Ruby, 50 chars, no stack overflow

Basically a direct rip of makapuf's Python solution:

def c(n)while n>1;n=n.odd?? n*3+1: n/2;p n end end

Ruby, 45 chars, will overflow

Basically a direct rip of the code provided in the question:

def c(n)p n;n.odd?? c(3*n+1):c(n/2)if n>1 end

C : 64 chars

main(x){for(scanf("%d",&x);x>=printf("%d,",x);x=x&1?3*x+1:x/2);}

With big integer support: 431 (necessary) chars

#include <stdlib.h>
#define B (w>=m?d=realloc(d,m=m+m):0)
#define S(a,b)t=a,a=b,b=t
main(m,w,i,t){char*d=malloc(m=9);for(w=0;(i=getchar()+2)/10==5;)
B,d[w++]=i%10;for(i=0;i<w/2;i++)S(d[i],d[w-i-1]);for(;;w++){
while(w&&!d[w-1])w--;for(i=w+1;i--;)putchar(i?d[i-1]+48:10);if(
w==1&&*d==1)break;if(*d&1){for(i=w;i--;)d[i]*=3;*d+=1;}else{
for(i=w;i-->1;)d[i-1]+=d[i]%2*10,d[i]/=2;*d/=2;}B,d[w]=0;for(i=0
;i<w;i++)d[i+1]+=d[i]/10,d[i]%=10;}}

Note: Do not remove #include <stdlib.h> without at least prototyping malloc/realloc, as doing so will not be safe on 64-bit platforms (64-bit void* will be converted to 32-bit int).

This one hasn't been tested vigorously yet. It could use some shortening as well.

Previous versions:

main(x){for(scanf("%d",&x);printf("%d,",x),x-1;x=x&1?3*x+1:x/2);} // 66

(removed 12 chars because no one follows the output format... :| )

Haskell : 50

c 1=[1];c n=n:(c$if odd n then 3*n+1 else n`div`2)

Perl : 31 chars

perl -nE 'say$_=$_%2?$_*3+1:$_/2while$_>1'
#         123456789 123456789 123456789 1234567

Edited to remove 2 unnecessary spaces.

Edited to remove 1 unnecessary space.

Python:

def collatz(n):
if (n%2) == 0:
return n/2
else:
return 3*n+1
def do_collatz(n):
while n > 1:
print n
n = collatz(n)
print n
do_collatz(int(input("Start number: ")))

Not vulnerable to stack overflows, but does not terminate on a sequence that does not converge on 1. (edit: forgot the input part)

Perl, 59 characters:

sub c{print my$x="@_\n";@_=$x&1?$x*3+1:$x/2,goto&c if$x!=1}

Mathematica, 45 50 chars

c=NestWhileList[If[OddQ@#,3#+1,#/2]&,#,#>1&]&

Python - 95 64 51 46 char

Obviously does not produce a stack overflow.

n=input()
while n>1:n=(n/2,n*3+1)[n%2];print n

x86 assembly, 1337 characters

;
; To assemble and link this program, just run:
;
; >> $ nasm -f elf collatz.asm && gcc -o collatz collatz.o
;
; You can then enjoy its output by passing a number to it on the command line:
;
; >> $ ./collatz 123
; >> 123 --> 370 --> 185 --> 556 --> 278 --> 139 --> 418 --> 209 --> 628 --> 314
; >> --> 157 --> 472 --> 236 --> 118 --> 59 --> 178 --> 89 --> 268 --> 134 --> 67
; >> --> 202 --> 101 --> 304 --> 152 --> 76 --> 38 --> 19 --> 58 --> 29 --> 88
; >> --> 44 --> 22 --> 11 --> 34 --> 17 --> 52 --> 26 --> 13 --> 40 --> 20 --> 10
; >> --> 5 --> 16 --> 8 --> 4 --> 2 --> 1
;
; There's even some error checking involved:
; >> $ ./collatz
; >> Usage: ./collatz NUMBER
;
section .text
global main
extern printf
extern atoi


main:


cmp dword [esp+0x04], 2
jne .usage


mov ebx, [esp+0x08]
push dword [ebx+0x04]
call atoi
add esp, 4


cmp eax, 0
je .usage


mov ebx, eax
push eax
push msg


.loop:
mov [esp+0x04], ebx
call printf


test ebx, 0x01
jz .even


.odd:
lea ebx, [1+ebx*2+ebx]
jmp .loop


.even:


shr ebx, 1
cmp ebx, 1
jne .loop


push ebx
push end
call printf


add esp, 16
xor eax, eax
ret


.usage:
mov ebx, [esp+0x08]
push dword [ebx+0x00]
push usage
call printf
add esp, 8
mov eax, 1
ret


msg db "%d --> ", 0
end db "%d", 10, 0
usage db "Usage: %s NUMBER", 10, 0

Haskell, 62 chars 63 76 83, 86, 97, 137

c 1=[1]
c n=n:c(div(n`mod`2*(5*n+2)+n)2)
main=readLn>>=print.c

User input, printed output, uses constant memory and stack, works with arbitrarily big integers.

