_TWO_WAY_LIST provides an elegant illustration of how to implement
			   a lazy scheme in Eiffel through the use of object-oriented data structures.
			 ]"
	warning        : "[
			   The formatting (<lang>) specifications for Eiffel in RosettaCode are slightly obsolete:
			   `note' and other newer keywords not supported, red color for manifest strings.
			   This should be fixed soon.
		         ]"

class
	APPLICATION

create
	make

feature {NONE} -- Initialization

	make
			-- Print first 20 Hamming numbers, in order, and the 1691-st one.
		local
			Hammings: like hamming
				-- List of Hamming numbers, up to 1691-st one.
		do
			Hammings := hamming (1691)
			across 1 |..| 20 as i loop
				io.put_natural (Hammings.i_th (i.item)); io.put_string (" ")
			end
			io.put_new_line; io.put_natural (Hammings.i_th (1691)); io.put_new_line
		end

feature -- Basic operations

	hamming (n: INTEGER): ARRAYED_LIST [NATURAL]
			-- First `n' elements (in order) of the Hamming sequence,
			-- or as many of them as will not produce overflow.
		local
			sl: SORTED_TWO_WAY_LIST [NATURAL]
			overflow: BOOLEAN
			first, next: NATURAL
		do
			create Result.make (n); create sl.make
			sl.extend (1); sl.start
			across 1 |..| n as i invariant
				-- "The numbers output so far are the first `i' - 1 Hamming numbers, in order".
				-- "Result.first is the `i'-th Hamming number."
			until sl.is_empty loop
				first := sl.first; sl.start
				Result.extend (first); sl.remove
				across << 2, 3, 5 >> as multiplier loop
					next := multiplier.item * first
					overflow := overflow or next <= first
					if not overflow and then not sl.has (next) then sl.extend (next) end
				end
			end
		end
end

{{out}}

1
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000

Elixir

defmodule Hamming do
  def generater do
    queues = [{2, queue}, {3, queue}, {5, queue}]
    Stream.unfold({1, queues}, fn {n, q} -> next(n, q) end)
  end

  defp next(n, queues) do
    queues = Enum.map(queues, fn {m, queue} -> {m, push(queue, m*n)} end)
    min    = Enum.map(queues, fn {_, queue} -> top(queue) end) |> Enum.min
    queues = Enum.map(queues, fn {m, queue} ->
               {m, (if min==top(queue), do: erase_top(queue), else: queue)}
             end)
    {n, {min, queues}}
  end

  defp queue, do: {[], []}

  defp push({input, output}, term), do: {[term | input], output}

  defp top({input, []}), do: List.last(input)
  defp top({_, [h|_]}), do: h

  defp erase_top({input, []}), do: erase_top({[], Enum.reverse(input)})
  defp erase_top({input, [_|t]}), do: {input, t}
end

IO.puts "first twenty Hamming numbers:"
IO.inspect Hamming.generater |> Enum.take(20)
IO.puts "1691st Hamming number:"
IO.puts Hamming.generater |> Enum.take(1691) |> List.last
IO.puts "one millionth Hamming number:"
IO.puts Hamming.generater |> Enum.take(1_000_000) |> List.last

{{out}}

first twenty Hamming numbers:
[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
1691st Hamming number:
2125764000
one millionth Hamming number:
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

ERRE

For bigger numbers, you have to use an external program, like MULPREC.R

PROGRAM HAMMING

!$DOUBLE

DIM H[2000]

PROCEDURE HAMMING(L%->RES)
      LOCAL I%,J%,K%,N%,M,X2,X3,X5
      H[0]=1
      X2=2  X3=3  X5=5
      FOR N%=1 TO L%-1 DO
        M=X2
        IF M>X3 THEN M=X3 END IF
        IF M>X5 THEN M=X5 END IF
        H[N%]=M
        IF M=X2 THEN I%+=1  X2=2*H[I%]  END IF
        IF M=X3 THEN J%+=1  X3=3*H[J%]  END IF
        IF M=X5 THEN K%+=1  X5=5*H[K%]  END IF
      END FOR
      RES=H[L%-1]
END PROCEDURE

