

A200526


a(n) = gcd(t(n), t(3n1)), where t = A200217.


0



2156316023, 211148507797805, 392841376460687116573, 13886731309220741899538675431, 1359801885649216204023955447726829, 2529908688645864568558938918274367865293, 89430911052730984787593270943984417274689212615
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OFFSET

2,1


COMMENTS

Successive maxima of the GCD in A200217 occur between A200217(n) and A200217(3n1) terms. Conjecture: All terms have same set of prime divisors, that can be used to complete prime factorization of every term in this sequence by the GCD method. All prime divisors with exception 19 are of the form 4k+1. The integer 19 divides a(3n+1) for n=0,1,2,3,...


LINKS



MATHEMATICA

ff = {}; Do[AppendTo[ff, GCD[15/8 Fibonacci[15 (1 + 2 n)]  9/20 Fibonacci[30 (1 + 2 n)] + 1/40 Fibonacci[45 (1 + 2 n)], 15/8 Fibonacci[15 (1 + 2 (3 n  1))]  9/20 Fibonacci[30 (1 + 2 (3 n  1))] + 1/40 Fibonacci[45 (1 + 2 (3 n  1))]]], {n, 2, 10}]; ff


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