A sample run of this code, given an 80 digit number of all '1's (!) as input, is pretty fun to look at.

Original, function only version:

Haskell 51 chars

f n=n:[[],f([n`div`2,3*n+1]!!(n`mod`2))]!!(1`mod`n)

Who the @&^# needs conditionals, anyway?

(edit: I was being "clever" and used fix. Without it, the code dropped to 54 chars. edit2: dropped to 51 by factoring out f())

Befunge

&>:.:1-|
>3*^ @
|%2: <
v>2/>+

MS Excel, 35 chars

=IF(A1/2=ROUND(A1/2,0),A1/2,A1*3+1)

Taken straight from Wikipedia:

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

It only took copy/pasting the formula 111 times to get the result for a starting number of 1000. ;)

Golfscript : 20 chars

  ~{(}{3*).1&5*)/}/1+`
#
# Usage: echo 21 | ruby golfscript.rb collatz.gs

This is equivalent to

stack<int> s;
s.push(21);
while (s.top() - 1) {
int x = s.top();
int numerator = x*3+1;
int denominator = (numerator&1) * 5 + 1;
s.push(numerator/denominator);
}
s.push(1);
return s;

Ruby, 43 characters

bignum supported, with stack overflow susceptibility:

def c(n)p n;n%2>0?c(3*n+1):c(n/2)if n>1 end

...and 50 characters, bignum supported, without stack overflow:

def d(n)while n>1 do p n;n=n%2>0?3*n+1:n/2 end end

Kudos to Jordan. I didn't know about 'p' as a replacement for puts.

F#, 90 characters

let c=Seq.unfold(function|n when n<=1->None|n when n%2=0->Some(n,n/2)|n->Some(n,(3*n)+1))


> c 21;;
val it : seq<int> = seq [21; 64; 32; 16; ...]

Or if you're not using F# interactive to display the result, 102 characters:

let c=Seq.unfold(function|n when n<=1->None|n when n%2=0->Some(n,n/2)|n->Some(n,(3*n)+1))>>printf"%A"

MATLAB 7.8.0 (R2009a): 58 characters

n=input('');while n>1,n=n/2+rem(n,2)*(n*5+2)/2;disp(n);end

Test case:

>> n=input('');while n>1,n=n/2+rem(n,2)*(n*5+2)/2;disp(n);end
21
64
32
16
8
4
2
1

C#: 216 Characters

using C=System.Console;class P{static void Main(){var p="start:";System.Action<object> o=C.Write;o(p);ulong i;while(ulong.TryParse(C.ReadLine(),out i)){o(i);while(i > 1){i=i%2==0?i/2:i*3+1;o(" -> "+i);}o("\n"+p);}}}

in long form:

using C = System.Console;
class P
{
static void Main()
{
var p = "start:";
System.Action<object> o = C.Write;
o(p);
ulong i;
while (ulong.TryParse(C.ReadLine(), out i))
{
o(i);
while (i > 1)
{
i = i % 2 == 0 ? i / 2 : i * 3 + 1;
o(" -> " + i);
}
o("\n" + p);
}
}
}

New Version, accepts one number as input provided through the command line, no input validation. 173 154 characters.

using System;class P{static void Main(string[]a){Action<object>o=Console.Write;var i=ulong.Parse(a[0]);o(i);while(i>1){i=i%2==0?i/2:i*3+1;o(" -> "+i);}}}

in long form:

using System;
class P
{
static void Main(string[]a)
{
Action<object>o=Console.Write;
var i=ulong.Parse(a[0]);
o(i);
while(i>1)
{
i=i%2==0?i/2:i*3+1;
o(" -> "+i);
}
}
}

I am able to shave a few characters by ripping off the idea in this answer to use a for loop rather than a while. 150 characters.

using System;class P{static void Main(string[]a){Action<object>o=Console.Write;for(var i=ulong.Parse(a[0]);i>1;i=i%2==0?i/2:i*3+1)o(i+" -> ");o(1);}}

Common Lisp, 141 characters:

(defun c ()
(format t"Number: ")
(loop for n = (read) then (if(oddp n)(+ 1 n n n)(/ n 2))
until (= n 1)
do (format t"~d -> "n))
(format t"1~%"))

Test run:

Number: 171
171 -> 514 -> 257 -> 772 -> 386 -> 193 -> 580 -> 290 -> 145 -> 436 ->
218 -> 109 -> 328 -> 164 -> 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

Based on ar's code, here's a perl version which actually conforms to the output requirements

perl -E 'print"Number: ";$_=<STDIN>;chomp;print"Results: $_";$_=$_%2?$_*3+1:$_/2,print" -> ",$_ while$_!=1;say""'

Length: 114 counting the perl invocation and quotes, 104 withouit

I'm sure some experienced golfer can reduce this crude version further.

