A066803 a(n) = gcd(2^n + 1, 3^n + 1).

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, 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, 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

Offset

1,2

Comments

a(n) divides a(k*n) if k is odd. - Robert Israel, Nov 15 2015

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

Links

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000 (first 1000 terms from Harry J. Smith)

Carlos Rivera, Puzzle 1064. GCD(2^p+1,3^p+1), The Prime Puzzles and Problems Connection.

Formula

a(n) = gcd(A000051(n), A034472(n)). - Michel Marcus, Nov 15 2015

Maple

seq(igcd(2^n+1, 3^n+1), n=1..100); # Robert Israel, Nov 15 2015

Mathematica

Table[GCD[3^n+1, 2^n+1], {n, 90}] (* Harvey P. Dale, Dec 03 2012 *)

Prog

(PARI) a(n) = gcd(3^n + 1, 2^n + 1); \\ Harry J. Smith, Mar 28 2010
(Python)
from math import gcd
def a(n): return gcd(2**n + 1, 3**n + 1)
print([a(n) for n in range(1, 84)]) # Michael S. Branicky, Nov 20 2021

Crossrefs

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

Sequence in context: A333751 A323921 A293235 * A089608 A250131 A100615

Adjacent sequences: A066800 A066801 A066802 * A066804 A066805 A066806

Keyword

nonn

Author

Benoit Cloitre, Jan 18 2002

Status

approved