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4, 8, 15, 16, 23, 42

Write a program that outputs this sequence of numbers infinitely. However, The Numbers must not appear in your source code anywhere.

The following is not a valid Java program to output The Numbers because The Numbers appear in its source code:

class TheNumbers {
    public static void main(String[] args) {
        for(int n = 0;;) System.out.println(
            n == 4 ? n = 8 :
            n == 8 ? n = 15 :
            n == 15 ? n = 16 :
            n == 16 ? n = 23 :
            n == 23 ? n = 42 : (n = 4)
        );
    }
}

The definition of "The Numbers must not appear in your source code" is as follows:

  • You must not use the numeral 4.
  • You must not use the numeral 8.
  • You must not use the numeral 1 followed by the numeral 5.
  • You must not use the numeral 1 followed by the numeral 6.
  • You must not use the numeral 2 followed by the numeral 3.

If your language ignores certain characters that can be placed between the numerals, it's not a valid substitution. So for example if your language interprets the literal 1_5 as 15, this would count as the numeral 1 followed by the numeral 5.

Alternative bases are included in the restriction, so for example:

  • Binary 100 can't be used as a substitute for 4.
  • Octal 10 can't be used as a substitute for 8.
  • Hexadecimal F can't be used as a substitute for 15.

Therefore, the following is a valid (but not very inspired) Java program to output The Numbers because The Numbers do not appear in its source code:

class TheNumbers {
    public static void main(String[] args) {
        for(int n = '*';;) {
            System.out.println(n -= '&');
            System.out.println(n *= 2);
            System.out.println(n += 7);
            System.out.println(++n);
            System.out.println(n += 7);
            System.out.println(n += 19);
        }
    }
}

Note that in that program, '*' and '&' are substituted for the integers 42 and 38, because otherwise the numerals 4 and 8 would appear in its source code.

The definition of "outputs the sequence infinitely" is open to interpretation. So, for example, a program that outputs glyphs getting smaller until they are "infinitely" small would be valid.

Kudos if you are able to generate the sequence in some way that's not basically hard-coding each number.

This is a popularity contest, so be creative. The answer with the most up votes on March 26th is the winner.

share|improve this question
8  
I can count 6 downvotes but no comments :/ –  Vereos Mar 12 at 13:52
9  
@Vereos, "This is a stupid question" isn't very constructive, which might be why no-one posted it as a comment. –  Peter Taylor Mar 12 at 15:26
14  
There are 11 types of people in this world: those that watched Lost, those that didn't, and those that don't understand binary. –  squeamish ossifrage Mar 12 at 16:26
7  
@PeterTaylor For sure, but newcomers mostly will not get that and leave the site instead of trying to improve their future questions. I guess that This isn't an interesting question, IMHO, since the solution is pretty trivial. Please post in the sandbox next time. would be way better than This is a stupid question., but that's just my personal opinion. –  Vereos Mar 12 at 16:57
3  
I notice the question does not prohibit outputting other numbers. So at least according to infinite-monkey-theory an unadulterated pseudo-random number generator should do the trick. –  kojiro Mar 13 at 13:21

108 Answers 108

up vote 219 down vote accepted

Java

I decided to add another entry since this is completely different from my first one (which was more like an example).

This program calculates the average of an array entered by the user...

import java.util.Scanner;

public class Numbers {
    public static double getSum(int[] nums) {
        double sum = 0;
        if(nums.length > 0) {
            for(int i = 0; i <= nums.length; i++) {
                sum += nums[i];
            }
        }

        return sum;
    }

    public static double getAverage(int[] nums) { return getSum(nums) / nums.length; }
    public static long roundAverage(int[] nums) { return Math.round(getAverage(nums)); }

    private static void beginLoop(int[] nums) {
        if(nums == null) {
            return;
        }

        long avg = roundAverage(nums);
        System.out.println("enter nums for average");
        System.out.println("example:");
        System.out.print("array is " + nums[0]);
        for(int i = 1; i <= nums.length; i++) {
            System.out.print(", " + nums[i]);
        }

        System.out.println();
        System.out.println("avg is " + avg);
    }

    private static int[] example = { 1, 2, 7, 9, };

    public static void main(String[] args) {
        boolean done = false;
        while(!done) {
            try {
                int[] nums = example;
                beginLoop(nums);

                nums = getInput();
                if(nums == null) {
                    done = true;
                } else {
                    System.out.println("avg is " + getAverage(nums));
                }
            } catch(Exception e) {
                e.printStackTrace();
            }
        }
    }

    static int[] getInput() {
        Scanner in = new Scanner(System.in);
        System.out.print("enter length of array to average or 0 to exit: ");
        int length = in.nextInt();
        if(length == 0) {
            return null;

        } else {
            int[] nums = new int[length];
            for(int i = 0; i <= nums.length; i++) {
                System.out.print("enter number for index " + i + ": ");
                nums[i] = in.nextInt();
            }
            return nums;
        }
    }
}

...or does it?

