...

2158 1079 3238 1619 4858 2429 7288 3644 1822

...

16 8 4 2 1

112

'''Number below 100,000 with the longest hailstone sequence, and the length of that sequence:'''


```beeswax>
@:  q pf1_#
>%"d3~@.Pqf#{g?` `{gpK@~BP9~5@P@q'M<
d~2~pL~1Ff<         <            >?d
    >zAg?MM@1~y@~gLpz2~yg@~3~hAg?M d
                   >?~fz1~y?yg@hhAg?Mb

Output:

77031 351

## Befunge


```befunge
93*:.    v
> :2%v  >
v+1*3_2/
>" ",:.v   v<
<v v-1:< <
+1\_$1+v^ \
v .,+94<>^>::v
>" "03pv  :* p
v67:" "<  0: 1
>p78p25  *^*p0
  v!-1:  <<*^<
9$_:0\  ^-^< v
v01g00:< 1   4
>g"@"*+`v^  <+
v01/"@":_ $ ^,
>p"@"%00p\$:^.
vg01g00  ,+49<
>"@"*+.@

27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
112
77031
351

Bracmat

(
  ( hailstone
  =   L len
    .   !arg:?L
      &   whl
        ' ( !arg:~1
          & (!arg*1/2:~/|3*!arg+1):?arg
          & !arg !L:?L
          )
      & (!L:? [?len&!len.!L)
  )
& ( reverse
  =   L e
    .   :?L
      & whl'(!arg:%?e ?arg&!e !L:?L)
      & !L
  )
& hailstone$27:(?len.?list)
& reverse$!list:?first4 [4 ? [-5 ?last4
& put$"Hailstone sequence starting with "
& put$!first4
& put$(str$(" has " !len " elements and ends with "))
& put$(!last4 \n)
& 1:?N
& 0:?max:?Nmax
&   whl
  ' ( !N+1:<100000:?N
    &   hailstone$!N
      : (   >!max:?max&!N:?Nmax
          | ?
        . ?
        )
    )
&   out
  $ ( str
    $ ( "The number <100000 with the longest hailstone sequence is "
        !Nmax
        " with "
        !max
        " elements."
      )
    )
);

=={{header|Brainfuck}}== Prints the number of terms required to map the input to 1. Does not count the first term of the sequence. >,[ [ ----------[ >>>[>>>>]+[[-]+<[->>>>++>>>>+[>>>>]++[->+<<<<<]]<<<] ++++++[>------<-]>--[>>[->>>>]+>+[<<<<]>-],< ]> ]>>>++>+>>[ <<[>>>>[-]+++++++++<[>-<-]+++++++++>[-[<->-]+[<<<<]]<[>+<-]>] >[>[>>>>]+[[-]<[+[->>>>]>+<]>[<+>[<<<<]]+<<<<]>>>[->>>>]+>+[<<<<]] >[[>+>>[<<<<+>>>>-]>]<<<<[-]>[-<<<<]]>>>>>>> ]>>+[[-]++++++>>>>]<<<<[[<++++++++>-]<.[-]<[-]<[-]<]<, ]

<lang Brainfuck>27
111

Brat

hailstone = { num |
  sequence = [num]
  while { num != 1 }
    { true? num % 2 == 0
      { num = num / 2 }
      { num = num * 3 + 1 }
      sequence << num
    }

  sequence
}

#Check sequence for 27
seq = hailstone 27
true? (seq[0,3] == [27 82 41 124] && seq[-1, -4] == [8 4 2 1])
  { p "Sequence for 27 is correct" }
  { p "Sequence for 27 is not correct!" }

#Find longest sequence for numbers < 100,000
longest = [number: 0 length: 0]

1.to 99999 { index |
    seq = hailstone index
    true? seq.length > longest[:length]
      { longest[:length] = seq.length
        longest[:number] = index
        p "Longest so far: #{index} @ #{longest[:length]} elements"
      }

    index = index + 1
  }

p "Longest was starting from #{longest[:number]} and was of length #{longest[:length]}"

Sequence for 27 is correct
Longest so far: 1 @ 1 elements
Longest so far: 2 @ 2 elements
Longest so far: 3 @ 8 elements
...
Longest so far: 52527 @ 340 elements
Longest so far: 77031 @ 351 elements
Longest was starting from 77031 and was of length 351

Burlesque

blsq ) 27{^^^^2.%{3.*1.+}\/{2./}\/ie}{1!=}w!bx{\/+]}{\/isn!}w!L[
112

C

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

int hailstone(int n, int *arry)
{
    int hs = 1;

    while (n!=1) {
        hs++;
        if (arry) *arry++ = n;
        n = (n&1) ? (3*n+1) : (n/2);
    }
    if (arry) *arry++ = n;
    return hs;
}

int main()
{
    int j, hmax = 0;
    int jatmax, n;
    int *arry;

    for (j=1; j<100000; j++) {
       n = hailstone(j, NULL);
       if (hmax < n) {
           hmax = n;
           jatmax = j;
       }
    }
    n = hailstone(27, NULL);
    arry = malloc(n*sizeof(int));
    n = hailstone(27, arry);

    printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
        arry[0],arry[1],arry[2],arry[3],
        arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
    printf("Max %d at j= %d\n", hmax, jatmax);
    free(arry);

    return 0;
}

[ 27, 82, 41, 124, ...., 8, 4, 2, 1] len= 112
Max 351 at j= 77031

With caching

Much faster if you want to go over a million or so.

#include <stdio.h>

#define N 10000000
#define CS N	/* cache size */

typedef unsigned long ulong;
ulong cache[CS] = {0};

ulong hailstone(ulong n)
{
	int x;
	if (n == 1) return 1;
	if (n < CS && cache[n]) return cache[n];

	x = 1 + hailstone((n & 1) ? 3 * n + 1 : n / 2);
	if (n < CS) cache[n] = x;
	return x;
}

int main()
{
	int i, l, max = 0, mi;
	for (i = 1; i < N; i++) {
		if ((l = hailstone(i)) > max) {
			max = l;
			mi = i;
		}
	}
	printf("max below %d: %d, length %d\n", N, mi, max);
	return 0;
}

C#

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace Hailstone
{
    class Program
    {
        public static List<int> hs(int n,List<int> seq)
        {
            List<int> sequence = seq;
            sequence.Add(n);
            if (n == 1)
            {
                return sequence;
            }else{
                int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
                return hs(newn, sequence);
            }
        }

        static void Main(string[] args)
        {
            int n = 27;
            List<int> sequence = hs(n,new List<int>());
            Console.WriteLine(sequence.Count + " Elements");
            List<int> start = sequence.GetRange(0, 4);
            List<int> end = sequence.GetRange(sequence.Count - 4, 4);
            Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
            int number = 0, longest = 0;
            for (int i = 1; i < 100000; i++)
            {
                int count = (hs(i, new List<int>())).Count;
                if (count > longest)
                {
                    longest = count;
                    number = i;
                }
            }
            Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
       }
    }
}

112 Elements
Starting with : 27,82,41,124 and ending with : 8,4,2,1
Number < 100000 with longest Hailstone seq.: 77031 with length of 351

With caching

As with the [[#C|C example]], much faster if you want to go over a million or so.

using System;
using System.Collections.Generic;

namespace ConsoleApplication1
{
    class Program
    {
        public static void Main()
        {
            int longestChain = 0, longestNumber = 0;

            var recursiveLengths = new Dictionary<int, int>();

            const int maxNumber = 100000;

            for (var i = 1; i <= maxNumber; i++)
            {
                var chainLength = Hailstone(i, recursiveLengths);
                if (longestChain >= chainLength)
                    continue;

                longestChain = chainLength;
                longestNumber = i;
            }
            Console.WriteLine("max below {0}: {1} ({2} steps)", maxNumber, longestNumber, longestChain);
        }

        private static int Hailstone(int num, Dictionary<int, int> lengths)
        {
            if (num == 1)
                return 1;

            while (true)
            {
                if (lengths.ContainsKey(num))
                    return lengths[num];

                lengths[num] = 1 + ((num%2 == 0) ? Hailstone(num/2, lengths) : Hailstone((3*num) + 1, lengths));
            }
        }
    }
}

max below 100000: 77031 (351 steps)

C++

#include <iostream>
#include <vector>
#include <utility>

std::vector<int> hailstone(int i)
{
    std::vector<int> v;
    while(true){
        v.push_back(i);
        if (1 == i) break;
        i = (i % 2) ? (3 * i + 1) : (i / 2);
    }
    return v;
}

std::pair<int,int> find_longest_hailstone_seq(int n)
{
    std::pair<int, int> maxseq(0, 0);
    int l;
    for(int i = 1; i < n; ++i){
        l = hailstone(i).size();
        if (l > maxseq.second) maxseq = std::make_pair(i, l);
    }
    return maxseq;
}

int main () {

// Use the routine to show that the hailstone sequence for the number 27
    std::vector<int> h27;
    h27 = hailstone(27);
// has 112 elements
    int l = h27.size();
    std::cout << "length of hailstone(27) is " << l;
// starting with 27, 82, 41, 124 and
    std::cout << " first four elements of hailstone(27) are ";
    std::cout << h27[0] << " " << h27[1] << " "
              << h27[2] << " " << h27[3] << std::endl;
// ending with 8, 4, 2, 1
    std::cout << " last four elements of hailstone(27) are "
              << h27[l-4] << " " << h27[l-3] << " "
              << h27[l-2] << " " << h27[l-1] << std::endl;

    std::pair<int,int> m = find_longest_hailstone_seq(100000);

    std::cout << "the longest hailstone sequence under 100,000 is " << m.first
              << " with " << m.second << " elements." <<std::endl;

    return 0;
}

 length of hailstone(27) is 112 first four elements of hailstone(27) are 27 82 41 124
 last four elements of hailstone(27) are 8 4 2 1
 the longest hailstone sequence under 100,000 is 77031 with 351 elements.

