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{{task}} The [http://planetmath.org/properdivisor proper divisors] of a positive integer '''N''' are those numbers, other than '''N''' itself, that divide '''N''' without remainder.

For '''N''' > 1 they will always include 1, but for '''N''' == 1 there are no proper divisors.

;Examples: The proper divisors of 6 are 1, 2, and 3.

The proper divisors of 100 are 1, 2, 4, 5, 10, 20, 25, and 50.

# Find a number in the range 1 to 20,000 with the most proper divisors. Show the number and just the count of how many proper divisors it has.

Show all output here.

• [[Amicable pairs]]
• [[Abundant, deficient and perfect number classifications]]
• [[Aliquot sequence classifications]]
• [[Factors of an integer]]
• [[Prime decomposition]]

## 360 Assembly

{{trans|Rexx}} This program uses two ASSIST macros (XDECO, XPRNT) to keep the code as short as possible.

```*        Proper divisors           14/06/2016
PROPDIV  CSECT
USING  PROPDIV,R13        base register
B      72(R15)            skip savearea
DC     17F'0'             savearea
STM    R14,R12,12(R13)    prolog
ST     R13,4(R15)         "
ST     R15,8(R13)         "
LR     R13,R15            "
LA     R10,1              n=1
LOOPN1   C      R10,=F'10'         do n=1 to 10
BH     ELOOPN1
LR     R1,R10             n
BAL    R14,PDIV           pdiv(n)
ST     R0,NN              nn=pdiv(n)
MVC    PG,PGT             init buffer
LA     R11,PG             pgi=0
XDECO  R10,XDEC           edit n
MVC    0(3,R11),XDEC+9    output n
LA     R11,7(R11)         pgi=pgi+7
L      R1,NN              nn
XDECO  R1,XDEC            edit nn
MVC    0(3,R11),XDEC+9    output nn
LA     R11,20(R11)        pgi=pgi+20
LA     R5,1               i=1
LOOPNI   C      R5,NN              do i=1 to nn
BH     ELOOPNI
LR     R1,R5              i
SLA    R1,2               *4
L      R2,TDIV-4(R1)      tdiv(i)
XDECO  R2,XDEC            edit tdiv(i)
MVC    0(3,R11),XDEC+9    output tdiv(i)
LA     R11,3(R11)         pgi=pgi+3
LA     R5,1(R5)           i=i+1
B      LOOPNI
ELOOPNI  XPRNT  PG,80              print buffer
LA     R10,1(R10)         n=n+1
B      LOOPN1
ELOOPN1  SR     R0,R0              0
ST     R0,M               m=0
LA     R10,1              n=1
LOOPN2   C      R10,=F'20000'      do n=1 to 20000
BH     ELOOPN2
LR     R1,R10             n
BAL    R14,PDIV           nn=pdiv(n)
C      R0,M               if nn>m
BNH    NNNHM
ST     R10,II             ii=n
ST     R0,M               m=nn
NNNHM    LA     R10,1(R10)         n=n+1
B      LOOPN2
ELOOPN2  MVC    PG,PGR             init buffer
L      R1,II              ii
XDECO  R1,XDEC            edit ii
MVC    PG(5),XDEC+7       output ii
L      R1,M               m
XDECO  R1,XDEC            edit m
MVC    PG+9(4),XDEC+8     output m
XPRNT  PG,80              print buffer
L      R13,4(0,R13)       epilog
LM     R14,R12,12(R13)    "
XR     R15,R15            "
BR     R14                exit
*------- pdiv   --function(x)----->number of divisors---
PDIV     ST     R1,X               x
C      R1,=F'1'           if x=1
BNE    NOTONE
LA     R0,0               return(0)
BR     R14
NOTONE   LR     R4,R1              x
N      R4,=X'00000001'    mod(x,2)
LA     R4,1(R4)           +1
ST     R4,ODD             odd=mod(x,2)+1
LA     R8,1               ia=1
LA     R0,1               1
ST     R0,TDIV            tdiv(1)=1
SR     R9,R9              ib=0
L      R7,ODD             odd
LA     R7,1(R7)           j=odd+1
LOOPJ    LR     R5,R7              do j=odd+1 by odd
MR     R4,R7              j*j
C      R5,X               while j*j<x
BNL    ELOOPJ
L      R4,X               x
SRDA   R4,32              .
DR     R4,R7              /j
LTR    R4,R4              if mod(x,j)=0
BNZ    ITERJ
LA     R8,1(R8)           ia=ia+1
LR     R1,R8              ia
SLA    R1,2               *4 (F)
ST     R7,TDIV-4(R1)      tdiv(ia)=j
LA     R9,1(R9)           ib=ib+1
L      R4,X               x
SRDA   R4,32              .
DR     R4,R7              j
LR     R2,R9              ib
SLA    R2,2               *4 (F)
ST     R5,TDIVB-4(R2)     tdivb(ib)=x/j
ITERJ    A      R7,ODD             j=j+odd
B      LOOPJ
ELOOPJ   LR     R5,R7              j
MR     R4,R7              j*j
C      R5,X               if j*j=x
BNE    JTJNEX
LA     R8,1(R8)           ia=ia+1
LR     R1,R8              ia
SLA    R1,2               *4 (F)
ST     R7,TDIV-4(R1)      tdiv(ia)=j
JTJNEX   LA     R1,TDIV(R1)        @tdiv(ia+1)
LA     R2,TDIVB-4(R2)     @tdivb(ib)
LTR    R6,R9              do i=ib to 1 by -1
BZ     ELOOPI
LOOPI    MVC    0(4,R1),0(R2)      tdiv(ia)=tdivb(i)
LA     R8,1(R8)           ia=ia+1
LA     R1,4(R1)           r1+=4
SH     R2,=H'4'           r2-=4
BCT    R6,LOOPI           i=i-1
ELOOPI   LR     R0,R8              return(ia)
*        ----   ----------------------------------------
TDIV     DS     80F
TDIVB    DS     40F
M        DS     F
NN       DS     F
II       DS     F
X        DS     F
ODD      DS     F
PGT      DC     CL80'... has .. proper divisors:'
PGR      DC     CL80'..... has ... proper divisors.'
PG       DC     CL80' '
XDEC     DS     CL12
YREGS
END    PROPDIV
```

{{out}}

```
1 has  0 proper divisors:
2 has  1 proper divisors:  1
3 has  1 proper divisors:  1
4 has  2 proper divisors:  1  2
5 has  1 proper divisors:  1
6 has  3 proper divisors:  1  2  3
7 has  1 proper divisors:  1
8 has  3 proper divisors:  1  2  4
9 has  2 proper divisors:  1  3
10 has  3 proper divisors:  1  2  5
15120 has  79 proper divisors.

```

The first part of the task is to ''create a routine to generate a list of the proper divisors''. To ease the re-use of this routine for other tasks, such as ''Abundant, Deficient and Perfect Number Classification'' [[http://rosettacode.org/wiki/Abundant,_deficient_and_perfect_number_classifications#Ada]], ''Abundant Odd Number'' [[http://rosettacode.org/wiki/Abundant_odd_numbers#Ada]], and ''Amicable Pairs'' [[http://rosettacode.org/wiki/Amicable_pairs#Ada]], we define this routine as a function of a generic package:

```generic
type Result_Type (<>) is limited private;
None: Result_Type;
with function One(X: Positive) return Result_Type;
with function Add(X, Y: Result_Type) return Result_Type
is <>;
package Generic_Divisors is

function Process
(N: Positive; First: Positive := 1) return Result_Type is
(if First**2 > N or First = N then None
elsif (N mod First)=0 then
(if First = 1 or First*First = N
else Process(N, First+1));

end Generic_Divisors;
```

Now we instantiate the ''generic package'' to solve the other two parts of the task. Observe that there are two different instantiations of the package: one to generate a list of proper divisors, another one to count the number of proper divisors without actually generating such a list:

```with Ada.Text_IO, Ada.Containers.Generic_Array_Sort, Generic_Divisors;

procedure Proper_Divisors is

begin
-- show the proper divisors of the numbers 1 to 10 inclusive.
declare
type Pos_Arr is array(Positive range <>) of Positive;
subtype Single_Pos_Arr is Pos_Arr(1 .. 1);
Empty: Pos_Arr(1 .. 0);

function Arr(P: Positive) return Single_Pos_Arr is ((others => P));

package Divisor_List is new Generic_Divisors
(Result_Type => Pos_Arr, None => Empty, One => Arr, Add =>  "&");

(Positive, Positive, Pos_Arr);
begin
for I in 1 .. 10 loop
declare
List: Pos_Arr := Divisor_List.Process(I);
begin
(Positive'Image(I) & " has" &
Natural'Image(List'Length) & " proper divisors:");
Sort(List);
for Item of List loop
end loop;
end;
end loop;
end;

-- find a number 1 .. 20,000 with the most proper divisors
declare
Number: Positive := 1;
Number_Count: Natural := 0;
Current_Count: Natural;

function Cnt(P: Positive) return Positive is (1);

package Divisor_Count is new Generic_Divisors
(Result_Type => Natural, None => 0, One => Cnt, Add =>  "+");

begin
for Current in 1 .. 20_000 loop
Current_Count := Divisor_Count.Process(Current);
if Current_Count > Number_Count then
Number := Current;
Number_Count := Current_Count;
end if;
end loop;
(Positive'Image(Number) & " has the maximum number of" &
Natural'Image(Number_Count) & " proper divisors.");
end;
end Proper_Divisors;
```

{{out}}

``` 1 has 0 proper divisors:
2 has 1 proper divisors: 1
3 has 1 proper divisors: 1
4 has 2 proper divisors: 1 2
5 has 1 proper divisors: 1
6 has 3 proper divisors: 1 2 3
7 has 1 proper divisors: 1
8 has 3 proper divisors: 1 2 4
9 has 2 proper divisors: 1 3
10 has 3 proper divisors: 1 2 5

15120 has the maximum number of 79 proper divisors.
```

## ALGOL 68

{{works with|ALGOL 68G|Any - tested with release 2.8.3.win32}}

```# MODE to hold an element of a list of proper divisors            #
MODE DIVISORLIST = STRUCT( INT divisor, REF DIVISORLIST next );

# end of divisor list value                                       #
REF DIVISORLIST nil divisor list = REF DIVISORLIST(NIL);

# resturns a DIVISORLIST containing the proper divisors of n      #
# if n = 1, 0 or -1, we return no divisors                        #
PROC proper divisors = ( INT n )REF DIVISORLIST:
BEGIN
REF DIVISORLIST result   := nil divisor list;
REF DIVISORLIST end list := result;
INT abs n  = ABS n;
IF abs n > 1 THEN
# build the list of divisors backeards, so they are  #
# returned in ascending order                        #
INT root n = ENTIER sqrt( abs n );
FOR d FROM root n BY -1 TO 2 DO
IF abs n MOD d = 0 THEN
# found another divisor                      #
result := HEAP DIVISORLIST
:= DIVISORLIST( d, result );
IF end list IS nil divisor list THEN
# first result                           #
end list := result
FI;
IF d * d /= n THEN
# add the other divisor to the end of    #
# the list                               #
next OF end list := HEAP DIVISORLIST
:= DIVISORLIST( abs n OVER d, nil divisor list );
end list         := next OF end list
FI
FI
OD;
# 1 is always a proper divisor of numbers > 1        #
result := HEAP DIVISORLIST
:= DIVISORLIST( 1, result )
FI;
result
END # proper divisors # ;

# returns the number of divisors in a DIVISORLIST                 #
PROC count divisors = ( REF DIVISORLIST list )INT:
BEGIN
INT result := 0;
REF DIVISORLIST divisors := list;
WHILE divisors ISNT nil divisor list DO
result +:= 1;
divisors := next OF divisors
OD;
result
END # count divisors # ;

# find the proper divisors of 1 : 10                              #
FOR n TO 10 DO
REF DIVISORLIST divisors := proper divisors( n );
print( ( "Proper divisors of: ", whole( n, -2 ), ": " ) );
WHILE divisors ISNT nil divisor list DO
print( ( " ", whole( divisor OF divisors, 0 ) ) );
divisors := next OF divisors
OD;
print( ( newline ) )
OD;

# find the first/only number in 1 : 20 000 with the most divisors  #
INT max number         = 20 000;
INT max divisors      :=      0;
INT has max divisors  :=      0;
INT with max divisors :=      0;
FOR d TO max number DO
INT divisor count = count divisors( proper divisors( d ) );
IF divisor count > max divisors THEN
# found a number with more divisors than the previous max  #
max divisors       := divisor count;
has max divisors   := d;
with max divisors  := 1
ELIF divisor count = max divisors THEN
# found another number with that many divisors             #
with max divisors +:= 1
FI
OD;
print( ( whole( has max divisors, 0 )
, " is the "
, IF with max divisors < 2 THEN "only" ELSE "first" FI
, " number upto "
, whole( max number, 0 )
, " with "
, whole( max divisors, 0 )
, " divisors"
, newline
) )
```

{{out}}

```
Proper divisors of:  1:
Proper divisors of:  2:  1
Proper divisors of:  3:  1
Proper divisors of:  4:  1 2
Proper divisors of:  5:  1
Proper divisors of:  6:  1 2 3
Proper divisors of:  7:  1
Proper divisors of:  8:  1 2 4
Proper divisors of:  9:  1 3
Proper divisors of: 10:  1 2 5
15120 is the first number upto 20000 with 79 divisors

```

=={{header|Algol-M}}== Algol-M's maximum allowed integer value of 16,383 prevented searching up to 20,000 for the number with the most divisors, so the code here searches only up to 10,000.

```
BEGIN

% COMPUTE P MOD Q %
INTEGER FUNCTION MOD (P, Q);
INTEGER P, Q;
BEGIN
MOD := P - Q * (P / Q);
END;

% COUNT, AND OPTIONALLY DISPLAY, PROPER DIVISORS OF N %
INTEGER FUNCTION DIVISORS(N, DISPLAY);
INTEGER N, DISPLAY;
BEGIN
INTEGER I, LIMIT, COUNT, START, DELTA;
IF MOD(N, 2) = 0 THEN
BEGIN
START := 2;
DELTA := 1;
END
ELSE  % ONLY NEED TO CHECK ODD DIVISORS %
BEGIN
START := 3;
DELTA := 2;
END;
% 1 IS A DIVISOR OF ANY NUMBER > 1 %
IF N > 1 THEN COUNT := 1 ELSE COUNT := 0;
IF (DISPLAY <> 0) AND (COUNT <> 0) THEN WRITEON(1);
% CHECK REMAINING POTENTIAL DIVISORS %
I := START;
LIMIT := N / START;
WHILE I <= LIMIT DO
BEGIN
IF MOD(N, I) = 0 THEN
BEGIN
IF DISPLAY <> 0 THEN WRITEON(I);
COUNT := COUNT + 1;
END;
I := I + DELTA;
IF COUNT = 1 THEN LIMIT := N / I;
END;
DIVISORS := COUNT;
END;

COMMENT MAIN PROGRAM BEGINS HERE;
INTEGER I, NDIV, TRUE, FALSE, HIGHDIV, HIGHNUM;
TRUE := -1;
FALSE := 0;

WRITE("PROPER DIVISORS OF FIRST TEN NUMBERS:");
FOR I := 1 STEP 1 UNTIL 10 DO
BEGIN
WRITE(I, " : ");
NDIV := DIVISORS(I, TRUE);
END;

WRITE("SEARCHING FOR NUMBER UP TO 10000 WITH MOST DIVISORS ...");
HIGHDIV := 1;
HIGHNUM := 1;
FOR I := 1 STEP 1 UNTIL 10000 DO
BEGIN
NDIV := DIVISORS(I, FALSE);
IF NDIV > HIGHDIV THEN
BEGIN
HIGHDIV := NDIV;
HIGHNUM := I;
END;
END;
WRITE("THE NUMBER IS:", HIGHNUM);
WRITE("IT HAS", HIGHDIV, " DIVISORS");

END