BEGIN
      FOR H%=1 TO 20 DO
        HAMMING(H%->RES)
        PRINT("H(";H%;")=";RES)
      END FOR
      HAMMING(1691->RES)
      PRINT("H(1691)=";RES)
END PROGRAM

{{out}}

H( 1 )= 1
H( 2 )= 2
H( 3 )= 3
H( 4 )= 4
H( 5 )= 5
H( 6 )= 6
H( 7 )= 8
H( 8 )= 9
H( 9 )= 10
H( 10 )= 12
H( 11 )= 15
H( 12 )= 16
H( 13 )= 18
H( 14 )= 20
H( 15 )= 24
H( 16 )= 25
H( 17 )= 27
H( 18 )= 30
H( 19 )= 32
H( 20 )= 36
H(1691)= 2125764000

F#

This version implements Dijkstra's merge solution, so is closely related to the Haskell classic version.

type LazyList<'a> = Cons of 'a * Lazy<LazyList<'a>> // ': fix colouring

let rec hammings() =
  let rec (-|-) (Cons(x, nxf) as xs) (Cons(y, nyf) as ys) =
    if x < y then Cons(x, lazy(nxf.Value -|- ys))
    elif x > y then Cons(y, lazy(xs -|- nyf.Value))
    else Cons(x, lazy(nxf.Value -|- nyf.Value))
  let rec inf_map f (Cons(x, nxf)) =
    Cons(f x, lazy(inf_map f nxf.Value))
  Cons(1I, lazy(let x = inf_map ((*) 2I) hamming
                let y = inf_map ((*) 3I) hamming
                let z = inf_map ((*) 5I) hamming
                x -|- y -|- z))

// testing...
[<EntryPoint>]
let main args =
  let rec iterLazyListFor f n (Cons(v, rf)) =
    if n > 0 then f v; iterLazyListFor f (n - 1) rf.Value
  let rec nthLazyList n ((Cons(v, rf)) as ll) =
    if n <= 1 then v else nthLazyList (n - 1) rf.Value
  printf "( "; iterLazyListFor (printf "%A ") 20 (hammings()); printfn ")"
  printfn "%A" (hammings() |> nthLazyList 1691)
  printfn "%A" (hammings() |> nthLazyList 1000000)
  0

The above code memory residency is quite high as it holds the entire lazy sequence in memory due to the reference preventing garbage collection as the sequence is consumed,

The following code reduces that high memory residency by making the routine a function and using internal local stream references for the intermediate streams so that they can be collected as the stream is consumed as long as no reference is held to the main results stream (which is not in the sample test functions); it also avoids duplication of factors by successively building up streams and further reduces memory use by ordering of the streams so that the least dense are determined first: {{trans|Haskell}}

type LazyList<'a> = Cons of 'a * Lazy<LazyList<'a>> // ': fix colouring

let hamming() =
  let rec merge (Cons(x, f) as xs) (Cons(y, g) as ys) =
    if x < y then Cons(x, lazy(merge (f.Force()) ys))
    else Cons(y, lazy(merge xs (g.Force())))
  let rec smult m (Cons(x, rxs)) =
    Cons(m * x, lazy(smult m (rxs.Force())))
  let rec first = smult 5I (Cons(1I, lazy first))
  let u s n =
    let rec r = merge s (smult n (Cons(1I, lazy r))) in r
  (Cons(1I, lazy(Seq.fold u first [| 3I; 2I |])))

The above code can by used just by substituting it for the "hamming" binding and substituting "hamming()" for "hamming" in the main testing function calls (three places).

