A308090 a(n) = gcd(2^n + n!, 3^n + n!, n+1).

1, 1, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, 1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, 1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, 1, 1, 53, 1, 1, 1, 1, 1, 59, 1, 61, 1, 1, 1, 1, 1, 67, 1, 1, 1, 71, 1, 73, 1, 1, 1, 1, 1, 79, 1, 1, 1, 83, 1, 1, 1, 1, 1, 89, 1, 1, 1, 1, 1, 1, 1, 97, 1, 1, 1

Offset: 1

Author

Pedro Caceres, May 11 2019

Keywords

• nonn

Comments

From observation: For n > 3, if n+1 is prime, then a(n) = n+1.

This implies that (2^n + n!)= 0 mod (n+1) iff (n+1) is prime, and (3^n + n!)= 0 mod (n+1) iff (n+1) is prime.

Conjecture: Conversely, if gcd(2^n + n!, 3^n + n!, n+1) = n+1, then n+1 is prime.

Appears to be the same as A090585(n) except at n=2. - R. J. Mathar, Jul 22 2021

Examples

			a(4) = gcd(2^4 + 4!, 3^4 + 4!, 5) = gcd(40, 105, 5) = 5.
a(5) = gcd(2^5 + 5!, 3^5 + 5!, 6) = gcd(152, 363, 6) = 1.
		

Crossrefs

Cf. A007611, A249945.

Programs

Mathematica

Table[GCD[2^n+n!,3^n+n!,n+1],{n,100}] (* Harvey P. Dale, Aug 27 2020 *)

PARI

a(n) = gcd([2^n + n!, 3^n + n!, n+1]); \\ Michel Marcus, May 12 2019

Formula

a(n) = gcd(A007611(n), A249945(n), n+1).