```

{{out}}

```
PROPER DIVISORS OF FIRST TEN NUMBERS:
1 :
2 :      1
3 :      1
4 :      1     2
5 :      1
6 :      1     2     3
7 :      1
8 :      1     2     4
9 :      1     3
10 :      1     2     5
SEARCHING FOR NUMBER UP TO 10000 WITH MOST DIVISORS:
THE NUMBER IS:  7560
IT HAS    63 DIVISORS

```

## AppleScript

{{Trans|JavaScript}}

```-- PROPER DIVISORS -----------------------------------------------------------

-- properDivisors :: Int -> [Int]
on properDivisors(n)
if n = 1 then
{1}
else
set realRoot to n ^ (1 / 2)
set intRoot to realRoot as integer
set blnPerfectSquare to intRoot = realRoot

-- isFactor :: Int -> Bool
script isFactor
on |λ|(x)
n mod x = 0
end |λ|
end script

-- Factors up to square root of n,
set lows to filter(isFactor, enumFromTo(1, intRoot))

-- and quotients of these factors beyond the square root,

-- integerQuotient :: Int -> Int
script integerQuotient
on |λ|(x)
(n / x) as integer
end |λ|
end script

-- excluding n itself (last item)
items 1 thru -2 of (lows & map(integerQuotient, ¬
items (1 + (blnPerfectSquare as integer)) thru -1 of reverse of lows))
end if
end properDivisors

-- TEST ----------------------------------------------------------------------
on run
-- numberAndDivisors :: Int -> [Int]
script numberAndDivisors
on |λ|(n)
{num:n, divisors:properDivisors(n)}
end |λ|
end script

-- maxDivisorCount :: Record -> Int -> Record
script maxDivisorCount
on |λ|(a, n)
set intDivisors to length of properDivisors(n)

if intDivisors ≥ divisors of a then
{num:n, divisors:intDivisors}
else
a
end if
end |λ|
end script

{oneToTen:map(numberAndDivisors, ¬
enumFromTo(1, 10)), mostDivisors:foldl(maxDivisorCount, ¬
{num:0, divisors:0}, enumFromTo(1, 20000))} ¬

end run

-- GENERIC FUNCTIONS ---------------------------------------------------------

-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if m > n then
set d to -1
else
set d to 1
end if
set lst to {}
repeat with i from m to n by d
set end of lst to i
end repeat
return lst
end enumFromTo

-- filter :: (a -> Bool) -> [a] -> [a]
on filter(f, xs)
tell mReturn(f)
set lst to {}
set lng to length of xs
repeat with i from 1 to lng
set v to item i of xs
if |λ|(v, i, xs) then set end of lst to v
end repeat
return lst
end tell
end filter

-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl

-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map

-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
```

{{Out}}

```{oneToTen:{{num:1, divisors:{1}}, {num:2, divisors:{1}}, {num:3, divisors:{1}},
{num:4, divisors:{1, 2}}, {num:5, divisors:{1}}, {num:6, divisors:{1, 2, 3}},
{num:7, divisors:{1}}, {num:8, divisors:{1, 2, 4}}, {num:9, divisors:{1, 3}},
{num:10, divisors:{1, 2, 5}}},
mostDivisors:{num:18480, divisors:79}}
```

## Arc

```

;; Given num, return num and the list of its divisors
(= divisor (fn (num)
(= dlist '())
(when (is 1 num) (= dlist '(1 0)))
(when (is 2 num) (= dlist '(2 1)))
(unless (or (is 1 num) (is 2 num))
(up i 1 (+ 1 (/ num 2))
(if (is 0 (mod num i))
(push i dlist)))
(= dlist (cons num dlist)))
dlist))

;; Find out what number has the most divisors between 2 and 20,000.
;; Print a list of the largest known number's divisors as it is found.
(= div-lists (fn (cnt (o show 0))
(= tlist '()) (= clist tlist)
(when (> show 0) (prn tlist))
(up i 1 cnt
(divisor i)
(when (is 1 show) (prn dlist))
(when (>= (len dlist) (len tlist))
(= tlist dlist)
(when (is show 2) (prn tlist))
(let c (- (len dlist) 1)
(push (list i c) clist))))

(= many-divisors (list ((clist 0) 1)))
(for n 0 (is ((clist n) 1) ((clist 0) 1)) (= n (+ 1 n))
(push ((clist n) 0) many-divisors))
(= many-divisors (rev many-divisors))
(prn "The number with the most divisors under " cnt
" has " (many-divisors 0) " divisors.")
(prn "It is the number "
(if (> 2 (len many-divisors)) (cut (many-divisors) 1)
(many-divisors 1)) ".")
(prn "There are " (- (len many-divisors) 1) " numbers"
" with this trait, and they are "
(map [many-divisors _] (range 1 (- (len many-divisors) 1))))
(prn (map [divisor _] (cut many-divisors 1)))
many-divisors))

(div-lists 10 1)
(div-lists 20000)
;; This took about 10 minutes on my machine.

```

{{Out}}

```
(1 0)
(2 1)
(3 1)
(4 2 1)
(5 1)
(6 3 2 1)
(7 1)
(8 4 2 1)
(9 3 1)
(10 5 2 1)
The number with the most divisors under 10 has 3 divisors.
It is the number 10.
There are 3 numbers with this trait, and they are (10 8 6)
((10 5 2 1) (8 4 2 1) (6 3 2 1))
'(3 10 8 6)

The number with the most divisors under 20000 has 79 divisors.
It is the number 18480.
There are 2 numbers with this trait, and they are (18480 15120)

```

## AWK

```
# syntax: GAWK -f PROPER_DIVISORS.AWK
BEGIN {
show = 0 # show divisors: 0=no, 1=yes
print("    N  cnt  DIVISORS")
for (i=1; i<=20000; i++) {
divisors(i)
if (i <= 10 || i == 100) { # including 100 as it was an example in task description
printf("%5d  %3d  %s\n",i,Dcnt,Dstr)
}
if (Dcnt < max_cnt) {
continue
}
if (Dcnt > max_cnt) {
rec = ""
max_cnt = Dcnt
}
rec = sprintf("%s%5d  %3d  %s\n",rec,i,Dcnt,show?Dstr:"divisors not shown")
}
printf("%s",rec)
exit(0)
}
function divisors(n,  i) {
if (n == 1) {
Dcnt = 0
Dstr = ""
return
}
Dcnt = Dstr = 1
for (i=2; i<n; i++) {
if (n % i == 0) {
Dcnt++
Dstr = sprintf("%s %s",Dstr,i)
}
}
return
}

```

output:

```
N  cnt  DIVISORS
1    0
2    1  1
3    1  1
4    2  1 2
5    1  1
6    3  1 2 3
7    1  1
8    3  1 2 4
9    2  1 3
10    3  1 2 5
100    8  1 2 4 5 10 20 25 50
15120   79  divisors not shown
18480   79  divisors not shown

```

## BaCon

```
FUNCTION ProperDivisor(nr, show)

LOCAL probe, total

FOR probe = 1 TO nr-1
IF MOD(nr, probe) = 0 THEN
IF show THEN PRINT " ", probe;
INCR total
END IF
NEXT

END FUNCTION

FOR x = 1 TO 10
PRINT x, ":";
IF ProperDivisor(x, 1) = 0 THEN PRINT " 0";
PRINT
NEXT

FOR x = 1 TO 20000
DivisorCount = ProperDivisor(x, 0)
IF DivisorCount > MaxDivisors THEN
MaxDivisors = DivisorCount
MagicNumber = x
END IF
NEXT

PRINT "Most proper divisors for number in the range 1-20000: ", MagicNumber, " with ", MaxDivisors, " divisors."

```

{{out}}

```
1: 0
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
Most proper divisors for number in the range 1-20000: 15120 with 79 divisors.

```

## C

### Brute Force

C has tedious boilerplate related to allocating memory for dynamic arrays, so we just skip the problem of storing values altogether.

```
#include <stdio.h>
#include <stdbool.h>

int proper_divisors(const int n, bool print_flag)
{
int count = 0;

for (int i = 1; i < n; ++i) {
if (n % i == 0) {
count++;
if (print_flag)
printf("%d ", i);
}
}

if (print_flag)
printf("\n");

return count;
}

int main(void)
{
for (int i = 1; i <= 10; ++i) {
printf("%d: ", i);
proper_divisors(i, true);
}

int max = 0;
int max_i = 1;

for (int i = 1; i <= 20000; ++i) {
int v = proper_divisors(i, false);
if (v >= max) {
max = v;
max_i = i;
}
}

printf("%d with %d divisors\n", max_i, max);
return 0;
}

```

{{out}}

```
1:
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
18480 with 79 divisors

```

### Number Theoretic

There is no need to go through all the divisors if only the count is needed, this implementation refines the brute force approach by solving the second part of the task via a Number Theory formula. The running time is noticeably faster than the brute force method above. Output is same as the above.

```
#include <stdio.h>
#include <stdbool.h>

int proper_divisors(const int n, bool print_flag)
{
int count = 0;

for (int i = 1; i < n; ++i) {
if (n % i == 0) {
count++;
if (print_flag)
printf("%d ", i);
}
}

if (print_flag)
printf("\n");

return count;
}

int countProperDivisors(int n){
int prod = 1,i,count=0;

while(n%2==0){
count++;
n /= 2;
}

prod *= (1+count);

for(i=3;i*i<=n;i+=2){
count = 0;

while(n%i==0){
count++;
n /= i;
}

prod *= (1+count);
}

if(n>2)
prod *= 2;

return prod - 1;
}

int main(void)
{
for (int i = 1; i <= 10; ++i) {
printf("%d: ", i);
proper_divisors(i, true);
}

int max = 0;
int max_i = 1;

for (int i = 1; i <= 20000; ++i) {
int v = countProperDivisors(i);
if (v >= max) {
max = v;
max_i = i;
}
}

printf("%d with %d divisors\n", max_i, max);
return 0;
}

```

## C#

```namespace RosettaCode.ProperDivisors
{
using System;
using System.Collections.Generic;
using System.Linq;

internal static class Program
{
private static IEnumerable<int> ProperDivisors(int number)
{
return
Enumerable.Range(1, number / 2)
.Where(divisor => number % divisor == 0);
}

private static void Main()
{
foreach (var number in Enumerable.Range(1, 10))
{
Console.WriteLine("{0}: {{{1}}}", number,
string.Join(", ", ProperDivisors(number)));
}

var record = Enumerable.Range(1, 20000).Select(number => new
{
Number = number,
Count = ProperDivisors(number).Count()
}).OrderByDescending(currentRecord => currentRecord.Count).First();
Console.WriteLine("{0}: {1}", record.Number, record.Count);
}
}
}
```

{{out}}

```1: {}
2: {1}
3: {1}
4: {1, 2}
5: {1}
6: {1, 2, 3}
7: {1}
8: {1, 2, 4}
9: {1, 3}
10: {1, 2, 5}
15120: 79
```

## C++

```#include <vector>
#include <iostream>
#include <algorithm>

std::vector<int> properDivisors ( int number ) {
std::vector<int> divisors ;
for ( int i = 1 ; i < number / 2 + 1 ; i++ )
if ( number % i == 0 )
divisors.push_back( i ) ;
return divisors ;
}

int main( ) {
std::vector<int> divisors ;
unsigned int maxdivisors = 0 ;
int corresponding_number = 0 ;
for ( int i = 1 ; i < 11 ; i++ ) {
divisors =  properDivisors ( i ) ;
std::cout << "Proper divisors of " << i << ":\n" ;
for ( int number : divisors ) {
std::cout << number << " " ;
}
std::cout << std::endl ;
divisors.clear( ) ;
}
for ( int i = 11 ; i < 20001 ; i++ ) {
divisors =  properDivisors ( i ) ;
if ( divisors.size( ) > maxdivisors ) {
maxdivisors = divisors.size( ) ;
corresponding_number = i ;
}
divisors.clear( ) ;
}

std::cout << "Most divisors has " << corresponding_number <<
" , it has " << maxdivisors << " divisors!\n" ;
return 0 ;
}

```

{{out}}

```
Proper divisors of 1:

Proper divisors of 2:
1
Proper divisors of 3:
1
Proper divisors of 4:
1 2
Proper divisors of 5:
1
Proper divisors of 6:
1 2 3
Proper divisors of 7:
1
Proper divisors of 8:
1 2 4
Proper divisors of 9:
1 3
Proper divisors of 10:
1 2 5
Most divisors has 15120 , it has 79 divisors!

```

## Ceylon

```shared void run() {

function divisors(Integer int) =>
if(int <= 1)
then {}
else (1..int / 2).filter((Integer element) => element.divides(int));

for(i in 1..10) {
print("``i`` => ``divisors(i)``");
}

value start = 1;
value end = 20k;

value mostDivisors =
map {for(i in start..end) i->divisors(i).size}
.inverse()
.max(byKey(byIncreasing(Integer.magnitude)));

print("the number(s) with the most divisors between ``start`` and ``end`` is/are:
``mostDivisors?.item else "nothing"`` with ``mostDivisors?.key else "no"`` divisors");
}
```

{{out}}

```1 => []
2 => { 1 }
3 => { 1 }
4 => { 1, 2 }
5 => { 1 }
6 => { 1, 2, 3 }
7 => { 1 }
8 => { 1, 2, 4 }
9 => { 1, 3 }
10 => { 1, 2, 5 }
the number(s) with the most divisors between 1 and 20000 is/are:
[15120, 18480] with 79 divisors
```

## Clojure

```(ns properdivisors
(:gen-class))

(defn proper-divisors [n]
" Proper divisors of n"
(if (= n 1)
[]
(filter #(= 0 (rem n %)) (range 1 n))))

;; Property divisors of numbers 1 to 20,000 inclusive
(def data (for [n (range 1 (inc 20000))]
[n (proper-divisors n)]))

;; Find Max
(defn maximal-key [k x & xs]
" Normal max-key only finds one key that produces maximum, while this function finds them all "
(reduce (fn [ys x]
(let [c (compare (k x) (k (peek ys)))]
(cond
(pos? c) [x]
(neg? c) ys
:else    (conj ys x))))
[x]
xs))

(println "n\tcnt\tPROPER DIVISORS")
(doseq [n (range 1 11)]
(let [factors (proper-divisors n)]
(println n "\t" (count factors) "\t" factors)))

(def max-data (apply maximal-key (fn [[i pd]] (count pd)) data))

(doseq [[n factors] max-data]
(println n " has " (count factors) " divisors"))

```

{{Output}}

```
n	cnt	PROPER DIVISORS
1 	 0 	 []
2 	 1 	 (1)
3 	 1 	 (1)
4 	 2 	 (1 2)
5 	 1 	 (1)
6 	 3 	 (1 2 3)
7 	 1 	 (1)
8 	 3 	 (1 2 4)
9 	 2 	 (1 3)
10 	 3 	 (1 2 5)
15120  has  79  divisors
18480  has  79  divisors

