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%I #11 Sep 06 2017 05:34:15
%S 0,0,12,560,15504,346104,6906900,129024512,2310796740,40226003064,
%T 686392118544,11543525003120,192052217662812,3169185696976320,
%U 51968632068982524,848016349271816384,13784507849163060240,223382961205435729512,3611184426083530971300,58264040214444951056384
%N a(n) = (1/10)*(2^(4*n-1)-5^n*L(2*n)+L(4*n)), where 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.
%F a(n) = Sum_{k = 0..floor((n-3)/5)} binomial(4n, 2n-10k-5).
%F O.g.f.: -4*x^3*(3+26*x+5*x^2)/((-1+16*x)*(1-15*x+25*x^2)*(1-7*x+x^2)) = -(1/20)+(1/10)*(-2+15*x)/(1-15*x+25*x^2)-(1/20)/(-1+16*x)+(1/10)*(2-7*x)/(1-7*x+x^2) . - _R. J. Mathar_, Nov 28 2007
%o (PARI) a(n) = sum(k=0, (n-3)\5, binomial(4*n, 2*n-10*k-5)); \\ _Michel Marcus_, Sep 06 2017
%K nonn
%O 1,3
%A _N. J. A. Sloane_, Nov 27 2007
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