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%I #11 Apr 17 2014 06:40:55
%S 1,13,110,765,4746,27314,149052,781725,3975730,19730150,95973956,
%T 459145778,2165937060,10095323460,46566906872,212857023069,
%U 965208806082,4345780250270,19442667426420,86489687956518
%N Convolution of Catalan numbers A000108 with A038845.
%C Also convolution of A018218(n+1), n >= 0, with A000302 (powers of 4); also convolution of A000346 with A002697.
%H Fung Lam, <a href="/A042941/b042941.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = binomial(n+3, 2)*(4^(n+1) - A000984(n+3)/A000984(2)) / 2.
%F G.f.: c(x)/(1-4*x)^3, where c(x) is the g.f. for Catalan numbers.
%F Recurrence: (n+1)*a(n) = 128*(1-2*n)*a(n-4) + 32*(8*n-1)*a(n-3) - 24*(4*n+1)*a(n-2) + 2*(8*n+5)*a(n-1). - _Fung Lam_, Apr 13 2014
%F a(n) ~ 2^(2*n)*(n^2 - 8*n^(3/2)/(3*sqrt(Pi))). - _Fung Lam_, Apr 13 2014
%F Recurrence: n*(n+1)*a(n) = 2*n*(4*n+9)*a(n-1) - 8*(n+2)*(2*n+3)*a(n-2). - _Vaclav Kotesovec_, Apr 16 2014
%t CoefficientList[Series[(1-Sqrt[1-4*x])/(2*x*(1-4*x)^3), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Apr 16 2014 *)
%K easy,nonn
%O 0,2
%A _Wolfdieter Lang_
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