Both codes output the same results as follows but the second is over three times faster: {{output}}

( 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 )
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Both codes are over 10 times slower as compared to Haskell (or Kotline or Scala or Clojure) when all are written in exactly the same style, perhaps due in some small degree to the BigInteger implementation being much slower for these operations than GMP and the JVM's implementation of BigInteger, but is also about twice as slow as the same algorithm written in C#. This seems to show up one of F#'s "warts" in that the implementation of closure functions enclosing free variables necessary to implement lazy as in the LazyList seem to be particularly inefficient even as compared to C#'s lambda functions (which are also less efficient than the others mentioned).

Fast somewhat imperative sequence version using logarithms

Since the above pure functional approach isn't very efficient, a more imperative approach using "growable" arrays which are "drained" of unnecessary older values in blocks once the back pointer indices are advanced is used in the following code. The code also implements an algorithm to avoid duplicate calculations and thus does the same number of operations as the above code but faster due to using integer and floating point operations rather an BigInteger ones. Due to the "draining" the memory use is the same as the above by a constant factor. However, note that other than the contents of these arrays, pure functional code using immutable values is used. Note that the implementation of IEnumerable using sequences in F# is also not very efficient and a "roll-your-own" IEnumerable implementation would likely be two or three times faster:

open System.Numerics

let hammingsLog() = // imperative arrays, eliminates the BigInteger operations...
  let lb3 = 1.5849625007211561814537389439478 // Math.Log(3) / Math.Log(2);
  let lb5 = 2.3219280948873623478703194294894 // Math.Log(5) / Math.Log(2);
  let inline mul2 (lg, x2, x3, x5) = (lg + 1.0, x2 + 1u, x3, x5)
  let inline mul3 (lg, x2, x3, x5) = (lg + lb3, x2, x3 + 1u, x5)
  let inline mul5 (lg, x2, x3, x5) = (lg + lb5, x2, x3, x5 + 1u)
  let one = (0.0, 0u, 0u, 0u)
  let s532, mrg = one |> mul2, one |> mul3
  let s53 = one |> mul3 |> mul3 // equivalent to 9 for advance step
  let s5 = one |> mul5
  let h = ResizeArray<_>()
  let m = ResizeArray<_>()
  let inline drplg (_, x2, x3, x5) = (x2, x3, x5)
  let inline nontriv() = Seq.unfold (fun (i, j, s532, mrg, s53, s5) -> // THIS STILL IS PATTERN MATCHING!!!!!
    let inline (<) (lga, _, _, _) (lgb, _, _, _) = lga < lgb
    let nv, ni, nj, ns532, nmrg, ns53, ns5 =
      if s532 < mrg then h.Add(s532)
                         s532, i + 1, j, h.[i] |> mul2, mrg, s53, s5
      else if s53 < s5 then h.Add(mrg); m.Add(s53)
                            mrg, i, j + 1, s532, s53, m.[j] |> mul3, s5
           else h.Add(mrg); m.Add(s5)
                mrg, i, j, s532, s5, s53, s5 |> mul5
    let nj = if nj >= m.Capacity / 2 then m.RemoveRange(0, nj); 0 else nj
    let ni = if ni >= h.Capacity / 2 then h.RemoveRange(0, ni); 0 else ni
    Some (drplg nv, (ni, nj, ns532, nmrg, ns53, ns5))) (0,0,s532,mrg,s53,s5)
  seq { yield drplg one; yield! nontriv() } // this is very slow

let trival (x2, x3, x5) = // convert trival to BigInteger
  let rec loop n mlt rslt =
    if n <= 0u then rslt
    else loop (n - 1u) mlt (mlt * rslt)
  loop x2 2I 1I |> loop x3 3I |> loop x5 5I

[<EntryPoint>]
let main argv =
  printf "( "; hammingsLog() |> Seq take 20 |> Seq.iter (printf "%A " << trival); printfn ")"
  printfn "%A" (hammingsLog() |> Seq.nth (1691 - 1))
  let strt = System.DateTime.Now.Ticks

  let rslt = (hammingsLog()) |> Seq.nth (1000000 - 1)

  let stop = System.DateTime.Now.Ticks

  printfn "%A" (rslt |> trival)
  printfn "\r\nFound this last up to %A in %A milliseconds." topNum ((stop - strt) / 10000L)

  printf "\r\nPress any key to exit:"
  System.Console.ReadKey(true) |> ignore
  printfn ""
  0 // return an integer exit code

The above code produces the same outputs as above, but takes only about 300 milliseconds rather than over three seconds.