java.lang.ArrayIndexOutOfBoundsException: 4
    at Numbers.getSum(Numbers.java:8)
    at Numbers.getAverage(Numbers.java:15)
    at Numbers.roundAverage(Numbers.java:16)
    at Numbers.beginLoop(Numbers.java:23)
    at Numbers.main(Numbers.java:42)
java.lang.ArrayIndexOutOfBoundsException: 4
    at Numbers.getSum(Numbers.java:8)
    at Numbers.getAverage(Numbers.java:15)
    at Numbers.roundAverage(Numbers.java:16)
    at Numbers.beginLoop(Numbers.java:23)
    at Numbers.main(Numbers.java:42)
java.lang.ArrayIndexOutOfBoundsException: 4
    at Numbers.getSum(Numbers.java:8)
    ...
share|improve this answer
15  
This is great! I would not have thought of something like that. –  Jordon Biondo Mar 12 at 21:48
2  
Wow, beautiful ! Great idea ;) –  Pierre Arlaud Mar 13 at 8:25
5  
Genius! Though the output is a bit verbose, but I guess that has to do with the language you chose here. ;) –  Pieter Witvoet Mar 13 at 12:30
3  
Just when I thought the Python "lizt=Lost plot" one couldn't be topped... –  Dave Mar 14 at 14:27
3  
@justhalf Actually it bugs me this was the top answer for a little while there. It's no fun to win my own question. –  Radiodef Mar 20 at 6:59
up vote 183 down vote
+250

Python

#!/usr/bin/python
lizt = ["SPOI",
        "LERS: Lo",
        "st begins with ",
        "a plane crash on",
        "a desert island and end",
        "s with its viewers stuck in limbo forever."
        ]

while True:
    for item in lizt:
        print len(item)

Edit: As per nneonneo's suggestion, script now includes no digits.

share|improve this answer
2  
So simple, and yet so good. –  xfix Mar 13 at 13:22
4  
Whether or not this gets my vote depends entirely on the answer to this question: is the spelling of "lizt" an "Arzt" reference? EDIT: Who am I kidding, it gets my vote anyhow. –  Plutor Mar 13 at 14:11
6  
I would write while True: so that your answer contains no digits at all. –  nneonneo Mar 14 at 1:07
2  
while True: is more common. –  queueoverflow Mar 17 at 14:09
1  
Doesn't that spoil the "no alternative bases" rule? Basically, it's just an array of base-1 numbers :-) –  Daniel Mar 24 at 22:15

Perl

There is nothing hidden in the source code. Nope. If the code doesn't work, type use re "eval"; before it (required in Perl 5.18).

''=~('('.'?'.('{').(
'`'|'%').('['^'-').(
"\`"| '!').('`'|',')
.'"'. '\\' .'@'.('`'
|'.') .'=' .'('.('^'
^('`'       |"\*")).
','.("\:"& '=').','.
('^'^('`'| ('/'))).(
'^'^("\`"| '+')).','
.('^'^('`'|('/'))).(
'^'^('`'|'(')).','.(
'^'^('`'|',')).('^'^
("\`"|     '-')).','
.('^' ^('`' |'*')).(
'^'^( "\`"| (','))).
(')').     ';'.('['^
','). ('`'| ('(')).(
"\`"| ')'). ('`'|','
).('`'     |'%').'('
.'\\'.'$'.'|'."\=".(
'^'^('`'|'/'))."\)".
'\\'.'{'.'\\'."\$".(
"\["^ '/')       .((
'=')  ).+( '^'^('`'|
'.' ) ).(( (';'))).(
"\`"| '&').     ('`'
|'/') .('['^')') .((
'(')) .''. '\\'. '@'
.+(     '`'     |'.'
).')'.'\\'.'{'.('['^
'(').('`'|',').('`'|
'%').('`'|'%').('['^
'+'). '\\'.     '$'.
'_'.  '-'. '\\'. '$'
.+( ( '[') ^'/').';'
.'\\' .'$'      .''.
('['^ '/') .'='. (((
'\\') )).+ "\$". '_'
.((     ';'     )).+
'\\'.'$'.'_'.'='.'='
.('^'^('`'|'*')).'|'
.'|'.('['^'+').('['^
')'     ).(     '`'|
(( ')')) ) .('`' |((
'.'))).( '['^'/' ).+
(((     (((     '\\'
)) )))).'"'.('{' ^((
(( '[')))) ).''. (((
((       ((     '\\'
))))))).'"'.';'.('['
^'+').('['^')').('`'
|')').('`'|'.').('['
^+ '/').''.     '\\'
.+ '}'. +( "\["^ '+'
). ('[' ^"\)").( '`'
|+       ((     ')')
)).('`' |+ '.').('['
^'/').( (( '{'))^'['
).'\\'. ((       '"'
)).('!'^'+').('\\').
'"'.'\\'.'}'.(('!')^
'+').'"'.'}'.')');$:
='.'#madebyxfix#'.'=
^'~';$~='@'|"\(";#;#

Explanation in spoiler.

