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
Keywords
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.
Programs
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Maple
seq(igcd(2^n+1, 3^n+1), n=1..100); # Robert Israel, Nov 15 2015
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Mathematica
Table[GCD[3^n+1,2^n+1],{n,90}] (* Harvey P. Dale, Dec 03 2012 *) -
PARI
a(n) = gcd(3^n + 1, 2^n + 1); \\ Harry J. Smith, Mar 28 2010
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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
Comments