

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

Table of n, a(n) for n=2..8.


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

Cf. A200217.
Sequence in context: A096561 A096560 A011581 * A015383 A016872 A016920
Adjacent sequences: A200523 A200524 A200525 * A200527 A200528 A200529


KEYWORD

nonn


AUTHOR

Artur Jasinski, Nov 18 2011


STATUS

approved