```

## Common Lisp

Ideally, the smallest-divisor function would only try prime numbers instead of odd numbers.

```(defun proper-divisors-recursive (product &optional (results '(1)))
"(int,list)->list::Function to find all proper divisors of a +ve integer."

(defun smallest-divisor (x)
"int->int::Find the smallest divisor of an integer > 1."
(if (evenp x) 2
(do ((lim (truncate (sqrt x)))
(sd 3 (+ sd 2)))
((or (integerp (/ x sd)) (> sd lim)) (if (> sd lim) x sd)))))

(defun pd-rec (fac)
"(int,int)->nil::Recursive function to find proper divisors of a +ve integer"
(when (not (member fac results))
(push fac results)
(let ((hifac (/ fac (smallest-divisor fac))))
(pd-rec hifac)
(pd-rec (/ product hifac)))))

(pd-rec product)
(butlast (sort (copy-list results) #'<)))

(defun task (method &optional (n 1) (most-pds '(0)))
(dotimes (i 19999)
(let ((npds (length (funcall method (incf n))))
(hiest (car most-pds)))
(when (>= npds hiest)
(if (> npds hiest)
(setf most-pds (list npds (list n)))
(setf most-pds (list npds (cons n (second most-pds))))))))
most-pds)

(defun main ()
(format t "Task 1:Proper Divisors of [1,10]:~%")
(dotimes (i 10) (format t "~A:~A~%" (1+ i) (proper-divisors-recursive (1+ i))))
(format t "Task 2:Count & list of numbers <=20,000 with the most Proper Divisors:~%~A~%"
```

{{out}}

```CL-USER(10): (main)
1:NIL
2:(1)
3:(1)
4:(1 2)
5:(1)
6:(1 2 3)
7:(1)
8:(1 2 4)
9:(1 3)
10:(1 2 5)
Task 2:Count & list of numbers <=20,000 with the most Proper Divisors:
(79 (18480 15120))
NIL
```

## Component Pascal

{{Works with|Black Box Component Builder}}

```
MODULE RosettaProperDivisor;
IMPORT StdLog;

PROCEDURE Pd*(n: LONGINT;OUT r: ARRAY OF LONGINT):LONGINT;
VAR
i,j: LONGINT;
BEGIN
i := 1;j := 0;
IF n >  1 THEN
WHILE (i < n) DO
IF (n MOD i) = 0 THEN
IF (j < LEN(r)) THEN r[j] := i END; INC(j)
END;
INC(i)
END;
END;
RETURN j
END Pd;

PROCEDURE Do*;
VAR
r: ARRAY 128 OF LONGINT;
i,j,found,max,idxMx: LONGINT;
mx: ARRAY 128 OF LONGINT;
BEGIN
FOR i := 1 TO 10 DO
found := Pd(i,r);
IF found > LEN(r) THEN (* Error. more pd than r can admit *) HALT(1) END;
StdLog.Int(i);StdLog.String("[");StdLog.Int(found);StdLog.String("]:> ");
FOR j := 0 TO found - 1 DO
StdLog.Int(r[j]);StdLog.Char(' ');
END;
StdLog.Ln
END;

max := 0;idxMx := 0;
FOR i := 1 TO 20000 DO
found := Pd(i,r);
IF found > max THEN
idxMx:= 0;mx[idxMx] := i;max := found
ELSIF found = max THEN
INC(idxMx);mx[idxMx] := i
END;
END;
StdLog.String("Found: ");StdLog.Int(idxMx + 1);
StdLog.String(" Numbers with the longest proper divisors [");
StdLog.Int(max);StdLog.String("]: ");StdLog.Ln;
FOR i := 0 TO idxMx DO
StdLog.Int(mx[i]);StdLog.Ln
END
END Do;

END RosettaProperDivisor.

^Q RosettaProperDivisor.Do~

```

{{out}}

```
1[ 0]:>
2[ 1]:>  1
3[ 1]:>  1
4[ 2]:>  1  2
5[ 1]:>  1
6[ 3]:>  1  2  3
7[ 1]:>  1
8[ 3]:>  1  2  4
9[ 2]:>  1  3
10[ 3]:>  1  2  5
Found:  2 Numbers with the longest proper divisors [ 79]:
15120
18480

```

## D

{{trans|Python}} Currently the lambda of the filter allocates a closure on the GC-managed heap.

```void main() /*@safe*/ {
import std.stdio, std.algorithm, std.range, std.typecons;

immutable properDivs = (in uint n) pure nothrow @safe /*@nogc*/ =>
iota(1, (n + 1) / 2 + 1).filter!(x => n % x == 0 && n != x);

iota(1, 11).map!properDivs.writeln;
iota(1, 20_001).map!(n => tuple(properDivs(n).count, n)).reduce!max.writeln;
}
```

{{out}}

```[[], [1], [1], [1, 2], [1], [1, 2, 3], [1], [1, 2, 4], [1, 3], [1, 2, 5]]
Tuple!(uint, int)(79, 18480)
```

The Run-time is about 0.67 seconds with the ldc2 compiler.

## Dyalect

{{trans|Swift}}

```func properDivs(n) {
if n == 1 {
return
}
for x in 1..(n-1) {
if n % x == 0 {
yield x
}
}
}

for i in 1..10 {
print("\(i): \(properDivs(i).toArray())")
}

var (num, max) = (0,0)

for i in 1..20000 {
const count = properDivs(i).len()
if count > max {
set (num, max) = (i, count)
}
}

print("\(num): \(max)")
```

{{out}}

```1: []
2: [1]
3: [1]
4: [1, 2]
5: [1]
6: [1, 2, 3]
7: [1]
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120: 79
```

## EchoLisp

```
(lib 'list) ;; list-delete

;; let n = product p_i^a_i , p_i prime
;; number of divisors = product (a_i + 1) - 1
(define (numdivs n)
(1- (apply * (map (lambda(g) (1+ (length g))) (group (prime-factors n))))))

(remember 'numdivs)

;; prime powers
;; input : a list g of grouped prime factors ( 3 3 3 ..)
;; returns (1 3 9 27 ...)
(define (ppows g (mult 1))
(for/fold (ppows '(1)) ((a g))
(set! mult (* mult a))
(cons mult ppows)))

;; proper divisors
;; decomp n into ((2 2 ..) ( 3 3 ..)  ) prime factors groups
;; combines (1 2 4 8 ..) (1 3 9 ..) lists
;; remove n from the list

(define (divs n)
(if (<= n 1) null
(list-delete
(for/fold (divs'(1)) ((g (map  ppows (group (prime-factors n)))))
(for*/list ((a divs) (b g)) (* a b)))
n )))

;; find number(s) with max # of proper divisors
;; returns list of (n . maxdivs)  for n in range 2..N

(define (most-proper N)
(define maxdivs 1)
(define ndivs 0)
(for/fold (most-proper null) ((n (in-range 2 N)))
(set! ndivs (numdivs n))
#:continue (< ndivs maxdivs)
(when (> ndivs maxdivs)
(set!-values (most-proper maxdivs) (values null ndivs)))
(cons (cons n maxdivs) most-proper)))

```

{{out}}

```
(for ((i (in-range 1 11))) (writeln i (divs i)))
1     null
2     (1)
3     (1)
4     (2 1)
5     (1)
6     (2 3 1)
7     (1)
8     (4 2 1)
9     (3 1)
10     (2 5 1)

(most-proper 20000)
→ ((18480 . 79) (15120 . 79))
(most-proper 1_000_000)
→ ((997920 . 239) (982800 . 239) (942480 . 239) (831600 . 239) (720720 . 239))

(lib 'bigint)
(numdivs 95952222101012742144)  → 666 ;; 🎩

```

## Eiffel

```
class
APPLICATION

create
make

feature

make
-- Test the feature proper_divisors.
local
count, number: INTEGER
do
across
1 |..| 10 as c
loop
list := proper_divisors (c.item)
io.put_string (c.item.out + ": ")
across
list as l
loop
io.put_string (l.item.out + " ")
end
io.new_line
end
across
1 |..| 20000 as c
loop
list := proper_divisors (c.item)
if list.count > count then
count := list.count
number := c.item
end
end
io.put_string (number.out + " has with " + count.out + " divisors the highest number of proper divisors.")
end

-- Proper divisors of 'n'.
do
create Result.make
across
1 |..| (n - 1) as c
loop
if n \\ c.item = 0 then
Result.extend (c.item)
end
end
end

end

```

{{out}}

```
1:
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
15120 has with 79 divisors the highest number of proper divisors.

```

## Elixir

{{trans|Erlang}}

```defmodule Proper do
def divisors(1), do: []
def divisors(n), do: [1 | divisors(2,n,:math.sqrt(n))] |> Enum.sort

defp divisors(k,_n,q) when k>q, do: []
defp divisors(k,n,q) when rem(n,k)>0, do: divisors(k+1,n,q)
defp divisors(k,n,q) when k * k == n, do: [k | divisors(k+1,n,q)]
defp divisors(k,n,q)                , do: [k,div(n,k) | divisors(k+1,n,q)]

def most_divisors(limit) do
{length,nums} = Enum.group_by(1..limit, fn n -> length(divisors(n)) end)
|> Enum.max_by(fn {length,_nums} -> length end)
IO.puts "With #{length}, Number #{inspect nums} has the most divisors"
end
end

Enum.each(1..10, fn n ->
IO.puts "#{n}: #{inspect Proper.divisors(n)}"
end)
Proper.most_divisors(20000)
```

{{out}}

```
1: []
2: [1]
3: [1]
4: [1, 2]
5: [1]
6: [1, 2, 3]
7: [1]
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
With 79, Number [18480, 15120] has the most divisors

```

```-module(properdivs).
-export([divs/1,sumdivs/1,longest/1]).

divs(0) -> [];
divs(1) -> [];
divs(N) -> lists:sort([1] ++ divisors(2,N,math:sqrt(N))).

divisors(K,_N,Q) when K > Q -> [];
divisors(K,N,Q) when N rem K =/= 0 ->
divisors(K+1,N,Q);
divisors(K,N,Q) when K * K  == N ->
[K] ++ divisors(K+1,N,Q);
divisors(K,N,Q) ->
[K, N div K] ++ divisors(K+1,N,Q).

sumdivs(N) -> lists:sum(divs(N)).

longest(Limit) -> longest(Limit,0,0,1).

longest(L,Current,CurLeng,Acc) when Acc >= L ->
io:format("With ~w, Number ~w has the most divisors~n", [CurLeng,Current]);
longest(L,Current,CurLeng,Acc) ->
A = length(divs(Acc)),
if A > CurLeng ->
longest(L,Acc,A,Acc+1);
true -> longest(L,Current,CurLeng,Acc+1)
end.
```

{{out}}

```
1> [io:format("X: ~w, N: ~w~n", [N,properdivs:divs(N)]) ||  N <- lists:seq(1,10)].
X: 1, N: []
X: 2, N: [1]
X: 3, N: [1]
X: 4, N: [1,2]
X: 5, N: [1]
X: 6, N: [1,2,3]
X: 7, N: [1]
X: 8, N: [1,2,4]
X: 9, N: [1,3]
X: 10, N: [1,2,5]
[ok,ok,ok,ok,ok,ok,ok,ok,ok,ok]

2> properdivs:longest(20000).
With 79, Number 15120 has the most divisors

```

## F#

```
let mutable a=0
let mutable b=0
let mutable c=0
let mutable d=0
let mutable e=0
let mutable f=0
for k=1 to 10 do
b <- 0
f <- k/2
printf "divisor "
for l=1 to f do
if k%l=0 then
b <- b+1
printf " %i," l
printf "no of divisor %i" b
printfn ""
for i=1 to 20000 do
b <- 0
f <- i/2
for j=1 to f do
if i%j=0 then
b <- b+1
if b=c then
d <- 0
d <- i
if c<b then
c <- b

printfn "%i has %i divisor" d c

```

A purely functional approach.

```
// the simple function with the answer
let propDivs n = [1..n/2] |> List.filter (fun x->n % x = 0)

// to cache the result length; helpful for a long search
let propDivDat n = propDivs n |> fun xs -> n, xs.Length, xs

// UI: always the longest and messiest
let show (n,count,divs) =
let showCount = count |> function | 0-> "no proper divisors" | 1->"1 proper divisor" | _-> sprintf "%d proper divisors" count
let showDiv = divs |> function | []->"" | x::[]->sprintf ": %d" x | _->divs |> Seq.map string |> String.concat "," |> sprintf ": %s"
printfn "%d has %s%s" n showCount showDiv

// generate output
[1..10] |> List.iter (propDivDat >> show)

// use a sequence: we don't really need to hold this data, just iterate over it
Seq.init 20000 ( ((+) 1) >> propDivDat)
|> Seq.fold (fun a b ->match a,b with | (_,c1,_),(_,c2,_) when c2 > c1 -> b | _-> a) (0,0,[])
|> fun (n,count,_) -> (n,count,[]) |> show

```

{{out}}

```
1 has no proper divisors
2 has 1 proper divisor: 1
3 has 1 proper divisor: 1
4 has 2 proper divisors: 1,2
5 has 1 proper divisor: 1
6 has 3 proper divisors: 1,2,3
7 has 1 proper divisor: 1
8 has 3 proper divisors: 1,2,4
9 has 2 proper divisors: 1,3
10 has 3 proper divisors: 1,2,5
15120 has 79 proper divisors

```

## Factor

```
USING: math.primes.factors math.ranges ;
10 [1,b] [ divisors but-last ] map [ 1 + pprint bl . ] each-index
20000 [1,b] [ divisors but-last length ] map dup supremum
swap dupd index 1 + pprint " with " write pprint " divisors." print

```

{{out}}

```
1 { }
2 { 1 }
3 { 1 }
4 { 1 2 }
5 { 1 }
6 { 1 2 3 }
7 { 1 }
8 { 1 2 4 }
9 { 1 3 }
10 { 1 2 5 }
15120 with 79 divisors.