=== {{libheader|Qt}} === Templated solution works for all of Qt's sequential container classes (QLinkedList, QList, QVector).

#include <QDebug>
#include <QVector>

template <class T>
T hailstone(typename T::value_type n)
{
    T seq;
    for (seq << n; n != 1; seq << n) {
        n = (n&1) ? (3*n)+1 : n/2;
    }
    return seq;
}

template <class T>
T longest_hailstone_seq(typename T::value_type n)
{
    T maxSeq;
    for (; n > 0; --n) {
        const auto seq = hailstone<T>(n);
        if (seq.size() > maxSeq.size()) {
            maxSeq = seq;
        }
    }
    return maxSeq;
}

int main(int, char *[]) {
    const auto seq = hailstone<QVector<uint_fast16_t>>(27);
    qInfo() << "hailstone(27):";
    qInfo() << "  length:" << seq.size() << "elements";
    qInfo() << "  first 4 elements:" << seq.mid(0,4);
    qInfo() << "  last 4 elements:" << seq.mid(seq.size()-4);

    const auto max = longest_hailstone_seq<QVector<uint_fast32_t>>(100000);
    qInfo() << "longest sequence with starting element under 100000:";
    qInfo() << "  length:" << max.size() << "elements";
    qInfo() << "  starting element:" << max.first();
}

hailstone(27):
  length: 112 elements
  first 4 elements: QVector(27, 82, 41, 124)
  last 4 elements: QVector(8, 4, 2, 1)
longest sequence with starting element under 100000:
  length: 351 elements
  starting element: 77031

Ceylon

shared void run() {

	{Integer*} hailstone(variable Integer n) {
		variable [Integer*] stones = [n];
		while(n != 1) {
			n = if(n.even) then n / 2 else 3 * n + 1;
			stones = stones.append([n]);
		}
		return stones;
	}

	value hs27 = hailstone(27);
	print("hailstone sequence for 27 is ``hs27.take(3)``...``hs27.skip(hs27.size - 3).take(3)`` with length ``hs27.size``");

	variable value longest = hailstone(1);
	for(i in 2..100k - 1) {
		value current = hailstone(i);
		if(current.size > longest.size) {
			longest = current;
		}
	}
	print("the longest sequence under 100,000 starts with ``longest.first else "what?"`` and has length ``longest.size``");
}

CLIPS

(deftemplate longest
  (slot bound)             ; upper bound for the range of values to check
  (slot next (default 2))  ; next value that needs to be checked
  (slot start (default 1)) ; starting value of longest sequence
  (slot len (default 1))   ; length of longest sequence
)

(deffacts startup
  (query 27)
  (longest (bound 100000))
)

(deffunction hailstone-next
  (?n)
  (if (evenp ?n)
    then (div ?n 2)
    else (+ (* 3 ?n) 1)
  )
)

(defrule extend-sequence
  ?hail <- (hailstone $?sequence ?tail&:(> ?tail 1))
  =>
  (retract ?hail)
  (assert (hailstone ?sequence ?tail (hailstone-next ?tail)))
)

(defrule start-query
  (query ?num)
  =>
  (assert (hailstone ?num))
)

(defrule result-query
  (query ?num)
  (hailstone ?num $?sequence 1)
  =>
  (bind ?sequence (create$ ?num ?sequence 1))
  (printout t "Hailstone sequence starting with " ?num ":" crlf)
  (bind ?len (length ?sequence))
  (printout t "  Length: " ?len crlf)
  (printout t "  First four: " (implode$ (subseq$ ?sequence 1 4)) crlf)
  (printout t "  Last four: " (implode$ (subseq$ ?sequence (- ?len 3) ?len)) crlf)
  (printout t crlf)
)

(defrule longest-create-next-hailstone
  (longest (bound ?bound) (next ?next))
  (test (<= ?next ?bound))
  (not (hailstone ?next $?))
  =>
  (assert (hailstone ?next))
)

(defrule longest-check-next-hailstone
  ?longest <- (longest (bound ?bound) (next ?next) (start ?start) (len ?len))
  (test (<= ?next ?bound))
  ?hailstone <- (hailstone ?next $?sequence 1)
  =>
  (retract ?hailstone)
  (bind ?thislen (+ 2 (length ?sequence)))
  (if (> ?thislen ?len) then
    (modify ?longest (start ?next) (len ?thislen) (next (+ ?next 1)))
    else
    (modify ?longest (next (+ ?next 1)))
  )
)

(defrule longest-finished
  (longest (bound ?bound) (next ?next) (start ?start) (len ?len))
  (test (> ?next ?bound))
  =>
  (printout t "The number less than " ?bound " that has the largest hailstone" crlf)
  (printout t "sequence is " ?start " with a length of " ?len "." crlf)
  (printout t crlf)
)

The number less than 100000 that has the largest hailstone
sequence is 77031 with a length of 351.

Hailstone sequence starting with 27:
  Length: 112
  First four: 27 82 41 124
  Last four: 8 4 2 1

Clojure

(defn hailstone-seq [n]
  {:pre [(pos? n)]}
  (lazy-seq
   (cond (= n 1)   '(1)
         (even? n) (cons n (hailstone-seq (/ n 2)))
         :else     (cons n (hailstone-seq (+ (* n 3) 1))))))

(let [hseq (hailstone-seq 27)]
  (->  hseq count      (= 112)            assert)
  (->> hseq (take 4)   (= [27 82 41 124]) assert)
  (->> hseq (drop 108) (= [8 4 2 1])      assert))

(let [{max-i :num, max-len :len}
      (reduce #(max-key :len %1 %2)
              (for [i (range 1 100000)]
                {:num i, :len (count (hailstone-seq i))}))]
  (println "Maximum length" max-len "was found for hailstone(" max-i ")."))

COBOL

Testing with GnuCOBOL

       identification division.
       program-id. hailstones.
       remarks. cobc -x hailstones.cob.

       data division.
       working-storage section.
       01 most                 constant as 1000000.
       01 coverage             constant as 100000.
       01 stones               usage binary-long.
       01 n                    usage binary-long.
       01 storm                usage binary-long.

       01 show-arg             pic 9(6).
       01 show-default         pic 99 value 27.
       01 show-sequence        usage binary-long.
       01 longest              usage binary-long occurs 2 times.

       01 filler.
          05 hail              usage binary-long
                               occurs 0 to most depending on stones.
       01 show                 pic z(10).
       01 low-range            usage binary-long.
       01 high-range           usage binary-long.
       01 range                usage binary-long.


       01 remain               usage binary-long.
       01 unused               usage binary-long.

       procedure division.
       accept show-arg from command-line
       if show-arg less than 1 or greater than coverage then
           move show-default to show-arg
       end-if
       move show-arg to show-sequence

       move 1 to longest(1)
       perform hailstone varying storm
                         from 1 by 1 until storm > coverage
       display "Longest at: " longest(2) " with " longest(1) " elements"
       goback.

      *> **************************************************************
       hailstone.
       move 0 to stones
       move storm to n
       perform until n equal 1
           if stones > most then
               display "too many hailstones" upon syserr
               stop run
           end-if

           add 1 to stones
           move n to hail(stones)
           divide n by 2 giving unused remainder remain
           if remain equal 0 then
               divide 2 into n
           else
               compute n = 3 * n + 1
           end-if
       end-perform
       add 1 to stones
       move n to hail(stones)

       if stones > longest(1) then
           move stones to longest(1)
           move storm to longest(2)
       end-if

       if storm equal show-sequence then
           display show-sequence ": " with no advancing
           perform varying range from 1 by 1 until range > stones
               move 5 to low-range
               compute high-range = stones - 4
               if range < low-range or range > high-range then
                   move hail(range) to show
                   display function trim(show) with no advancing
                   if range < stones then
                       display ", " with no advancing
                   end-if
               end-if
               if range = low-range and stones > 8 then
                   display "..., " with no advancing
               end-if
           end-perform
           display ": " stones " elements"
       end-if
       .

       end program hailstones.

prompt$ cobc -x hailstones.cob
prompt$ ./hailstones
+0000000027: 27, 82, 41, 124, ..., 8, 4, 2, 1: +0000000112 elements
Longest at: +0000077031 with +0000000351 elements
prompt$ ./hailstones 42
+0000000042: 42, 21, 64, 32, ..., 8, 4, 2, 1: +0000000009 elements
Longest at: +0000077031 with +0000000351 elements

CoffeeScript

Recursive version:

hailstone = (n) ->
  if n is 1
    [n]

  else if n % 2 is 0
    [n].concat hailstone n/2

  else
    [n].concat hailstone (3*n) + 1

h27 = hailstone 27
console.log "hailstone(27) = #{h27[0..3]} ... #{h27[-4..]} (length: #{h27.length})"

maxlength = 0
maxnums = []

for i in [1..100000]
  seq = hailstone i

  if seq.length is maxlength
    maxnums.push i
  else if seq.length > maxlength
    maxlength = seq.length
    maxnums = [i]

console.log "Max length: #{maxlength}; numbers generating sequences of this length: #{maxnums}"

hailstone(27) = 27,82,41,124 ... 8,4,2,1 (length: 112)
Max length: 351; numbers generating sequences of this length: 77031

