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a(n) = 4^n*binomial(2*n+1, n).
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%I #33 Jan 16 2024 01:38:27

%S 1,12,160,2240,32256,473088,7028736,105431040,1593180160,24216338432,

%T 369849532416,5671026163712,87246556364800,1346089726771200,

%U 20819521107394560,322702577164615680,5011381198321090560,77954818640550297600,1214454016715941478400

%N a(n) = 4^n*binomial(2*n+1, n).

%H G. C. Greubel, <a href="/A098400/b098400.txt">Table of n, a(n) for n = 0..820</a>

%F G.f.: (1-sqrt(1-16*x))/(8*x*sqrt(1-16*x)).

%F E.g.f.: a(n) = n! * [x^n] exp(8*x)*(BesselI(0, 8*x) + BesselI(1, 8*x)). - _Peter Luschny_, Aug 25 2012

%F (n+1)*a(n) - 8*(2*n+1)*a(n-1) = 0. - _R. J. Mathar_, Nov 26 2012

%F a(n) = 4^n*(2*n+1)*Hypergeometric2F1([1-n,-n],[2],1). - _Peter Luschny_, Sep 22 2014

%F From _G. C. Greubel_, Dec 27 2023: (Start)

%F a(n) = 4^n * A001700(n).

%F a(n) = 4^n * (2*n+1) * A000108(n).

%F a(n) = (2*n+1) * A151403(n). (End)

%F From _Amiram Eldar_, Jan 16 2024: (Start)

%F Sum_{n>=0} 1/a(n) = 8/15 + 128*arcsin(1/4)/(15*sqrt(15)).

%F Sum_{n>=0} (-1)^n/a(n) = 8/17 + 128*arcsinh(1/4)/(17*sqrt(17)). (End)

%t Table[4^n Binomial[2n+1,n],{n,0,20}] (* _Harvey P. Dale_, Jan 22 2019 *)

%o (PARI) a(n)=binomial(2*n+1,n)<<(2*n) \\ _Charles R Greathouse IV_, Oct 23 2023

%o (Magma) [4^n*(2*n+1)*Catalan(n): n in [0..30]]; // _G. C. Greubel_, Dec 27 2023

%o (SageMath) [4^n*binomial(2*n+1,n) for n in range(31)] # _G. C. Greubel_, Dec 27 2023

%Y Cf. A000108, A001700, A069720, A069723, A098399, A151403.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Sep 06 2004