```

## Fortran

Compiled using G95 compiler, run on x86 system under Puppy Linux

```

function icntprop(num  )
icnt=0
do i=1 , num-1
if (mod(num , i)  .eq. 0)  then
icnt = icnt + 1
if (num .lt. 11) print *,'    ',i
end if
end do
icntprop =  icnt
end function

limit = 20000
maxcnt = 0
print *,'N   divisors'
do j=1,limit,1
if (j .lt. 11) print *,j
icnt = icntprop(j)

if (icnt .gt. maxcnt) then
maxcnt = icnt
maxj = j
end if

end do

print *,' '
print *,' from 1 to ',limit
print *,maxj,' has max proper divisors: ',maxcnt
end

```

{{out}}

```

N   divisors
1
2
1
3
1
4
1
2
5
1
6
1
2
3
7
1
8
1
2
4
9
1
3
10
1
2
5

from 1 to  20000
15120  has max proper divisors:  79

```

## FreeBASIC

```
' FreeBASIC v1.05.0 win64

Sub ListProperDivisors(limit As Integer)
If limit < 1 Then Return
For i As Integer = 1 To limit
Print Using "##"; i;
Print " ->";
If i = 1 Then
Print " (None)"
Continue For
End if
For j As Integer = 1 To i \ 2
If i Mod j = 0 Then Print " "; j;
Next j
Print
Next i
End Sub

Function CountProperDivisors(number As Integer) As Integer
If number < 2 Then Return 0
Dim count As Integer = 0
For i As Integer = 1 To number \ 2
If number Mod i = 0 Then count += 1
Next
Return count
End Function

Dim As Integer n, count, most = 1, maxCount = 0

Print "The proper divisors of the following numbers are :"
Print
ListProperDivisors(10)

For n As Integer = 2 To 20000
count = CountProperDivisors(n)
If count > maxCount Then
maxCount = count
most = n
EndIf
Next

Print
Print Str(most); " has the most proper divisors, namely"; maxCount
Print
Print "Press any key to exit the program"
Sleep
End

```

{{out}}

```
The proper divisors of the following numbers are :

1 -> (None)
2 ->  1
3 ->  1
4 ->  1  2
5 ->  1
6 ->  1  2  3
7 ->  1
8 ->  1  2  4
9 ->  1  3
10 ->  1  2  5

15120 has the most proper divisors, namely 79

```

## Frink

Frink's built-in factorization routines efficiently find factors of arbitrary-sized integers.

```
for n = 1 to 10
println["\$n\t" + join[" ", properDivisors[n]]]

println[]

d = new dict
for n = 1 to 20000
{
c = length[properDivisors[n]]
}

most = max[keys[d]]
println[d@most + " have \$most factors"]

properDivisors[n] := allFactors[n, true, false, true]

```

{{out}}

```
1
2       1
3       1
4       1 2
5       1
6       1 2 3
7       1
8       1 2 4
9       1 3
10      1 2 5

[15120, 18480] have 79 factors

```

## GFA Basic

OPENW 1 CLEARW 1 ' ' Array f% is used to hold the divisors DIM f%(SQR(20000)) ! cannot redim arrays, so set size to largest needed ' ' 1. Show proper divisors of 1 to 10, inclusive ' FOR i%=1 TO 10 num%=@proper_divisors(i%) PRINT "Divisors for ";i%;":"; FOR j%=1 TO num% PRINT " ";f%(j%); NEXT j% PRINT NEXT i% ' ' 2. Find (smallest) number <= 20000 with largest number of proper divisors ' result%=1 ! largest so far number%=0 ! its number of divisors FOR i%=1 TO 20000 num%=@proper_divisors(i%) IF num%>number% result%=i% number%=num% ENDIF NEXT i% PRINT "Largest number of divisors is ";number%;" for ";result% ' ~INP(2) CLOSEW 1 ' ' find the proper divisors of n%, placing results in f% ' and return the number found ' FUNCTION proper_divisors(n%) LOCAL i%,root%,count% ' ARRAYFILL f%(),0 count%=1 ! index of next slot in f% to fill ' IF n%>1 f%(count%)=1 count%=count%+1 root%=SQR(n%) FOR i%=2 TO root% IF n% MOD i%=0 f%(count%)=i% count%=count%+1 IF i%*i%<>n% ! root% is an integer, so check if i% is actual squa- lists:seq(1,10)]. X: 1, N: [] X: 2, N: [1] X: 3, N: [1] X: 4, N: [1,2] X: 5, N: [1] X: 6, N: [1,2,3] X: 7, N: [1] X: 8, N: [1,2,4] X: 9, N: [1,3] X: 10, N: [1,2,5] [ok,ok,ok,ok,ok,ok,ok,ok,ok,ok]

2> properdivs:longest(20000). With 79, Number 15120 has the most divisors re root of n% f%(count%)=n%/i% count%=count%+1 ENDIF ENDIF NEXT i% ENDIF ' RETURN count%-1 ENDFUNC

```

Output is:

```txt

Divisors for 1:
Divisors for 2: 1
Divisors for 3: 1
Divisors for 4: 1 2
Divisors for 5: 1
Divisors for 6: 1 2 3
Divisors for 7: 1
Divisors for 8: 1 2 4
Divisors for 9: 1 3
Divisors for 10: 1 2 5
Largest number of divisors is 79 for 15120

```

## Go

{{trans|Kotlin}}

```package main

import (
"fmt"
"strconv"
)

func listProperDivisors(limit int) {
if limit < 1 {
return
}
width := len(strconv.Itoa(limit))
for i := 1; i <= limit; i++ {
fmt.Printf("%*d -> ", width, i)
if i == 1 {
fmt.Println("(None)")
continue
}
for j := 1; j <= i/2; j++ {
if i%j == 0 {
fmt.Printf(" %d", j)
}
}
fmt.Println()
}
}

func countProperDivisors(n int) int {
if n < 2 {
return 0
}
count := 0
for i := 1; i <= n/2; i++ {
if n%i == 0 {
count++
}
}
return count
}

func main() {
fmt.Println("The proper divisors of the following numbers are :\n")
listProperDivisors(10)
fmt.Println()
maxCount := 0
most := []int{1}
for n := 2; n <= 20000; n++ {
count := countProperDivisors(n)
if count == maxCount {
most = append(most, n)
} else if count > maxCount {
maxCount = count
most = most[0:1]
most[0] = n
}
}
fmt.Print("The following number(s) <= 20000 have the most proper divisors, ")
fmt.Println("namely", maxCount, "\b\n")
for _, n := range most {
fmt.Println(n)
}
}
```

{{out}}

```
The proper divisors of the following numbers are :

1 -> (None)
2 ->  1
3 ->  1
4 ->  1 2
5 ->  1
6 ->  1 2 3
7 ->  1
8 ->  1 2 4
9 ->  1 3
10 ->  1 2 5

The following number(s) <= 20000 have the most proper divisors, namely 79

15120
18480

```

```import Data.Ord
import Data.List

divisors :: (Integral a) => a -> [a]
divisors n = filter ((0 ==) . (n `mod`)) [1 .. (n `div` 2)]

main :: IO ()
main = do
putStrLn "divisors of 1 to 10:"
mapM_ (print . divisors) [1 .. 10]
putStrLn "a number with the most divisors within 1 to 20000 (number, count):"
print \$ maximumBy (comparing snd)
[(n, length \$ divisors n) | n <- [1 .. 20000]]
```

{{out}}

```divisors of 1 to 10:
[]
[1]
[1]
[1,2]
[1]
[1,2,3]
[1]
[1,2,4]
[1,3]
[1,2,5]
a number with the most divisors within 1 to 20000 (number, count):
(18480,79)
```

Or, for a little more efficiency, filtering only up to the root, and deriving the higher proper divisors from the lower ones, as quotients:

```import Data.List (maximumBy)
import Data.Ord (comparing)
import Data.Bool (bool)

properDivisors
:: Integral a
=> a -> [a]
properDivisors n =
let root = (floor . sqrt . fromIntegral) n
lows = filter ((0 ==) . rem n) [1 .. root]
in init (lows ++ bool id tail (n == root * root) (reverse (quot n <\$> lows)))

main :: IO ()
main = do
putStrLn "Proper divisors of 1 to 10:"
mapM_ (print . properDivisors) [1 .. 10]
mapM_
putStrLn
[ ""
, "A number in the range 1 to 20,000 with the most proper divisors,"
, "as (number, count of proper divisors):"
, ""
]
print \$
maximumBy (comparing snd) \$
(,) <*> (length . properDivisors) <\$> [1 .. 20000]
```

{{Out}}

```Proper divisors of 1 to 10:
[]
[1]
[1]
[1,2]
[1]
[1,2,3]
[1]
[1,2,4]
[1,3]
[1,2,5]

A number in the range 1 to 20,000 with the most proper divisors,
as (number, count of proper divisors):

(18480,79)
```

## J

The proper divisors of an integer are the [[Factors of an integer]] without the integer itself.

So, borrowing from [[Factors of an integer#J|the J implementation]] of that related task:

```factors=: [: /:~@, */&>@{@((^ i.@>:)&.>/)@q:~&__
properDivisors=: factors -. ]
```

Proper divisors of numbers 1 through 10:

```   (,&": ' -- ' ,&": properDivisors)&>1+i.10
1 --
2 -- 1
3 -- 1
4 -- 1 2
5 -- 1
6 -- 1 2 3
7 -- 1
8 -- 1 2 4
9 -- 1 3
10 -- 1 2 5
```

Number(s) not exceeding 20000 with largest number of proper divisors (and the count of those divisors):

```   (, #@properDivisors)&> 1+I.(= >./) #@properDivisors@> 1+i.20000
15120 79
18480 79
```

Note that it's a bit more efficient to simply count factors here, when selecting the candidate numbers.

```      (, #@properDivisors)&> 1+I.(= >./) #@factors@> 1+i.20000
15120 79
18480 79
```

We could also arbitrarily toss either 15120 or 18480 (keeping the other number), if it were important that we produce only one result.

## Java

{{works with|Java|1.5+}}

```import java.util.Collections;
import java.util.List;

public class Proper{
public static List<Integer> properDivs(int n){
if(n == 1) return divs;
for(int x = 2; x < n; x++){
if(n % x == 0) divs.add(x);
}

Collections.sort(divs);

return divs;
}

public static void main(String[] args){
for(int x = 1; x <= 10; x++){
System.out.println(x + ": " + properDivs(x));
}

int x = 0, count = 0;
for(int n = 1; n <= 20000; n++){
if(properDivs(n).size() > count){
x = n;
count = properDivs(n).size();
}
}
System.out.println(x + ": " + count);
}
}
```

{{out}}

```1: []
2: [1]
3: [1]
4: [1, 2]
5: [1]
6: [1, 2, 3]
7: [1]
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120: 79
```

## JavaScript

### ES5

```(function () {

// Proper divisors
function properDivisors(n) {
if (n < 2) return [];
else {
var rRoot = Math.sqrt(n),
intRoot = Math.floor(rRoot),

lows = range(1, intRoot).filter(function (x) {
return (n % x) === 0;
});

return lows.concat(lows.slice(1).map(function (x) {
return n / x;
}).reverse().slice((rRoot === intRoot) | 0));
}
}

// [m..n]
function range(m, n) {
var a = Array(n - m + 1),
i = n + 1;
while (i--) a[i - 1] = i;
return a;
}

var tblOneToTen = [
['Number', 'Proper Divisors', 'Count']
].concat(range(1, 10).map(function (x) {
var ds = properDivisors(x);

return [x, ds.join(', '), ds.length];
})),

dctMostBelow20k = range(1, 20000).reduce(function (a, x) {
var lng = properDivisors(x).length;

return lng > a.divisorCount ? {
n: x,
divisorCount: lng
} : a;
}, {
n: 0,
divisorCount: 0
});

// [[a]] -> bool -> s -> s
return '{| class="wikitable" ' + (
strStyle ? 'style="' + strStyle + '"' : ''
) + lstRows.map(function (lstRow, iRow) {
var strDelim = ((blnHeaderRow && !iRow) ? '!' : '|');

return '\n|-\n' + strDelim + ' ' + lstRow.map(function (v) {
return typeof v === 'undefined' ? ' ' : v;
}).join(' ' + strDelim + strDelim + ' ');
}).join('') + '\n|}';
}

return wikiTable(
tblOneToTen,
true
) + '\n\nMost proper divisors below 20,000:\n\n  ' + JSON.stringify(
dctMostBelow20k
);

})();
```

{{out}}

{| class="wikitable" |- ! Number !! Proper Divisors !! Count |- | 1 || || 0 |- | 2 || 1 || 1 |- | 3 || 1 || 1 |- | 4 || 1, 2 || 2 |- | 5 || 1 || 1 |- | 6 || 1, 2, 3 || 3 |- | 7 || 1 || 1 |- | 8 || 1, 2, 4 || 3 |- | 9 || 1, 3 || 2 |- | 10 || 1, 2, 5 || 3 |}

Most proper divisors below 20,000:

{"n":15120,"divisorCount":79}

### ES6

```(() => {

// properDivisors :: Int -> [Int]
let properDivisors = n => {
let rRoot = Math.sqrt(n),
intRoot = Math.floor(rRoot),
blnPerfectSquare = rRoot === intRoot,

lows = range(1, intRoot)
.filter(x => (n % x) === 0);

// for perfect squares, we can drop
// the head of the 'highs' list
return lows.concat(lows
.map(x => n / x)
.reverse()
.slice(blnPerfectSquare | 0)
)
.slice(0, -1); // except n itself
},

// range :: Int -> Int -> [Int]
range = (m, n) => Array.from({
length: (n - m) + 1
}, (_, i) => m + i);

return {
properDivisorsOf1to10: range(1, 10)
.reduce((a, x) => (
a[x.toString()] = properDivisors(x),
a
), {}),

intMaxDivisorsUnder20k: range(1, 20000)
.reduce((a, x) => {
let intDivisors = properDivisors(x)
.length;

return intDivisors >= a.divisors ? {
max: x,
divisors: intDivisors
} : a;

}, {
max: 0,
divisors: 0
})
};

})();
```

{{Out}}

```{
"properDivisorsOf1to10":{
"1":[], "2":[1], "3":[1], "4":[1, 2], "5":[1],
"6":[1, 2, 3], "7":[1], "8":[1, 2, 4], "9":[1, 3], "10":[1, 2, 5]
},
"intMaxDivisorsUnder20k":{"max":18480, "divisors":79}
}
```

## jq

{{works with|jq|1.4}} In the following, proper_divisors returns a stream. In order to count the number of items in the stream economically, we first define "count(stream)":

```def count(stream): reduce stream as \$i (0; . + 1);

# unordered
def proper_divisors:
. as \$n
| if \$n > 1 then 1,
( range(2; 1 + (sqrt|floor)) as \$i
| if (\$n % \$i) == 0 then \$i,
((\$n / \$i) | if . == \$i then empty else . end)
else empty
end)
else empty
end;

# The first integer in 1 .. n inclusive
# with the maximal number of proper divisors in that range:
def most_proper_divisors(n):
reduce range(1; n+1) as \$i
( [null, 0];
count( \$i | proper_divisors ) as \$count
| if \$count > .[1] then [\$i, \$count] else . end);
```

```"The proper divisors of the numbers 1 to 10 inclusive are:",
(range(1;11) as \$i | "\(\$i): \( [ \$i | proper_divisors] )"),
"",
"The pair consisting of the least number in the range 1 to 20,000 with",
"the maximal number proper divisors together with the corresponding",
"count of proper divisors is:",
most_proper_divisors(20000)
```

{{out}}

```\$ jq -n -c -r -f /Users/peter/jq/proper_divisors.jq
The proper divisors of the numbers 1 to 10 inclusive are:
1: []
2: [1]
3: [1]
4: [1,2]
5: [1]
6: [1,2,3]
7: [1]
8: [1,2,4]
9: [1,3]
10: [1,2,5]

The pair consisting of the least number in the range 1 to 20,000 with
the maximal number proper divisors together with the corresponding
count of proper divisors is:
[15120,79]
```

## Julia

Use `factor` to obtain the prime factorization of the target number. I adopted the argument handling style of `factor` in my `properdivisors` function.

