|
%I #9 Mar 07 2024 12:55:26
%S 2,19,245,3631,58121,973843,16773677,293759095,5196109073,92455824667,
%T 1650850175669,29537478199039,529130102195225,9485447592486691,
%U 170110949757514301,3051485664370912903,54745886982174938657
%N a(n)=(1/(4n))*sum(k=1,n,F(6k)*B(2n-2k)*binomial(2n,2k)) where F=Fibonacci numbers and B=Bernoulli numbers.
%C Linear recurrence and empirical g.f. confirmed by more terms. - _Ray Chandler_, Mar 07 2024
%H Ray Chandler, <a href="/A124125/b124125.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (35, -383, 1465, -1516, 80).
%F a(n)=(1/4)*(F(6n-3)+4^n*F(2n-1)+2*5^(n-1))
%F Empirical G.f.: -x*(68*x^4-597*x^3+346*x^2-51*x+2) / ((5*x-1)*(x^2-18*x+1)*(16*x^2-12*x+1)). [_Colin Barker_, Dec 01 2012]
%o (PARI) a(n)=(1/4)*(fibonacci(6*n-3)+4^n*fibonacci(2*n-1)+2*5^(n-1))
%Y Cf. A111262.
%K nonn
%O 1,1
%A _Benoit Cloitre_, Nov 29 2006
|
