A253368 a(n) = F(12*n)/(12^2) with the Fibonacci numbers F = A000045.
1, 322, 103683, 33385604, 10750060805, 3461486193606, 1114587804280327, 358893811492071688, 115562692712642803209, 37210828159659490561610, 11981771104717643318035211, 3858093084890921488916776332, 1242293991563772001787883943693
Offset: 1
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..399
- Index entries for linear recurrences with constant coefficients, signature (322, -1).
Programs
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Mathematica
Table[Fibonacci[12 n]/144, {n, 11}] (* Michael De Vlieger, Apr 03 2015 *) LinearRecurrence[{322, -1},{1, 322},11] (* Ray Chandler, Aug 12 2015 *)
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PARI
Vec(x/(x^2-322*x+1) + O(x^20)) \\ Colin Barker, Dec 31 2014
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PARI
vector(20, n, fibonacci(12*n)/(9*2^4)) \\ Altug Alkan, Nov 11 2015
Formula
G.f.: x / (x^2 - 322*x + 1). - Colin Barker, Dec 31 2014
From Peter Bala, Apr 03 2015: (Start)
For integer k, 1 + k*(36 - k)*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + k/8*Sum_{n >= 1} Fibonacci(6*n)*x^n )*( 1 + k/8*Sum_{n >= 1} Fibonacci(6*n)*(-x)^n ).
1 + 64*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + Sum_{n >= 1} Fibonacci(6*n+3)*x^n )*( 1 + Sum_{n >= 1} Fibonacci(6*n+3)*(-x)^n ).
1 + 320*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + Sum_{n >= 1} Lucas(6*n)*x^n )*( 1 + Sum_{n >= 1} Lucas(6*n)*(-x)^n ).
(End)
a(n) = (((161+72*sqrt(5))^(-n)*(-1+(161+72*sqrt(5))^(2*n))))/(144*sqrt(5)). - Colin Barker, Jun 02 2016
Extensions
Errors in name, data and formula corrected by Colin Barker, Dec 31 2014
Edited: numbers and name changed, formula and programs adjusted by Wolfdieter Lang, Jan 20 2015
Name simplified using "12" as the common number.Peter M. Chema, Mar 31 2016
Comments