```
function properdivisors{T<:Integer}(n::T)
0 < n || throw(ArgumentError("number to be factored must be ≥ 0, got \$n"))
1 < n || return T[]
!isprime(n) || return T[one(T), n]
f = factor(n)
d = T[one(T)]
for (k, v) in f
c = T[k^i for i in 0:v]
d = d*c'
d = reshape(d, length(d))
end
sort!(d)
return d[1:end-1]
end

lo = 1
hi = 10
println("List the proper divisors for ", lo, " through ", hi, ".")
for i in lo:hi
println(@sprintf("%4d", i), " ", properdivisors(i))
end

hi = 2*10^4
println("\nFind the numbers within [", lo, ",", hi, "] having the most divisors.")

maxdiv = 0
nlst = Int[]

for i in lo:hi
ndiv = length(properdivisors(i))
if ndiv > maxdiv
maxdiv = ndiv
nlst = [i]
elseif ndiv == maxdiv
push!(nlst, i)
end
end

println(nlst, " have the maximum proper divisor count of ", maxdiv, ".")

```

{{out}}

```
List the proper divisors for 1 through 10.
1 []
2 [1,2]
3 [1,3]
4 [1,2]
5 [1,5]
6 [1,2,3]
7 [1,7]
8 [1,2,4]
9 [1,3]
10 [1,2,5]

Find the numbers within [1,20000] having the most divisors.
[15120,18480] have the maximum proper divisor count of 79.

```

## Kotlin

```// version 1.0.5-2

fun listProperDivisors(limit: Int) {
if (limit < 1) return
for(i in 1..limit) {
if (i == 1) {
println("(None)")
continue
}
(1..i/2).filter{ i % it == 0 }.forEach { print(" \$it") }
println()
}
}

fun countProperDivisors(n: Int): Int {
if (n < 2) return 0
return (1..n/2).count { (n % it) == 0 }
}

fun main(args: Array<String>) {
println("The proper divisors of the following numbers are :\n")
listProperDivisors(10)
println()
var count: Int
var maxCount = 0
val most: MutableList<Int> = mutableListOf(1)
for (n in 2..20000) {
count = countProperDivisors(n)
if (count == maxCount)
else if (count > maxCount) {
maxCount = count
most.clear()
}
}
println("The following number(s) have the most proper divisors, namely " + maxCount + "\n")
for (n in most) println(n)
}
```

{{out}}

```
The proper divisors of the following numbers are :

1 -> (None)
2 ->  1
3 ->  1
4 ->  1 2
5 ->  1
6 ->  1 2 3
7 ->  1
8 ->  1 2 4
9 ->  1 3
10 ->  1 2 5

The following number(s) have the most proper divisors, namely 79

15120
18480

```

## Lua

```-- Return a table of the proper divisors of n
function propDivs (n)
if n < 2 then return {} end
local divs, sqr = {1}, math.sqrt(n)
for d = 2, sqr do
if n % d == 0 then
table.insert(divs, d)
if d ~= sqr then table.insert(divs, n/d) end
end
end
table.sort(divs)
return divs
end

-- Show n followed by all values in t
function show (n, t)
io.write(n .. ":\t")
for _, v in pairs(t) do io.write(v .. " ") end
print()
end

-- Main procedure
local mostDivs, numDivs, answer = 0
for i = 1, 10 do show(i, propDivs(i)) end
for i = 1, 20000 do
numDivs = #propDivs(i)
if numDivs > mostDivs then
mostDivs = numDivs
end
end
print(answer .. " has " .. mostDivs .. " proper divisors.")
```

{{out}}

```1:
2:      1
3:      1
4:      1 2
5:      1
6:      1 2 3
7:      1
8:      1 2 4
9:      1 3
10:     1 2 5
15120 has 79 proper divisors.
```

A Function that yields the proper divisors of an integer n:

```ProperDivisors[n_Integer /; n > 0] := Most@Divisors@n;
```

Proper divisors of n from 1 to 10:

```Grid@Table[{n, ProperDivisors[n]}, {n, 1, 10}]
```

{{out}}

```1	{}
2	{1}
3	{1}
4	{1,2}
5	{1}
6	{1,2,3}
7	{1}
8	{1,2,4}
9	{1,3}
10	{1,2,5}
```

The number with the most divisors between 1 and 20,000:

```Fold[
Last[SortBy[{#1, {#2, Length@ProperDivisors[#2]}}, Last]] &,
{0, 0},
Range[20000]]
```

{{out}}

```{18480, 79}
```

An alternate way to find the number with the most divisors between 1 and 20,000:

```Last@SortBy[
Table[
{n, Length@ProperDivisors[n]},
{n, 1, 20000}],
Last]
```

{{out}}

```{15120, 79}
```

## Matlab

```
function D=pd(N)
K=1:ceil(N/2);
D=K(~(rem(N, K)));

```

{{out}}

```
for I=1:10
disp([num2str(I) ' : ' num2str(pd(I))])
end
1 : 1
2 : 1
3 : 1
4 : 1  2
5 : 1
6 : 1  2  3
7 : 1
8 : 1  2  4
9 : 1  3
10 : 1  2  5

maxL=0; maxI=0;
for I=1:20000
L=length(pd(I));
if L>maxL
maxL=L; maxI=I;
end
end
maxI

maxI =

15120

maxL

maxL =

79

```

```MODULE ProperDivisors;
FROM FormatString IMPORT FormatString;

PROCEDURE WriteInt(n : INTEGER);
VAR buf : ARRAY[0..15] OF CHAR;
BEGIN
FormatString("%i", buf, n);
WriteString(buf)
END WriteInt;

PROCEDURE proper_divisors(n : INTEGER; print_flag : BOOLEAN) : INTEGER;
VAR count,i : INTEGER;
BEGIN
count := 0;
FOR i:=1 TO n-1 DO
IF n MOD i = 0 THEN
INC(count);
IF print_flag THEN
WriteInt(i);
WriteString(" ")
END
END
END;
IF print_flag THEN WriteLn END;
RETURN count;
END proper_divisors;

VAR
buf : ARRAY[0..63] OF CHAR;
i,max,max_i,v : INTEGER;
BEGIN
FOR i:=1 TO 10 DO
WriteInt(i);
WriteString(": ");
proper_divisors(i, TRUE)
END;

max := 0;
max_i := 1;

FOR i:=1 TO 20000 DO
v := proper_divisors(i, FALSE);
IF v>= max THEN
max := v;
max_i := i
END
END;

FormatString("%i with %i divisors\n", buf, max_i, max);
WriteString(buf);

END ProperDivisors.
```

## Objeck

```use Collection;

class Proper{
function : Main(args : String[]) ~ Nil {
for(x := 1; x <= 10; x++;) {
Print(x, ProperDivs(x));
};

x := 0;
count := 0;

for(n := 1; n <= 20000; n++;) {
if(ProperDivs(n)->Size() > count) {
x := n;
count := ProperDivs(n)->Size();
};
};
"{\$x}: {\$count}"->PrintLine();
}

function : ProperDivs(n : Int) ~ IntVector {
divs := IntVector->New();

if(n = 1) {
return divs;
};

for(x := 2; x < n; x++;) {
if(n % x = 0) {
};
};
divs->Sort();

return divs;
}

function : Print(x : Int, result : IntVector) ~ Nil {
"{\$x}: "->Print();
result->ToArray()->ToString()->PrintLine();
}
}

```

Output:

```
1: []
2: [1]
3: [1]
4: [1,2]
5: [1]
6: [1,2,3]
7: [1]
8: [1,2,4]
9: [1,3]
10: [1,2,5]
15120: 79

```

```
MODULE ProperDivisors;
IMPORT
Out;

CONST
initialSize = 128;
TYPE
Result* = POINTER TO ResultDesc;
ResultDesc = RECORD
found-: LONGINT; (* number of slots in pd *)
pd-: POINTER TO ARRAY OF LONGINT;
cap: LONGINT;   (* Capacity *)
END;

VAR
i,found,max,idxMx: LONGINT;
mx: ARRAY 32 OF LONGINT;
rs: Result;

PROCEDURE (r: Result) Init(size: LONGINT);
BEGIN
r.found := 0;
r.cap := size;
NEW(r.pd,r.cap);
END Init;

BEGIN
(* Out.String("--->");Out.LongInt(n,0);Out.String(" At: ");Out.LongInt(r.found,0);Out.Ln; *)
IF (r.found < LEN(r.pd^) - 1) THEN
r.pd[r.found] := n;
ELSE
(* expand pd for more room *)
END;
INC(r.found);

PROCEDURE (r:Result) Show();
VAR
i: LONGINT;
BEGIN
Out.String("(Result:");Out.LongInt(r.found + 1,0);(* Out.String("/");Out.LongInt(r.cap,0);*)
Out.String("-");
IF r.found > 0 THEN
FOR i:= 0 TO r.found - 1 DO
Out.LongInt(r.pd[i],0);
IF i = r.found - 1 THEN Out.Char(')') ELSE Out.Char(',') END
END
END;
Out.Ln
END Show;

PROCEDURE (r:Result) Reset();
BEGIN
r.found := 0;
END Reset;

PROCEDURE GetFor(n: LONGINT;VAR rs: Result);
VAR
i: LONGINT;
BEGIN
IF n > 1 THEN
WHILE (i < n) DO
IF (n MOD i) = 0 THEN rs.Add(i) END;
INC(i)
END
END;
END GetFor;

BEGIN
NEW(rs);rs.Init(initialSize);
FOR i := 1 TO 10 DO
Out.LongInt(i,4);Out.Char(':');
GetFor(i,rs);
rs.Show();
rs.Reset();
END;
Out.LongInt(100,4);Out.Char(':');GetFor(100,rs);rs.Show();rs.Reset();
max := 0;idxMx := 0;found := 0;
FOR i := 1 TO 20000 DO
GetFor(i,rs);
IF rs.found > max THEN
idxMx:= 0;mx[idxMx] := i;max := rs.found
ELSIF rs.found = max THEN
INC(idxMx);mx[idxMx] := i
END;
rs.Reset()
END;
Out.String("Found: ");Out.LongInt(idxMx + 1,0);
Out.String(" Numbers with most proper divisors ");
Out.LongInt(max,0);Out.String(": ");Out.Ln;
FOR i := 0 TO idxMx DO
Out.LongInt(mx[i],0);Out.Ln
END
END ProperDivisors.

```

{{out}}

```
1:(Result:1-
2:(Result:2-1)
3:(Result:2-1)
4:(Result:3-1,2)
5:(Result:2-1)
6:(Result:4-1,2,3)
7:(Result:2-1)
8:(Result:4-1,2,4)
9:(Result:3-1,3)
10:(Result:4-1,2,5)
100:(Result:9-1,2,4,5,10,20,25,50)
Found: 2 Numbers with most proper divisors 79:
15120
18480

```

## Oforth

```Integer method: properDivs  self 2 / seq filter(#[ self swap mod 0 == ]) }

10 seq apply(#[ dup print " : " print properDivs println ])
20000 seq map(#[ dup properDivs size Pair new ]) reduce(#maxKey) println
```

{{out}}

```
1 : []
2 : [1]
3 : [1]
4 : [1, 2]
5 : [1]
6 : [1, 2, 3]
7 : [1]
8 : [1, 2, 4]
9 : [1, 3]
10 : [1, 2, 5]
[79, 15120]

```

## PARI/GP

```proper(n)=if(n==1, [], my(d=divisors(n)); d[2..#d]);
apply(proper, [1..10])
r=at=0; for(n=1,20000, t=numdiv(n); if(t>r, r=t; at=n)); [at, numdiv(t)-1]
```

{{out}}

```%1 = [[], [2], [3], [2, 4], [5], [2, 3, 6], [7], [2, 4, 8], [3, 9], [2, 5, 10]]
%2 = [15120, 7]
```

## Pascal

{{works with|Free Pascal}} Using prime factorisation

```{\$IFDEF FPC}{\$MODE DELPHI}{\$ELSE}{\$APPTYPE CONSOLE}{\$ENDIF}
uses
sysutils;
const
MAXPROPERDIVS = 1920;
type
tRes = array[0..MAXPROPERDIVS] of LongWord;
tPot = record
potPrim,
potMax :LongWord;
end;

tprimeFac = record
pfPrims : array[1..10] of tPot;
pfCnt,
pfNum   : LongWord;
end;
tSmallPrimes = array[0..6541] of longWord;

var
SmallPrimes: tSmallPrimes;

procedure InitSmallPrimes;
var
pr,testPr,j,maxprimidx: Longword;
isPrime : boolean;
Begin
maxprimidx := 0;
SmallPrimes[0] := 2;
pr := 3;
repeat
isprime := true;
j := 0;
repeat
testPr := SmallPrimes[j];
IF testPr*testPr > pr then
break;
If pr mod testPr = 0 then
Begin
isprime := false;
break;
end;
inc(j);
until false;

if isprime then
Begin
inc(maxprimidx);
SmallPrimes[maxprimidx]:= pr;
end;
inc(pr,2);
until pr > 1 shl 16 -1;
end;

procedure PrimeFacOut(primeDecomp:tprimeFac);
var
i : LongWord;
begin
with primeDecomp do
Begin
write(pfNum,' = ');
For i := 1 to pfCnt-1 do
with pfPrims[i] do
If potMax = 1 then
write(potPrim,'*')
else
write(potPrim,'^',potMax,'*');
with pfPrims[pfCnt] do
If potMax = 1 then
write(potPrim)
else
write(potPrim,'^',potMax);
end;
end;

procedure PrimeDecomposition(n:LongWord;var res:tprimeFac);
var
i,pr,cnt,quot{to minimize divisions} : LongWord;
Begin
res.pfNum := n;
res.pfCnt:= 0;
i := 0;
cnt := 0;
repeat
pr := SmallPrimes[i];
IF pr*pr>n then
Break;

quot := n div pr;
IF pr*quot = n then
with res do
Begin
inc(pfCnt);
with pfPrims[pfCnt] do
Begin
potPrim := pr;
potMax := 0;
repeat
n := quot;
quot := quot div pr;
inc(potMax);
until pr*quot <> n;
end;
end;
inc(i);
until false;
//a big prime left over?
IF n <> 1 then
with res do
Begin
inc(pfCnt);
with pfPrims[pfCnt] do
Begin
potPrim := n;
potMax := 1;
end;
end;
end;

function CntProperDivs(const primeDecomp:tprimeFac):LongWord;
//count of proper divisors
var
i: LongWord;
begin
result := 1;
with primeDecomp do
For i := 1 to pfCnt do
result := result*(pfPrims[i].potMax+1);
//remove
dec(result);
end;

function findProperdivs(n:LongWord;var res:TRes):LongWord;
//simple trial division to get a sorted list of all proper divisors
var
i,j: LongWord;
Begin
result := 0;
i := 1;
j := n;
while j>i do
begin
j := n DIV i;
IF i*j = n then
Begin
//smaller factor part at the beginning upwards
res[result]:= i;
IF i <> j then
//bigger factor at the end downwards
res[MAXPROPERDIVS-result]:= j
else
//n is square number
res[MAXPROPERDIVS-result]:= 0;
inc(result);
end;
inc(i);
end;

If result>0 then
Begin
//move close together
i := result;
j := MAXPROPERDIVS-result+1;
result := 2*result-1;
repeat
res[i] := res[j];
inc(j);
inc(i);
until i > result;

if res[result-1] = 0 then
dec(result);
end;
end;

procedure AllFacsOut(n: Longword);
var
res:TRes;
i,k,j:LongInt;
Begin
j := findProperdivs(n,res);
write(n:5,' : ');
For k := 0 to j-2 do write(res[k],',');
IF j>=1 then
write(res[j-1]);
writeln;
end;

var
primeDecomp: tprimeFac;
rs : tRes;
i,j,max,maxcnt: LongWord;
BEGIN
InitSmallPrimes;
For i := 1 to 10 do
AllFacsOut(i);
writeln;
max    := 0;
maxCnt := 0;
For i := 1 to 20*1000 do
Begin
PrimeDecomposition(i,primeDecomp);
j := CntProperDivs(primeDecomp);
IF j> maxCnt then
Begin
maxcnt := j;
max := i;
end;
end;
PrimeDecomposition(max,primeDecomp);
j := CntProperDivs(primeDecomp);

PrimeFacOut(primeDecomp);writeln('  ',j:10,' factors'); writeln;
//https://en.wikipedia.org/wiki/Highly_composite_number <= HCN
//http://wwwhomes.uni-bielefeld.de/achim/highly.txt the first 1200 HCN
max := 3491888400;
PrimeDecomposition(max,primeDecomp);
j := CntProperDivs(primeDecomp);
PrimeFacOut(primeDecomp);writeln('  ',j:10,' factors'); writeln;
END.
```

{{Output}}

```
1 :
2 : 1
3 : 1
4 : 1,2
5 : 1
6 : 1,2,3
7 : 1
8 : 1,2,4
9 : 1,3
10 : 1,2,5

15120 = 2^4*3^3*5*7          79 factors

3491888400 = 2^4*3^3*5^2*7*11*13*17*19        1919 factors

real    0m0.004s
```

## Perl

### Using a module for divisors

```use ntheory qw/divisors/;
sub proper_divisors {
my \$n = shift;
# Like Pari/GP, divisors(0) = (0,1) and divisors(1) = ()
return 1 if \$n == 0;
my @d = divisors(\$n);
pop @d;  # divisors are in sorted order, so last entry is the input
@d;
}
say "\$_: ", join " ", proper_divisors(\$_) for 1..10;
# 1. For the max, we can do a traditional loop.
my(\$max,\$ind) = (0,0);
for (1..20000) {
my \$nd = scalar proper_divisors(\$_);
(\$max,\$ind) = (\$nd,\$_) if \$nd > \$max;
}
say "\$max \$ind";
# 2. Or we can use List::Util's max with decoration (this exploits its implementation)
{
use List::Util qw/max/;
no warnings 'numeric';
say max(map { scalar(proper_divisors(\$_)) . " \$_" } 1..20000);
}
```

{{out}}

```1:
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
79 15120
79 18480
```

Note that the first code will choose the first max, while the second chooses the last.

## Perl 6

{{Works with|rakudo|2018.10}} Once your threshold is over 1000, the maximum proper divisors will always include 2, 3 and 5 as divisors, so only bother to check multiples of 2, 3 and 5.

There really isn't any point in using concurrency for a limit of 20_000. The setup and bookkeeping drowns out any benefit. Really doesn't start to pay off until the limit is 50_000 and higher. Try swapping in the commented out race map iterator line below for comparison.