===Extremely fast non-enumerating version sorting values in error band===

If one is willing to forego sequences and just calculate the nth Hamming number, then some reading on the relationship between the size of numbers to the sequence numbers is helpful (Wikipedia: regular number). One finds that there is a very distinct relationship and that it quite quickly reduces to quite a small error band proportional to the log of the output value for larger ranges. Thus, the following code just scans for logarithmic representations to insert into a sequence for this top error band and extracts the correct nth representation from that band. It reduces time complexity to O(n^(2/3)) from O(n) for the sequence versions, but even more amazingly, reduces memory requirements to O(n^(1/3)) from O(n^(2/3)) and thus makes it possible to calculate very large values in the sequence on common personal computers. The code is as follows: {{trans|Haskell}}

let nthHamming n =
  if n < 1UL then failwith "nthHamming; argument must be > 0!"
  if n < 2UL then 0u, 0u, 0u else // trivial case for first value of one
  let lb3 = 1.5849625007211561814537389439478 // Math.Log(3) / Math.Log(2);
  let lb5 = 2.3219280948873623478703194294894 // Math.Log(5) / Math.Log(2);
  let fctr = 6.0 * lb3 * lb5
  let crctn = 2.4534452978042592646620291867186 // Math.Log(Math.sqrt(30.0)) / Math.Log(2.0)
  let lbest = (fctr * double n) ** (1.0/3.0) - crctn // from WP formula
  let lbhi = lbest + 1.0 / lbest
  let lblo = 2.0 * lbest - lbhi // upper and lower bound of upper "band"
  let klmt = uint32 (lbhi / lb5)
  let rec loopk k kcnt kbnd =
    if k > klmt then kcnt, kbnd else
      let p = lbhi - double k * lb5
      let jlmt = uint32 (p / lb3)
      let rec loopj j jcnt jbnd =
        if j > jlmt then loopk (k + 1u) jcnt jbnd else
          let q = p - double j * lb3
          let i = uint32 q
          let lg = lbhi - q + double i // current log 2 value (estimated)
          let nbnd = if lg >= lblo then (lg, (uint32 i, j, k)) :: jbnd else jbnd
          loopj (j + 1u) (jcnt + uint64 i + 1UL) nbnd in loopj 0u kcnt kbnd
  let count, bnd = loopk 0u 0UL [] // 64-bit value so doesn't overflow
  if n > count then failwith "nthHamming:  band high estimate is too low!"
  let ndx = int (count - n)
  if ndx >= bnd.Length then failwith "NthHamming.findNth:  band low estimate is too high!"
  let sbnd = bnd |> List.sortBy (fun (lg, _) -> -lg) // sort in decending order
  let _, rslt = sbnd.[ndx]
  rslt

[<EntryPoint>]
let main argv =
  let topNum = 1000000UL
  printf "( "; {1..20} |> Seq.iter (printf "%A " << trival << nthHamming << uint64); printfn ")"
  printfn "%A" (nthHamming 1691UL |> trival)
  let rslt = nthHammingx topNum
  let strt = System.DateTime.Now.Ticks

  let rslt = nthHamming topNum

  let stop = System.DateTime.Now.Ticks

  let x2, x3, x5 = rslt
  printfn "2**%A times 3**%A times 5**%A" x2 x3 x5
  let lgrthm = log10 2.0 * (double x2 + (double x3 * log 3.0 + double x5 * log 5.0) / log 2.0)
  let exp = floor lgrthm |> int
  let mntsa = 10.0 ** (lgrthm - double exp)
  printfn "Approximately %AE+%A" mntsa exp
  let s = trival rslt |> string
  let lngth = s.Length
  printfn "Digits:  %A" lngth
  if lngth <= 10000 then
    {0..100..lngth-1}
      |> Seq.iter (fun i ->
        printfn "%s" (s.Substring(i, if i + 100 < lngth then 100 else lngth - i)))

  printfn "\r\nFound this last up to %A in %A milliseconds." topNum ((stop - strt) / 10000L)

  printf "\r\nPress any key to exit:"
  System.Console.ReadKey(true) |> ignore
  printfn ""
  0 // return an integer exit code