This is a simple Perl program which makes use of multiple bitwise operations, and evaluates the regular expression using =~ operator. The regex begins with (?{ and ends with }). In Perl, this runs code while evaluating regular expression - this lets me use eval without actually using it. Normally, however, re "eval" is required, for security reasons, when evaluating regular expressions from strings (some older programs actually took regular expressions from the user) - but it turns out that before Perl 5.18 there was a bug causing constant folded expressions to work even without this pragma - if you are using Perl 5.18, type use re "eval"; before the code to make it work. Other than that, there is not much else to this code.

share|improve this answer
22  
+1 for the design. Look at it from a distance. –  tomsmeding Mar 12 at 16:01
7  
I'm starting to look like this but I still don't see it.. –  rdurand Mar 12 at 16:55
64  
@xfix "This is a simple Perl program" - if that's the case, I'd hate to see a complicated one. –  mikeTheLiar Mar 12 at 19:24
7  
Hey look, it's a schooner. –  roippi Mar 12 at 20:31
6  
Protip: copy/paste to Notepad++ and zoom all the way out. –  mikeTheLiar Mar 14 at 18:59

Brainfuck

I'm so bad at ASCII art !

++        ++++++++    +[>+>++    ++>++++
+<        <<-]>++>    >-           --<
<<        +[    >>    >.<.>++      ++.
<.        >-    --    ----.++      ++.
<.>---    -.+++++.         <.      >--
-/-./+    .<.>+.-/    -.++<<.      </]

Test it here : http://ideone.com/kh3DYI

share|improve this answer
    
This is a really nice solution :) –  gilberto.agostinho.f Mar 13 at 15:14

Unix C

There are lots of places to find numeric constants.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <errno.h>
#include <limits.h>
#include <signal.h>
#include <fcntl.h>
#include <pwd.h>
#include <netdb.h>

int main(void)
{
  int thenumbers[] = {
    S_IRGRP|S_IXGRP|S_IWOTH,
    ntohs(getservbyname("telnet", "tcp")->s_port),
    exp(M_E)-cos(M_PI),
    SIGTERM,
    CHAR_BIT,
    strlen(getpwuid(EXIT_SUCCESS)->pw_name)
  }, i=sizeof(thenumbers)/sizeof(*thenumbers);
  while(i--)
    printf("%d\n", thenumbers[i]);
  return main();
}
share|improve this answer
10  
The obfuscation here is pretty supreme for being simple substitution. –  Radiodef Mar 13 at 1:14
1  
Doesn't it run into stack overflow due to recursion? –  skyjur Mar 13 at 16:52
    
@Skirmantas I assume all compilers will use tail-recursion for this (eg. replace the call to main with a jump to main). –  Tyilo Mar 13 at 22:31

C#

Formula "stolen" from https://oeis.org/A130826 : a(n) is the smallest number such that twice the number of divisors of (a(n)-n)/3 gives the n-th term in the first differences of the sequence produced by the Flavius-Josephus sieve.

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

public static class LostNumberCalculator
{
    public static int GetNumber(int i)
    {
        int a = GetPairwiseDifferences(GetFlaviusJosephusSieveUpTo(100)).ElementAt(i);
        int b = FindSmallestNumberWithNDivisors(a / 2);
        return b * 3 + i + 1;
    }

    public static IEnumerable<int> GetFlaviusJosephusSieveUpTo(int max)
    {
        List<int> numbers = Enumerable.Range(1, max).ToList();

        for (int d = 2; d < max; d++)
        {
            List<int> newNumbers = new List<int>();
            for (int i = 0; i < numbers.Count; i++)
            {
                bool deleteNumber = (i + 1) % d == 0;
                if (!deleteNumber)
                {
                    newNumbers.Add(numbers[i]);
                }
            }
            numbers = newNumbers;
        }

        return numbers;
    }

    public static IEnumerable<int> GetPairwiseDifferences(IEnumerable<int> numbers)
    {
        var list = numbers.ToList();
        for (int i = 0; i < list.Count - 1; i++)
        {
            yield return list[i + 1] - list[i];
        }
    }

    public static int FindSmallestNumberWithNDivisors(int n)
    {
        for (int i = 1; i <= int.MaxValue; i++)
        {
            if (CountDivisors(i) == n)
            {
                return i;
            }
        }
        throw new ArgumentException("n is too large");
    }

    public static int CountDivisors(int number)
    {
        int divisors = 0;
        for (int i = 1; i <= number; i++)
        {
            if (number % i == 0)
            {
                divisors++;
            }
        }
        return divisors;
    }
}

class Program
{
    static void Main(string[] args)
    {
        while (true)
        {
            for (int i = 0; i < 6; i++)
            {
                int n = LostNumberCalculator.GetNumber(i);
                Console.WriteLine(n);
            }
        }
    }
}
share|improve this answer
9  
+1 For someone that actually went to oeis.org in order to research a formula that fits the sequence :) –  MrPaulch Mar 17 at 6:57
    
a(i)=a(i-1)+a(i-3)+a(i-5) really seems like an easier solution –  Cruncher Mar 20 at 20:40
1  
@Cruncher That formula requires you to predefine the first 5 terms (including 4, 8 and 15), which is both boring and against the rules. –  Sebastian Negraszus Mar 20 at 21:01

C#

Using the fact that any sequence of N elements can be generated by an N-1 polynomial and entering the numbers involved a lot of beeps and boops. For reference, the polynomial I derived is

( -9(X^5) +125(X^4) -585(X^3) +1075(X^2) -446(X) +160 ) / 40

I assigned the factors to the variables named for the numbers, for simplicity ;)

First version:

int BEEP,
// Magic numbers, do not touch.
four = -9,
eight = 125,
fifteen = -117*5, 
sixteen = 1075,
twenty_three = (-1-1337) /3,
forty_two = 320/2;


for(BEEP=0;;BEEP=++BEEP%6)
{
    Console.WriteLine( 0.025* (
        four *BEEP*BEEP*BEEP*BEEP*BEEP+ 
        eight *BEEP*BEEP*BEEP*BEEP+ 
        fifteen *BEEP*BEEP*BEEP+
        sixteen *BEEP*BEEP+
        twenty_three *BEEP+ 
        forty_two ));
}

I liked the implication of rising tension as the number of BEEPs decreases after each number.

