A070832 a(n) = Sum_{k=0..n} binomial(8*n,8*k).

1, 2, 12872, 1470944, 622116992, 125858012672, 36758056208384, 8793364151263232, 2334899414608412672, 586347560750962049024, 151652224498623981289472, 38612725801339748322639872, 9913426188311626771400228864

Offset: 0

Author

Sebastian Gutierrez and Sarah Kolitz (skolitz(AT)mit.edu), May 15 2002

Keywords

• easy
• nonn

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..415
• Index entries for linear recurrences with constant coefficients, signature (136,32880,-552704,-65536).

Crossrefs

Sum_{k=0..n} binomial(b*n,b*k): A000079 (b=1), A081294 (b=2), A007613 (b=3), A070775 (b=4), A070782 (b=5), A070967 (b=6), A094211 (b=7), this sequence (b=8), A094213 (b=9), A070833 (b=10).

Programs

Mathematica

Table[Sum[Binomial[8n,8k],{k,0,n}],{n,0,15}] (* Harvey P. Dale, Nov 25 2020 *)

PARI

a(n)=sum(k=0,n,binomial(8*n,8*k)); \\ Benoit Cloitre, May 27 2004

PARI

Vec((1 - 134*x - 20280*x^2 + 207296*x^3 + 8192*x^4) / ((1 - 16*x)*(1 - 256*x)*(1 + 136*x + 16*x^2)) + O(x^15)) \\ Colin Barker, May 27 2019

Formula

Let b(n) = a(n)-2^(8*n)/8 then b(n)+120*b(n-1)-2160*b(n-2)-256*b(n-3)=0. - Benoit Cloitre, May 27 2004

a(n) = 1/4*16^n + 1/8*256^n + 1/4*(-68 + 48*sqrt(2))^n + 1/4*(-68-48*sqrt(2))^n.

From Colin Barker, May 27 2019: (Start)

G.f.: (1 - 134*x - 20280*x^2 + 207296*x^3 + 8192*x^4) / ((1 - 16*x)*(1 - 256*x)*(1 + 136*x + 16*x^2)).

a(n) = 21*a(n-1) + 353*a(n-2) - 32*a(n-3) for n>4.

(End)

Extensions

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 15 2007