Scheme: 72

(define(c n)(if(= n 1)`(1)(cons n(if(odd? n)(c(+(* n 3)1))(c(/ n 2))))))

This uses recursion, but the calls are tail-recursive so I think they'll be optimized to iteration. In some quick testing, I haven't been able to find a number for which the stack overflows anyway. Just for example:

(c 9876543219999999999000011234567898888777766665555444433332222 7777777777777777777777777777777798797657657651234143375987342987 5398709812374982529830983743297432985230985739287023987532098579 058095873098753098370938753987)

...runs just fine. [that's all one number -- I've just broken it to fit on screen.]

import java.math.BigInteger;
public class SortaJava {


static final BigInteger THREE = new BigInteger("3");
static final BigInteger TWO = new BigInteger("2");


interface BiFunc<R, A, B> {
R call(A a, B b);
}


interface Cons<A, B> {
<R> R apply(BiFunc<R, A, B> func);
}


static class Collatz implements Cons<BigInteger, Collatz> {
BigInteger value;
public Collatz(BigInteger value) { this.value = value; }
public <R> R apply(BiFunc<R, BigInteger, Collatz> func) {
if(BigInteger.ONE.equals(value))
return func.call(value, null);
if(value.testBit(0))
return func.call(value, new Collatz((value.multiply(THREE)).add(BigInteger.ONE)));
return func.call(value, new Collatz(value.divide(TWO)));
}
}


static class PrintAReturnB<A, B> implements BiFunc<B, A, B> {
boolean first = true;
public B call(A a, B b) {
if(first)
first = false;
else
System.out.print(" -> ");
System.out.print(a);
return b;
}
}


public static void main(String[] args) {
BiFunc<Collatz, BigInteger, Collatz> printer = new PrintAReturnB<BigInteger, Collatz>();
Collatz collatz = new Collatz(new BigInteger(args[0]));
while(collatz != null)
collatz = collatz.apply(printer);
}
}

C#, 88 chars I think. Recursive

void T(int i,string s){Console.Write("{0}{1}",s,i);if(i!=1)T(i%2==0?i/2:(i*3)+ 1,"->");}

Plus this initial call

T(171, "");

Here is a non-recursive method, 107 chars I think

void T2(int i){string s="";while(i>=1){Console.Write("{0}{1}",s,i);i=i==1?-1:i=i%2==0?i/2:(i*3)+1;s="->";}}

VB.Net, about 180 characters

Sub Main()
Dim q = New Queue(Of Integer)
q.Enqueue(CInt(Console.ReadLine))
Do
q.Enqueue(CInt(If(q.Peek Mod 2 = 0, q.Dequeue / 2, q.Dequeue * 3 + 1)))
Console.WriteLine(q.Peek)
Loop Until q.Peek = 1
End Sub

funny thing is converting this code into c# create more characters

to make it work in an empty .vb file (about 245 characters)

Imports System.Collections.Generic
Imports System
Module m
Sub Main()
Dim q = New Queue(Of Integer)
q.Enqueue(CInt(Console.ReadLine))
Do
q.Enqueue(CInt(If(q.Peek Mod 2 = 0, q.Dequeue / 2, q.Dequeue * 3 + 1)))
Console.WriteLine(q.Peek)
Loop Until q.Peek = 1
End Sub
End Module

not the shortest, but an elegant clojure solution

(defn collatz [n]
(print n "")
(if (> n 1)
(recur
(if (odd? n)
(inc (* 3 n))
(/ n 2)))))

bc 41 chars

I guess this kind of problems is what bc was invented for:

for(n=read();n>1;){if(n%2)n=n*6+2;n/=2;n}

Test:

bc1 -q collatz.bc
21
64
32
16
8
4
2
1

Proper code:

for(n=read();n>1;){if(n%2)n=n*3+1else n/=2;print n,"\n"}

bc handles numbers with up to INT_MAX digits

Edit: The Wikipedia article mentions this conjecture has been checked for all values up to 20x2⁵⁸ (aprox. 5.76e18). This program:

c=0;for(n=2^20000+1;n>1;){if(n%2)n=n*6+2;n/=2;c+=1};n;c

tests 2^20,000+1 (aprox. 3.98e6,020) in 68 seconds, 144,404 cycles.

C++ 113 100 95

#include <iostream>
int main(int i){for(std::cin>>i;i>1;i=i&1?i*3+1:i/2)std::cout<<i<<" -> ";}

Miranda (101 characters)

c n=" 1",if n=1
=" "++shownum(n)++c(n*3+1),if n mod 2=1
=" "++shownum(n)++c(n div 2),otherwise

(whitespace is syntactically important)

Using an extension of HQ9+ (that I haven't written yet), called HQ9+C where C prints the Collatz Sequence on a number taken from stdin.