Common Lisp

(defun hailstone (n)
  (cond ((= n 1) '(1))
	((evenp n) (cons n (hailstone (/ n 2))))
	(t (cons n (hailstone (+ (* 3 n) 1))))))

(defun longest (n)
  (let ((k 0) (l 0))
    (loop for i from 1 below n do
	 (let ((len (length (hailstone i))))
	   (when (> len l) (setq l len k i)))
	 finally (format t "Longest hailstone sequence under ~A for ~A, having length ~A." n k l))))

Sample session:

ROSETTA> (length (hailstone 27))
112
ROSETTA> (subseq (hailstone 27) 0 4)
(27 82 41 124)
ROSETTA> (last (hailstone 27) 4)
(8 4 2 1)
ROSETTA> (longest-hailstone 100000)
Longest hailstone sequence under 100000 for 77031, having length 351.
NIL

Crystal

def hailstone(n)
    seq = [n]
    until n == 1
        n = n.even? ? n / 2 : n * 3 + 1
        seq << n
    end
    seq
end

max_len = (1...100_000).max_by{|n| hailstone(n).size }
max = hailstone(max_len)
puts ([max_len, max.size, max.max, max.first(4), max.last(4)])
# => [77031, 351, 21933016, [77031, 231094, 115547, 346642], [8, 4, 2, 1]]

twenty_seven = hailstone(27)
puts ([twenty_seven.size, twenty_seven.first(4), max.last(4)])
# => [112, [27, 82, 41, 124], [8, 4, 2, 1]]

D

Basic Version

import std.stdio, std.algorithm, std.range, std.typecons;

auto hailstone(uint n) pure nothrow {
  auto result = [n];
  while (n != 1) {
    n = (n & 1) ? (n * 3 + 1) : (n / 2);
    result ~= n;
  }
  return result;
}

void main() {
  enum M = 27;
  immutable h = M.hailstone;
  writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
  writeln("Length hailstone(", M, ")= ", h.length);

  enum N = 100_000;
  immutable p = iota(1, N)
                .map!(i => tuple(i.hailstone.length, i))
                .reduce!max;
  writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}

hailstone(27)= [27, 82, 41, 124] ... [8, 4, 2, 1]
Length hailstone(27)= 112
Longest sequence in [1,100000]= 77031 with len 351

Lazy Version

Same output.

import std.stdio, std.algorithm, std.typecons, std.range;

auto hailstone(uint m) pure nothrow @nogc {
    return m
           .recurrence!q{ a[n - 1] & 1 ? a[n - 1] * 3 + 1 : a[n - 1]/2}
           .until!q{ a == 1 }(OpenRight.no);
}

void main() {
  enum M = 27;
  immutable h = M.hailstone.array;
  writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
  writeln("Length hailstone(", M, ")= ", h.length);

  enum N = 100_000;
  immutable p = iota(1, N)
                .map!(i => tuple(i.hailstone.walkLength, i))
                .reduce!max;
  writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}

Faster Lazy Version

Same output.

struct Hailstone {
  uint n;
  bool empty() const pure nothrow @nogc { return n == 0; }
  uint front() const pure nothrow @nogc { return n; }
  void popFront() pure nothrow @nogc {
    n = n == 1 ? 0 : (n & 1 ? (n * 3 + 1) : n / 2);
  }
}

void main() {
  import std.stdio, std.algorithm, std.range, std.typecons;

  enum M = 27;
  immutable h = M.Hailstone.array;
  writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
  writeln("Length hailstone(", M, ")= ", h.length);

  enum N = 100_000;
  immutable p = iota(1, N)
                .map!(i => tuple(i.Hailstone.walkLength, i))
                .reduce!max;
  writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}

Lazy Version With Caching

Faster, same output.

import std.stdio, std.algorithm, std.range, std.typecons;

struct Hailstone(size_t cacheSize = 500_000) {
  size_t n;
  __gshared static size_t[cacheSize] cache;

  bool empty() const pure nothrow @nogc { return n == 0; }
  size_t front() const pure nothrow @nogc { return n; }

  void popFront() nothrow {
    if (n >= cacheSize) {
      n = n == 1 ? 0 : (n & 1 ? n*3 + 1 : n/2);
    } else if (cache[n]) {
      n = cache[n];
    } else {
      immutable n2 = n == 1 ? 0 : (n & 1 ? n*3 + 1 : n/2);
      n = cache[n] = n2;
    }
  }
}

void main() {
  enum M = 27;
  const h = M.Hailstone!().array;
  writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
  writeln("Length hailstone(", M, ")= ", h.length);

  enum N = 100_000;
  immutable p = iota(1, N)
                .map!(i => tuple(i.Hailstone!().walkLength, i))
                .reduce!max;
  writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}

Generator Range Version

import std.stdio, std.algorithm, std.range, std.typecons, std.concurrency;

auto hailstone(size_t n) {
    return new Generator!size_t({
        yield(n);
        while (n > 1) {
            n = (n & 1) ? (3 * n + 1) : (n / 2);
            yield(n);
        }
    });
}

void main() {
  enum M = 27;
  const h = M.hailstone.array;
  writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
  writeln("Length hailstone(", M, ")= ", h.length);

  enum N = 100_000;
  immutable p = iota(1, N)
                .map!(i => tuple(i.hailstone.walkLength, i))
                .reduce!max;
  writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}

Dart

 hailstone(int n) {
  if(n<=0) {
    throw new IllegalArgumentException("start value must be >=1)");
  }
  Queue<int> seq=new Queue<int>();
  seq.add(n);
  while(n!=1) {
    n=n%2==0?(n/2).toInt():3*n+1;
    seq.add(n);
  }
  return new List<int>.from(seq);
}

// apparently List is missing toString()
String iterableToString(Iterable seq) {
  String str="[";
  Iterator i=seq.iterator();
  while(i.hasNext()) {
    str+=i.next();
    if(i.hasNext()) {
      str+=",";
    }
  }
  return str+"]";
}

main() {
  for(int i=1;i<=10;i++) {
    print("h($i)="+iterableToString(hailstone(i)));
  }
  List<int> h27=hailstone(27);
  List<int> first4=h27.getRange(0,4);
  print("first 4 elements of h(27): "+iterableToString(first4));
  Expect.listEquals([27,82,41,124],first4);

  List<int> last4=h27.getRange(h27.length-4,4);
  print("last 4 elements of h(27): "+iterableToString(last4));
  Expect.listEquals([8,4,2,1],last4);

  print("length of sequence h(27): "+h27.length);
  Expect.equals(112,h27.length);

  int seq,max=0;
  for(int i=1;i<=100000;i++) {
    List<int> h=hailstone(i);
    if(h.length>max) {
      max=h.length;
      seq=i;
    }
  }
  print("up to 100000 the sequence h($seq) has the largest length ($max)");
}

h(1)=[1]
h(2)=[2,1]
h(3)=[3,10,5,16,8,4,2,1]
h(4)=[4,2,1]
h(5)=[5,16,8,4,2,1]
h(6)=[6,3,10,5,16,8,4,2,1]
h(7)=[7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1]
h(8)=[8,4,2,1]
h(9)=[9,28,14,7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1]
h(10)=[10,5,16,8,4,2,1]
first 4 elements of h(27): [27,82,41,124]
last 4 elements of h(27): [8,4,2,1]
length of sequence h(27): 112
up to 100000 the sequence h(77031) has the largest length (351)

Dc

Firstly, this code takes the value from the stack, computes and prints the corresponding Hailstone sequence, and the length of the sequence. The q procedure is for counting the length of the sequence. The e and o procedure is for even and odd number respectively. The x procedure is for overall control.

27
[[--: ]nzpq]sq
[d 2/ p]se
[d 3*1+ p]so
[d2% 0=e d1=q d2% 1=o d1=q lxx]dsxx

82
41
124
62
(omitted)
8
4
2
1
--: 112

Then we could wrap the procedure x with a new procedure s, and call it with l which is loops the value of t from 1 to 100000, and cleaning up the stack after each time we finish up with a number. Register L for the length of the longest sequence and T for the corresponding number. Also, procedure q is slightly modified for storing L and T if needed, and all printouts in procedure e and o are muted.

