A358318 For n >= 5, a(n) is the number of zeros that need to be inserted to the left of the ones digit of the n-th prime so that the result is composite.

2, 2, 2, 4, 1, 1, 1, 2, 3, 1, 1, 3, 3, 2, 3, 5, 1, 2, 1, 3, 5, 1, 1, 1, 3, 3, 1, 3, 3, 1, 4, 1, 1, 1, 3, 1, 2, 2, 3, 1, 2, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 5, 2, 2, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 6, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 2, 3, 1, 2, 1, 1, 1, 1, 3

Offset: 5

Author

Rida Hamadani, Nov 09 2022

Keywords

• nonn
• base

Comments

Conjecture: the sequence is bounded.

Examples

			For n = 8, prime(8) is 19. 109, 1009, and 10009 are all primes, while 100009 is not, thus a(8) = 4.
For n = 30, prime(30) is 113. 1103 and 11003 are prime, while 110003 is not, thus a(30) = 3.
		

Crossrefs

Cf. A000040, A344637.

Programs

Mathematica

a[n_] := Module[{p = Prime[n], c = 1, q, r}, r = Mod[p, 10]; q = 10*(p - r); While[PrimeQ[q + r], q *= 10; c++]; c]; Array[a, 100, 5] (* Amiram Eldar, Nov 27 2022 *)

PARI

a(n) = n=prime(n); for(i=1,oo, isprime(n=10*n-n%10*9) || return(i)); \\ Kevin Ryde, Dec 08 2022

Python

from sympy import isprime, prime
def a(n):
    s, c = str(prime(n)), 1
    while isprime(int(s[:-1] + '0'*c + s[-1])): c += 1
    return c
print([a(n) for n in range(5, 92)]) # Michael S. Branicky, Nov 09 2022

Formula

If a(n) > 1 then a(pi(k)) = a(n) - 1 where k = 100*floor(p/10) + p mod 10 and p = prime(n) (i.e., k is the result when a single 0 is inserted to the left of the ones digit of p).