A345456 a(n) = Sum_{k=0..n} binomial(5*n+2,5*k).

1, 22, 859, 25773, 843756, 26789257, 859595529, 27481113638, 879683351911, 28146676447417, 900729032983924, 28822936611339453, 922338323835136341, 29514778095285204502, 944473434343229560419, 30223143962480773595093, 967140672636207153780796

Offset

0,2

Links

Seiichi Manyama, Table of n, a(n) for n = 0..500

Index entries for linear recurrences with constant coefficients, signature (21,353,-32).

Formula

G.f.: (1 + x + 44*x^2) / ((1 - 32*x)*(1 + 11*x - x^2)).

a(n) = 21*a(n-1) + 353*a(n-2) - 32*a(n-3) for n>2.

a(n) = A139398(5*n+2).

a(n) = 2^(5*n + 3)/10 + ((-295 + 131*sqrt(5))/phi^(5*n) + (115 - 49*sqrt(5))*(-1)^n*phi^(5*n)) / (10*(41*sqrt(5)-90)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Jun 20 2021

Mathematica

a[n_] := Sum[Binomial[5*n + 2, 5*k], {k, 0, n}]; Array[a, 17, 0] (* Amiram Eldar, Jun 20 2021 *)
LinearRecurrence[{21, 353, -32}, {1, 22, 859}, 20] (* Harvey P. Dale, Aug 25 2022 *)

Prog

(PARI) a(n) = sum(k=0, n, binomial(5*n+2, 5*k));
(PARI) my(N=20, x='x+O('x^N)); Vec((1+x+44*x^2)/((1-32*x)*(1+11*x-x^2)))

Crossrefs

Sum_{k=0..n} binomial(b*n+c,b*k): A070782 (b=5,c=0), A345455 (b=5,c=1), this sequence (b=5,c=2), A345457 (b=5,c=3), A345458 (b=5,c=4).

Cf. A139398.

Sequence in context: A237605 A231159 A095696 * A322930 A200575 A102995

Adjacent sequences: A345453 A345454 A345455 * A345457 A345458 A345459

Keyword

nonn

Author

Seiichi Manyama, Jun 20 2021

Status

approved