C

:P

PHP

function Collatz($n)
{
$i = 0;
while($n>1)
{
if($n % 2)
{
$n = (3*$n) + 1;
$i++;
echo "step $i:  $n <br/>";
}


else
{
$n = $n/2;
$i++;
echo "step $i:  $n <br/>";
}
}


}

Powershell : 80 characters

One-liner:

"Results: $(for($x=read-host Number;1%$x;$x=@($x/2;3*$x+1)[$x%2]){""$x ->""}) 1"

Pretty-printed:

"Results: $( for( $x = read-host Number; 1%$x; $x = @( $x/2; 3*$x+1 )[ $x%2 ] )
{
""$x ->""
}
) 1"

Without input prompt and output formatting - 44 characters:

for($x=read-host;1%$x;$x=@($x/2;3*$x+1)[$x%2]){$x}1

The program frm Jerry Coffin has integer over flow, try this one:

#include <iostream>


int main(unsigned long long i)
{
int j = 0;
for(  std::cin>>i; i>1; i = i&1? i*3+1:i/2, ++j)
std::cout<<i<<" -> ";


std::cout<<"\n"<<j << " iterations\n";
}

tested with

The number less than 100 million with the longest total stopping time is 63,728,127, with 949 steps.

The number less than 1 billion with the longest total stopping time is 670,617,279, with 986 steps.

Bash, 130 including spaces and newlines:

#!/bin/bash
if [ $1 == 1 ]; then echo $1
else if [ $(($1%2)) == 0 ]; then n=$(($1/2))
else n=$(($1*3+1))
fi
echo "$1 -> `c $n`"
fi

This assumes that c is the name of the script file, and it is in the path of the user who's running the script.

VBScript: 105 Characters

Apparently I'm a glutton for punishment.

c(InputBox("?"))
Public Sub c(i)
msgbox(i)
If i>1 Then
If i mod 2=0 Then
c(i/2)
Else
c(3*i+1)
End If
End If
End Sub

Python 45 Char

Shaved a char off of makapuf's answer.

n=input()
while~-n:n=(n/2,n*3+1)[n%2];print n

dc - 24 chars 25 28

dc is a good tool for this sequence:

?[d5*2+d2%*+2/pd1<L]dsLx

dc -f collatz.dc
21
64
32
16
8
4
2
1

Also 24 chars using the formula from the Golfscript entry:

?[3*1+d2%5*1+/pd1<L]dsLx

57 chars to meet the specs:

[Number: ]n?[Results: ]ndn[d5*2+d2%*+2/[ -> ]ndnd1<L]dsLx

dc -f collatz-spec.dc
Number: 3
Results: 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1

Scala + Scalaz

import scalaz._
import Scalaz._
val collatz =
(_:Int).iterate[Stream](a=>Seq(a/2,3*a+1)(a%2)).takeWhile(1<) // This line: 61 chars

And in action:

scala> collatz(7).toList
res15: List[Int] = List(7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2)

Scala 2.8

val collatz =
Stream.iterate(_:Int)(a=>Seq(a/2,3*a+1)(a%2)).takeWhile(1<) :+ 1

This also includes the trailing 1.

scala> collatz(7)
res12: scala.collection.immutable.Stream[Int] = Stream(7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1)

With the following implicit

implicit def intToEven(i:Int) = new {
def ~(even: Int=>Int, odd: Int=>Int) = {
if (i%2==0) { even(i) } else { odd(i) }
}
}

this can be shortened to

val collatz = Stream.iterate(_:Int)(_~(_/2,3*_+1)).takeWhile(1<) :+ 1

Edit - 58 characters (including input and output, but not including initial number)

var n=readInt;while(n>1){n=Seq(n/2,n*3+1)(n%2);println(n)}

Could be reduced by 2 if you don't need newlines...

Perl

I decided to be a little anticompetitive, and show how you would normally code such problem in Perl.
_{There is also a 46 (total) char code-golf entry at the end.}

These first three examples all start out with this header.

#! /usr/bin/env perl
use Modern::Perl;
# which is the same as these three lines:
# use 5.10.0;
# use strict;
# use warnings;


while( <> ){
chomp;
last unless $_;
Collatz( $_ );
}

• Simple recursive version

  use Sub::Call::Recur;
  sub Collatz{
  my( $n ) = @_;
  $n += 0; # ensure that it is numeric
  die 'invalid value' unless $n > 0;
  die 'Integer values only' unless $n == int $n;
  say $n;
  given( $n ){
  when( 1 ){}
  when( $_ % 2 != 0 ){ # odd
  recur( 3 * $n + 1 );
  }
  default{ # even
  recur( $n / 2 );
  }
  }
  }
  

• Simple iterative version

  sub Collatz{
  my( $n ) = @_;
  $n += 0; # ensure that it is numeric
  die 'invalid value' unless $n > 0;
  die 'Integer values only' unless $n == int $n;
  say $n;
  while( $n > 1 ){
  if( $n % 2 ){ # odd
  $n = 3 * $n + 1;
  } else { #even
  $n = $n / 2;
  }
  say $n;
  }
  }
  

