%I #22 Oct 16 2024 09:23:34
%S 1,2,30,500,8750,157500,2887500,53625000,1005468750,18992187500,
%T 360851562500,6888984375000,132038867187500,2539208984375000,
%U 48970458984375000,946762207031250000,18343517761230468750,356080050659179687500,6923778762817382812500
%N a(n) = (4*0^n + 5^n*binomial(2*n,n))/5.
%C (1 + (k-1)*sqrt(1-4*k*x))/(k*sqrt(1-4*k*x)) is the g.f. for ((k-1)*0^n + k^n*binomial(2*n,n))/k.
%H G. C. Greubel, <a href="/A099046/b099046.txt">Table of n, a(n) for n = 0..760</a>
%F G.f.: (1 + 4*sqrt(1-20*x))/(5*sqrt(1-20*x)).
%F n*a(n) +10*(-2*n+1)*a(n-1)=0. - _R. J. Mathar_, Nov 24 2012
%F E.g.f.: (4 + exp(10*x) * BesselI(0,10*x)) / 5. - _Ilya Gutkovskiy_, Nov 17 2021
%F a(n) = Integral_{x = 0..20} x^n * w(x) dx for n >= 1, where w(x) = 1/( 5*Pi*sqrt(x*(20 - x)) ) is positive on the interval (0, 20). The weight function w(x) is singular at x = 0 and at x = 20 and is the solution of the Hausdorff moment problem. - _Peter Bala_, Oct 12 2024
%t CoefficientList[Series[(1+4Sqrt[1-20x])/(5Sqrt[1-20x]),{x,0,20}],x] (* _Harvey P. Dale_, Mar 30 2011 *)
%t Join[{1}, Table[5^(n - 1)*Binomial[2*n, n], {n,1,50}]] (* _G. C. Greubel_, Dec 31 2017 *)
%o (Magma) [(4*0^n + 5^n*Binomial(2*n, n))/5: n in [ 0..30]]; // _G. C. Greubel_, Dec 31 2017
%o (PARI) for(n=0,30, print1((4*0^n + 5^n*binomial(2*n,n))/5, ", ")) \\ _G. C. Greubel_, Dec 31 2017
%Y Cf. A069723, A088218, A099044, A099045.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Sep 24 2004