A098400 a(n) = 4^n*binomial(2*n+1, n).
1, 12, 160, 2240, 32256, 473088, 7028736, 105431040, 1593180160, 24216338432, 369849532416, 5671026163712, 87246556364800, 1346089726771200, 20819521107394560, 322702577164615680, 5011381198321090560, 77954818640550297600, 1214454016715941478400
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..820
Programs
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Magma
[4^n*(2*n+1)*Catalan(n): n in [0..30]]; // G. C. Greubel, Dec 27 2023
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Mathematica
Table[4^n Binomial[2n+1,n],{n,0,20}] (* Harvey P. Dale, Jan 22 2019 *) -
PARI
a(n)=binomial(2*n+1,n)<<(2*n) \\ Charles R Greathouse IV, Oct 23 2023
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SageMath
[4^n*binomial(2*n+1,n) for n in range(31)] # G. C. Greubel, Dec 27 2023
Formula
G.f.: (1-sqrt(1-16*x))/(8*x*sqrt(1-16*x)).
E.g.f.: a(n) = n! * [x^n] exp(8*x)*(BesselI(0, 8*x) + BesselI(1, 8*x)). - Peter Luschny, Aug 25 2012
(n+1)*a(n) - 8*(2*n+1)*a(n-1) = 0. - R. J. Mathar, Nov 26 2012
a(n) = 4^n*(2*n+1)*Hypergeometric2F1([1-n,-n],[2],1). - Peter Luschny, Sep 22 2014
From G. C. Greubel, Dec 27 2023: (Start)
a(n) = 4^n * A001700(n).
a(n) = 4^n * (2*n+1) * A000108(n).
a(n) = (2*n+1) * A151403(n). (End)
From Amiram Eldar, Jan 16 2024: (Start)
Sum_{n>=0} 1/a(n) = 8/15 + 128*arcsin(1/4)/(15*sqrt(15)).
Sum_{n>=0} (-1)^n/a(n) = 8/17 + 128*arcsinh(1/4)/(17*sqrt(17)). (End)