A Belphegor number is a number of the form \$(10^{n+3}+666)*10^{n+1}+1\$ (1{n zeroes}666{n zeroes}1) where \$n\$ is an non-negative integer. A Belphegor prime is a Belphegor number that is also prime.

The \$n\$ values of the first few Belphegor primes are 0, 13, 42, 506 (A232448)


Write a program that either:

  • takes no input and outputs all Belphegor primes.
  • takes a input \$k\$ and outputs the first \$k\$ Belphegor primes.

A reference python implementation can be found here.


  • You may output the \$n\$ value for the Belphegor prime instead of the prime itself
  • You may use probabilistic primality tests as long as there is no known counter case.


This is so shortest bytes wins.

Inspired by The Most Evil Number - Numberphile

  • 2
    \$\begingroup\$ what do you mean by "All"? there might be infinitely many... \$\endgroup\$
    – ZaMoC
    Commented May 27, 2020 at 16:33
  • \$\begingroup\$ @J42161217 by "All" I mean to write an program that does not stop and will eventually output all Belphegor primes. \$\endgroup\$ Commented May 27, 2020 at 16:37
  • \$\begingroup\$ Do you mean "\$n\text{-th}\$ value" by "\$n\$ value"? \$\endgroup\$ Commented May 27, 2020 at 22:48
  • 2
    \$\begingroup\$ To be pedantic (and reading the fine print), the only known Belphegor primes are 16661 and 1000000000000066600000000000001. The rest of the numbers in the sequence are only probable primes. \$\endgroup\$
    – Abigail
    Commented May 28, 2020 at 10:48
  • 1
    \$\begingroup\$ I find the criterion as long as there is no known counter case a bit shaky. It sounds like "it's fine as long as we can't tell". (For example, even if we don't know a counter case for a strong Baillie-PSW primality test to date, it is conjectured that there are infinitely many of them.) \$\endgroup\$
    – Arnauld
    Commented May 29, 2020 at 8:06

11 Answers 11


05AB1E, 14 10 9 bytes


Outputs the infinite sequence.
Extremely slow due to the prime-check on large numbers, so times out before it even reaches the n=13 Belphegor prime on TIO..

Try it online or verify the numbers without the prime-check filter.


∞          # Push an infinite positive list: [1,2,3,4,5,...]
 °         # Take 10 to the power each: [10,100,1000,10000,100000,...]
  66+      # Add 66 to each: [76,166,1066,10066,100066,...]
     €û    # Palindromize each: [767,16661,1066601,100666001,10006660001,...]
       ʒ   # Filter the list by:
        p  #  Check whether it's a prime number
           #  (minor note: 767 is NOT a prime)
           # (after which the resulting list is output implicitly)

Wolfram Language (Mathematica), 51 bytes

outputs the n value
"...program that does not stop and will eventually output all..."


Try it online!

thanks to @DanTheMan for saving 4 bytes
and also to @mypronoun -7 bytes

  • \$\begingroup\$ Using Do[...,∞] would be shorter than using While. Additionally, If can use infix syntax. \$\endgroup\$
    – DanTheMan
    Commented May 28, 2020 at 0:32
  • \$\begingroup\$ 53 bytes: Try it online! \$\endgroup\$ Commented May 28, 2020 at 5:30
  • 1
    \$\begingroup\$ 51 bytes by rearranging 10^(2c+2) to 100^++c: Try it online! \$\endgroup\$ Commented May 28, 2020 at 10:53
  • \$\begingroup\$ @mypronounismonicareinstate good job! \$\endgroup\$
    – ZaMoC
    Commented May 28, 2020 at 13:26

Pyth, 17 bytes


Try it online!

Takes k as input and outputs the n corresponding to the first k Belphegor primes.


.f             Q0    Find the first k values of Z where the following is true,
                     starting at 0 and counting upwards.
            ^TZ      Raise 10 to the power of Z
           `         Convert to a string
         _B          Pair with reversal
     j666            Join with 666 in the middle
    s                Convert to number
  P_                 Check for primality.

Ruby -rprime, 45 bytes

Prime.map{|p|p p if"#{p}"=~/^1(0*)666(\1)1$/}

Try it online!

Stutters, then checks all primes against a Belphegor prime regex. Very slow.

(Edit: Kudos to @Abigail, whose earlier Perl answer used a similar regex. I didn't notice it until after I posted my answer.)


Jelly, 19 bytes


A full program which prints the Belphegor primes.

Try it online!


+ؽṬ6×1;ŒḄḌṄẒ¡×0µ1# - Main Link: no arguments (implicit input = 0)
                µ1# - count up, from n = 0, finding the first n for which
                      this yields a truthy value:
 ؽ                 -   [1,2]
+                   -   add to n -> [n+1, n+2]
   Ṭ                -   un-truth -> [0]*n+[1,1]  (e.g. n = 3: [0,0,0,1,1])
    6×              -   multiply by six -> [0]*n+[6,6]
      1;            -   prefix with a one -> [1]+[0]*n+[6,6]
        ŒḄ          -   bounce -> [1]+[0]*n+[6,6,6]+[0*n]+[1]
          Ḍ         -   from base 10 -> 100...0066600...001
             ¡      -   repeat...
            Ẓ       -   ...number of times?: is prime?
           Ṅ        -   ...action?: print it and a newline character
              ×0    -   multiply the result by 0 (forcing an infinite loop)

Python 3, 80 bytes

while 1:n=100**b+10**~-b*666+1;all(n%m for m in range(2,n))and print(n);b+=1

Try it online!

