A098520 E.g.f. exp(x)*BesselI(1,4*x)/2.

0, 1, 2, 15, 52, 285, 1206, 6027, 27560, 134073, 633130, 3062279, 14676828, 71045845, 343195230, 1665555075, 8084777040, 39343835505, 191627687250, 934855945215, 4565076327300, 22318461756045, 109211684822790, 534907610833275, 2621997452787192, 12862364386480425, 63140696801700986

Offset

0,3

Comments

Binomial transform of e.g.f. BesselI(1,4x)/2, or {0,1,0,12,0,160,0,2240,0,32256,0,...} with g.f. 2x/(1-16x^2+sqrt(1-16x^2)). The binomial transform of e.g.f. BesselI(1,2*sqrt(r)x)/sqrt(r) with g.f. 2x/(1-(2*sqrt(r)x)^2+sqrt(1-(2*sqrt(r)x)^2)) has g.f. 2x/(1-2x-((2*sqrt(r))^2-1)x^2+(1-x)*sqrt(1-2x-((2*sqrt(r))^2-1)x^2)).

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..300

Formula

G.f.: 2*x/(1-2*x-15*x^2+(1-x)*sqrt(1-2*x-15*x^2)).

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

Conjecture: (n-1)*(n+1)*a(n) - n*(2n-1)*a(n-1) - 15*n*(n-1)*a(n-2) = 0. - R. J. Mathar, Dec 11 2011

a(n) ~ 5^(n+1/2)/(4*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 15 2012

a(n) = 2^(n-1)*GegenbauerC(n-1, -n, -1/4). - Peter Luschny, May 08 2016

Maple

a := n -> simplify(2^(n-1)*GegenbauerC(n-1, -n, -1/4)):
seq(a(n), n=0..26); # Peter Luschny, May 08 2016

Mathematica

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

Prog

(PARI) x='x+O('x^66); concat([0], Vec(2*x/(1-2*x-15*x^2+(1-x)*sqrt(1-2*x-15*x^2)))) \\ Joerg Arndt, May 11 2013

Crossrefs

Sequence in context: A248544 A344819 A015520 * A216333 A056078 A142861

Adjacent sequences: A098517 A098518 A098519 * A098521 A098522 A098523

Keyword

easy,nonn

Author

Paul Barry, Sep 12 2004

Status

approved