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A098520 E.g.f. exp(x)*BesselI(1,4*x)/2. 3

%I #19 May 23 2021 02:42:04

%S 0,1,2,15,52,285,1206,6027,27560,134073,633130,3062279,14676828,

%T 71045845,343195230,1665555075,8084777040,39343835505,191627687250,

%U 934855945215,4565076327300,22318461756045,109211684822790,534907610833275,2621997452787192,12862364386480425,63140696801700986

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

%C 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)).

%H Vincenzo Librandi, <a href="/A098520/b098520.txt">Table of n, a(n) for n = 0..300</a>

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

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

%F 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

%F a(n) ~ 5^(n+1/2)/(4*sqrt(2*Pi*n)). - _Vaclav Kotesovec_, Oct 15 2012

%F a(n) = 2^(n-1)*GegenbauerC(n-1, -n, -1/4). - _Peter Luschny_, May 08 2016

%p a := n -> simplify(2^(n-1)*GegenbauerC(n-1, -n, -1/4)):

%p seq(a(n), n=0..26); # _Peter Luschny_, May 08 2016

%t 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 *)

%o (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

%K easy,nonn

%O 0,3

%A _Paul Barry_, Sep 12 2004

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