```sub propdiv (\x) {
my @l = 1 if x > 1;
(2 .. x.sqrt.floor).map: -> \d {
unless x % d { @l.push: d; my \y = x div d; @l.push: y if y != d }
}
@l
}

put "\$_ [{propdiv(\$_)}]" for 1..10;

my @candidates;
loop (my int \$c = 30; \$c <= 20_000; \$c += 30) {
#(30, *+30 …^ * > 500_000).race.map: -> \$c {
my \mx = +propdiv(\$c);
next if mx < @candidates - 1;
@candidates[mx].push: \$c
}

say "max = {@candidates - 1}, candidates = {@candidates.tail}";
```

{{out}}

```1 []
2 [1]
3 [1]
4 [1 2]
5 [1]
6 [1 2 3]
7 [1]
8 [1 2 4]
9 [1 3]
10 [1 2 5]
max = 79, candidates = 15120 18480
```

## Phix

The factors routine is an auto-include. The actual implementation of it, from builtins\pfactors.e is

```global function factors(atom n, integer include1=0)
-- returns a list of all integer factors of n
--  if include1 is 0 (the default), result does not contain either 1 or n
--  if include1 is 1, and n>1, the result contains 1 and n
--  if include1 is -1, and n>1, the result contains 1 but not n
sequence lfactors = {}, hfactors = {}
atom hfactor
integer p = 2,
lim = floor(sqrt(n))

if n!=1 and include1!=0 then
lfactors = {1}
if include1=1 then
hfactors = {n}
end if
end if
while p<=lim do
if remainder(n,p)=0 then
lfactors = append(lfactors,p)
hfactor = n/p
if hfactor=p then exit end if
hfactors = prepend(hfactors,hfactor)
end if
p += 1
end while
return lfactors & hfactors
end function
```

The compiler knows where to find that, so the main program is just:

```for i=1 to 10 do
?{i,factors(i,-1)}
end for

integer maxd = 0, k
sequence candidates = {}

for i=1 to 20000 do
k = length(factors(i,-1))
if k>=maxd then
if k=maxd then
candidates &= i
else
candidates = {i}
maxd = k
end if
end if
end for

printf(1,"%d divisors:", maxd)
?candidates
{} = wait_key()
```

{{out}}

```
{1,{}}
{2,{1}}
{3,{1}}
{4,{1,2}}
{5,{1}}
{6,{1,2,3}}
{7,{1}}
{8,{1,2,4}}
{9,{1,3}}
{10,{1,2,5}}
79 divisors:{15120,18480}

```

## PHP

```<?php
function ProperDivisors(\$n) {
yield 1;
\$large_divisors = [];
for (\$i = 2; \$i <= sqrt(\$n); \$i++) {
if (\$n % \$i == 0) {
yield \$i;
if (\$i*\$i != \$n) {
\$large_divisors[] = \$n / \$i;
}
}
}
foreach (array_reverse(\$large_divisors) as \$i) {
yield \$i;
}
}

assert([1, 2, 4, 5, 10, 20, 25, 50] ==
iterator_to_array(ProperDivisors(100)));

foreach (range(1, 10) as \$n) {
echo "\$n =>";
foreach (ProperDivisors(\$n) as \$divisor) {
echo " \$divisor";
}
echo "\n";
}

\$divisorsCount = [];
for (\$i = 1; \$i < 20000; \$i++) {
\$divisorsCount[sizeof(iterator_to_array(ProperDivisors(\$i)))][] = \$i;
}
ksort(\$divisorsCount);

echo "Numbers with most divisors: ", implode(", ", end(\$divisorsCount)), ".\n";
echo "They have ", key(\$divisorsCount), " divisors.\n";

```

Outputs:

```1 => 1
2 => 1
3 => 1
4 => 1 2
5 => 1
6 => 1 2 3
7 => 1
8 => 1 2 4
9 => 1 3
10 => 1 2 5
Numbers with most divisors: 15120, 18480.
They have 79 divisors.
```

## PicoLisp

```# Generate all proper divisors.
(de propdiv (N)
'((X) (=0 (% N X)))
(range 1 N) )) )

# Obtaining the values from 1 to 10 inclusive.
(mapcar propdiv (range 1 10))
# Output:
# (NIL (1) (1) (1 2) (1) (1 2 3) (1) (1 2 4) (1 3) (1 2 5))
```

===Brute-force===

```(de propdiv (N)
(cdr
(rot
(make
(for I N
(and (=0 (% N I)) (link I)) ) ) ) ) )
(de countdiv (N)
(let C -1
(for I N
(and (=0 (% N I)) (inc 'C)) )
C ) )
(let F (-5 -8)
(tab F "N" "LIST")
(for I 10
(tab F
I
(glue " + " (propdiv I)) ) ) )
(println
(maxi
countdiv
(range 1 20000) ) )
```

### Factorization

```(de accu1 (Var Key)
(if (assoc Key (val Var))
(con @ (inc (cdr @)))
(push Var (cons Key 2)) )
Key )
(de factor (N)
(let
(R NIL
D 2
L (1 2 2 . (4 2 4 2 4 6 2 6 .))
M (sqrt N) )
(while (>= M D)
(if (=0 (% N D))
(setq M
(sqrt (setq N (/ N (accu1 'R D)))) )
(inc 'D (pop 'L)) ) )
(accu1 'R N)
(dec (apply * (mapcar cdr R))) ) )
(bench
(println
(maxi
factor
(range 1 20000) )
@@ ) )
```

Output:

```
15120 79
0.081 sec

```

## PL/I

```*process source xref;
(subrg):
cpd: Proc Options(main);
p9a=time();
Dcl (p9a,p9b) Pic'(9)9';
Dcl cnt(3) Bin Fixed(31) Init((3)0);
Dcl x Bin Fixed(31);
Dcl pd(300) Bin Fixed(31);
Dcl sumpd   Bin Fixed(31);
Dcl npd     Bin Fixed(31);
Dcl hi      Bin Fixed(31) Init(0);
Dcl (xl(10),xi) Bin Fixed(31);
Dcl i       Bin Fixed(31);
Do x=1 To 10;
Call proper_divisors(x,pd,npd);
Put Edit(x,' -> ',(pd(i) Do i=1 To npd))(Skip,f(2),a,10(f(2)));
End;
xi=0;
Do x=1 To 20000;
Call proper_divisors(x,pd,npd);
Select;
When(npd>hi) Do;
xi=1;
xl(1)=x;
hi=npd;
End;
When(npd=hi) Do;
xi+=1;
xl(xi)=x;
End;
Otherwise;
End;
End;
Put Edit(hi,' -> ',(xl(i) Do i=1 To xi))(Skip,f(3),a,10(f(6)));

x= 166320; Call proper_divisors(x,pd,npd);
Put Edit(x,' -> ',npd)(Skip,f(8),a,f(4));
x=1441440; Call proper_divisors(x,pd,npd);
Put Edit(x,' -> ',npd)(Skip,f(8),a,f(4));

p9b=time();
Put Edit((p9b-p9a)/1000,' seconds elapsed')(Skip,f(6,3),a);
Return;

proper_divisors: Proc(n,pd,npd);
Dcl (n,pd(300),npd) Bin Fixed(31);
Dcl (d,delta)       Bin Fixed(31);
npd=0;
If n>1 Then Do;
If mod(n,2)=1 Then  /* odd number  */
delta=2;
Else                /* even number */
delta=1;
Do d=1 To n/2 By delta;
If mod(n,d)=0 Then Do;
npd+=1;
pd(npd)=d;
End;
End;
End;
End;

End;
```

{{out}}

```
1 ->
2 ->  1
3 ->  1
4 ->  1 2
5 ->  1
6 ->  1 2 3
7 ->  1
8 ->  1 2 4
9 ->  1 3
10 ->  1 2 5
79 ->  15120 18480
166320 ->  159
1441440 ->  287
0.530 seconds elapsed
```

## PowerShell

### version 1

```
function proper-divisor (\$n) {
if(\$n -ge 2) {
\$lim = [Math]::Floor([Math]::Sqrt(\$n))
\$less, \$greater = @(1), @()
for(\$i = 2; \$i -lt \$lim; \$i++){
if(\$n%\$i -eq 0) {
\$less += @(\$i)
\$greater = @(\$n/\$i) + \$greater
}
}
if((\$lim -ne 1) -and (\$n%\$lim -eq 0)) {\$less += @(\$lim)}
\$(\$less + \$greater)
} else {@()}
}
"\$(proper-divisor 100)"
"\$(proper-divisor 496)"
"\$(proper-divisor 2048)"

```

### version 2

```
function proper-divisor (\$n) {
if(\$n -ge 2) {
\$lim = [Math]::Floor(\$n/2)+1
\$proper = @(1)
for(\$i = 2; \$i -lt \$lim; \$i++){
if(\$n%\$i -eq 0) {
\$proper += @(\$i)
}
}
\$proper
} else {@()}
}
"\$(proper-divisor 100)"
"\$(proper-divisor 496)"
"\$(proper-divisor 2048)"

```

### version 3

```
function eratosthenes (\$n) {
if(\$n -gt 1){
\$prime = @(0..\$n| foreach{\$true})
\$m = [Math]::Floor([Math]::Sqrt(\$n))
function multiple(\$i) {
for(\$j = \$i*\$i; \$j -le \$n; \$j += \$i) {
\$prime[\$j] = \$false
}
}
multiple 2
for(\$i = 3; \$i -le \$m; \$i += 2) {
if(\$prime[\$i]) {multiple \$i}
}
2
for(\$i = 3; \$i -le \$n; \$i += 2) {
if(\$prime[\$i]) {\$i}
}

} else {
Write-Error "\$n is not greater than 1"
}
}
function prime-decomposition (\$n) {
\$array = eratosthenes \$n
\$prime = @()
foreach(\$p in \$array) {
while(\$n%\$p -eq 0) {
\$n /= \$p
\$prime += @(\$p)
}
}
\$prime
}
function proper-divisor (\$n) {
if(\$n -ge 2) {
\$array = prime-decomposition \$n
\$lim = \$array.Count
function state(\$res, \$i){
if(\$i -lt \$lim) {
state (\$res) (\$i + 1)
state (\$res*\$array[\$i]) (\$i + 1)
} elseif(\$res -lt \$n) {\$res}
}
state 1 0 | sort -Unique
} else {@()}
}
"\$(proper-divisor 100)"
"\$(proper-divisor 496)"
"\$(proper-divisor 2048)"

```

Output:

```
1 2 4 5 10 20 25 50
1 2 4 8 16 31 62 124 248
1 2 4 8 16 32 64 128 256 512 1024

```

## Prolog

{{works with | SWI-Prolog 7}}

Taking a cue from [http://stackoverflow.com/a/171779 an SO answer]:

```divisor(N, Divisor) :-
UpperBound is round(sqrt(N)),
between(1, UpperBound, D),
0 is N mod D,
(
Divisor = D
;
LargerDivisor is N/D,
LargerDivisor =\= D,
Divisor = LargerDivisor
).

proper_divisor(N, D) :-
divisor(N, D),
D =\= N.

%

proper_divisors(N, Ds) :-
setof(D, proper_divisor(N, D), Ds).

%

show_proper_divisors_of_range(Low, High) :-
findall( N:Ds,
( between(Low, High, N),
proper_divisors(N, Ds) ),
Results ),
maplist(writeln, Results).

%

proper_divisor_count(N, Count) :-
proper_divisors(N, Ds),
length(Ds, Count).

find_most_proper_divisors_in_range(Low, High, Result) :-
aggregate_all( max(Count, N),
( between(Low, High, N),
proper_divisor_count(N, Count) ),
max(MaxCount, Num) ),
Result = (num(Num)-divisor_count(MaxCount)).
```

Output:

```?- show_proper_divisors_of_range(1,10).
2:[1]
3:[1]
4:[1,2]
5:[1]
6:[1,2,3]
7:[1]
8:[1,2,4]
9:[1,3]
10:[1,2,5]
true.