Then I figured I could calculate the factors using beep and boops, too:

int BEEEP=0, BEEP=++BEEEP ,BOOP=++BEEP,BLEEP=++BOOP+BEEP,

four = BOOP*-BOOP,
eight = BLEEP*BLEEP*BLEEP,
fifteen = BOOP*-(BOOP+(BEEP*BLEEP))*BLEEP*BOOP,
sixteen = BLEEP*BLEEP*(BOOP+(BLEEP*BEEP*BEEP*BEEP)),
twenty_three = BEEP*-((BLEEP*BOOP*BLEEP*BOOP)-BEEP),
forty_two = BEEP*BEEP*BEEP*BEEP*BEEP*BLEEP;

Went a little overboard after that...

int BEEEP=default(int), BEEP=++BEEEP ,BOOP=++BEEP,BLEEP=++BOOP+BEEP;

for(--BEEEP;;BEEEP=++BEEEP%(BEEP*BOOP))
{
    Console.WriteLine(

    BOOP*(                       (BOOP*-BOOP)*BEEEP    *BEEEP*BEEEP*BEEEP    *BEEEP+(BLEEP*BLEEP*
    BLEEP)                       *BEEEP*      BEEEP*    BEEEP*                     BEEEP+
    (BOOP*                       -(BOOP+      (BEEP*    BLEEP)                    )*BLEEP
    *BOOP)                       *BEEEP*      BEEEP*    BEEEP+(BLEEP*BLEEP        *(BOOP+
    (BLEEP*                       BEEP*        BEEP*                 BEEP)))       *BEEEP*
    BEEEP+                       (BEEP*-(     (BLEEP                 *BOOP*         BLEEP
    *BOOP)                       -BEEP))      *BEEEP+                (BEEP*         BEEP*
    BEEP*BEEP*BEEP*BLEEP))/     (BEEP*((BEEP*BEEP*BEEP  *BEEP*BEEP*BEEP)-(        BEEP+BEEP))));
}

Using the default operator in C# for value types allows initialization of BEEEP to zero. This way no numeric literals are used in the code. The basic algorithm is the same. but the factors are calculated inline.

share|improve this answer
    
@kódfodrász thanks for the suggested edit! –  Rik Mar 14 at 11:44
6  
I see a numeral 8 in there, you bad person you –  Thebluefish Mar 15 at 14:20
    
@Thebluefish I am ashamed. –  Rik Mar 17 at 8:11

D

Not allowed to use the numbers 4, 8, 15, 16, 23, or 42 in my code? No problem, then I won't use numbers at all!

import std.stdio;

void main()
{
    while( true )
    {
        ( ',' - '('  ).writeln;
        ( '/' - '\'' ).writeln;
        ( '/' - ' '  ).writeln;
        ( '_' - 'O'  ).writeln;
        ( '^' - 'G'  ).writeln;
        ( '~' - 'T'  ).writeln;
    }
}
share|improve this answer
6  
ASCII arithmetic is best arithmetic. –  Pharap Mar 15 at 17:42
2  
So after C, came a language called D? –  cegprakash Mar 20 at 9:01

Javascript + HTML

Anti-golf!