  • \$\begingroup\$ By "you may use probabilistic primality tests as long as there is no known counter case." i meant any counter case not just counter cases that are belphegor primes. And fermat primality tests with base 2 has lot of counter cases oeis.org/A001567 \$\endgroup\$ Commented May 28, 2020 at 3:31
  • \$\begingroup\$ @Mukadan I see. So what kind of probabilistic primality test doesn't have a known counter case? \$\endgroup\$ Commented May 28, 2020 at 3:47
  • \$\begingroup\$ Baillie–PSW primality test or miller-rabin with the bases 2, 325, 9375, 28178, 450775, 9780504, 1795265022. Both of them are deterministic upto 2^64 and have no known counter cases \$\endgroup\$ Commented May 28, 2020 at 3:56
  • \$\begingroup\$ 79 bytes with Wilson‘s Theorem, outputs n. \$\endgroup\$
    – ovs
    Commented May 28, 2020 at 6:50
  • \$\begingroup\$ 77 bytes, because 4 is (obviously) not a Belphegor number. \$\endgroup\$
    – ovs
    Commented May 28, 2020 at 8:42

perl -M5.010 -Mbigint -Mexperimental=signatures, 62 70 bytes

sub p($m=3){$m>=$_||($_%$m&&p($m+2))}$_=16661;{p&&say;s/6+/0$&0/;redo}

Try it online!

@ikegami pointed out the original solution doesn't work, because .. doesn't work well with bigints. So we replaced it with recursive function which checks whether a number is a prime (by checking whether it isn't evenly divisible by any odd digit less than the tested number (excecpt 1)). We also no longer iterating over all numbers, instead, we're just checking all the Belphegor numbers; we can easily make the next one from the previous by replacing 666 by 06660.

It's still slow, because of the rather dumb primeness checking. Running it on TIO doesn't actually produce any output (it seems to run at most one minute). Running it from the command line quickly produces 16661, but I couldn't bother waiting for it to reach 1000000000000066600000000000001, the next Belphegor prime. It is likely to die from memory exhaustion when trying to determine one of the Belphegor numbers is prime, before finding 1000000000000066600000000000001 anyway.

  • \$\begingroup\$ This doesn't work. .. requires native ints, so your bigints are being converted to native ints. Your program will die with Range iterator outside integer range long before finding 1000000000000066600000000000001. (Try perl -Mbigint -e'$x = 1000000000000066600000000000001; grep 1, 2..$x-2'.) \$\endgroup\$
    – ikegami
    Commented May 28, 2020 at 8:18

Pyth, 22 bytes


Try it online!

Implements the formula provided in the question. Prints the n values rather than the primes themselves.

Since this version (not surprisingly) times out on TIO, here is a version that prints all n values lower than the input: Try it online!

  • \$\begingroup\$ Alternate 22 byte solution by doing string addition \$\endgroup\$ Commented May 28, 2020 at 3:17

JavaScript (Node.js), 71 bytes

A full program that prints Belphegor primes forever ... and takes forever to print them.


Try it online!


for(k = 10n;;)            // outer loop: start with k = 10 and loop forever
  for(                    //   inner loop:
    d = n =               //     start with d = n =
      666n * k +          //       666 * k +
      (k *= 10n) * k +    //       (10 * k)² +
      1n;                 //       1
                          //     and update k to 10 * k
      n % --d ||          //     decrement d until it divides n
        d < 2n &&         //     if d is less than 2:
          console.log(n); //       n is prime --> print it
  );                      //

JavaScript (Node.js), 176 bytes (non-competing)

A much faster version that uses a single iteration of the Miller-Rabin primality test.


Try it online!

I guess it doesn't comply with the challenge rules since the test is likely to produce false-positives. It does however find the same 5 first terms as other answers.


Vyxal, 108 bitsv2, 13.5 bytes


Try it Online!

Using the T flag sets the interpreter to time out after 1 minute :P

  • \$\begingroup\$ Porting my 05AB1E answer, is 11.375 bytes (91 bits), and possibly even shorter if you know how to prevent casting to/from strings to numbers for the palindromizing: try it online. \$\endgroup\$ Commented Oct 2, 2023 at 15:05

Python, 220 164 bytes

def a(k,s=set()):
 for i in range(k):
  for d in range(1,int(n**.5//1/2)):
 return s

Simple prime search by checking modulus below the square root; fastened by skipping every even divisor.

Likely there's room for improvement, as it becomes incredibly slow for k > 10.

Edit: thanks to @JonathanAllan and @mathjunkie for ideas and sources. This update has heavy use of tweaks and bit-operations.

  • 1
    \$\begingroup\$ I doubt that altering the golfiest Python prime identifying program would even print out the first one without stupid amounts of resources :) \$\endgroup\$ Commented May 27, 2020 at 22:35
  • 1
    \$\begingroup\$ True->1; inline the ifs; use p*=n%((d*2)+1)>0 (maybe even p*=n%-~(d*2)>0?); (i+1)->-~i \$\endgroup\$ Commented May 27, 2020 at 22:38
  • 1
    \$\begingroup\$ from math import* appears to be shorter. \$\endgroup\$ Commented May 27, 2020 at 22:46
  • \$\begingroup\$ @JonathanAllan can you explain -~i? I'm not familiar with bitwise operators that much. How negating the bitwise negation of i equals +1? \$\endgroup\$ Commented May 27, 2020 at 23:33
  • 2
    \$\begingroup\$ @ZoltánSchmidt Take a look at this Python golfing tip \$\endgroup\$ Commented May 27, 2020 at 23:36

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