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

%I #36 Dec 07 2021 11:17:54

%S 1,5,1,1,1,5,1,1,19,25,1,1,1,145,1,1,1,5,1,1,43,5,1,97,1,265,19,1,1,

%T 25,1,1,67,5,1,1,1,5,1,1,1,145,1,1,19,5,1,1,1,12625,307,1,1,5,1,1,1,5,

%U 1,241,1,5,817,1,1,5,1,1,139,725,1,55969,1,745,1,1,1,265,1,1,3097,5,499

%N a(n) = gcd(2^n + 1, 3^n + 1).

%C a(n) divides a(k*n) if k is odd. - _Robert Israel_, Nov 15 2015

%C Conjecture: a(2^k) = 1 for k != 1. That is to say, there is no prime p > 5 such that ord(2,p) and ord(3,p) is the same power of 2, where ord(a,p) is the multiplicative order of a modulo p. - _Jianing Song_, Nov 20 2021

%H Jon E. Schoenfield, <a href="/A066803/b066803.txt">Table of n, a(n) for n = 1..10000</a> (first 1000 terms from Harry J. Smith)

%H Carlos Rivera, <a href="https://www.primepuzzles.net/puzzles/puzz_1064.htm">Puzzle 1064. GCD(2^p+1,3^p+1)</a>, The Prime Puzzles and Problems Connection.

%F a(n) = gcd(A000051(n), A034472(n)). - _Michel Marcus_, Nov 15 2015

%p seq(igcd(2^n+1, 3^n+1), n=1..100); # _Robert Israel_, Nov 15 2015

%t Table[GCD[3^n+1,2^n+1],{n,90}] (* _Harvey P. Dale_, Dec 03 2012 *)

%o (PARI) a(n) = gcd(3^n + 1, 2^n + 1); \\ _Harry J. Smith_, Mar 28 2010

%o (Python)

%o from math import gcd

%o def a(n): return gcd(2**n + 1, 3**n + 1)

%o print([a(n) for n in range(1, 84)]) # _Michael S. Branicky_, Nov 20 2021

%Y Cf. A000051 (2^n+1), A034472 (3^n+1), A260674, A349722.

%K nonn

%O 1,2

%A _Benoit Cloitre_, Jan 18 2002