A345455 a(n) = Sum_{k=0..n} binomial(5*n+1,5*k).

1, 7, 474, 12393, 427351, 13333932, 430470899, 13733091643, 439924466026, 14072420067757, 450374698997499, 14411355379952868, 461170414282959151, 14757375158697584607, 472236871202375365274, 15111570273013075344193, 483570355262634763462351

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 - 14*x - 26*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+1).

a(n) = 2^(5*n + 2)/10 + ((-475 + 213*sqrt(5))/phi^(5*n) - ( 65 - 33*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 + 1, 5*k], {k, 0, n}]; Array[a, 17, 0] (* Amiram Eldar, Jun 20 2021 *)
LinearRecurrence[{21, 353, -32}, {1, 7, 474}, 20] (* Harvey P. Dale, Jul 20 2021 *)

Prog

(PARI) a(n) = sum(k=0, n, binomial(5*n+1, 5*k));
(PARI) my(N=20, x='x+O('x^N)); Vec((1-14*x-26*x^2)/((1-32*x)*(1+11*x-x^2)))

Crossrefs

Sum_{k=0..n} binomial(b*n+c,b*k): A082311 (b=3,c=1), A090407 (b=4,c=1), A070782 (b=5,c=0), this sequence (b=5,c=1), A345456 (b=5,c=2), A345457 (b=5,c=3), A345458 (b=5,c=4).

Cf. A139398.

Sequence in context: A238164 A112949 A254966 * A261806 A332147 A278143

Adjacent sequences: A345452 A345453 A345454 * A345456 A345457 A345458

Keyword

nonn

Author

Seiichi Manyama, Jun 20 2021

Status

approved