A098519 E.g.f. exp(x)*BesselI(1,2*sqrt(3)*x)/sqrt(3).

0, 1, 2, 12, 40, 185, 726, 3157, 13112, 56331, 239230, 1028522, 4414224, 19045039, 82237442, 356104140, 1544056864, 6707220443, 29172892518, 127058629852, 554006070200, 2418204764451, 10565384173762, 46202462762837, 202207635999240, 885642000534925, 3881697614968706

Offset

0,3

Comments

Binomial transform of e.g.f. BesselI(1,2*sqrt(3)x)/sqrt(3), or {0,1,0,9,0,90,0,945,0,10206,0,...} with g.f. 2x/(1-12x^2+sqrt(1-12x^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-11*x^2+(1-x)*sqrt(1-2*x-11*x^2)).

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

Conjecture: (n-1)*(n+1)*a(n) - n*(2*n-1)*a(n-1) - 11*n*(n-1)*a(n-2) = 0. - R. J. Mathar, Nov 23 2011

a(n) ~ sqrt(6+sqrt(3))*(1+2*sqrt(3))^n/(6*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 15 2012

Mathematica

Table[SeriesCoefficient[2*x/(1-2*x-11*x^2+(1-x)*Sqrt[1-2*x-11*x^2]), {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 15 2012 *)
With[{nn=30}, CoefficientList[Series[Exp[x] BesselI[1, 2x Sqrt[3]]/Sqrt[3], {x, 0, nn}], x] Range[0, nn]!]//Simplify (* Harvey P. Dale, Apr 27 2016 *)

Prog

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

Crossrefs

Sequence in context: A110953 A003683 A188572 * A127725 A371357 A280174

Adjacent sequences: A098516 A098517 A098518 * A098520 A098521 A098522

Keyword

easy,nonn

Author

Paul Barry, Sep 12 2004

Status

approved