<!DOCTYPE html>
<html>
<head>
</head>
<body>
<pre>
/*hereIsTheDataPart~                    Es="5030000307000022
E2000000100000010000                    E5370000507000022200
E0010100001110000005                    E0337001010000102220
E0100010010111005033                    E7001010000102220010
E1010010111~33079900                    E1000111102221000001
E1110111~03037910100                    E0111102220010100001
E0111".replace(/~/g,                    E5);Zfillfillfillfil
Eqw=21;fq=2;fz=fq*2;                    Efl=fz*2;fm=fl*2;fw=
Efm+2; M=Math;functi                    Eon r(n,i,z){return 
Efunction(){l=i||'';                    E;for(m=0;m!=n;m++)l
E+=String.fromCharCo                    Ede(97+M.floor(M.ran
Edom()*26));return l                    E+(z||'')}};kb=r(fm,
E'/*','*'+'/');kc=r(                    Efw,'//');kd=r(20);Z
Eke=r(fw,'/*');kf=r(                    E20);kg=r(fw,'','*'+
E'/');kh=kf;ki=new Z                    EArray(21).join(' ')
E;x=[];for(n=35*ix;n                    E!=s.length;++n){x.Z
Epush(parseInt(s[n])                    E)};oo=function(){oZ
E+=z==1?kb():z==9?kc                    E():z==3?(ee.shift()
E||kd()):z==5?(y==0?                    Eke():(ee.shift()||Z
Ekf())):z==7?(y==(yl                    E-1)?kg():(ee.shift(
E)||kh())):z==0?ki:Z                    Epl.shift();}Ze=mc^2
EZthis=does*nothing;                    EZnor*does+this-haha
EZawkw0rd+space+fi11                    EZrunn1ng/out+of=stf
EZfjsddfkuhkarekhkhk                    777777777777777777*/
0;ix=typeof ix=="number"?(ix+1)%6:1;s=text();ee=[];pl=[];//2
0;q=function(n,m){return s.substr(n,m)};evl="";xl=20;yl=12//
0;while(s.length){c=s[0];m=1;if(c=='\n'){s=q(1);continue;}//
0;if(c=='E'){ev=q(0,xl);i=ev.indexOf('Z');ee.push(ev);//sd//
0;evl+=i==-1?ev.substr(1):ev.substr(1, i-1);}if(c=='0'){//sd
0;pl.push(q(0,xl*3),'','');m=3};s=q(xl*m);}eval(evl);o="";//
0;for(r=0;r!=5;++r){for(y=0;y!=yl;++y){for(n=0;n!=7;++n){//s
0;z=x[n+r*7];oo()}o+="\n"}}setTimeout(function(){text(o);//z
0;(function(){var space=' ____ ',garbage='asfdasr#@%$sdfgk';
0;var filler=space+garbage+space+garbage+space+garbage;//s//
0;})("test",1199119919191,new Date(),"xyz",30/11/1)//asdfsaf
0;eval(text());},1000);//askfdjlkasjhr,kajberksbhfsdmhbkjygk
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
                                        /*1111111111111111*/
</pre>
<script>
window.onload = function () {
setTimeout(function() {
    text = function (txt) { 
        pre = document.getElementsByTagName('pre')[0];
        if(!txt) {
            return pre.innerText;
        }
        pre.innerText = txt;
    }
    eval(text());
}, 1000);
}
</script>
</body>
</html>

The <pre> element displays a number in the sequence. It also contains all the code necessary to get to the next number in the sequence. So the <pre> is eval'd, which results in the text of the <pre> being updated to resemble the next number in the sequence. This process repeats endlessly.

Here it is in action!

share|improve this answer
2  
Plus one for ingenuity. Welcome to PPCG! –  Jonathan Van Matre Mar 13 at 23:27

C

Get your squinting goggles on :-)

main(         i){char*s     ="*)2;,5p   7ii*dpi*t1p+"
"={pi       7,i)?1!'p)(a! (ii(**+)(o(,( '(p-7rr)=pp="
"/(('       (^r9e   n%){1 !ii):   a;pin     7,p+"
"{*sp       ;'*p*   op=p) in,**             i+)s"
"pf/=       (t=2/   *,'i% f+)0f7i=*%a       (rpn"
"p(p;       )ri=}   niipp   +}(ipi%*ti(     !{pi"
"+)sa       tp;}*   s;}+%         *n;==     cw-}"
"9{ii       i*(ai   a5n(a +fs;i   *1'7",    *p=s-
1;while(p=('T'^i)?++p:s){ for(i=1;55!=*     p;p++
)i+=(' '!=*   p);printf     ("%d ",i/       2);}}
share|improve this answer
11  
As pretty as this may be, I count three 4s and two 8s in there. –  Geobits Mar 13 at 2:33
6  
@Geobits I obviously need a new pair of squinting goggles! Fixed now. –  squeamish ossifrage Mar 13 at 8:28

Haskell, 1 LoC

import Data.Char; main = putStr $ unwords $ map (show . (+)(-ord 'D') . ord) $ cycle "HLST[n" 

I've decided to go for a readable one-liner just to show how awesome Haskell is. Also, I've decided to avoid all digits, just in case.

Thanks to built-in lazy evaluation, Haskell can manipulate (map, split, join, filter...) infinitely long lists just fine. It even has multiple built-ins to create them. Since a string is just a list of characters, infinitely long strings are no mystery to Haskell either.

share|improve this answer
2  
I love the way Haskell and the like do functional programming :D –  Jwosty Mar 12 at 12:14
2  
fromEnum looks nicer than Data.Char.ord, and is somewhat shorter –  mniip Mar 12 at 13:43
1  
Whuh ... how? Could you explain? –  Pureferret Mar 13 at 14:57
1  
I just noticed the innocuous characters right at the end. I assume the have something to do with it? –  Pureferret Mar 13 at 17:01
1  
Ah. Yes, indeed, they do. –  Jan Dvorak Mar 13 at 17:20

Mathematica

We can answer the question by focusing on the repeating partial denominators of the periodic continued fraction shown below. They are what we need.

repeat cf

After all, they comprise the non-terminating sequence we are trying to produce : 4, 8, 15, 16, 23, 42, 4, 8, 15, 16, 23, 42 ...


In Mathematica one obtains the quadratic irrational corresponding to the periodic continued fraction by

FromContinuedFraction[{0, {4, 8, 15, 16, 23, 42}}]

quad irr 1

where the 0 refers to the implicit integer part.