?- find_most_proper_divisors_in_range(1,20000,Result).
Result = num(15120)-divisor_count(79).

```

## PureBasic

```
EnableExplicit

Procedure ListProperDivisors(Number, List Lst())
If Number < 2 : ProcedureReturn : EndIf
Protected i
For i = 1 To Number / 2
If Number % i = 0
Lst() = i
EndIf
Next
EndProcedure

Procedure.i CountProperDivisors(Number)
If Number < 2 : ProcedureReturn 0 : EndIf
Protected i, count = 0
For i = 1 To Number / 2
If Number % i = 0
count + 1
EndIf
Next
ProcedureReturn count
EndProcedure

Define n, count, most = 1, maxCount = 0
If OpenConsole()
PrintN("The proper divisors of the following numbers are : ")
PrintN("")
NewList lst()
For n = 1 To 10
ListProperDivisors(n, lst())
Print(RSet(Str(n), 3) + " -> ")
If ListSize(lst()) = 0
Print("(None)")
Else
ForEach lst()
Print(Str(lst()) + " ")
Next
EndIf
ClearList(lst())
PrintN("")
Next
For n = 2 To 20000
count = CountProperDivisors(n)
If count > maxCount
maxCount = count
most = n
EndIf
Next
PrintN("")
PrintN(Str(most) + " has the most proper divisors, namely " + Str(maxCount))
PrintN("")
PrintN("Press any key to close the console")
Repeat: Delay(10) : Until Inkey() <> ""
CloseConsole()
EndIf

```

{{out}}

```
The proper divisors of the following numbers are :

1 -> (None)
2 -> 1
3 -> 1
4 -> 1 2
5 -> 1
6 -> 1 2 3
7 -> 1
8 -> 1 2 4
9 -> 1 3
10 -> 1 2 5

15120 has the most proper divisors, namely 79

```

## Python

### Python: Literal

A very literal interpretation

``` def proper_divs2(n):
...     return {x for x in range(1, (n + 1) // 2 + 1) if n % x == 0 and n != x}
...
>>> [proper_divs2(n) for n in range(1, 11)]
[set(), {1}, {1}, {1, 2}, {1}, {1, 2, 3}, {1}, {1, 2, 4}, {1, 3}, {1, 2, 5}]
>>>
>>> n, length = max(((n, len(proper_divs2(n))) for n in range(1, 20001)), key=lambda pd: pd[1])
>>> n
15120
>>> length
79
>>>
```

### Python: From prime factors

I found [http://stackoverflow.com/a/171784/10562 a reference] on how to generate factors from all the prime factors and the number of times each prime factor goes into N - its multiplicity.

For example, given N having prime factors P(i) with associated multiplicity M(i}) then the factors are given by:

```
for m[0] in range(M(0) + 1):
for m[1] in range(M[1] + 1):
...
for m[i - 1] in range(M[i - 1] + 1):
mult = 1
for j in range(i):
mult *= P[j] ** m[j]
yield mult
```

This version is over an order of magnitude faster for generating the proper divisors of the first 20,000 integers; at the expense of simplicity.

```from math import sqrt
from functools import lru_cache, reduce
from collections import Counter
from itertools import product

MUL = int.__mul__

def prime_factors(n):
'Map prime factors to their multiplicity for n'
d = _divs(n)
d = [] if d == [n] else (d[:-1] if d[-1] == d else d)
pf = Counter(d)
return dict(pf)

@lru_cache(maxsize=None)
def _divs(n):
'Memoized recursive function returning prime factors of n as a list'
for i in range(2, int(sqrt(n)+1)):
d, m  = divmod(n, i)
if not m:
return [i] + _divs(d)
return [n]

def proper_divs(n):
'''Return the set of proper divisors of n.'''
pf = prime_factors(n)
pfactors, occurrences = pf.keys(), pf.values()
multiplicities = product(*(range(oc + 1) for oc in occurrences))
divs = {reduce(MUL, (pf**m for pf, m in zip(pfactors, multis)), 1)
for multis in multiplicities}
try:
divs.remove(n)
except KeyError:
pass
return divs or ({1} if n != 1 else set())

if __name__ == '__main__':
rangemax = 20000

print([proper_divs(n) for n in range(1, 11)])
print(*max(((n, len(proper_divs(n))) for n in range(1, 20001)), key=lambda pd: pd[1]))
```

{{out}}

```[set(), {1}, {1}, {1, 2}, {1}, {1, 2, 3}, {1}, {1, 2, 4}, {1, 3}, {1, 2, 5}]
15120 79
```

## R

{{Works with|R|3.3.2 and above}}

```
# Proper divisors. 12/10/16 aev
require(numbers);
V <- sapply(1:20000, Sigma, k = 0, proper = TRUE); ind <- which(V==max(V));
cat("  *** max number of divisors:", max(V), "\n"," *** for the following indices:",ind, "\n");

```

{{Output}}

```
*** max number of divisors: 79
*** for the following indices: 15120 18480

```

## Racket

### Short version

```#lang racket
(require math)
(define (proper-divisors n) (drop-right (divisors n) 1))
(for ([n (in-range 1 (add1 10))])
(printf "proper divisors of: ~a\t~a\n" n (proper-divisors n)))
(define most-under-20000
(for/fold ([best '(1)]) ([n (in-range 2 (add1 20000))])
(define divs (proper-divisors n))
(if (< (length (cdr best)) (length divs)) (cons n divs) best)))
(printf "~a has ~a proper divisors\n"
(car most-under-20000) (length (cdr most-under-20000)))
```

{{out}}

```proper divisors of: 1	()
proper divisors of: 2	(1)
proper divisors of: 3	(1)
proper divisors of: 4	(1 2)
proper divisors of: 5	(1)
proper divisors of: 6	(1 2 3)
proper divisors of: 7	(1)
proper divisors of: 8	(1 2 4)
proper divisors of: 9	(1 3)
proper divisors of: 10	(1 2 5)
15120 has 79 proper divisors
```

### Long version

The '''main''' module will only be executed when this file is executed. When used as a library, it will not be used.

```#lang racket/base
(provide fold-divisors ; name as per "Abundant..."
proper-divisors)

(define (fold-divisors v n k0 kons)
(cond
[(>= n1 (vector-length v))
(define rv (make-vector n1 k0))
(for* ([n (in-range 1 n1)] [m (in-range (* 2 n) n1 n)])
(vector-set! rv m (kons n (vector-ref rv m))))
rv]
[else v]))

(define proper-divisors
(let ([p.d-v (vector)])
(λ (n)
(set! p.d-v (reverse (fold-divisors p.d-v n null cons)))
(vector-ref p.d-v n))))

(module+ main
(for ([n (in-range 1 (add1 10))])
(printf "proper divisors of: ~a\t~a\n" n (proper-divisors n)))

(define count-proper-divisors
(let ([p.d-v (vector)])
(λ(n) (set! p.d-v (fold-divisors p.d-v n 0 (λ (d n) (add1 n))))
(vector-ref p.d-v n))))

(void (count-proper-divisors 20000))

(define-values [C I]
(for*/fold ([C 0] [I 1])
[c (in-value (count-proper-divisors i))]
#:when [> c C])
(values c i)))
(printf "~a has ~a proper divisors\n" I C))
```

The output is the same as the short version above.

## REXX

### version 1

```Call time 'R'
Do x=1 To 10
Say x '->' proper_divisors(x)
End

hi=1
Do x=1 To 20000
/* If x//1000=0 Then Say x */
npd=count_proper_divisors(x)
Select
When npd>hi Then Do
list.npd=x
hi=npd
End
When npd=hi Then
list.hi=list.hi x
Otherwise
Nop
End
End

Say hi '->' list.hi

Say ' 166320 ->' count_proper_divisors(166320)
Say '1441440 ->' count_proper_divisors(1441440)

Say time('E') 'seconds elapsed'
Exit

proper_divisors: Procedure
Parse Arg n
If n=1 Then Return ''
pd=''
/* Optimization reduces 37 seconds to 28 seconds */
If n//2=1 Then  /* odd number  */
delta=2
Else            /* even number */
delta=1
Do d=1 To n%2 By delta
If n//d=0 Then
pd=pd d
End
Return space(pd)

count_proper_divisors: Procedure
Parse Arg n
Return words(proper_divisors(n))
```

{{out}}

```1 ->
2 -> 1
3 -> 1
4 -> 1 2
5 -> 1
6 -> 1 2 3
7 -> 1
8 -> 1 2 4
9 -> 1 3
10 -> 1 2 5
79 -> 15120 18480
166320 -> 159
1441440 -> 287
28.342000 seconds elapsed
```

### version 2

The following REXX version is an adaptation of the ''optimized'' version for the REXX language example for ''Factors of an integer''.

This REXX version handles all integers (negative, zero, positive) and automatically adjusts the precision (decimal digits).

It also allows the specification of the ranges (for display and for finding the maximum), and allows for extra numbers to be

specified.

With the (function) optimization, it's over '''20''' times faster.

```/*REXX program finds proper divisors (and count) of integer ranges; finds the max count.*/
parse arg bot top inc range xtra                 /*obtain optional arguments from the CL*/
if   bot=='' |   bot==","  then    bot=     1    /*Not specified?  Then use the default.*/
if   top=='' |   top==","  then    top=    10    /* "      "         "   "   "     "    */
if   inc=='' |   inc==","  then    inc=     1    /* "      "         "   "   "     "    */
if range=='' | range==","  then  range= 20000    /* "      "         "   "   "     "    */
w= max( length(top), length(bot), length(range)) /*determine the biggest number of these*/
numeric digits max(9, w + 1)                     /*have enough digits for  //  operator.*/
@.= 'and'                                        /*a literal used to separate #s in list*/
do n=bot  to top  by inc                   /*process the first range specified.   */
q= Pdivs(n);    #= words(q)                /*get proper divs; get number of Pdivs.*/
if q=='∞'  then #= q                       /*adjust number of Pdivisors for zero. */
say right(n, max(20, w) )   'has'   center(#, 4)     "proper divisors: "    q
end   /*n*/
m=0                                              /*M ≡ maximum number of Pdivs (so far).*/
do r=1  for range;    q= Pdivs(r)          /*process the second range specified.  */
#= words(q);          if #<m  then iterate /*get proper divs; get number of Pdivs.*/
if #<m  then iterate                       /*Less then max?   Then ignore this #. */
@.#= @.#  @.  r;      m=#                  /*add this Pdiv to max list; set new M.*/
end   /*r*/                                /* [↑]   process 2nd range of integers.*/
say
say m  ' is the highest number of proper divisors in range 1──►'range,
", and it's for: "       subword(@.m, 3)
say                                              /* [↓]  handle any given extra numbers.*/
do i=1  for words(xtra);  n= word(xtra, i) /*obtain an extra number from XTRA list*/
w= max(w, 1 + length(n) )                  /*use maximum width for aligned output.*/
numeric digits max(9, 1 + length(n) )      /*have enough digits for  //  operator.*/
q= Pdivs(n);              #= words(q)      /*get proper divs; get number of Pdivs.*/
say  right(n, max(20, w) )     'has'     center(#, 4)      "proper divisors."
end   /*i*/                                /* [↑] support extra specified integers*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
Pdivs: procedure; parse arg x,b;  x= abs(x);   if x==1  then return ''          /*unity?*/
odd= x // 2;                            if x==0  then return '∞'         /*zero ?*/
a= 1                                      /* [↓]  use all, or only odd #s.    ___*/
do j=2+odd  by 1+odd  while j*j < x   /*divide by some integers up to    √ X */
if x//j==0  then do;  a=a j;  b=x%j b /*if ÷, add both divisors to α & ß.    */
end
end   /*j*/                           /* [↑]  %  is the REXX integer division*/
/* [↓]  adjust for a square.        ___*/
if j*j==x  then  return  a j b            /*Was  X  a square?    If so, add  √ X */
return  a   b            /*return the divisors  (both lists).   */
```

{{out|output|text= when using the following input: 0 10 1 20000 166320 1441440 11796480000 }}

```
0 has  ∞   proper divisors:  ∞
1 has  0   proper divisors:
2 has  1   proper divisors:  1
3 has  1   proper divisors:  1
4 has  2   proper divisors:  1 2
5 has  1   proper divisors:  1
6 has  3   proper divisors:  1 2 3
7 has  1   proper divisors:  1
8 has  3   proper divisors:  1 2 4
9 has  2   proper divisors:  1 3
10 has  3   proper divisors:  1 2 5

79 is the highest number of proper divisors in range 1──►20000, and it's for:  15120 and 18480

166320 has 159  proper divisors.
1441440 has 287  proper divisors.
11796480000 has 329  proper divisors.

```

### version 3

When factoring 20,000 integers, this REXX version is about '''10%''' faster than the REXX version 2.

When factoring 200,000 integers, this REXX version is about '''30%''' faster.

When factoring 2,000,000 integers, this REXX version is about '''40%''' faster.

When factoring 20,000,000 integers, this REXX version is about '''38%''' faster.

It accomplishes a faster speed by incorporating the calculation of an ''integer square root'' of an integer (without using any floating point arithmetic).

```/*REXX program finds proper divisors (and count) of integer ranges; finds the max count.*/
parse arg bot top inc range xtra                 /*obtain optional arguments from the CL*/
if   bot=='' |   bot==","  then    bot=    1     /*Not specified?  Then use the default.*/
if   top=='' |   top==","  then    top=   10     /* "      "         "   "   "     "    */
if   inc=='' |   inc==","  then    inc=    1     /* "      "         "   "   "     "    */
if range=='' | range==","  then  range=20000     /* "      "         "   "   "     "    */
w= max( length(top), length(bot), length(range)) /*determine the biggest number of these*/
numeric digits max(9, w + 1)                     /*have enough digits for  //  operator.*/
@.= 'and'                                        /*a literal used to separate #s in list*/
do n=bot  to top  by inc                   /*process the first range specified.   */
q= Pdivs(n);    #= words(q)                /*get proper divs; get number of Pdivs.*/
if q=='∞'  then #= q                       /*adjust number of Pdivisors for zero. */
say right(n, max(20, w) )   'has'   center(#, 4)     "proper divisors: "    q
end   /*n*/
m=0                                              /*M ≡ maximum number of Pdivs (so far).*/
do r=1  for range;    q= Pdivs(r)          /*process the second range specified.  */
#= words(q);          if #<m  then iterate /*get proper divs; get number of Pdivs.*/
if #<m  then iterate                       /*Less then max?   Then ignore this #. */
@.#= @.#  @.  r;      m=#                  /*add this Pdiv to max list; set new M.*/
end   /*r*/                                /* [↑]   process 2nd range of integers.*/
say
say m  ' is the highest number of proper divisors in range 1──►'range,
", and it's for: "       subword(@.m, 3)
say                                              /* [↓]  handle any given extra numbers.*/
do i=1  for words(xtra);  n= word(xtra, i) /*obtain an extra number from XTRA list*/
w= max(w, 1 + length(n) )                  /*use maximum width for aligned output.*/
numeric digits max(9, 1 + length(n) )      /*have enough digits for  //  operator.*/
q= Pdivs(n);              #= words(q)      /*get proper divs; get number of Pdivs.*/
say  right(n, max(20, w) )     'has'     center(#, 4)      "proper divisors."
end   /*i*/                                /* [↑] support extra specified integers*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
Pdivs: procedure; parse arg x 1 z,b;  x= abs(x);   if x==1  then return ''      /*unity?*/
odd= x // 2;                                if x==0  then return '∞'     /*zero ?*/
r= 0;         q= 1                        /* [↓] ══integer square root══     ___ */
do while q<=z; q=q*4; end            /*R:  an integer which will be    √ X  */
do while q>1;  q=q%4; _= z-r-q;  r=r%2;  if _>=0  then  do;  z=_;  r=r+q;  end
end   /*while q>1*/                  /* [↑]  compute the integer sqrt of  X.*/
a=1                                       /* [↓]  use all, or only odd #s.   ___ */
do j=2 +odd  by 1 +odd to r -(r*r==x) /*divide by some integers up to   √ X  */
if x//j==0  then do;  a=a j;  b=x%j b /*if ÷, add both divisors to α & ß.    */
end
end   /*j*/                           /* [↑]  %  is the REXX integer division*/
/* [↓]  adjust for a square.        ___*/
if j*j==x  then  return  a j b            /*Was  X  a square?    If so, add  √ X */
return  a   b            /*return the divisors  (both lists).   */
```

{{out|output|text= is identical to the 2nd REXX version when using the same inputs.}}

## Ring

```
# Project : Proper divisors

limit = 10
for n=1 to limit
if n=1
see "" + 1 + " -> (None)" + nl
loop
ok
see "" + n + " -> "
for m=1 to n-1
if n%m = 0
see " " + m
ok
next
see nl
next

```

Output:

