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%I #19 Jan 11 2022 22:06:42
%S 1,6,43,332,2661,21810,181455,1526040,12939145,110413406,947052723,
%T 8157680228,70518067309,611426078346,5315138311383,46308989294640,
%U 404274406256145,3535479068797110,30966952059306555,271616893912241532,2385412594943633781,20973327081776664546
%N a(n) = Sum_{k = 0..n} binomial(n,k) * binomial(n+1,k+1) * 4^k.
%C Fifth binomial transform of A098664.
%H Vincenzo Librandi, <a href="/A098665/b098665.txt">Table of n, a(n) for n = 0..200</a>
%F G.f.: ((1+3*x)-sqrt(1-10*x+9*x^2))/(8*x*sqrt(1-10*x+9*x^2)).
%F E.g.f.: exp(5x)*(BesselI(0, 4x)+BesselI(1, 4x)/2).
%F 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
%F a(n) ~ 9^(n+1)/(4*sqrt(2*Pi*n)). - _Vaclav Kotesovec_, Oct 15 2012
%F From _Peter Bala_, Jan 07 2022: (Start)
%F The following formulas assume an offset of 1:
%F a(n) = (1/4) * Sum_{k = 0..n} binomial(n,k)*A119259(k).
%F a(n) = (1/4) * Sum_{k = 0..n} binomial(n,k)*binomial(2*n-k-1,n-k)*3^k.
%F a(n) = (1/4) * [x^n] ((1 + 3*x)/(1 - x))^n.
%F 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)
%t 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 *)
%o (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
%Y Cf. A098664, A119259.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Sep 20 2004
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