We can check by inverting the operation:

enter image description here

{0, {4, 8, 15, 16, 23, 42}}


The 4' s and 8' s violate one of the rules of the challenge. The substring 15 is an additional violation. We can reformat the quadratic irrational to satisfy the rules.

c

{0, {4, 8, 15, 16, 23, 42}}


Now we grab the sequence of interest:

Last[c]

{4, 8, 15, 16, 23, 42}

And print the list forever …

While[True, Print@Row[ContinuedFraction[(-3220235/5+Sqrt[(10611930613350/25)])/(61630/2)],"\t"]]

table

share|improve this answer
    
Well, that is one nice math solution. I really like this one –  C5H8NNaO4 Mar 17 at 17:59
    
@C5H8NNaO4, Thanks, MSG. It was fun to figure out. –  David Carraher Mar 18 at 1:26
1  
+1 You edited to get rid of the 16 in the fraction while I was typing a comment about it. –  Geobits Mar 18 at 1:27
    
@Geobits. Funny about that. I thought I'd check whether I satisfied the rules; there were several violations that I since fixed. –  David Carraher Mar 18 at 3:00

Java

I can't find a pattern in that sequence. If there's no recognizable pattern, we might as well just throw a bunch of small primes together, cram them into Java's built-in RNG, and call it a day. I don't see how that could possibly go wrong, but then again, I'm an optimist :)

import java.util.Random;
public class LostNumbers {
    public static void main(String[] args) {
        long nut=2*((2*5*7)+1)*((2*2*3*((2*2*2*2*11)+3))+5)*
                   ((3*5*((2*3*3)+1)*((2*2*2*2*2*3)+1))+2L);
        int burner=2*2*2*5;
        while(true){
            Random dice = new Random(nut);
            for(int i=0;i<6;i++)
                System.out.print((dice.nextInt(burner)+3) + " "); // cross your fingers!
            System.out.println();
        }
    }
}
share|improve this answer
1  
Minus the while(true): ideone.com/1xaPdO –  Tim S. Mar 12 at 20:30
7  
There is a pattern, but it's not... very obvious: oeis.org/A130826 :) –  Sebastian Negraszus Mar 12 at 20:51

C / C++

Using only the characters L, O, S and T repeatedly in that order:

int main(){for(;;)printf("%d %d %d %d %d %d\n",

    'L'-     'O'*'S'    &'T','L'  &'O'+'S'*
    'T',    'L'^  'O'  |'S'*        'T'&
    'L',    'O'*  'S'    &'T'/      'L'+
    'O',    'S'^  'T'      &'L',    'O'*
    'S'&'T'   +'L'+    'O'^'S'+     'T')   ;}
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Bash one-liner

yes `curl -s "https://oeis.org/search?q=id:A$((130726+100))&fmt=text" |
grep %S | cut -d " " -f 3 | cut -d "," -f 1-6`

Line break added for readability. It (ab)uses the fact that these are the first six numbers of OEIS Sequence A130826.

share|improve this answer
    
You can also pipe awk -F"[ ,]" '/%S/ {for (i=3;i<=9;i++) printf $i" "}' to curl. –  fedorqui Mar 13 at 13:18
1  
You can remove the loop altogether with yes and drop the redirect to /dev/null with curl -s. Something like yes $(curl -s "https://oeis.org/search?q=id:A$((130726+100))&t=text" | awk -F"[ ,]" '/%S/ {for (i=3;i<9;i++) printf $i" "}') –  DigitalTrauma Mar 14 at 6:20
    
@DigitalTrauma: Thanks, I did not know about yes and curl -s -- I just shamelessly added this to my answer. :-) –  Heinzi Mar 14 at 6:33

C using no numbers at all and no character values

s(int x) { return x+x; }
p(int x) { return printf("%d ",x); }
main()
{
    for(;;){
    int a = s(p(s((s==s)+(p==p))));
    int b = a+s(a+p(a+a));
    putchar(b-s(p(b*a-b-s(p(s(s(p(b-(s==s))+p(b)))-(p==p))))));
    }
}
share|improve this answer

I like the idea of using the sequence

a[n+5] = a[n] + a[n+2] + a[n+4]

as in this answer. Found it through the OEIS Search as sequence A122115.

If we go through the sequence in reverse we will find a suitable initialization quintuple that doesn’t contain 4, 8, 15, 16 or 23.

Python3:

l = [3053, 937, -1396, -1757, -73]
while l[-1] != 66:
    l.append(l[-5] + l[-3] + l[-1])
while True:
    print(l[-6:-1])
share|improve this answer
    
very clever! Nice. –  David Carraher Mar 25 at 12:26

Python

b=a=True;b<<=a;c=b<<a;d=c<<a;e=d<<a;f=e<<a
while a: print c,d,e-a,e,e+d-a,f+d+b

Bitwise operators and some simple math.

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JavaScript

No numbers at all is a good move. But rather than print the sequence once per pass through the loop, only print once number per pass.

t = "....A...B......CD......E..................FEDCBA";
b = k = --t.length;
do {
    console.log(p = t.indexOf(t[k]));
} while (k-=!!(p-k)||(k-b));

The lower part of the string codes the numbers to print and the upper part of the string codes the next character to find. Where the two parts meet (a single F) codes resetting the cycle.