0dsLsT1st
[dsLltsT]sM
[[zdlL<M q]sq
[d 2/]se
[d 3*1+ ]so
[d2% 0=e d1=q d2% 1=o d1=q lxx]dsxx]ss
[lt1+dstlsxc lt100000>l]dslx
lTn[:]nlLp

{{out}} (Takes quite some time on a decent machine)

77031:351

DCL

$ n = f$integer( p1 )
$ i = 1
$ loop:
$  if p2 .nes. "QUIET" then $ s'i = n
$  if n .eq. 1 then $ goto done
$  i = i + 1
$  if .not. n
$  then
$   n = n / 2
$  else
$   if n .gt. 715827882 then $ exit  ! avoid overflowing
$   n = 3 * n + 1
$  endif
$  goto loop
$ done:
$ if p2 .nes. "QUIET"
$ then
$  penultimate_i = i - 1
$  antepenultimate_i = i - 2
$  preantepenultimate_i = i - 3
$  write sys$output "sequence has ", i, " elements starting with ", s1, ", ", s2, ", ", s3, ", ", s4, " and ending with ", s'preantepenultimate_i, ", ", s'antepenultimate_i, ", ", s'penultimate_i, ", ", s'i
$ endif
$ sequence_length == i

$ @hailstone 27
sequence has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1

$ limit = f$integer( p1 )
$ i = 1
$ max_so_far = 0
$ loop:
$  call hailstone 'i quiet
$  if sequence_length .gt. max_so_far
$  then
$   max_so_far = sequence_length
$   current_record_holder = i
$  endif
$  i = i + 1
$  if i .lt. limit then $ goto loop
$ write sys$output current_record_holder, " is the number less than ", limit, " which has the longest hailstone sequence which is ", max_so_far, " in length"
$ exit
$
$ hailstone: subroutine
$ n = f$integer( p1 )
$ i = 1
$ loop:
$  if p2 .nes. "QUIET" then $ s'i = n
$  if n .eq. 1 then $ goto done
$  i = i + 1
$  if .not. n
$  then
$   n = n / 2
$  else
$   if n .gt. 715827882 then $ exit  ! avoid overflowing
$   n = 3 * n + 1
$  endif
$  goto loop
$ done:
$ if p2 .nes. "QUIET"
$ then
$  penultimate_i = i - 1
$  antepenultimate_i = i - 2
$  preantepenultimate_i = i - 3
$  write sys$output "sequence has ", i, " elements starting with ", s1, ", ", s2, ", ", s3, ", ", s4, " and ending with ", s'preantepenultimate_i, ", ", s'antepenultimate_i, ", ", s'penultimate_i, ", ", s'i
$ endif
$ sequence_length == I
$ exit
$ endsubroutine

$ @longest_hailstone 100000
77031 is the number less than 100000 which has the longest hailstone sequence which is 351 in length

=={{header|Déjà Vu}}==

local hailstone:
	swap [ over ]
	while < 1 dup:
		if % over 2:
			#odd
			++ * 3
		else:
			#even
			/ swap 2
		swap push-through rot dup
	drop

if = (name) :(main):
	local :h27 hailstone 27
	!. = 112 len h27
	!. = 27 h27! 0
	!. = 82 h27! 1
	!. = 41 h27! 2
	!. = 124 h27! 3
	!. = 8 h27! 108
	!. = 4 h27! 109
	!. = 2 h27! 110
	!. = 1 h27! 111

	local :max 0
	local :maxlen 0
	for i range 1 99999:
		dup len hailstone i
		if < maxlen:
			set :maxlen
			set :max i
		else:
			drop
	!print( "number: " to-str max ", length: " to-str maxlen )
else:
	@hailstone

true
true
true
true
true
true
true
true
true
number: 77031, length: 351

Delphi

program ShowHailstoneSequence;

{$APPTYPE CONSOLE}

uses SysUtils, Generics.Collections;

procedure GetHailstoneSequence(aStartingNumber: Integer; aHailstoneList: TList<Integer>);
var
  n: Integer;
begin
  aHailstoneList.Clear;
  aHailstoneList.Add(aStartingNumber);
  n := aStartingNumber;

  while n <> 1 do
  begin
    if Odd(n) then
      n := (3 * n) + 1
    else
      n := n div 2;
    aHailstoneList.Add(n);
  end;
end;

var
  i: Integer;
  lList: TList<Integer>;
  lMaxSequence: Integer;
  lMaxLength: Integer;
begin
  lList := TList<Integer>.Create;
  try
    GetHailstoneSequence(27, lList);
    Writeln(Format('27: %d elements', [lList.Count]));
    Writeln(Format('[%d,%d,%d,%d ... %d,%d,%d,%d]',
      [lList[0], lList[1], lList[2], lList[3],
      lList[lList.Count - 4], lList[lList.Count - 3], lList[lList.Count - 2], lList[lList.Count - 1]]));
    Writeln;

    lMaxSequence := 0;
    lMaxLength := 0;
    for i := 1 to 100000 do
    begin
      GetHailstoneSequence(i, lList);
      if lList.Count > lMaxLength then
      begin
        lMaxSequence := i;
        lMaxLength := lList.Count;
      end;
    end;
    Writeln(Format('Longest sequence under 100,000: %d with %d elements', [lMaxSequence, lMaxLength]));
  finally
    lList.Free;
  end;

  Readln;
end.

27: 112 elements
[27 82 41 124 ... 8 4 2 1]

Longest sequence under 100,000: 77031 with 351 elements

EchoLisp

(lib 'hash)
(lib 'sequences)
(lib 'compile)

(define (hailstone n)
(when (> n 1)
	(if (even? n) (/ n 2) (1+ (* n 3)))))

(define H (make-hash))

;; (iterator/f seed f) returns seed, (f seed) (f(f seed)) ...

(define (hlength seed)
	(define collatz (iterator/f hailstone seed))
	(or
	   (hash-ref H seed) ;; known ?
	   (hash-set H seed
	      (for ((i (in-naturals)) (h collatz))
              ;; add length of subsequence if already known
	      #:break (hash-ref H h) => (+ i (hash-ref H h))
	      (1+ i)))))

(define (task (nmax 100000))
	(for ((n [1 .. nmax])) (hlength n)) ;; fill hash table

	(define hmaxlength (apply max (hash-values H)))
	(define hmaxseed (hash-get-key H hmaxlength))
	(writeln 'maxlength= hmaxlength 'for hmaxseed))

(define H27 (iterator/f hailstone 27))
(take H27 6)
   → (27 82 41 124 62 31)
(length H27)
   → 112
(list-tail (take H27 112) -6)
   → (5 16 8 4 2 1)

(task)
maxlength=     351     for     77031

;; more ...
(lib 'bigint)

(task 200000)
    maxlength=     383     for     156159
(task 300000)
    maxlength=     443     for     230631
(task 400000)
    maxlength=     443     for     230631
(task 500000)
    maxlength=     449     for     410011
(task 600000)
    maxlength=     470     for     511935
(task 700000)
    maxlength=     509     for     626331
(task 800000)
    maxlength=     509     for     626331
(task 900000)
    maxlength=     525     for     837799
(task 1000000)
    maxlength=     525     for     837799

Egel

import "prelude.eg"

namespace Hailstone (

    using System
    using List

    def even = [ N -> (N%2) == 0 ]

    def hailstone =
        [ 1 -> {1}
        | N -> if even N then cons N (hailstone (N/2))
               else cons N (hailstone (N * 3 + 1)) ]

    def hailpair =
        [ N -> (N, length (hailstone N)) ]

    def hailmax =
        [ (N, NMAX), (M, MMAX) -> if (NMAX < MMAX) then (M, MMAX) else (N, NMAX) ]

    def largest =
        [ 1 -> (1, 1)
        | N ->
            let M0 = hailpair N in
            let M1 = largest (N - 1) in
                hailmax M0 M1 ]
)

using System
using List
using Hailstone

def task0 = let H27 = hailstone 27 in length H27

def task1 =
    let H27 = hailstone 27 in
    let L   = length H27 in
        (take 4 H27, drop (L - 4) H27)

def task2 = largest 100000

def main = (task0, task1, task2)

Eiffel

class
	APPLICATION

create
	make

feature

	make
		local
			test: LINKED_LIST [INTEGER]
			count, number, te: INTEGER
		do
			create test.make
			test := hailstone_sequence (27)
			io.put_string ("There are " + test.count.out + " elements in the sequence for the number 27.")
			io.put_string ("%NThe first 4 elements are: ")
			across
				1 |..| 4 as t
			loop
				io.put_string (test [t.item].out + "%T")
			end
			io.put_string ("%NThe last 4 elements are: ")
			across
				(test.count - 3) |..| test.count as t
			loop
				io.put_string (test [t.item].out + "%T")
			end
			across
				1 |..| 99999 as c
			loop
				test := hailstone_sequence (c.item)
				te := test.count
				if te > count then
					count := te
					number := c.item
				end
			end
			io.put_string ("%NThe longest sequence for numbers below 100000 is " + count.out + " for the number " + number.out + ".")
		end

	hailstone_sequence (n: INTEGER): LINKED_LIST [INTEGER]
			-- Members of the Hailstone Sequence starting from 'n'.
		require
			n_is_positive: n > 0
		local
			seq: INTEGER
		do
			create Result.make
			from
				seq := n
			until
				seq = 1
			loop
				Result.extend (seq)
				if seq \\ 2 = 0 then
					seq := seq // 2
				else
					seq := ((3 * seq) + 1)
				end
			end
			Result.extend (seq)
		ensure
			sequence_terminated: Result.last = 1
		end

end

There are 112 elements in the sequence for the number 27.
The first 4 elements are: 27    82    41    124
The last 4 elements are: 8    4    2    1
The longest sequence for numbers below 100000 is 351 for the number 77031.

Elena

ELENA 4.x :

import system'collections;
import extensions;

const int maxNumber = 100000;

Hailstone(int n,Map<int,int> lengths)
{
    if (n == 1)
    {
        ^ 1
    };

    while (true)
    {
        if (lengths.containsKey(n))
        {
            ^ lengths[n]
        }
        else
        {
            if (n.isEven())
            {
                lengths[n] := 1 + Hailstone(n/2, lengths)
            }
            else
            {
                lengths[n] := 1 + Hailstone(3*n + 1, lengths)
            }
        }
    }
}

public program()
{
    int longestChain := 0;
    int longestNumber := 0;
    auto recursiveLengths := new Map<int,int>(4096,4096);

    for(int i := 1, i < maxNumber, i+=1)
    {
        var chainLength := Hailstone(i, recursiveLengths);
        if (longestChain < chainLength)
        {
               longestChain := chainLength;
               longestNumber := i
        }
    };

    console.printFormatted("max below {0}: {1} ({2} steps)", maxNumber, longestNumber, longestChain)
}

max bellow 100000: 77031 (351 steps)

Elixir

defmodule Hailstone do
  require Integer

  def step(1)                        , do: 0
  def step(n) when Integer.is_even(n), do: div(n,2)
  def step(n)                        , do: n*3 + 1

  def sequence(n) do
    Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
  end

  def run do
    seq27 = sequence(27)
    len27 = length(seq27)
    repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
    IO.puts "Hailstone(27) has #{len27} elements: #{repr}"

    {len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
    IO.puts "Longest sequence starting under 100000 begins with #{start} and has #{len} elements."
  end
end

Hailstone.run

Hailstone(27) has 112 elements: [27, 82, 41, 124, ..., 8, 4, 2, 1]
Longest sequence starting under 100000 begins with 77031 and has 351 elements.

