A098665 a(n) = Sum_{k = 0..n} binomial(n,k) * binomial(n+1,k+1) * 4^k.

1, 6, 43, 332, 2661, 21810, 181455, 1526040, 12939145, 110413406, 947052723, 8157680228, 70518067309, 611426078346, 5315138311383, 46308989294640, 404274406256145, 3535479068797110, 30966952059306555, 271616893912241532, 2385412594943633781, 20973327081776664546

Offset

0,2

Comments

Fifth binomial transform of A098664.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Formula

G.f.: ((1+3*x)-sqrt(1-10*x+9*x^2))/(8*x*sqrt(1-10*x+9*x^2)).

E.g.f.: exp(5x)*(BesselI(0, 4x)+BesselI(1, 4x)/2).

Recurrence: (n+1)*(2*n-1)*a(n) = 4*(5*n^2-2)*a(n-1) - 9*(n-1)*(2*n+1)*a(n-2). - Vaclav Kotesovec, Oct 15 2012

a(n) ~ 9^(n+1)/(4*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 15 2012

From Peter Bala, Jan 07 2022: (Start)

The following formulas assume an offset of 1:

a(n) = (1/4) * Sum_{k = 0..n} binomial(n,k)*A119259(k).

a(n) = (1/4) * Sum_{k = 0..n} binomial(n,k)*binomial(2*n-k-1,n-k)*3^k.

a(n) = (1/4) * [x^n] ((1 + 3*x)/(1 - x))^n.

The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p >= 3 and positive integers n and k. (End)

Mathematica

Table[SeriesCoefficient[((1+3*x)-Sqrt[1-10*x+9*x^2])/(8*x*Sqrt[1-10*x+9*x^2]), {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 15 2012 *)

Prog

(PARI) my(x='x+O('x^66)); Vec(((1+3*x)-sqrt(1-10*x+9*x^2))/(8*x*sqrt(1-10*x+9*x^2))) \\ Joerg Arndt, May 12 2013

Crossrefs

Cf. A098664, A119259.

Sequence in context: A091128 A349302 A025594 * A153397 A005786 A071541

Adjacent sequences: A098662 A098663 A098664 * A098666 A098667 A098668

Keyword

easy,nonn

Author

Paul Barry, Sep 20 2004

Status

approved