A247134 - OEIS

A247134

Least k such that 2n+1 = 2^k - p^m for some prime p and integer m > 0, or 0 if no such k exists.

2, 3, 3, 4, 4, 4, 4, 5, 6, 5, 5, 5, 5, 5, 5, 7, 6, 6, 6, 6, 6, 29, 6, 6, 7, 6, 6, 6, 6, 6, 6, 8, 8, 7, 7, 10, 9, 7, 8, 7, 7, 8, 7, 7, 8, 7, 8, 10, 7, 7, 7, 7, 7, 8, 7, 7, 10, 7, 7, 7, 7, 7, 7, 47, 8, 8, 9, 8, 10, 9, 10, 8, 9, 8, 8, 9, 8, 8, 15, 8, 10, 9, 9, 8

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OFFSET

0,1

COMMENTS

Since even numbers cannot be of this form unless they are powers of 2, only odd 2n+1 are considered.

Except for n=3, 11, 18, 19, 27, 39, 50, 51, ..., the value of k corresponds to m=1, i.e., 2^k - (2n+1) is prime.

a(935) >= 70000. - Hiroaki Yamanouchi, Nov 22 2014

LINKS

Hiroaki Yamanouchi, Table of n, a(n) for n = 0..934

FORMULA

a(n) = min{ k | exists m>1: 2n+1 = A000079(k) - A000961(m) } > log_2(n).

PROG