Erlang

-module(hailstone).
-import(io).
-export([main/0]).

hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].

max_length(Start, Stop) ->
    F = fun (N) -> {length(hailstone(N)), N} end,
    Lengths = lists:map(F, lists:seq(Start, Stop)),
    lists:max(Lengths).

main() ->
    io:format("hailstone(4): ~w~n", [hailstone(4)]),
    Seq27 = hailstone(27),
    io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
    io:format("hailstone(27) first 4: ~w~n",
              [lists:sublist(Seq27, 4)]),
    io:format("hailstone(27) last 4: ~w~n",
              [lists:nthtail(length(Seq27) - 4, Seq27)]),
    io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
    {Length, N} = max_length(1, 100000),
    io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).

Eshell V5.8.4  (abort with ^G)
1> c(hailstone).
{ok,hailstone}
2> hailstone:main().
hailstone(4): [4,2,1]
hailstone(27) length: 112
hailstone(27) first 4: [27,82,41,124]
hailstone(27) last 4: [8,4,2,1]
finding maximum hailstone(N) length for 1 <= N <= 100000... done.
hailstone(77031) length: 351
ok

'''Erlang 2'''

This version has one collatz function for just calculating totals (just for fun) and the second generating lists.

-module(collatz).
-export([main/0,collatz/1,coll/1,max_atz_under/1]).

collatz(1) -> 1;
collatz(N) when N rem 2 == 0 -> 1 + collatz(N div 2);
collatz(N) when N rem 2 > 0 -> 1 + collatz(3 * N +1).

max_atz_under(N) ->
  F = fun (X) -> {collatz(X), X} end,
  {_, Index} = lists:max(lists:map(F, lists:seq(1, N))),
  Index.

coll(1) -> [1];
coll(N) when N rem 2 == 0 -> [N|coll(N div 2)];
coll(N) -> [N|coll(3 * N + 1)].

main() ->
    io:format("collatz(4) non-list total: ~w~n", [collatz(4)]),
    io:format("coll(4) with lists ~w~n",  [coll(4)] ),
    Seq27 = coll(27),
    Seq1000 = coll(max_atz_under(100000)),
    io:format("coll(27) length: ~B~n", [length(Seq27)]),
    io:format("coll(27) first 4: ~w~n", [lists:sublist(Seq27, 4)]),
    io:format("collatz(27) last 4: ~w~n",
              [lists:nthtail(length(Seq27) - 4, Seq27)]),
    io:format("maximum  N <= 100000..."),
    io:format("Max: ~w~n", [max_atz_under(100000)]),
    io:format("Total: ~w~n", [ length( Seq1000 ) ] ).

'''Output'''

64> collatz:main().
collatz(4) non-list total: 3
coll(4) with lists [4,2,1]
coll(27) length: 112
coll(27) first 4: [27,82,41,124]
collatz(27) last 4: [8,4,2,1]
maximum  N <= 100000...Max: 77031
Total: 351
ok

ERRE

In Italy it's known also as "Ulam conjecture".

PROGRAM ULAM

!$DOUBLE

PROCEDURE HAILSTONE(X,PRT%->COUNT)
   COUNT=1
   IF PRT% THEN PRINT(X,) END IF
   REPEAT
      IF X/2<>INT(X/2) THEN
          X=X*3+1
        ELSE
          X=X/2
      END IF
      IF PRT% THEN PRINT(X,) END IF
      COUNT=COUNT+1
   UNTIL X=1
   IF PRT% THEN PRINT END IF
END PROCEDURE

BEGIN
   HAILSTONE(27,TRUE->COUNT)
   PRINT("Sequence length for 27:";COUNT)
   MAX_COUNT=2
   NMAX=2
   FOR I=3 TO 100000 DO
      HAILSTONE(I,FALSE->COUNT)
      IF COUNT>MAX_COUNT THEN NMAX=I MAX_COUNT=COUNT END IF
   END FOR
   PRINT("Max. number is";NMAX;" with";MAX_COUNT;"elements")
END PROGRAM

        27        82        41       124        62
        31        94        47       142        71
       214       107       322       161       484
       242       121       364       182        91
       274       137       412       206       103
       310       155       466       233       700
       350       175       526       263       790
       395      1186       593      1780       890
       445      1336       668       334       167
       502       251       754       377      1132
       566       283       850       425      1276
       638       319       958       479      1438
       719      2158      1079      3238      1619
      4858      2429      7288      3644      1822
       911      2734      1367      4102      2051
      6154      3077      9232      4616      2308
      1154       577      1732       866       433
      1300       650       325       976       488
       244       122        61       184        92
        46        23        70        35       106
        53       160        80        40        20
        10         5        16         8         4
         2         1

Sequence length for 27: 112
Max. number is 77031 with 351 elements

Euler Math Toolbox

>function hailstone (n) ...
$  v=[n];
$  repeat
$    if mod(n,2) then n=3*n+1;
$    else n=n/2;
$    endif;
$    v=v|n;
$    until n==1;
$  end;
$  return v;
$  endfunction
>hailstone(27), length(%)
 [ 27  82  41  124  62  31  94  47  142  71  214  107  322  161  484  242
 121  364  182  91  274  137  412  206  103  310  155  466  233  700
 350  175  526  263  790  395  1186  593  1780  890  445  1336  668
 334  167  502  251  754  377  1132  566  283  850  425  1276  638  319
 958  479  1438  719  2158  1079  3238  1619  4858  2429  7288  3644
 1822  911  2734  1367  4102  2051  6154  3077  9232  4616  2308  1154
 577  1732  866  433  1300  650  325  976  488  244  122  61  184  92
 46  23  70  35  106  53  160  80  40  20  10  5  16  8  4  2  1 ]
 112
>function hailstonelength (n) ...
$  v=zeros(1,n);
$  v[1]=4; v[2]=2;
$  loop 3 to n;
$    count=1;
$    n=#;
$    repeat
$      if mod(n,2) then n=3*n+1;
$      else n=n/2;
$      endif;
$      if n<=cols(v) and v[n] then
$        v[#]=v[n]+count;
$        break;
$      endif;
$      count=count+1;
$    end;
$  end;
$  return v;
$  endfunction
>h=hailstonelength(100000);
>ex=extrema(h); ex[3], ex[4]
 351
 77031

Euphoria

function hailstone(atom n)
    sequence s
    s = {n}
    while n != 1 do
        if remainder(n,2)=0 then
            n /= 2
        else
            n = 3*n + 1
        end if
        s &= n
    end while
    return s
end function

function hailstone_count(atom n)
    integer count
    count = 1
    while n != 1 do
        if remainder(n,2)=0 then
            n /= 2
        else
            n = 3*n + 1
        end if
        count += 1
    end while
    return count
end function

sequence s
s = hailstone(27)
puts(1,"hailstone(27) =\n")
? s
printf(1,"len = %d\n\n",length(s))

integer max,imax,count
max = 0
for i = 2 to 1e5-1 do
    count = hailstone_count(i)
    if count > max then
        max = count
        imax = i
    end if
end for

printf(1,"The longest hailstone sequence under 100,000 is %d with %d elements.\n",
    {imax,max})

hailstone(27) =
{27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,
91,274,137,412,206,103,310,155,466,233,700,350,175,526,263,790,395,
1186,593,1780,890,445,1336,668,334,167,502,251,754,377,1132,566,283,
850,425,1276,638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,
7288,3644,1822,911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,
577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,70,35,
106,53,160,80,40,20,10,5,16,8,4,2,1}
len = 112

The longest hailstone sequence under 100,000 is 77031 with 351 elements.

Excel

In cell '''A1''', place the starting number.
In cell '''A2''' enter this formula '''=IF(MOD(A1,2)=0,A1/2,A1*3+1)'''
Drag and copy the formula down until 4, 2, 1

Ezhil

Ezhil is a Tamil programming language, see [http://en.wikipedia.org/wiki/Ezhil_%28programming_language%29 | Wikipedia] entry.

