A167506 Number of m >= 0, m <= n such that 2^(n-m)*3^m + 1 or 2^(n-m)*3^m - 1 is prime.

2, 2, 3, 4, 5, 2, 6, 7, 6, 3, 5, 1, 10, 1, 3, 8, 10, 2, 7, 4, 3, 2, 9, 1, 5, 1, 5, 5, 6, 2, 13, 6, 3, 1, 9, 5, 10, 2, 5, 7, 13, 1, 11, 6, 4, 0, 12, 1, 8, 3, 7, 9, 11, 1, 7, 7, 4, 2, 11, 1, 11, 2, 9, 6, 6, 1, 13, 8, 8, 1, 9, 2, 13, 0, 5, 4, 12, 1, 11, 2, 10, 3, 13, 2, 8, 2, 4, 6, 9, 1, 6, 7, 4, 1, 8, 1, 9, 1

Offset: 1

Author

M. F. Hasler, Nov 19 2009

Keywords

• nonn

Comments

M. Underwood observed that for all primes p < 3187 we have a(p) > 1, and asks whether there is a prime such that a(p) = 0. (This is equivalent to A167504(p) = A167505(p) = 0.)

Links

• Robert Israel, Table of n, a(n) for n = 1..2000
• Mark Underwood, 2^a*3^b one away from a prime. Post to primenumbers group, Nov. 19, 2009.
• Mark Underwood and Jens Kruse Andersen, 2^a*3^b one away from a prime, digest of 3 messages in primenumbers Yahoo group, Nov 19, 2009.

Crossrefs

Cf. A167504, A167505.

Programs

Maple

g:= proc(n,m) local t; t:= 2^(n-m)*3^m; isprime(t+1) or isprime(t-1) end proc:
f:= proc(n) nops(select(m -> g(n,m), [$0..n])) end proc:
map(f, [$1..100]); # Robert Israel, Mar 11 2025

PARI

A167505(n)=sum( b=0,n, ispseudoprime(3^b<<(n-b)-1) || ispseudoprime(3^b<<(n-b)+1))

Formula

max { A167504(n), A167505(n) } <= a(n) <= A167504(n)+A167505(n).