• Optimized iterative version

  sub Collatz{
  my( $n ) = @_;
  $n += 0; # ensure that it is numeric
  die 'invalid value' unless $n > 0;
  die 'Integer values only' unless $n == int $n;
  #
  state @next;
  $next[1] //= 0; # sets $next[1] to 0 if it is undefined
  #
  # fill out @next until we get to a value we've already worked on
  until( defined $next[$n] ){
  say $n;
  #
  if( $n % 2 ){ # odd
  $next[$n] = 3 * $n + 1;
  } else { # even
  $next[$n] = $n / 2;
  }
  #
  $n = $next[$n];
  }
  say $n;
  # finish running until we get to 1
  say $n while $n = $next[$n];
  }
  

Now I'm going to show how you would do that last example with a version of Perl prior to v5.10.0

#! /usr/bin/env perl
use strict;
use warnings;


while( <> ){
chomp;
last unless $_;
Collatz( $_ );
}
{
my @next = (0,0); # essentially the same as a state variable
sub Collatz{
my( $n ) = @_;
$n += 0; # ensure that it is numeric
die 'invalid value' unless $n > 0;


# fill out @next until we get to a value we've already worked on
until( $n == 1 or defined $next[$n] ){
print $n, "\n";


if( $n % 2 ){ # odd
$next[$n] = 3 * $n + 1;
} else { # even
$next[$n] = $n / 2;
}
$n = $next[$n];
}
print $n, "\n";


# finish running until we get to 1
print $n, "\n" while $n = $next[$n];
}
}

Benchmark

First off the IO is always going to be the slow part. So if you actually benchmarked them as-is you should get about the same speed out of each one.

To test these then, I opened a file handle to /dev/null ($null), and edited every say $n to instead read say {$null} $n. This is to reduce the dependence on IO.

#! /usr/bin/env perl
use Modern::Perl;
use autodie;


open our $null, '>', '/dev/null';


use Benchmark qw':all';


cmpthese( -10,
{
Recursive => sub{ Collatz_r( 31 ) },
Iterative => sub{ Collatz_i( 31 ) },
Optimized => sub{ Collatz_o( 31 ) },
});


sub Collatz_r{
...
say {$null} $n;
...
}
sub Collatz_i{
...
say {$null} $n;
...
}
sub Collatz_o{
...
say {$null} $n;
...
}

After having run it 10 times, here is a representative sample output:

Rate Recursive Iterative Optimized
Recursive 1715/s        --      -27%      -46%
Iterative 2336/s       36%        --      -27%
Optimized 3187/s       86%       36%        --

Finally, a real code-golf entry:

perl -nlE'say;say$_=$_%2?3*$_+1:$_/2while$_>1'

46 chars total

If you don't need to print the starting value, you could remove 5 more characters.

perl -nE'say$_=$_%2?3*$_+1:$_/2while$_>1'

41 chars total
_{31 chars for the actual code portion, but the code won't work without the -n switch. So I include the entire example in my count.}

nroff¹

Run with nroff -U hail.g

.warn
.pl 1
.pso (printf "Enter a number: " 1>&2); read x; echo .nr x $x
.while \nx>1 \{\
.  ie \nx%2 .nr x \nx*3+1
.  el .nr x \nx/2
\nx
.\}

^{1. groff version}

F# 82 Chars

let rec f n=printfn "%A" n;if n>1I then if n%2I=0I then f(n/2I)else f(3I*n+1I)

Factor:

without golfing

USE: math
: body ( n -- n ) >integer dup . "->" . dup odd? = [ 3 * 1 + ] [ 2 / ] if ;
: hailstone ( n --  ) dup 1 > [ body hailstone ] [ . ] if  ;
21 hailstone

golfed:

21 [ dup 1 > ] [ >integer dup . "->" . dup 2 mod 1 = [ 3 * 1 + ] [ 2 / ] if ] while .

Output:

21
"->"
64
"->"
32
"->"
16
"->"
8
"->"
4
"->"
2
"->"
1

ruby, 43, possibly meeting the I/O requirement

Run with ruby -n hail

n=$_.to_i
(n=n%2>0?n*3+1: n/2
p n)while n>1

Fortran: 71 chars

n=1
1 if(n==1)read*,n
n=merge(n/2,3*n+1,mod(n,2)==0)
print*,n
goto1
end

Because someone had to do it :)

The count includes required newlines. Fully conformant Fortran 95 (and later) code. Includes full I/O, and performs as many times as you want!

Edit: one less char using a goto (points for style!)

PHP, 78 72 67 characters

I actually wrote this program about two years ago, after I read about the sequence in a Pickover book. I cleaned it up a bit, and this is the smallest I can make it and still have user input and a nice, readable output:

<?$n=fgets(STDIN);while($n!=1){$n=(($n&1)==0)?($n/2):(($n*3)+1);echo"$n\n";}?>

One has to assume short tags are enabled, and I'm not so sure that input will work on all consoles. But it works perfectly on my Windows machine.