      @( (எண்%2) == 1 )  ஆனால்
      	   hailstone( 3*எண் + 1)
          இல்லை
           hailstone( எண்/2 )
          முடி

## Factor


```factor
! rosetta/hailstone/hailstone.factor
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone

: hailstone ( n -- seq )
    [ 1vector ] keep
    [ dup 1 number= ]
    [
        dup even? [ 2 / ] [ 3 * 1 + ] if
        2dup swap push
    ] until
    drop ;

<PRIVATE
: main ( -- )
    27 hailstone dup dup
    "The hailstone sequence from 27:" print
    "  has length " write length .
    "  starts with " write 4 head [ unparse ] map ", " join print
    "  ends with " write 4 tail* [ unparse ] map ", " join print

    ! Maps n => { length n }, and reduces to longest Hailstone sequence.
    1 100000 [a,b)
    [ [ hailstone length ] keep 2array ]
    [ [ [ first ] bi@ > ] most ] map-reduce
    first2
    "The hailstone sequence from " write pprint
    " has length " write pprint "." print ;
PRIVATE>

MAIN: main

$ ./factor -run=rosetta.hailstone
Loading resource:work/rosetta/hailstone/hailstone.factor
The hailstone sequence from 27:
  has length 112
  starts with 27, 82, 41, 124
  ends with 8, 4, 2, 1
The hailstone sequence from 77031 has length 351.

[$1&$[%3*1+0~]?~[2/]?]n:
[[$." "$1>][n;!]#%]s:
[1\[$1>][\1+\n;!]#%]c:
27s;! 27c;!."
"
0m:0f:
1[$100000\>][$c;!$m;>[m:$f:0]?%1+]#%
f;." has hailstone sequence length "m;.

: hail-next ( n -- n )
  dup 1 and if 3 * 1+ else 2/ then ;
: .hail ( n -- )
  begin dup . dup 1 > while hail-next repeat drop ;
: hail-len ( n -- n )
  1 begin over 1 > while swap hail-next swap 1+ repeat nip ;

27 hail-len . cr
27 .hail cr

: longest-hail ( max -- )
  0 0 rot 1+ 1 do    ( n length )
    i hail-len 2dup < if
      nip nip i swap
    else drop then
  loop
  swap . ." has hailstone sequence length " . ;

100000 longest-hail

program Hailstone
  implicit none

  integer :: i, maxn
  integer :: maxseqlen = 0, seqlen
  integer, allocatable :: seq(:)

  call hs(27, seqlen)
  allocate(seq(seqlen))
  call hs(27, seqlen, seq)
  write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
  write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ",  &
                                     seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)

  do i = 1, 99999
    call hs(i, seqlen)
    if (seqlen > maxseqlen) then
      maxseqlen = seqlen
      maxn = i
    end if
  end do
  write(*,*)
  write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"

  deallocate(seq)

contains

subroutine hs(number, length, seqArray)
  integer, intent(in)  :: number
  integer, intent(out) :: length
  integer, optional, intent(inout) :: seqArray(:)
  integer :: n

  n = number
  length = 1
  if(present(seqArray)) seqArray(1) = n
  do while(n /= 1)
    if(mod(n,2) == 0) then
      n = n / 2
    else
      n = n * 3 + 1
    end if
    length = length + 1
    if(present(seqArray)) seqArray(length) = n
  end do
end subroutine

end program

Hailstone sequence for 27 has 112 elements
Sequence = 27, 82, 41, 124, ...., 8, 4, 2, 1

Longest sequence under 100000 is for 77031 with 351 elements

module Hailstone where

import Data.List (maximumBy)

hailstone :: Int -> [Int]
hailstone 1             = [1]
hailstone n | even n    = n : hailstone (n `div` 2)
            | otherwise = n : hailstone (n * 3 + 1)

withResult :: (t -> t1) -> t -> (t1, t)
withResult f x = (f x, x)

main :: IO ()
main = do
 let h27 = hailstone 27
 putStrLn $ show $ length h27
 let h4 = show $ take 4 h27
 let t4 = show $ drop (length h27 - 4) h27
 putStrLn ("hailstone 27: " ++ h4 ++ " ... " ++ t4)
 putStrLn $ show $ maximumBy (comparing fst) $ map (withResult (length . hailstone)) [1..100000]

112
hailstone 27: [27, 82, 41, 124] ... [8, 4, 2, 1]
(351, 77031)
runtime 0.969 wallclock seconds.

hailstone[n] :=
{
   results = new array

   while n != 1
   {
      results.push[n]
      if n mod 2 == 0    // n is even?
         n = n / 2
      else
         n = (3n + 1)
   }

   results.push[1]
   return results
}

longestLen = 0
longestN = 0
for n = 1 to 100000
{
   seq = hailstone[n]
   if length[seq] > longestLen
   {
      longestLen = length[seq]
      longestN = n
   }
}

println["$longestN has length $longestLen"]

let rec hailstone n = seq {
  match n with
  | 1                -> yield 1
  | n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
  | n                -> yield n; yield! hailstone (n * 3 + 1)
}

let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])

let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI

Maximum length 351 was found for hailstone(77031)

def
  hailstone( 1 ) = [1]
  hailstone( n ) = n # hailstone( if 2|n then n/2 else n*3 + 1 )

if _name_ == '-main-'
  h27 = hailstone( 27 )
  assert( h27.length() == 112 and h27.startsWith([27, 82, 41, 124]) and h27.endsWith([8, 4, 2, 1]) )

  val (n, len) = maxBy( snd, [(i, hailstone( i ).length()) | i <- 1:100000] )

  println( n, len )

77031, 351

fun hailstone_step(x: int): int =
  if (x % 2) == 0
  then x/2
  else (3*x) + 1

fun hailstone_seq(x: int): []int =
  let capacity = 100
  let i = 1
  let steps = replicate capacity (-1)
  let steps[0] = x
  loop ((capacity,i,steps,x)) = while x != 1 do
    let (steps, capacity) =
      if i == capacity then
        (concat steps (replicate capacity (-1)),
         capacity * 2)
      else (steps, capacity)
    let x = hailstone_step x
    let steps[i] = x
    in (capacity, i+1, steps, x)
  in #1 (split i steps)

fun hailstone_len(x: int): int =
  let i = 1
  loop ((i,x)) = while x != 1 do
    (i+1, hailstone_step x)
  in i

fun max (x: int) (y: int): int = if x < y then y else x

fun main (x: int) (n: int): ([]int, int) =
  (hailstone_seq x,
   reduce max 0 (map hailstone_len (map (1+) (iota (n-1)))))

CollatzSequence := function(n)
  local v;
  v := [ n ];
  while n > 1 do
    if IsEvenInt(n) then
      n := QuoInt(n, 2);
    else
      n := 3*n + 1;
    fi;
    Add(v, n);
  od;
  return v;
end;

CollatzLength := function(n)
  local m;
  m := 1;
  while n > 1 do
    if IsEvenInt(n) then
      n := QuoInt(n, 2);
    else
      n := 3*n + 1;
    fi;
    m := m + 1;
  od;
  return m;
end;

CollatzMax := function(a, b)
  local n, len, nmax, lmax;
  lmax := 0;
  for n in [a .. b] do
    len := CollatzLength(n);
    if len > lmax then
      nmax := n;
      lmax := len;
    fi;
  od;
  return [ nmax, lmax ];
end;

CollatzSequence(27);
# [ 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206,
#   103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502,
#   251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429,
#   7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300,
#   650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 ]
CollatzLength(27);
# 112

CollatzMax(1, 100);
# [ 97, 119 ]
CollatzMax(1, 1000);
# [ 871, 179 ]
CollatzMax(1, 10000);
# [ 6171, 262 ]
CollatzMax(1, 100000);
# [ 77031, 351 ]
CollatzMax(1, 1000000);
# [ 837799, 525 ]

package main

import "fmt"

// 1st arg is the number to generate the sequence for.
// 2nd arg is a slice to recycle, to reduce garbage.
func hs(n int, recycle []int) []int {
    s := append(recycle[:0], n)
    for n > 1 {
        if n&1 == 0 {
            n = n / 2
        } else {
            n = 3*n + 1
        }
        s = append(s, n)
    }
    return s
}

func main() {
    seq := hs(27, nil)
    fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
        len(seq), seq[0], seq[1], seq[2], seq[3],
        seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])

    var maxN, maxLen int
    for n := 1; n < 100000; n++ {
        seq = hs(n, seq)
        if len(seq) > maxLen {
            maxN = n
            maxLen = len(seq)
        }
    }
    fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}

hs(27): 112 elements: [27 82 41 124 ... 8 4 2 1]
hs(77031): 351 elements

package main

import "fmt"

// Task 1 implemented with a generator.  Calling newHg will "create
// a routine to generate the hailstone sequence for a number."
func newHg(n int) func() int {
    return func() (n0 int) {
        n0 = n
        if n&1 == 0 {
            n = n / 2
        } else {
            n = 3*n + 1
        }
        return
    }
}

func main() {
    // make generator for sequence starting at 27
    hg := newHg(27)
    // save first four elements for printing later
    s1, s2, s3, s4 := hg(), hg(), hg(), hg()
    // load next four elements in variables to use as shift register.
    e4, e3, e2, e1 := hg(), hg(), hg(), hg()
    // 4+4= 8 that we've generated so far
    ec := 8
    // until we get to 1, generate another value, shift, and increment.
    // note that intermediate elements--those shifted off--are not saved.
    for e1 > 1 {
        e4, e3, e2, e1 = e3, e2, e1, hg()
        ec++
    }
    // Complete task 2:
    fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
        ec, s1, s2, s3, s4, e4, e3, e2, e1)

    // Task 3:  strategy is to not store sequences, but just the length
    // of each sequence.  as soon as the sequence we're currently working on
    // dips into the range that we've already computed, we short-circuit
    // to the end by adding the that known length to whatever length
    // we've accumulated so far.

    var nMaxLen int // variable holds n with max length encounted so far
    // slice holds sequence length for each n as it is computed
    var computedLen [1e5]int
    computedLen[1] = 1
    for n := 2; n < 1e5; n++ {
        var ele, lSum int
        for hg := newHg(n); ; lSum++ {
            ele = hg()
            // as soon as we get an element in the range we have already