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Ruby

Generates the Numbers by embedding the equally mystical sequence 0, ∞, 9, 0, 36, 6, 6, 63;
No good can come from this.

(0..1/0.0).each{|i|puts"kw9ygp0".to_i(36)>>i%6*6&63}
share|improve this answer
    
All ruby code looks like it should just error and die; it shocks me to this day that any of it runs at all! –  a_programmer Mar 21 at 15:35

C (54 50 chars)

I'm posting a golf answer because golfing at least makes it fun.

main(a){while(printf("%d\n","gAELMT"[a++%6]-61));}
share|improve this answer
    
If you're golfing, you could (arguably) drop the a=0;. The only effect would be that you may start the sequence somewhere other than 4 (probably 8). Anyway, this will mess up the sequence when a overflows. It's technically undefined behavior, but the likely result is that you'll print garbage half the time. –  jerry Mar 13 at 17:25
    
Or just cycle the string to "gAELMT" :) –  orion Mar 13 at 18:53
    
Sure, unless someone invokes your program with arguments :) Still printing garbage half the time, though. –  jerry Mar 14 at 0:56
3  
If you give arguments to a program that doesn't need any, you pay the price :) –  orion Mar 14 at 9:42
1  
for doesn't help if there's no initialization. for(;;) is the same number of characters as while(). I interpreted the rules so that newlines have to be there... But I could use tail recursion with main... –  orion Mar 20 at 14:36

Haskell

main = mapM_ (print . round . go) [0..]
  where
    go n = 22 - 19.2*cos t + 6*cos (2*t) - 5.3*cos (3*t) + 0.5*cos (5*t)
      where t = fromInteger (n `mod` 6) / 6 * pi

http://ideone.com/erQfcd

Edit: What I used to generate the coefficients: https://gist.github.com/ion1/9578025

Edit: I really liked agrif’s program and ended up writing a Haskell equivalent while figuring it out. I picked a different base for the magic number.

import Data.Fixed
main = mapM_ print (go (369971733/5272566705 :: Rational))
  where go n = d : go m where (d,m) = divMod' (59*n) 1

http://ideone.com/kzL6AK

Edit: I also liked his second program and ended up writing a Haskell implementation of quadratic irrationals ;-). Using the library and agrif’s magic number, this program will print the sequence.

import qualified Data.Foldable as F
import Numeric.QuadraticIrrational

main = F.mapM_ print xs
  where (_, xs) = qiToContinuedFraction n
        n = qi (-16101175) 1 265298265333750 770375

This is how one could look for the magic number with the help of the library:

> continuedFractionToQI (0, Cyc [] 4 [8,15,16,23,42])
qi (-644047) 1 424477224534 30815

The printed value stands for the number (−644047 + 1 √424477224534)/30815. All you need to do is to find factors that get rid of disallowed digit sequences in the numbers while not changing the value of the expression.

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Welcome to the site =) –  Riot Mar 16 at 9:28

C#

var magicSeed = -1803706451;
var lottery = new Random(magicSeed);
var hurleysNumbers = new List<int>();
for (int i = 0; i < 6; i++) hurleysNumbers.Add(lottery.Next(43));
while (true) Console.WriteLine(String.Join(",", hurleysNumbers));

I found the seed after listening to some radio station in a flight over the pacific.

share|improve this answer
6  
There are 4s and 8s inside. –  zakk Mar 14 at 11:31

Python

import math

def periodic(x):
    three_cycle = abs(math.sin(math.pi * \
        (x/float(3) + (math.cos(float(2)/float(3)*x*math.pi)-1)/9)))
    two_cycle = abs(math.sin(math.pi * x / float(2)))
    six_cycle = three_cycle + 2*two_cycle
    return round(six_cycle, 2) # Correct for tiny floating point errors

def polynomial(x):
    numerator = (312+100)*(x**5) - 3000*x*(x**3) + (7775+100)*(x**3) - \
        (7955+1000)*(x**2) + (3997+1)*x + 120
    denominator = float(30)
    return float(numerator)/denominator

def print_lost_number(x):
    lost_number = polynomial(periodic(float(x)))
    print(int(lost_number)) # Get rid of ugly .0's at the end

i=0
while (1):
    print_lost_number(i)
    i += 1

While a lot of people used patterns taken from OEIS, I decided to create my own set of functions to represent the numbers.

The first function I created was periodic(). It is a function that repeats every six input numbers using the cyclical properties of the trig functions. It goes like this:

periodic(0) = 0
periodic(1) = 5/2
periodic(2) = 1
periodic(3) = 2
periodic(4) = 1/2
periodic(5) = 3
periodic(6) = 0
...

Then, I create polynomial(). That uses the following polynomial:

412x^5-3000x^4+7875x^3-8955x^2+3998x+120
----------------------------------------
                  30

(In my code, some of the coefficients are represented as sums because they contain the lost numbers as one of their digits.)

This polynomial converts the output of periodic() to its proper lost number, like this:

polynomial(0)   = 4
polynomial(5/2) = 8
polynomial(1)   = 15
polynomial(2)   = 16
polynomial(1/2) = 23
polynomial(3)   = 42

By constantly increasing i and passing it through both functions, I get the lost numbers repeating infinitely.

