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%I #15 Jul 13 2022 17:35:44
%S 0,0,1,91,3060,74613,1562275,30045016,548354601,9669627915,
%T 166514967388,2819214031645,47139484522131,780855182286336,
%U 12842348655153745,210042772449096763,3420451064885308740,55509625058510689221,898396209147305575171,14508414020570344661800
%N a(n) = (1/10)*(2^(4*n-3)-5^n*F(2*n-1)+L(4*n-2)), where F() = Fibonacci numbers A000045 and L() = Lucas numbers A000032.
%H H.-J. Seiffert, <a href="http://www.fq.math.ca/Problems/February2007advancednewversion.pdf">Problem H-651</a>, Fib. Quart., 45 (2007), 91.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (38,-483,2286,-3065,400).
%F a(n) = Sum_{k = 0..floor((n-3)/5)} binomial(4n-2, 2n-10k-6).
%F G.f.: -x^3*(85*x^2+53*x+1) / ((16*x-1)*(x^2-7*x+1)*(25*x^2-15*x+1)). - _Colin Barker_, Apr 09 2013
%t LinearRecurrence[{38,-483,2286,-3065,400},{0,0,1,91,3060},20] (* _Harvey P. Dale_, Jul 13 2022 *)
%o (PARI) a(n) = sum(k=0, (n-3)\5, binomial(4*n-2, 2*n-10*k-6)); \\ _Michel Marcus_, Sep 06 2017
%K nonn,easy
%O 1,4
%A _N. J. A. Sloane_, Nov 27 2007
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