A133306 a(n) = (1/n)*Sum_{i=0..n-1} C(n,i)*C(n,i+1)*5^i*6^(n-i), a(0)=1.

1, 6, 66, 906, 13926, 229326, 3956106, 70572066, 1291183806, 24095736726, 456879955026, 8776867331706, 170459895028566, 3341423256586206, 66023812564384026, 1313634856606430226, 26295597219228901806, 529199848207277494566, 10701116421278640683106, 217317899302044152030826

Offset

0,2

Comments

Sixth column of array A103209.

The Hankel transform of this sequence is 30^C(n+1,2). - Philippe Deléham, Oct 28 2007

Links

G. C. Greubel, Table of n, a(n) for n = 0..745

Formula

G.f.: (1-z-sqrt(z^2-22*z+1))/(10*z).

a(n) = Sum_{k, 0<=k<=n} A088617(n,k)*5^k.

a(n) = Sum_{k, 0<=k<=n} A060693(n,k)*5^(n-k).

a(n) = Sum_{k, 0<=k<=n} C(n+k, 2*k) 5^k*C(k), C(n) given by A000108.

a(0)=1, a(n) = a(n-1) + 5*Sum_{k=0..n-1} a(k)*a(n-1-k). - Philippe Deléham, Oct 23 2007

Conjecture: (n+1)*a(n) + 11*(-2*n+1)*a(n-1) + (n-2)*a(n-2) = 0. - R. J. Mathar, May 23 2014

G.f.: 1/(1 - 6*x/(1 - 5*x/(1 - 6*x/(1 - 5*x/(1 - 6*x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, May 10 2017

a(n) ~ 3^(1/4) * (11 + 2*sqrt(30))^(n + 1/2) / (10^(3/4) * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Nov 29 2021

Mathematica

CoefficientList[Series[(1-x-Sqrt[x^2-22*x+1])/(10*x), {x, 0, 50}], x] (* G. C. Greubel, Feb 10 2018 *)

Prog

(PARI) x='x+O('x^30); Vec((1-x-sqrt(x^2-22*x+1))/(10*x)) \\ G. C. Greubel, Feb 10 2018
(Magma) Q:=Rationals(); R<x>:=PowerSeriesRing(Q, 40); Coefficients(R!((1-x-Sqrt(x^2-22*x+1))/(10*x))) // G. C. Greubel, Feb 10 2018

Crossrefs

Cf. A000108, A060693, A103209, A103210, A103211.

Sequence in context: A130977 A191096 A151832 * A371681 A216636 A169715

Adjacent sequences: A133303 A133304 A133305 * A133307 A133308 A133309

Keyword

nonn

Author

Philippe Deléham, Oct 18 2007

Status

approved