            // computed, we're done...
            if ele < n {
                break
            }
        }
        // just add the sequence length already computed from this point.
        lSum += computedLen[ele]
        // save the sequence length for this n
        computedLen[n] = lSum
        // and note if it's the maximum so far
        if lSum > computedLen[nMaxLen] {
            nMaxLen = n
        }
    }
    fmt.Printf("hs(%d): %d elements\n", nMaxLen, computedLen[nMaxLen])
}

def hailstone = { long start ->
    def sequence = []
    while (start != 1) {
        sequence << start
        start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l
    }
    sequence << start
}

def sequence = hailstone(27)
assert sequence.size() == 112
assert sequence[0..3] == [27, 82, 41, 124]
assert sequence[-4..-1] == [8, 4, 2, 1]

def results = (1..100000).collect { [n:it, size:hailstone(it).size()] }.max { it.size }
println results

[n:77031, size:351]

import Data.List (maximumBy)
import Data.Ord (comparing)

collatz :: Int -> Int
collatz n
  | even n = n `div` 2
  | otherwise = 3 * n + 1

hailstone :: Int -> [Int]
hailstone = takeWhile (/= 1) . iterate collatz

longestChain :: Int
longestChain =
  fst $
  maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000]

--TEST -------------------------------------------------------------------------
main :: IO ()
main =
  mapM_
    putStrLn
    [ "Collatz sequence for 27: "
    , (show . hailstone) 27
    , "The number " ++ show longestChain
    , "has the longest hailstone sequence for any number less then 100000. "
    , "The sequence has length: " ++ (show . length . hailstone $ longestChain)
    ]

Collatz sequence for 27:
[27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,91,274,137,412,206,103,310,155,466,233,700,350,175,526,263,790,395,1186,593,1780,890,445,1336,668,334,167,502,251,754,377,1132,566,283,850,425,1276,638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,7288,3644,1822,911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,70,35,106,53,160,80,40,20,10,5,16,8,4,2]
The number 77031
has the longest hailstone sequence for any number less then 100000.
The sequence has length: 350

import Data.Ord (comparing)
import Data.List (maximumBy, intercalate)

hailstone :: Int -> [Int]
hailstone 1 = [1]
hailstone n
  | even n = n : hailstone (n `div` 2)
  | otherwise = n : hailstone (n * 3 + 1)

withResult :: (Int -> Int) -> Int -> (Int, Int)
withResult f x = (f x, x)

h27 :: [Int]
h27 = hailstone 27

main :: IO ()
main =
  mapM_
    putStrLn
    [ (show . length) h27
    , "hailstone 27: " ++
      intercalate " ... " (show <$> [take 4 h27, drop (length h27 - 4) h27])
    , show $
      maximumBy (comparing fst) $
      withResult (length . hailstone) <$> [1 .. 100000]
    ]

112
hailstone 27: [27,82,41,124] ... [8,4,2,1]
(351,77031)

import Data.List (unfoldr)

hailStones :: Int -> [Int]
hailStones =
  (++ [1]) .
  unfoldr
    (\x ->
        if x < 2
          then Nothing
          else Just
                 ( x
                 , if even x
                     then div x 2
                     else (3 * x) + 1))

mostStones :: Int -> (Int, Int)
mostStones n =
  foldr
    (\x (m, ml) ->
        let l = length (hailStones x)
        in if l > ml
             then (x, l)
             else (m, ml))
    (0, 0)
    [1 .. n]

-- GENERIC  -------------------------------------------------------------------
lastN_ :: Int -> [Int] -> [Int]
lastN_ = (foldr (const (drop 1)) <*>) . drop

-- TEST -----------------------------------------------------------------------
h27, start27, end27 :: [Int]
[h27, start27, end27] = [id, take 4, lastN_ 4] <*> [hailStones 27]

maxNum, maxLen :: Int
(maxNum, maxLen) = mostStones 100000

main :: IO ()
main =
  mapM_
    putStrLn
    [ "Sequence 27 length:"
    , show $ length h27
    , "Sequence 27 start:"
    , show start27
    , "Sequence 27 end:"
    , show end27
    , ""
    , "N with longest sequence where N <= 100000"
    , show maxNum
    , "length of this sequence:"
    , show maxLen
    ]

Sequence 27 length:
112
Sequence 27 start:
[27,82,41,124]
Sequence 27 end:
[8,4,2,1]

N with longest sequence where N <= 100000
77031
length of this sequence:
351

DIMENSION stones(1000)

H27 = hailstone(27)
ALIAS(stones,1, first4,4)
ALIAS(stones,H27-3,  last4,4)
WRITE(ClipBoard, Name) H27, first4, "...", last4

longest_sequence = 0
DO try = 1, 1E5
  elements = hailstone(try)
  IF(elements >= longest_sequence) THEN
      number = try
      longest_sequence = elements
      WRITE(StatusBar, Name) number, longest_sequence
  ENDIF
ENDDO
WRITE(ClipBoard, Name) number, longest_sequence
END

FUNCTION hailstone( n )
   USE : stones

   stones(1) = n
   DO i = 1, LEN(stones)
     IF(stones(i) == 1) THEN
         hailstone = i
         RETURN
     ELSEIF( MOD(stones(i),2) ) THEN
       stones(i+1) = 3*stones(i) + 1
     ELSE
       stones(i+1) = stones(i) / 2
     ENDIF
   ENDDO
END

procedure hailstone(n)
    while n > 1 do {
        suspend n
        n := if n%2 = 0 then n/2 else 3*n+1
        }
    suspend 1
end

procedure main(args)
    n := integer(!args) | 27
    every writes(" ",hailstone(n))
end

procedure hailstone(n)
    static cache
    initial {
        cache := table()
        cache[1] := [1]
        }
    /cache[n] := [n] ||| hailstone(if n%2 = 0 then n/2 else 3*n+1)
    return cache[n]
end

procedure main(args)
    n := integer(!args) | 27
    task2(n)
    write()
    task3()
end

procedure task2(n)
    count := 0
    every writes(" ",right(!(sequence := hailstone(n)),5)) do
        if (count +:= 1) % 15 = 0 then write()
    write()
    write(*sequence," value",(*sequence=1,"")|"s"," in the sequence.")
end

procedure task3()
    maxHS := 0
    every n := 1 to 100000 do {
        count := *hailstone(n)
        if maxHS <:= count then maxN := n
        }
    write(maxN," has a sequence of ",maxHS," values")
end

->hs
    27    82    41   124    62    31    94    47   142    71   214   107   322   161   484
   242   121   364   182    91   274   137   412   206   103   310   155   466   233   700
   350   175   526   263   790   395  1186   593  1780   890   445  1336   668   334   167
   502   251   754   377  1132   566   283   850   425  1276   638   319   958   479  1438
   719  2158  1079  3238  1619  4858  2429  7288  3644  1822   911  2734  1367  4102  2051
  6154  3077  9232  4616  2308  1154   577  1732   866   433  1300   650   325   976   488
   244   122    61   184    92    46    23    70    35   106    53   160    80    40    20
    10     5    16     8     4     2     1
112 values in the sequence.

77031 has a sequence of 351 values
->

makeItHail := method(n,
  stones := list(n)
  while (n != 1,
    if(n isEven,
      n = n / 2,
      n = 3 * n + 1
    )
    stones append(n)
  )
  stones
)

out := makeItHail(27)
writeln("For the sequence beginning at 27, the number of elements generated is ", out size, ".")
write("The first four elements generated are ")
for(i, 0, 3,
  write(out at(i), " ")
)
writeln(".")

write("The last four elements generated are ")
for(i, out size - 4, out size - 1,
  write(out at(i), " ")
)
writeln(".")

numOfElems := 0
nn := 3
for(x, 3, 100000,
  out = makeItHail(x)
  if(out size > numOfElems,
    numOfElems = out size
    nn = x
  )
)

writeln("For numbers less than or equal to 100,000, ", nn,
" has the longest sequence of ", numOfElems, " elements.")

For the sequence beginning at 27, the number of elements generated is 112.
The first four elements generated are 27 82 41 124 .
The last four elements generated are 8 4 2 1 .
For numbers less than or equal to 100,000, 77031 has the longest sequence of 351 elements.

collatz = method(n,
  n println
  unless(n <= 1,
    if(n even?, collatz(n / 2), collatz(n * 3 + 1)))
)

Home is a room.

To decide which list of numbers is the hailstone sequence for (N - number):
	let result be a list of numbers;
	add N to result;
	while N is not 1:
		if N is even, let N be N / 2;
		otherwise let N be (3 * N) + 1;
		add N to result;
	decide on result.

Hailstone length cache relates various numbers to one number.

To decide which number is the hailstone sequence length for (N - number):
	let ON be N;
	let length so far be 0;
	while N is not 1:
		if N relates to a number by the hailstone length cache relation:
			let result be length so far plus the number to which N relates by the hailstone length cache relation;
			now the hailstone length cache relation relates ON to result;
			decide on result;
		if N is even, let N be N / 2;
		otherwise let N be (3 * N) + 1;
		increment length so far;
	let result be length so far plus 1;
	now the hailstone length cache relation relates ON to result;
	decide on result.

To say first and last (N - number) entry/entries in (L - list of values of kind K):
	let length be the number of entries in L;
	if length <= N * 2:
		say L;
	else:
		repeat with M running from 1 to N:
			if M > 1, say ", ";
			say entry M in L;
		say " ... ";
		repeat with M running from length - N + 1 to length:
			say entry M in L;
			if M < length, say ", ".

When play begins:
	let H27 be the hailstone sequence for 27;
	say "Hailstone sequence for 27 has [number of entries in H27] element[s]: [first and last 4 entries in H27].";
	let best length be 0;
	let best number be 0;
	repeat with N running from 1 to 99999:
		let L be the hailstone sequence length for N;
		if L > best length:
			let best length be L;
			let best number be N;
	say "The number under 100,000 with the longest hailstone sequence is [best number] with [best length] element[s].";
