A346206 Primes p, with k digits, such that the Sum_{i=1..k} (p without its i-th digit)/(its i-th digit) is a prime.

11, 21673, 27367, 32611, 33311, 41141, 48821, 82781, 171263, 211441, 243433, 323443, 343243, 449699, 632623, 663661, 727271, 772127, 847871, 882881, 944969, 1129699, 1192699, 1193939, 1262633, 1334341, 1342433, 1343423, 1361441, 1388641, 1399193, 1461883, 1613441

Offset: 1

Author

Michel Marcus, Jul 10 2021

Keywords

• nonn
• base

Examples

			21673 gives 1673/2 + 2673/1 + 2173/6 + 2163/7 + 2167/3 = 4903; so 21673 is a term.
		

Links

• Michel Marcus, Table of n, a(n) for n = 1..441
• Carlos Rivera, Puzzle 1045. One nice puzzle from Paolo Lava, The Prime Puzzles and Problems Connection.

Crossrefs

Subsequence of A038618 (zeroless primes).

Programs

PARI

subs(d, j) = {my(x=""); for (k=1, #d, if (j != k, x = concat(x, d[k]));); eval(x);}
isok(p) = {my(d=digits(p), res);  if (isprime(p) && vecmin(d), res = sum(j=1, #d, subs(d, j)/d[j]); (denominator(res)==1) && isprime(res););}

Python

from sympy import isprime, primerange
from fractions import Fraction
def ok(p):
    s = str(p)
    if '0' in s or len(s) == 1: return False
    f = sum(Fraction(int(s[:i]+s[i+1:]), int(s[i])) for i in range(len(s)))
    return f.denominator == 1 and isprime(f.numerator)
def aupto(lim): return [p for p in primerange(1, lim+1) if ok(p)]
print(aupto(1620000)) # Michael S. Branicky, Jul 11 2021