```
1 -> (None)
2 ->  1
3 ->  1
4 ->  1 2
5 ->  1
6 ->  1 2 3
7 ->  1
8 ->  1 2 4
9 ->  1 3
10 ->  1 2 5

```

## Ruby

```require "prime"

class Integer
def proper_divisors
return [] if self == 1
primes = prime_division.flat_map{|prime, freq| [prime] * freq}
(1...primes.size).each_with_object([1]) do |n, res|
primes.combination(n).map{|combi| res << combi.inject(:*)}
end.flatten.uniq
end
end

(1..10).map{|n| puts "#{n}: #{n.proper_divisors}"}

size, select = (1..20_000).group_by{|n| n.proper_divisors.size}.max
select.each do |n|
puts "#{n} has #{size} divisors"
end
```

{{out}}

```
1: []
2: [1]
3: [1]
4: [1, 2]
5: [1]
6: [1, 2, 3]
7: [1]
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120 has 79 divisors
18480 has 79 divisors

```

### An Alternative Approach

```#Determine the integer within a range of integers that has the most proper divisors
#Nigel Galloway: December 23rd., 2014
require "prime"
n, g = 0
(1..20000).each{|i| e = i.prime_division.inject(1){|n,g| n * (g[1]+1)}
n, g = e, i if e > n}
puts "#{g} has #{n-1} proper divisors"
```

{{out}} In the range 1..200000

```
15120 has 79 proper divisors

```

and in the ranges 1..2000000 & 1..20000000

```
166320 has 159 proper divisors
1441440 has 287 proper divisors

```

## Rust

```trait ProperDivisors {
fn proper_divisors(&self) -> Option<Vec<u64>>;
}

impl ProperDivisors for u64 {
fn proper_divisors(&self) -> Option<Vec<u64>> {
if self.le(&1) {
return None;
}
let mut divisors: Vec<u64> = Vec::new();

for i in 1..*self {
if *self % i == 0 {
divisors.push(i);
}
}
Option::from(divisors)
}
}

fn main() {
for i in 1..11 {
println!("Proper divisors of {:2}: {:?}", i,
i.proper_divisors().unwrap_or(vec![]));
}

let mut most_idx: u64 = 0;
let mut most_divisors: Vec<u64> = Vec::new();
for i in 1..20_001 {
let divs = i.proper_divisors().unwrap_or(vec![]);
if divs.len() > most_divisors.len() {
most_divisors = divs;
most_idx = i;
}
}
println!("In 1 to 20000, {} has the most proper divisors at {}", most_idx,
most_divisors.len());
}

```

{{out}}

```Proper divisors of  1: []
Proper divisors of  2: [1]
Proper divisors of  3: [1]
Proper divisors of  4: [1, 2]
Proper divisors of  5: [1]
Proper divisors of  6: [1, 2, 3]
Proper divisors of  7: [1]
Proper divisors of  8: [1, 2, 4]
Proper divisors of  9: [1, 3]
Proper divisors of 10: [1, 2, 5]
In 1 to 20000, 15120 has the most proper divisors at 79

```

```
\$constant false = 0
\$constant true = FFFFH

rem - compute p mod q
function mod(p, q = integer) = integer
end = p - q * (p/q)

rem - count, and optionally display, proper divisors of n
function divisors(n, display = integer) = integer
var i, limit, count, start, delta = integer
if mod(n, 2) = 0 then
begin
start = 2
delta = 1
end
else
begin
start = 3
delta = 2
end
if n < 2 then count = 0 else count = 1
if display and (count = 1) then print using "#####"; 1;
i = start
limit = n / start
while i <= limit do
begin
if mod(n, i) = 0 then
begin
if display then print using "#####"; i;
count = count + 1
end
i = i + delta
if count = 1 then limit = n / i
end
if display then print
end = count

rem - main program begins here
var i, ndiv, highdiv, highnum = integer

print "Proper divisors of first 10 numbers:"
for i = 1 to 10
print using "### : "; i;
ndiv = divisors(i, true)
next i

print "Searching for number with most divisors ..."
highdiv = 1
highnum = 1
for i = 1 to 20000
ndiv = divisors(i, false)
if ndiv > highdiv then
begin
highdiv = ndiv
highnum = i
end
next i
print "Searched up to"; i
print highnum; " has the most divisors: "; highdiv

end

```

{{out}}

```
Proper divisors of first 10 numbers:
1 :
2 :     1
3 :     1
4 :     1     2
5 :     1
6 :     1     2    3
7 :     1
8 :     1     2    4
9 :     1     3
10 :     1     2    5
Searching for number with most divisors ...
Searched up to 20000
15120 has the most divisors:  79

```

## Scala

### Simple proper divisors

```def properDivisors(n: Int) = (1 to n/2).filter(i => n % i == 0)
def format(i: Int, divisors: Seq[Int]) = f"\$i%5d    \${divisors.length}%2d   \${divisors mkString " "}"

println(f"    n   cnt   PROPER DIVISORS")
val (count, list) = (1 to 20000).foldLeft( (0, List[Int]()) ) { (max, i) =>
val divisors = properDivisors(i)
if (i <= 10 || i == 100) println( format(i, divisors) )
if (max._1 < divisors.length) (divisors.length, List(i))
else if (max._1 == divisors.length) (divisors.length, max._2 ::: List(i))
else max
}

list.foreach( number => println(f"\$number%5d    \${properDivisors(number).length}") )
```

{{out}}

```    n   cnt   PROPER DIVISORS
1     0
2     1   1
3     1   1
4     2   1 2
5     1   1
6     3   1 2 3
7     1   1
8     3   1 2 4
9     2   1 3
10     3   1 2 5
100     8   1 2 4 5 10 20 25 50
15120    79
18480    79
```

### Proper divisors for integers for big integers

If ''Long''s are enough to you you can replace every ''BigInt'' with ''Long'' and the one ''BigInt(1)'' with ''1L''

```import scala.annotation.tailrec

def factorize(x: BigInt): List[BigInt] = {
@tailrec
def foo(x: BigInt, a: BigInt = 2, list: List[BigInt] = Nil): List[BigInt] = a * a > x match {
case false if x % a == 0 => foo(x / a, a, a :: list)
case false => foo(x, a + 1, list)
case true => x :: list
}

foo(x)
}

def properDivisors(n: BigInt): List[BigInt] = {
val factors = factorize(n)
val products = (1 until factors.length).flatMap(i => factors.combinations(i).map(_.product).toList).toList
(BigInt(1) :: products).filter(_ < n)
}
```

## Seed7

```\$ include "seed7_05.s7i";

const proc: writeProperDivisors (in integer: n) is func
local
var integer: i is 0;
begin
for i range 1 to n div 2 do
if n rem i = 0 then
write(i <& " ");
end if;
end for;
writeln;
end func;

const func integer: countProperDivisors (in integer: n) is func
result
var integer: count is 0;
local
var integer: i is 0;
begin
for i range 1 to n div 2 step succ(n rem 2) do
if n rem i = 0 then
incr(count);
end if;
end for;
end func;

const proc: main is func
local
var integer: i is 0;
var integer: v is 0;
var integer: max is 0;
var integer: max_i is 1;
begin
for i range 1 to 10 do
write(i <& ": ");
writeProperDivisors(i);
end for;
for i range 1 to 20000 do
v := countProperDivisors(i);
if v > max then
max := v;
max_i := i;
end if;
end for;
writeln(max_i <& " with " <& max <& " divisors");
end func;
```

{{out}}

```
1:
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
15120 with 79 divisors

```

## Sidef

{{trans|Perl 6}}

```func propdiv (n) {
n.divisors.slice(0, -2)
}

{|i| printf("%2d: %s\n", i, propdiv(i)) } << 1..10

var max = 0
var candidates = []

for i in (1..20_000) {
var divs = propdiv(i).len
if (divs > max) {
candidates = []
max = divs
}
candidates << i if (divs == max)
}

say "max = #{max}, candidates = #{candidates}"
```

{{out}}

```
1: []
2: [1]
3: [1]
4: [1, 2]
5: [1]
6: [1, 2, 3]
7: [1]
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
max = 79, candidates = [15120, 18480]

```

## Swift

Simple function:

```func properDivs1(n: Int) -> [Int] {

return filter (1 ..< n) { n % \$0 == 0 }
}
```

More efficient function:

```import func Darwin.sqrt

func sqrt(x:Int) -> Int { return Int(sqrt(Double(x))) }

func properDivs(n: Int) -> [Int] {

if n == 1 { return [] }

var result = [Int]()

for div in filter (1 ... sqrt(n), { n % \$0 == 0 }) {

result.append(div)

if n/div != div && n/div != n { result.append(n/div) }
}

return sorted(result)

}
```

```for i in 1...10 {
println("\(i): \(properDivs(i))")
}

var (num, max) = (0,0)

for i in 1...20_000 {

let count = properDivs(i).count
if (count > max) { (num, max) = (i, count) }
}

println("\(num): \(max)")
```

{{out}}

```1: []
2: [1]
3: [1]
4: [1, 2]
5: [1]
6: [1, 2, 3]
7: [1]
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120: 79

```

## tbas

```
dim _proper_divisors(100)

sub proper_divisors(n)
dim i
dim _proper_divisors_count = 0
if n <> 1 then
for i = 1 to (n \ 2)
if n %% i = 0 then
_proper_divisors_count = _proper_divisors_count + 1
_proper_divisors(_proper_divisors_count) = i
end if
next
end if
return _proper_divisors_count
end sub

sub show_proper_divisors(n, tabbed)
dim cnt = proper_divisors(n)
print str\$(n) + ":"; tab(4);"(" + str\$(cnt) + " items) ";
dim j
for j = 1 to cnt
if tabbed then
print str\$(_proper_divisors(j)),
else
print str\$(_proper_divisors(j));
end if
if (j < cnt) then print ",";
next
print
end sub

dim i
for i = 1 to 10
show_proper_divisors(i, false)
next

dim c
dim maxindex = 0
dim maxlength = 0
for t = 1 to 20000
c = proper_divisors(t)
if c > maxlength then
maxindex = t
maxlength = c
end if
next

print "A maximum at ";
show_proper_divisors(maxindex, false)

```
```
>tbas proper_divisors.bas
1:  (0 items)
2:  (1 items) 1
3:  (1 items) 1
4:  (2 items) 1,2
5:  (1 items) 1
6:  (3 items) 1,2,3
7:  (1 items) 1
8:  (3 items) 1,2,4
9:  (2 items) 1,3
10: (3 items) 1,2,5
A maximum at 15120:(79 items) 1,2,3,4,5,6,7,8,9,10,12,14,15,16,18,20,21,24,27,28,30,
35,36,40,42,45,48,54,56,60,63,70,72,80,84,90,105,108,112,120,126,135,
140,144,168,180,189,210,216,240,252,270,280,315,336,360,378,420,432,
504,540,560,630,720,756,840,945,1008,1080,1260,1512,1680,1890,2160,
2520,3024,3780,5040,7560

```

## Tcl

Note that if a number, $k$, greater than 1 divides $n$ exactly, both $k$ and $n/k$ are proper divisors. (The raw answers are not sorted; the pretty-printer code sorts.)

```proc properDivisors {n} {
if {\$n == 1} return
set divs 1
for {set i 2} {\$i*\$i <= \$n} {incr i} {
if {!(\$n % \$i)} {
lappend divs \$i
if {\$i*\$i < \$n} {
lappend divs [expr {\$n / \$i}]
}
}
}
return \$divs
}

for {set i 1} {\$i <= 10} {incr i} {
puts "\$i => {[join [lsort -int [properDivisors \$i]] ,]}"
}
set maxI [set maxC 0]
for {set i 1} {\$i <= 20000} {incr i} {
set c [llength [properDivisors \$i]]
if {\$c > \$maxC} {
set maxI \$i
set maxC \$c
}
}
puts "max: \$maxI => (...\$maxC…)"
```

{{out}}

```
1 => {}
2 => {1}
3 => {1}
4 => {1,2}
5 => {1}
6 => {1,2,3}
7 => {1}
8 => {1,2,4}
9 => {1,3}
10 => {1,2,5}
max: 15120 => (...79...)

```

## VBA

```Public Sub Proper_Divisor()
Dim t() As Long, i As Long, l As Long, j As Long, c As Long
For i = 1 To 10
Debug.Print "Proper divisor of " & i & " : " & Join(S(i), ", ")
Next
For i = 2 To 20000
l = UBound(S(i)) + 1
If l > c Then c = l: j = i
Next
Debug.Print "Number in the range 1 to 20,000 with the most proper divisors is : " & j
Debug.Print j & " count " & c & " proper divisors"
End Sub

Private Function S(n As Long) As String()
'returns the proper divisors of n
Dim j As Long, t() As String, c As Long
't = list of proper divisor of n
If n > 1 Then
For j = 1 To n \ 2
If n Mod j = 0 Then
ReDim Preserve t(c)
t(c) = j
c = c + 1
End If
Next
End If
S = t
End Function
```

{{out}}

```Proper divisor of 1 :
Proper divisor of 2 : 1
Proper divisor of 3 : 1
Proper divisor of 4 : 1, 2
Proper divisor of 5 : 1
Proper divisor of 6 : 1, 2, 3
Proper divisor of 7 : 1
Proper divisor of 8 : 1, 2, 4
Proper divisor of 9 : 1, 3
Proper divisor of 10 : 1, 2, 5
Number in the range 1 to 20,000 with the most proper divisors is : 15120
15120 count 79 proper divisors
```

## Visual Basic .NET

{{trans|C#}}

```Module Module1

Function ProperDivisors(number As Integer) As IEnumerable(Of Integer)
Return Enumerable.Range(1, number / 2).Where(Function(divisor As Integer) number Mod divisor = 0)
End Function

Sub Main()
For Each number In Enumerable.Range(1, 10)
Console.WriteLine("{0}: {{{1}}}", number, String.Join(", ", ProperDivisors(number)))
Next

Dim record = Enumerable.Range(1, 20000).Select(Function(number) New With {.Number = number, .Count = ProperDivisors(number).Count()}).OrderByDescending(Function(currentRecord) currentRecord.Count).First()
Console.WriteLine("{0}: {1}", record.Number, record.Count)
End Sub

End Module
```

{{out}}

```1: {}
2: {1}
3: {1}
4: {1, 2}
5: {1}
6: {1, 2, 3}
7: {1}
8: {1, 2, 4}
9: {1, 3}
10: {1, 2, 5}
15120: 79
```

## zkl

{{trans|D}} This is the simple version :

```fcn properDivs(n){ [1.. (n + 1)/2 + 1].filter('wrap(x){ n%x==0 and n!=x }) }
```

This version is MUCH faster (the output isn't ordered however):

```fcn properDivs(n){
if(n==1) return(T);
( pd:=[1..(n).toFloat().sqrt()].filter('wrap(x){ n%x==0 }) )
.pump(pd,'wrap(pd){ if(pd!=1 and (y:=n/pd)!=pd ) y else Void.Skip })
}
```
```[1..10].apply(properDivs).println();
[1..20_001].apply('wrap(n){ T(properDivs(n).len(),n) })
.reduce(fcn([(a,_)]ab, [(c,_)]cd){ a>c and ab or cd },T(0,0))
.println();
```

{{out}}

```
L(L(),L(1),L(1),L(1,2),L(1),L(1,2,3),L(1),L(1,2,4),L(1,3),L(1,2,5))
L(79,18480)

```