Update: By cheating on the math just a little, we can shave off some characters:

<?$n=fgets(STDIN);while($n!=1){$n=(($n&1)==0)?$n/2:$n*3+1;echo"$n\n";}?>

Update:

• Given that $n&1 returns either 1 or 0, we can take advantage of PHP's loose typedness and remove a couple more characters.
• Also, incorporating Christian's comment below (with a minor alteration to prevent infinite looping), we can remove one more.
• Finally, since PHP scripts don't need a terminating ?>, we can get rid of yet two more characters:

The end result:

<?$n=fgets(STDIN);while($n>1){$n=(!($n&1))?$n/2:$n*3+1;echo"$n\n";}

J, 45 characters

(-: * 0&=@(2&|)) + (1 + 3&*) * -.@(0&=@(2&|))

I'm no expert in J. Since the function for mean is +/%#, I'm sure this can be made shorter.

Microsoft Small Basic

TextWindow.Write( "Number: " )
n = TextWindow.ReadNumber()
TextWindow.Write( "Results: " )
While ( n > 1 )
TextWindow.Write( n + " -> " )
If Math.Remainder( n, 2 ) = 0  Then
n = n / 2
Else
n = n * 3 + 1
EndIf
EndWhile
TextWindow.WriteLine(1)

You can run it at: http://smallbasic.com/program/?ZZR544

Another assembler version. This one is not limited to 32 bit numbers, it can handle numbers up to 10⁶⁵⁵³⁴ although the ".com" format MS-DOS uses is limited to 80 digit numbers. Written for A86 assembler and requires a Win-XP DOS box to run. Assembles to 180 bytes:

    mov ax,cs
mov si,82h
add ah,10h
mov es,ax
mov bh,0
mov bl,byte ptr [80h]
cmp bl,1
jbe ret
dec bl
mov cx,bx
dec bl
xor di,di
p1:lodsb
sub al,'0'
cmp al,10
jae ret
stosb
loop p1
xor bp,bp
push es
pop ds
p2:cmp byte ptr ds:[bp],0
jne p3
inc bp
jmp p2
ret
p3:lea si,[bp-1]
cld
p4:inc si
mov dl,[si]
add dl,'0'
mov ah,2
int 21h
cmp si,bx
jne p4
cmp bx,bp
jne p5
cmp byte ptr [bx],1
je ret
p5:mov dl,'-'
mov ah,2
int 21h
mov dl,'>'
int 21h
test byte ptr [bx],1
jz p10
;odd
mov si,bx
mov di,si
mov dx,3
dec bp
std
p6:lodsb
mul dl
add al,dh
aam
mov dh,ah
stosb
cmp si,bp
jnz p6
or dh,dh
jz p7
mov al,dh
stosb
dec bp
p7:mov si,bx
mov di,si
p8:lodsb
inc al
xor ah,ah
aaa
stosb
or ah,ah
jz p9
cmp si,bp
jne p8
mov al,1
stosb
jmp p2
p9:inc bp
jmp p2
p10:mov si,bp
mov di,bp
xor ax,ax
p11:lodsb
test ah,1
jz p12
add al,10
p12:mov ah,al
shr al,1
cmp di,bx
stosb
jne p11
jmp p2

PHP. 69 Characters

Thanks to Danko Durbić for the fgets(STDIN), hope you dont mind :)

<?$i=fgets(STDIN);while($i!=1){echo ($i=$i%2?$i*3+1:$i/=2),"\r\n";}?>

bash 57/61/60

Another bash entry. Does not perform infinite precision math, and it may overflow.

#!/bin/bash
x=$1;echo $x;((x>1))&&$0 $((x%2?x*3+1:x/2))

A version that should not overflow could be

#!/bin/bash
x=$1;echo $x;((x>1))&&exec $0 $((x%2?x*3+1:x/2))

(edit) An iterative version as well:

#!/bin/bash
for((x=$1;x>1;x=x%2?x*3+1:x/2));do echo $x;done

JavaScript, 61 70 character with input

Iterative, precision depends on JS limits

var i=prompt('');while(i>1){console.log(i);i=(i%2)?3*i+1:i/2}

TI-BASIC

Not the shortest, but a novel approach. Certain to slow down considerably with large sequences, but it shouldn't overflow.

PROGRAM:COLLATZ
:ClrHome
:Input X
:Lbl 1
:While X≠1
:If X/2=int(X/2)
:Then
:Disp X/2→X
:Else
:Disp X*3+1→X
:End
:Goto 1
:End

Javascript, 67 56 chars

for(a=[i=prompt()];i-1;a.push(i=i%2?i*3+1:i/2));alert(a)

Ruby, 41 chars

n=gets.to_i
p n=[n/2,n*3+1][n%2]while n>1

C: 63 chars

main(x){scanf("%d",&x);while(x>printf("%d ",x=x&1?3*x+1:x/2));}

This is based on an answer from KennyTM. The for loop was chaged to a while loop and the code brought inside the while.

