Find all Belphegor primes

Question

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)

Task

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.

Rules

• 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.

Scoring

This is code-golf so shortest bytes wins.

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Inspired by The Most Evil Number - Numberphile

Answer 1

05AB1E, 14 10 9 bytes

∞°66+€ûʒp

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.

Explanation:

∞          # 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)

Answer 2

Wolfram Language (Mathematica), 51 bytes

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

PrimeQ[10^c*666+1+100^++c]~If~Print[c-2]~Do~{c,∞}     

Try it online!

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

Answer 3

Pyth, 17 bytes

.fP_sj666_B`^TZQ0

Try it online!

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

Explanation:

.fP_sj666_B`^TZQ0
.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.

Answer 4

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.)

Answer 5

Jelly, 19 bytes

+ؽṬ6×1;ŒḄḌṄẒ¡×0µ1#

A full program which prints the Belphegor primes.

Try it online!

How?

+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)

Answer 6

Python 3, 80 bytes

b=2
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!

Answer 7

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.

Answer 8

Pyth, 22 bytes

.V0IP_h*+^T+3b666^Thbb

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!

Answer 9

JavaScript (Node.js), 71 bytes

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

for(k=10n;;)for(d=n=666n*k+(k*=10n)*k+1n;n%--d||d<2n&&console.log(n););

Try it online!

Commented

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
  );                      //

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

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

for(k=10n;;)(n=666n*k+(k*=10n)*k+1n,~-(x=(g=(d,r,a)=>d?g(d/2n,d&1n?r*a%n:r,a*a%n):r)(d=n/(~-n&1n-n),1n,2n))&&~x+n?(g=d=>~d+n?~-(x=x*x%n)?~x+n&&g(d+d):1:1)(d):0)||console.log(n)

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.

Answer 10

Vyxal, 108 bits^v2, 13.5 bytes

Þ:'3+↵666+n›↵*›æ

Try it Online!

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

Answer 11

Python, 220 164 bytes

def a(k,s=set()):
 for i in range(k):
  p=1;n=(10**(i+3)+666)*10**-~i+1
  for d in range(1,int(n**.5//1/2)):
   p*=n%-~(d*2)>0
   if~-p:break
  s.add(p*n)
 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.

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