{{output}}

( 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 )
2125764000
2**55u times 3**47u times 5**64u
Approximately 5.193127804E+83
Digits:  84
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Found this last up to 1000000UL in 0L milliseconds.

Even though the above code is implemented in a completely functional style using immutable bindings and (non-lazy) lists (without closures), it is about as fast as implementations in the fastest of languages. It is faster than the Haskell version due to that version using lazy lists with the overhead of creating the requisite "thunks".

It takes too short a time to be measured to calculate the millionth Hamming number, the billionth number in the sequence can be calculated in just about 15 milliseconds, the trillionth in about one second, the thousand trillionth in about a hundred seconds, and it should be possible to calculate the 10^19th value in less than a day (untested) on common personal computers. The (2^64 - 1)th value (18446744073709551615) cannot be calculated due to a slight overflow problem as it approaches that limit.

'''Enhancement to by able to find Hamming numbers beyond the ten trillionth one'''

Due to the limited 53-bit mantissa of 64-bit double floating piint numbers, the above code can't properly sort the error band for input arguments somewhere above 10**13; the following code makes the sort accurate by using a multi-precision logarithm representation of sufficient precision so that the sort is accurate for arguments well beyond the uint64 input argument range, at about a doubling in cost in execution speed: {{trans|Haskell}}

let nthHamming n =
  if n < 1UL then failwith "nthHamming:  argument must be > 0!"
  if n < 2UL then 0u, 0u, 0u else // trivial case for first value of one
  let lb3 = 1.5849625007211561814537389439478 // Math.Log(3) / Math.Log(2);
  let lb5 = 2.3219280948873623478703194294894 // Math.Log(5) / Math.Log(2);
  let fctr = 6.0 * lb3 * lb5
  let crctn = 2.4534452978042592646620291867186 // Math.Log(Math.sqrt(30.0)) / Math.Log(2.0)
  let lbest = (fctr * double n) ** (1.0/3.0) - crctn // from WP formula
  let lbhi = lbest + 1.0/lbest
  let lblo = 2.0 * lbest - lbhi // upper and lower bound of upper "band"
  let bglb2 = 1267650600228229401496703205376I
  let bglb3 = 2009178665378409109047848542368I
  let bglb5 = 2943393543170754072109742145491I
  let klmt = uint32 (lbhi / lb5)
  let rec loopk k kcnt kbnd =
    if k > klmt then kcnt, kbnd else
      let p = lbhi - double k * lb5
      let jlmt = uint32 (p / lb3)
      let rec loopj j jcnt jbnd =
        if j > jlmt then loopk (k + 1u) jcnt jbnd else
          let q = p - double j * lb3
          let i = uint32 q
          let lg = lbhi - q + double i // current log 2 value (estimated)
          let nbnd = if lg < lblo then jbnd else
                       let bglg = bglb2 * bigint i + bglb3 * bigint j + bglb5 * bigint k in
                       (bglg, (uint32 i, j, k)) :: jbnd
          loopj (j + 1u) (jcnt + uint64 i + 1UL) nbnd in loopj 0u kcnt kbnd
  let count, bnd = loopk 0u 0UL [] // 64-bit value so doesn't overflow
  if n > count then failwith "nthHamming:  band high estimate is too low!"
  let ndx = int (count - n)
  if ndx >= bnd.Length then failwith "NthHamming.findNth:  band low estimate is too high!"
  let sbnd = bnd |> List.sortBy (fun (lg, _) -> -lg) // sort in decending order
  let _, rslt = sbnd.[ndx]
  rslt