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%I #19 Jan 30 2020 21:29:15
%S 1,2,24,128,1096,7632,60864,461568,3648096,28551872,226695424,
%T 1799989248,14380907776,115126211072,924791445504,7444100947968,
%U 60057602459136,485388465196032,3929580292706304,31858982479331328
%N Expansion of 1/sqrt(1-4x-36x^2).
%C Define Q(n,x)=sum{k=0..floor(n/2), binomial(n,k)binomial(2(n-k),n)x^(n-2k)}. Then a(n)=3^n*Q(n,1/3). A084770(n) is 2^n*Q(n,1/2). Central coefficient of (1+2x+10x^2)^n.
%H Vincenzo Librandi, <a href="/A098455/b098455.txt">Table of n, a(n) for n = 0..200</a>
%H Hacène Belbachir, Abdelghani Mehdaoui, László Szalay, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL22/Szalay/szalay42.html">Diagonal Sums in the Pascal Pyramid, II: Applications</a>, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
%F E.g.f.: exp(2x)*BesselI(0, 2*sqrt(10)*x).
%F a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*binomial(2(n-k), n)*9^k.
%F D-finite with recurrence: n*a(n) = 2*(2*n-1)*a(n-1) + 36*(n-1)*a(n-2). - _Vaclav Kotesovec_, Oct 15 2012
%F a(n) ~ sqrt(50+5*sqrt(10))*(2+2*sqrt(10))^n/(10*sqrt(Pi*n)). - _Vaclav Kotesovec_, Oct 15 2012
%t Table[SeriesCoefficient[1/Sqrt[1-4*x-36*x^2],{x,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 15 2012 *)
%o (PARI) x='x+O('x^66); Vec(1/sqrt(1-4*x-36*x^2)) \\ _Joerg Arndt_, May 11 2013
%K easy,nonn
%O 0,2
%A _Paul Barry_, Sep 08 2004
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