(Note: I use float() a lot in the code. This is so Python does floating-point division instead of i.e. saying 2/3=0.)

share|improve this answer
1  
Easy to fix, but you still have a 4 in polynomial. –  Geobits Mar 13 at 2:28
    
@Geobits whoops, didn't notice that. Thanks. –  Andrew Soutar Mar 13 at 2:45

Emacs Lisp 73 chars

The best way to loop forever? A cyclic list!

(let((a'(?\^D?\^H?\^O?\^P?\^W?*)))(setcdr(last a)a)(while(print(pop a))))

But wait, there's more!

?\^D is the nice way to insert the char for EOT, however if I was just submitting a file I wouldn't need the literal "\^D" I could just insert a '?' followed by an actual EOT character, thus taking the real number of needed chars down to: 63

Edit

I've been working on "gel" which is not a real language yet, but is basically series of emacs lisp macros for code golf. In "gel" this would be the solution:

(m a(~o ?\^D?\^H?\^O?\^P?\^W?*)(@(<^(^ a))(...)))

and without the waiting:

(m a(~o ?\^D?\^H?\^O?\^P?\^W?*)(@(<^(^ a))))

44 chars with nice character entry. Would be 34 if not for it being a web submission.

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Julia

By researching a while i found a mathematical way to express the sequence by other sequences without using any of the numbers (or tricky ways to use them):

L(n)=n==0?2:n==1?1:L(n-1)+L(n-2) #Lucas numbers.
O(n)=int(n*(n+1)*(n+2)/6)
S(n)=n in [O(i) for i=1:50]?0:1 #A014306
T(n)=begin k=ifloor(n/2);sum([L(i)*S(n+1-i) for i=1:k]) end #A025097

lost(n)=n>5?lost(n-1)+lost(n-3)+lost(n-5):(n+3)>5?T(n+3):-T(n+3) #A122115

[lost(i-2) for i=5:10]

Output:

6-element Array{Int64,1}:
  4
  8
 15
 16
 23
 42
share|improve this answer

C++

A nice clean language like C++ can permit you to lay out your source in a neat and highly readable way, and has the advantage of being easy to copy out by hand with minimum ambiguity.

Here the solution is reached using only the number 1.

#include <iostream>

typedef long int lI;
auto &VV = std::cout;
std::string vv = " ";

int main() {
  for(lI UU; UU --> UU;) {
    lI l1=1l+1l;lI 
    ll=1l << l1;VV 
    << ll << vv;lI 
    Il=ll*l1;VV << 
    Il << vv;VV <<
    ll*ll-1l << vv;
    lI II=ll*ll;VV 
    << II << vv;VV 
    <<(II += Il-1l)
    << vv;VV << l1
    * (II-l1)<< vv;
  }
}

Test: http://ideone.com/fuOdem

share|improve this answer

Scheme (Guile)

(let l ((x 179531901/2199535975))
  (let* ((b (* x 51)) (f (floor b)))
    (format #t "~a " f)
    (l (- b f))))

http://ideone.com/QBzuBC

Arguably this breaks the "don't encode the numbers in other bases" rule, but I think it's obscure enough that it doesn't count. As evidence of this obscurity, those two magic numbers in base 51 are:

26:27:21:9:18 / 6:19:6:19:6:19

Edit: Same trick, different representation. I actually like this one more, since it does not depend on an arbitrarily-chosen base. However, it requires a scheme implementation with infinite-accuracy support for quadratic irrationals, which (AFAIK) doesn't exist. You could implement it in something like Mathematica, though.

(let l ((x (/ (+ -16101175 (sqrt 265298265333750)) 770375)))
  (let* ((b (/ 1 x)) (f (floor b)))
    (format #t "~a " f)
    (l (- b f))))
share|improve this answer
    
Welcome to the site =) –  Riot Mar 16 at 9:27
    
+1 for "it requires a scheme implementation with infinite-accuracy support for quadratic irrationals, which (AFAIK) doesn't exist." –  Oxinabox Mar 18 at 11:21

PHP

I thought it was time someone submited a php answer, not the best but a fun one anyway

while(true)
{
    $lost = array(
    "Aaah",
    "Aaaaaaah",
    "Aaaaaaaaaaaaaah",
    "Aaaaaaaaaaaaaaah",
    "Aaaaaaaaaaaaaaaaaaaaaah",
    "Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaah");
    foreach ($lost as $a)
    {
        echo strlen($a).'
        ';
    }
}

the Ahs are the screams of the passengers as the plane crashes

share|improve this answer

Perl

#!/usr/bin/perl
use Math::Trig;

$alt = 2600;
$m   = 10 x 2;
$ip  = 1 - pi/100;
@candidates = (
    "Locke",
    "Hugo",
    "Sawyer",
    "Sayid Jarrah",
    "Jack Sh.",
    "Jin-Soo Kwon"
);

@lost = map {map{ $a+=ord; $a-=($a>$alt)?($r=$m,$m=-$ip*$m,$r):$z; }/./g; $a/100 }@candidates;
for(;;) {
    printf "%d\n",$_ for @lost;
}
share|improve this answer

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