	end the story.

Hailstone sequence for 27 has 112 elements: 27, 82, 41, 124 ... 8, 4, 2, 1.
The number under 100,000 with the longest hailstone sequence is 77031 with 351 elements.

hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0

   # hailseq 27                 NB. sequence length
112
   4 _4 {."0 1 hailseq 27       NB. first & last 4 numbers in sequence
27 82 41 124
 8  4  2   1
   (>:@(i. >./) , >./) #@hailseq }.i. 1e5  NB. number < 100000 with max seq length & its seq length
77031 351

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Hailstone {

  public static List<Long> getHailstoneSequence(long n) {
    if (n <= 0)
      throw new IllegalArgumentException("Invalid starting sequence number");
    List<Long> list = new ArrayList<Long>();
    list.add(Long.valueOf(n));
    while (n != 1) {
      if ((n & 1) == 0)
        n = n / 2;
      else
        n = 3 * n + 1;
      list.add(Long.valueOf(n));
    }
    return list;
  }

  public static void main(String[] args) {
    List<Long> sequence27 = getHailstoneSequence(27);
    System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);

    long MAX = 100000;
    // Simple way
    {
      long highestNumber = 1;
      int highestCount = 1;
      for (long i = 2; i < MAX; i++) {
        int count = getHailstoneSequence(i).size();
        if (count > highestCount) {
          highestCount = count;
          highestNumber = i;
        }
      }
      System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
    }

    // More memory efficient way
    {
      long highestNumber = 1;
      int highestCount = 1;
      for (long i = 2; i < MAX; i++) {
        int count = 1;
        long n = i;
        while (n != 1) {
          if ((n & 1) == 0)
            n = n / 2;
          else
            n = 3 * n + 1;
          count++;
        }
        if (count > highestCount) {
          highestCount = count;
          highestNumber = i;
        }
      }
      System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
    }

    // Efficient for analyzing all sequences
    {
      long highestNumber = 1;
      long highestCount = 1;
      Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
      sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));

      List<Long> currentList = new ArrayList<Long>();
      for (long i = 2; i < MAX; i++) {
        currentList.clear();
        Long n = Long.valueOf(i);
        Integer count = null;
        while ((count = sequenceMap.get(n)) == null) {
          currentList.add(n);
          long nValue = n.longValue();
          if ((nValue & 1) == 0)
            n = Long.valueOf(nValue / 2);
          else
            n = Long.valueOf(3 * nValue + 1);
        }
        int curCount = count.intValue();
        for (int j = currentList.size() - 1; j >= 0; j--)
          sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
        if (curCount > highestCount) {
          highestCount = curCount;
          highestNumber = i;
        }
      }
      System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
    }
    return;
  }
}

Sequence for 27 has 112 elements: [27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1]
Method 1, number 77031 has the longest sequence, with a length of 351
Method 2, number 77031 has the longest sequence, with a length of 351
Method 3, number 77031 has the longest sequence, with a length of 351

function hailstone (n) {
    var seq = [n];
    while (n > 1) {
        n = n % 2 ? 3 * n + 1 : n / 2;
        seq.push(n);
    }
    return seq;
}

// task 2: verify the sequence for n = 27
var h = hailstone(27), hLen = h.length;
print("sequence 27 is (" + h.slice(0, 4).join(", ") + " ... "
    + h.slice(hLen - 4, hLen).join(", ") + "). length: " + hLen);

// task 3: find the longest sequence for n < 100000
for (var n, max = 0, i = 100000; --i;) {
    var seq = hailstone(i), sLen = seq.length;
    if (sLen > max) {
        n = i;
        max = sLen;
    }
}
print("longest sequence: " + max + " numbers for starting point " + n);

sequence 27 is (27, 82, 41, 124 ... 8, 4, 2, 1). length: 112
longest sequence: 351 numbers for starting point 77031

(function () {

  // Hailstone Sequence
  // n -> [n]
  function hailstone(n) {
    return n === 1 ? [1] : (
      [n].concat(
        hailstone(n % 2 ? n * 3 + 1 : n / 2)
      )
    )
  }

  var lstCollatz27 = hailstone(27);

  return {
    length: lstCollatz27.length,
    sequence: lstCollatz27
  };

})();

{"length":112,"sequence":[27,82,41,124,62,31,94,47,142,71,214,
107,322,161,484,242,121,364,182,91,274,137,412,206,103,310,155,466,233,700,350,
175,526, 263,790,395,1186,593,1780,890,445,1336,668,334,167,502,251,754,377,
1132,566,283,850,425,1276,638,319,958,479,1438,719,2158,1079,3238,1619,4858,
2429,7288,3644,1822,911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,577,
1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,70,35,106,53,160,80,
40,20,10,5,16,8,4,2,1]}

(function () {

  function memoizedHailstone() {
    var dctMemo = {};

    return function hailstone(n) {
      var value = dctMemo[n];

      if (typeof value === "undefined") {
        dctMemo[n] = value = (n === 1) ?
          [1] : ([n].concat(hailstone(n % 2 ? n * 3 + 1 : n / 2)));
      }
      return value;
    }
  }

  // Derived a memoized version of the function,
  // which can reuse previously calculated paths
  var fnCollatz = memoizedHailstone();

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

  // Fold/reduce over an array to find the maximum length
  function longestBelow(n) {
    return range(1, n).reduce(
      function (a, x, i) {
        var lng = fnCollatz(x).length;

        return lng > a.l ? {
          n: i + 1,
          l: lng
        } : a

      }, {
        n: 0,
        l: 0
      }
    )
  }

  return longestBelow(100000);

})();

// Number, length of sequence
{"n":77031, "l":351}

(function (n) {

  var dctMemo = {};

  // Length only of hailstone sequence
  // n -> n
  function collatzLength(n) {
    var i = 1,
      a = n,
      lng;

    while (a !== 1) {
      lng = dctMemo[a];
      if ('u' === (typeof lng)[0]) {
        a = (a % 2 ? 3 * a + 1 : a / 2);
        i++;
      } else return lng + i - 1;
    }
    return i;
  }

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

  // Fold/reduce over an array to find the maximum length
  function longestBelow(n) {

    return range(1, n).reduce(
      function (a, x) {

        var lng = dctMemo[x] || (dctMemo[x] = collatzLength(x));

        return lng > a.l ? {
          n: x,
          l: lng
        } : a

      }, {
        n: 0,
        l: 0
      }
    )
  }

  return [100000, 1000000, 10000000].map(longestBelow);

})();

[
  {"n":77031, "l":351},   // 100,000
  {"n":837799, "l":525},  // 1,000,000
  {"n":8400511, "l":686}  // 10,000,000
]

longestBelow(100000000)
-> {"n":63728127, "l":950}

(() => {

    // hailstones :: Int -> [Int]
    const hailstones = x => {
        const collatz = memoized(n =>
            even(n) ? div(n, 2) : (3 * n) + 1);
        return reverse(until(
            xs => xs[0] === 1,
            xs => cons(collatz(xs[0]), xs), [x]
        ));
    };

    // collatzLength :: Int -> Int
    const collatzLength = n =>
        until(
            xi => xi[0] === 1,
            ([x, i]) => [(x % 2 ? 3 * x + 1 : x / 2), i + 1], //
            [n, 1]
        )[1];

    // GENERIC FUNCTIONS -----------------------------------------------------

    // comparing :: (a -> b) -> (a -> a -> Ordering)
    const comparing = f =>
        (x, y) => {
            const
                a = f(x),
                b = f(y);
            return a < b ? -1 : (a > b ? 1 : 0);
        };

    // cons :: a -> [a] -> [a]
    const cons = (x, xs) => [x].concat(xs);

    // div :: Int -> Int -> Int
    const div = (x, y) => Math.floor(x / y);

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

    // even :: Int -> Bool
    const even = n => n % 2 === 0;

    // fst :: (a, b) -> a
    const fst = pair => pair.length === 2 ? pair[0] : undefined;

    // map :: (a -> b) -> [a] -> [b]
    const map = (f, xs) => xs.map(f);

    // maximumBy :: (a -> a -> Ordering) -> [a] -> a
    const maximumBy = (f, xs) =>
        xs.length > 0 ? (
            xs.slice(1)
            .reduce((a, x) => f(x, a) > 0 ? x : a, xs[0])
        ) : undefined;

    // memoized :: (a -> b) -> (a -> b)
    const memoized = f => {
        const dctMemo = {};
        return x => {
            const v = dctMemo[x];
            return v !== undefined ? v : (dctMemo[x] = f(x));
        };
    };

    // reverse :: [a] -> [a]
    const reverse = xs =>
        xs.slice(0)
        .reverse();

    // unlines :: [String] -> String
    const unlines = xs => xs.join('\n');

    // until :: (a -> Bool) -> (a -> a) -> a -> a
    const until = (p, f, x) => {
        let v = x;
        while (!p(v)) v = f(v);
        return v;
    };

    // MAIN ------------------------------------------------------------------
    const
        // ceiling :: Int
        ceiling = 100000,

        // (maxLen, maxNum) :: (Int, Int)
        [maxLen, maxNum] =
        maximumBy(
            comparing(fst),
            map(i => [collatzLength(i), i], enumFromTo(1, ceiling))
        );
    return unlines([
        'Collatz sequence for 27: ',
        `${hailstones(27)}`,
        '',
        `The number ${maxNum} has the longest hailstone sequence`,
        `for any starting number under ${ceiling}.`,
        '',
        `The length of that sequence is ${maxLen}.`
    ]);
})();