A242775 - OEIS

A242775

Let b_k=3...3 consist of k>=1 3's. Then a(n) is the smallest k such that the concatenation b_k and prime(n) is prime, or a(n)=0 if there is no such prime.

0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 1, 1, 4, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 3, 3, 2, 1, 2, 7, 3, 1, 3, 2, 2, 8, 1, 1, 7, 2, 1, 1, 5, 3, 2, 2, 2, 3, 1, 3, 8, 5, 1, 1, 4, 3, 1, 4, 5, 3, 6, 1, 2, 1, 2, 1, 3, 1, 2, 2, 1, 3, 1, 6, 3, 1, 3, 4, 2, 3, 8, 4, 1, 3, 34, 1

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OFFSET

1,8

COMMENTS

Conjecture: for n>=4, a(n)>0.

Records >=1: 1,2,4,7,8,34,... correspond to primes 7,19,41,127,157,443,...

LINKS

Peter J. C. Moses, Table of n, a(n) for n = 1..2000

EXAMPLE

For n<=3, a(n) = 0, because 3..32, 3..33 and 3..35 can never be prime, whatever the number of 3's that are concatenated.

For n=4, prime(n)=7, 37 is prime. So a(4)=1.

PROG