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A133415
a(n) = (1/10)*(2^(4*n-1)-5^n*L(2*n)+L(4*n)), where L() = Lucas numbers A000032.
0
0, 0, 12, 560, 15504, 346104, 6906900, 129024512, 2310796740, 40226003064, 686392118544, 11543525003120, 192052217662812, 3169185696976320, 51968632068982524, 848016349271816384, 13784507849163060240, 223382961205435729512, 3611184426083530971300, 58264040214444951056384
OFFSET
1,3
LINKS
H.-J. Seiffert, Problem H-651, Fib. Quart., 45 (2007), 91.
FORMULA
a(n) = Sum_{k = 0..floor((n-3)/5)} binomial(4n, 2n-10k-5).
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
MATHEMATICA
Table[Sum[Binomial[4n, 2n-10k-5], {k, 0, Floor[(n-3)/5]}], {n, 20}] (* or *) LinearRecurrence[{38, -483, 2286, -3065, 400}, {0, 0, 12, 560, 15504}, 20] (* James C. McMahon, Mar 25 2025 *)
PROG
(PARI) a(n) = sum(k=0, (n-3)\5, binomial(4*n, 2*n-10*k-5)); \\ Michel Marcus, Sep 06 2017
CROSSREFS
Sequence in context: A193381 A224539 A210816 * A192602 A227143 A274259
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Nov 27 2007
STATUS
approved