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

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%I A308090 #31 Jul 22 2021 10:32:08
%S A308090 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,
%T A308090 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,
%U A308090 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
%N A308090 a(n) = gcd(2^n + n!, 3^n + n!, n+1).
%C A308090 From observation: For n > 3, if n+1 is prime, then a(n) = n+1.
%C A308090 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.
%C A308090 Conjecture: Conversely, if gcd(2^n + n!, 3^n + n!, n+1) = n+1, then n+1 is prime.
%C A308090 Appears to be the same as A090585(n) except at n=2. - _R. J. Mathar_, Jul 22 2021
%F A308090 a(n) = gcd(A007611(n), A249945(n), n+1).
%e A308090 a(4) = gcd(2^4 + 4!, 3^4 + 4!, 5) = gcd(40, 105, 5) = 5.
%e A308090 a(5) = gcd(2^5 + 5!, 3^5 + 5!, 6) = gcd(152, 363, 6) = 1.
%t A308090 Table[GCD[2^n+n!,3^n+n!,n+1],{n,100}] (* _Harvey P. Dale_, Aug 27 2020 *)
%o A308090 (PARI) a(n) = gcd([2^n + n!, 3^n + n!, n+1]); \\ _Michel Marcus_, May 12 2019
%Y A308090 Cf. A007611, A249945.
%K A308090 nonn
%O A308090 1,4
%A A308090 _Pedro Caceres_, May 11 2019