C# : 659 chars with BigInteger support

using System.Linq;using C=System.Console;class Program{static void Main(){var v=C.ReadLine();C.Write(v);while(v!="1"){C.Write("->");if(v[v.Length-1]%2==0){v=v.Aggregate(new{s="",o=0},(r,c)=>new{s=r.s+(char)((c-48)/2+r.o+48),o=(c%2)*5}).s.TrimStart('0');}else{var q=v.Reverse().Aggregate(new{s="",o=0},(r, c)=>new{s=(char)((c-48)*3+r.o+(c*3+r.o>153?c*3+r.o>163?28:38:48))+r.s,o=c*3+r.o>153?c*3+r.o>163?2:1:0});var t=(q.o+q.s).TrimStart('0').Reverse();var x=t.First();q=t.Skip(1).Aggregate(new{s=x>56?(x-57).ToString():(x-47).ToString(),o=x>56?1:0},(r,c)=>new{s=(char)(c-48+r.o+(c+r.o>57?38:48))+r.s,o=c+r.o>57?1:0});v=(q.o+q.s).TrimStart('0');}C.Write(v);}}}

Ungolfed

using System.Linq;
using C = System.Console;
class Program
{
static void Main()
{
var v = C.ReadLine();
C.Write(v);
while (v != "1")
{
C.Write("->");
if (v[v.Length - 1] % 2 == 0)
{
v = v
.Aggregate(
new { s = "", o = 0 },
(r, c) => new { s = r.s + (char)((c - 48) / 2 + r.o + 48), o = (c % 2) * 5 })
.s.TrimStart('0');
}
else
{
var q = v
.Reverse()
.Aggregate(
new { s = "", o = 0 },
(r, c) => new { s = (char)((c - 48) * 3 + r.o + (c * 3 + r.o > 153 ? c * 3 + r.o > 163 ? 28 : 38 : 48)) + r.s, o = c * 3 + r.o > 153 ? c * 3 + r.o > 163 ? 2 : 1 : 0 });
var t = (q.o + q.s)
.TrimStart('0')
.Reverse();
var x = t.First();
q = t
.Skip(1)
.Aggregate(
new { s = x > 56 ? (x - 57).ToString() : (x - 47).ToString(), o = x > 56 ? 1 : 0 },
(r, c) => new { s = (char)(c - 48 + r.o + (c + r.o > 57 ? 38 : 48)) + r.s, o = c + r.o > 57 ? 1 : 0 });
v = (q.o + q.s)
.TrimStart('0');
}
C.Write(v);
}
}
}

Go, 130 characters

package main
import(."os"
."strconv")
func main(){n,_:=Atoi(Args[1])
println(n)
for n>1{if n%2!=0{n=n*3+1}else{n/=2}
println(n)}}

Example

./collatz 3
3
10
5
16
8
4
2
1

Fortran - 60 Chars

read*,n
1 if(n/2*2<n)n=6*n+2
n=n/2
print*,n
if(n>1)goto1
end

GW-BASIC - 54 chars

1INPUT N
2N=(N+(N*5+2)*(N MOD 2))/2:?N:IF N>1GOTO 2

Windows cmd - 68 chars

@set/pd=
:l
@set/ad=(d+d%%2*(d*5+2))/2&echo %d%&if %d% NEQ 1 goto:l

JavaScript - 68 chars

Unlike the other JS (and most other languages) this one actually adheres to the -> in the output.

for(s='',c=' -> ',i=readline();i>1;i=i%2?i*3+1:i/2)s+=i+c
print(s+1)

If we avoid that, this is a 53 char alternative, prints one number per line:

for(p=print,i=readline(),p(i);i>1;)p(i=i%2?i*3+1:i/2)

Meant to run with SpiderMonkey:

echo 21 | js thisfile.js


21 -> 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1

LOLCODE: 406 CHARAKTERZ

HAI
BTW COLLATZ SOUNDZ JUS LULZ


CAN HAS STDIO?


I HAS A NUMBAR
BTW, I WANTS UR NUMBAR
GIMMEH NUMBAR


VISIBLE NUMBAR


IM IN YR SEQUENZ
MOD OF NUMBAR AN 2
BOTH SAEM IT AN 0, O RLY?
YA RLY, NUMBAR R QUOSHUNT OF NUMBAR AN 2
NO WAI, NUMBAR R SUM OF PRODUKT OF NUMBAR AN 3 AN 1
OIC
VISIBLE NUMBAR
DIFFRINT 2 AN SMALLR OF 2 AN NUMBAR, O RLY?
YA RLY, GTFO
OIC
IM OUTTA YR SEQUENZ


KTHXBYE

TESTD UNDR JUSTIN J. MEZA'S INTERPRETR. KTHXBYE!

Erlang, 120 chars

-module (f).
-export ([f/1]).
f(1)->1;
f(N)->
io:format("~p ",[N]),
if N rem 2 =:= 0
->f(trunc(N/2));
true->f(3*N+1)
end.

test:

f:f(171).


171 514 257 772 386 193 580 290 145 436 218 109 328 164 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

Josl - 58 characters

This version will not stack overflow due to tail recursion.

c dup println dup 1 > if dup odd? if 3 * 1+ else 2 / end c

Use:

main 21 c

Or, other examples:

main
21 c
63,728,127 c

Clojure - 70 chars

((fn[n](prn n)(if(> n 1)(recur(if(odd? n)(+(* 3 n)1)(/ n 2)))))(read))

Or, with proper whitespace and indentation:

((fn [n]
(prn n)
(if (> n 1)
(recur
(if (odd? n)
(+ (* 3 n) 1)
(/ n 2)))))
(read))

recur forces Clojure to use tail call recursion, so no stack overflows. Works with arbitrarily big numbers. Includes input and output, although it will crash if you enter a non-number :).


