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A200526
a(n) = gcd(t(n), t(3n-1)), where t = A200217.
0
2156316023, 211148507797805, 392841376460687116573, 13886731309220741899538675431, 1359801885649216204023955447726829, 2529908688645864568558938918274367865293, 89430911052730984787593270943984417274689212615
OFFSET
2,1
COMMENTS
Successive maxima of the GCD in A200217 occur between A200217(n) and A200217(3n-1) 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,...
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
KEYWORD
nonn
AUTHOR
Artur Jasinski, Nov 18 2011
STATUS
approved