Let's assign the numbers 0 through 94 to the 95 printable ASCII characters:


Space is 0, ! is 1, and so on until ~ is 94. We'll also assign 95 to tab (\t) and 96 to newline (\n).

Now consider the infinite string whose Nth character is the character above that the Nth prime number, modulo 97, has been assigned to. We'll call this string S.

For example, the first prime number is 2, and 2 mod 97 is 2, and 2 is assigned to ", so the first character of S is ". Similarly, the 30th prime number is 113, and 113 mod 97 is 16, and 16 is assigned to 0, so the 30th character of S is 0.

The first 1000 characters of S are as follows:

"#%'+-137=?EIKOU[]cgiosy $&*,0>BHJTV\bflrt~
#%1=ACGMOY_ekmswy"046:HNXZ^dlrx|!)-5?AKMSW]eiko{"&.28DFX^hntv|%+139?CEQ[]agmo{  $,6>HPV\`hnrz~+5ACMOSU_mqsw$(*.BFNX`djp~!'-5;GKQS]_eoq{}"48:>DJRX^tv
'17=EQU[aciu    026<>DHJNZ\b#)/7ISaegkqy}   $0:<@BFLXdlx~!'/3;?MQWY]ceku(.24LPR\hjt|!'-?EIKWamu$28<>BDNZ`fxz)+AGOUY[_gmwy"0:@LNRT^jl|~#')3;Meiow&(,4DFJRX^bnp%+-37=KQUW]agsy    ,06BJPTn
)15;=CYegw  ".<FHLTZ`dfjpx|~#-/9AES]ikquw&48>FLPbjtz
%139EUW[aosu&>HNPZ\fhrxz#%/5=[egqy  (:@LXZlrv|!35?MSWY]uw"(8@FL^nptz|!'17COacim &>BDHNP\`n+5;GU[eqsw}$*46:HNTX^`jl|'/AEKWY_ek&,:>FPXdvz|
7CIK[agu    ,0NTZ`hnrt
%)+1GMOSegkwy   "<BHLT^~-/59;?AKY_cku{.24:X\dntz!'37=?EIOQ[]ms&*6D`fz~/7=AGU[akmw"*46@HT^vx|#)-5GQW]_eo{}&,28@FPVX^djt|39OQcgoy6>PTV`fhnr#+7IY_ams} (*0:HLdfvx!#-AEGKScioq},48>\^hjptz
'-1=CKW[iu  6<HNPfn

Stack Exchange turns tabs into spaces, so here's a PasteBin with the tabs intact.


Find a substring of S that is a valid program in your language of choice that outputs the first M prime numbers, one per line, in order, for some positive integer M.

For example, 2 is a substring of S (it occurs in multiple places but any will do), and 2 is a valid CJam program whose output is


which is the first M = 1 prime numbers, one per line, in order.

Similarly, the string 2N3N5 may be a substring of S somewhere, and 2N3N5 is a valid CJam program that outputs


which is the first M = 3 prime numbers, one per line, in order.


The submission with the highest M wins. Tie breaker goes to the submission posted first.


  • There should be no additional output besides the single primes on each line, except for an optional trailing newline after the last line. There is no input.

  • The substring may be any length as long as it's finite.

  • The substring may occur anywhere within S. (And S may contain it in multiple places.)

  • The program must be a full-fledged program. You may not assume it is run in a REPL environment.

  • The program must run and terminate in a finite amount of time without errors.

  • "Newline" may be interpreted as any common newline representation necessary for your system/interpreter/etc. Just treat it as one character.

You must give the index of S where your substring starts, as well as the length of the substring if not the substring itself. You may not only show that the substring must exist.

Related: Looking for programs in a huge Boggle board

  • 1
    \$\begingroup\$ Can you give the code to produce that big string upto any number of characters ? (I suppose you already have one) \$\endgroup\$
    – Optimizer
    Feb 22, 2015 at 13:42
  • \$\begingroup\$ If there are 95 printable ASCII characters then why are you doing modulo 97? Ah nevermind, you also use tab and newline. \$\endgroup\$ Feb 22, 2015 at 15:19
  • \$\begingroup\$ Considering that 0 mod 97 can only occur once, the lack of space really hurts... \$\endgroup\$
    – Sp3000
    Feb 23, 2015 at 0:09
  • \$\begingroup\$ @Sp3000 Shoot, that didn't occur to me. :/ \$\endgroup\$ Feb 23, 2015 at 0:20

2 Answers 2


Lenguage, M = ∞

All of the programs start at the beginning of the string. The following poorly written Python program calculates how many characters are needed for a given M.

def program_length(n):
    PLUS, MINUS, DOT = '000', '001', '100'
    i = 1
    s = ''
    while n > 0:
        i += 1
        if all(i%f for f in range(2,i)): 
            s += str(i) + '\n'
            n -= 1
    out = '110111'
    ch = 0
    for c in s:
        dif = ord(c) - ch
        if dif > 0: out += PLUS * dif
        else: out += MINUS * -dif
        out += DOT
        ch = ord(c)
    return int(out, 2)

For example, for M = 5, the program is the first 2458595061728800486379873255763299470031450306332287344758771914371767127738856987726323081746207100511846413417615836995266879023298634729597739072625027450872641123623948113460334798483696686473335593598924642330139401455349473945729379748942060643508071340354553446024108199659348217846094898762753583206697609445347611002385321978831186831089882700897165873209445730704069057276108988230177356 characters.

  • \$\begingroup\$ If in doubt, there's a BF variant that'll do it for you. \$\endgroup\$
    – ymbirtt
    Feb 22, 2015 at 17:04
  • 3
    \$\begingroup\$ It's funny how Lenguage was inspired by another challenge of mine. It's like I'm bringing about my own downfall. \$\endgroup\$ Feb 22, 2015 at 22:09

CJam, M = 2

Short and sweet:


This sequence starts at position 54398, using 1-indexing of the string. You can test it online here.

I attempted to search for a few possible variations, but this was the first solution I found.

I am currently trying to find an M = 3 version, but I don't expect to find one within a reasonable period of time. If the sequence is uniformly random (an approximation), then the starting index for a length 5 sequence could be on the order of 10^9.


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