Note: shortly after posting my answer I noticed another Clojure implementation with pretty much the same algorithm. However, since that one doesn't attempt to be short, I'll leave my answer here, for what it's worth.

Grooovy - 59 chars

int n=args[0] as int
while(n>1){println n=n%2==0?n/2:n*3+1}

Example

$ ./collatz.groovy 5
16
8
4
2
1

With prettier output (66 chars)

int n=args[0] as int
while(n>1){print " -> ${n=n%2==0?n/2:n*3+1}"}

Example

$ ./collatz.groovy 5
-> 16 -> 8 -> 4 -> 2 -> 1

J, 31 characters

-:`(>:@(3&*))`1:@.(1&=+2&|)^:a:

Usage:

-:`(>:@(3&*))`1:@.(1&=+2&|)^:a: 9
9 28 14 7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1

Common Lisp, 76 74 chars

(defun c(n)(if (eql n 1)'(1)(cons n(if(oddp n)(c(1+(* 3 n)))(c(/ n 2))))))

or, written properly,

(defun collatz (n)
(if (eql n 1)
'(1)
(cons n (if (oddp n)
(collatz (1+ (* 3 n)))
(collatz (/ n 2))))))

Smalltalk, 103 characters

[:n||l|l:=OrderedCollection with:1.[n>1]whileTrue:[l addLast:n.n:=n odd ifTrue:[3*n+1]ifFalse:[n/2]].l]

Call by sending it the message #value: with the desired parameter:

[:n||l|l:=OrderedCollection with:1.[n>1]whileTrue:[l addLast:n.n:=n odd ifTrue:[3*n+1]ifFalse:[n/2]].l] value: 123

Or, more sanely:

[:n | | result |
result := OrderedCollection with: 1.
[n > 1] whileTrue: [
result addLast: n.
n := n odd ifTrue: [3*n + 1] ifFalse: [n / 2]].
result] value: 123

(The Proper Way would be to define the above as a method on Integer so you'd say "123 collatz", rather than as an anonymous closure.)

Since there seems to be a bit of interest with the LOLCODE solution, I thought I'd compare implementations of the two solution approaches (iterative and tail-recursive) in this language.

First, there is the iterative solution at 203 characters:

HAI 1.2
I HAS A n
GIMMEH n
IM IN YR l
VISIBLE n
BOTH SAEM 1 BIGGR OF 1 n
O RLY?
YA RLY
GTFO
OIC
MOD OF n 2
WTF?
OMG 0
n R QUOSHUNT OF n 2
GTFO
OMG 1
n R SUM OF 1 PRODUKT OF 3 n
OIC
IM OUTTA YR l
KTHXBYE

To give you the gist of what's going on:

• Input is read from STDIN using the GIMMEH keyword
• Looping is done between the IM IN YR <loopname> and IM OUTTA YR <loopname> statements
• VISIBLE is used to print to STDOUT
• The O RLY?, YA RLY, and OIC statements handle the conditional If/Then/Else logic
• The WTF?, OMG <expression>, and OIC statements handle the conditional Switch/Case logic
• Assignment is performed using <variable> R <value>
• GTFO breaks from a loop or conditional Switch/Case

And then there is the tail-recursive solution which manages to shave off 2 additional characters for a final count of 201:

HAI 1.2
HOW DUZ I c YR n
VISIBLE n
DIFFRINT 1 BIGGR OF 1 n
O RLY?
YA RLY
MOD OF n 2
WTF?
OMG 0
c QUOSHUNT OF n 2
GTFO
OMG 1
c SUM OF 1 PRODUKT OF 3 n
OIC
OIC
IF U SAY SO
I HAS A i
GIMMEH i
c i
KTHXBYE

Differences here are the definition of a function between the HOW DUZ I <funcname> YR <args> and IF U SAY SO statements. Functions are called with <funcname> <args>.

Ruby 55 chars

n=gets.to_i
while(n>1) do n=((n%2)==1)?3*n+1:n/2;p n end

Perl, 59 Chars

$n=shift;for($i=1;$n>1;$i++){$n=$n%2==0?$n/2:$n*3+1;printf"%002s: %s\n",$i,$n;}

However, I like this version at 79 chars (not counting whitespace) better because it prints the line number and the iteration value:

$n = shift; for($i = 1; $n > 1; $i++){ $n = $n % 2 == 0 ? $n / 2 : $n*3 + 1; printf "%002s: %s\n", $i, $n;}


$n = shift;


for($i = 1; $n > 1; $i++){
$n = $n % 2 == 0 ? $n / 2 : $n*3 + 1;
printf "%002s: %s\n", $i, $n;
}