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%I #28 Sep 08 2016 18:04:18
%S 1,2,4,12,36,120,400,1400,4900,17640,63504,232848,853776,3171168,
%T 11778624,44169840,165636900,625739400,2363904400,8982836720,
%U 34134779536,130332794592,497634306624,1907598175392,7312459672336,28124844893600,108172480360000
%N a(n) = binomial(n,floor(n/2))*binomial(n+1,floor(n/2+1/2))*(1+floor(n/2))/(1+2*floor(n/2)).
%H Alois P. Heinz, <a href="/A241530/b241530.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = ((8*n-4)*a(n-1)+16*(n-1)*(n-2)*a(n-2))/(n*(n+1)) for n>=2, a(n) = 2^n for n<2. - _Alois P. Heinz_, Apr 25 2014
%F G.f.: ((1+4*x)*K(4*x) - E(4*x))/(2*Pi*x), where K and E are the complete elliptic integrals of the first and second kind, respectively, with modulus k = 4*x. - _Benedict W. J. Irwin_, Aug 15 2016
%F From _Wolfdieter Lang_, Sep 06 2016 (Start):
%F The preceding g.f. can be rewritten as ((1+4*x)*F(1/2,1/2;1;(4*x)^2) -
%F F(-1/2,1/2;1;(4*x)^2))/(4*x), where F is the hypergyometric function F(a,b;c;z).
%F This leads to the bisection a(2*k) = ((2*k)!)^2/k!^4 = A002894(k) and a(2*k+1) = 2*(2*k)!*(2*k+1)!/((k+1)*k!^4) = 2*A000894(k), for k >= 0.
%F (End)
%p A241530 := n -> binomial(n,iquo(n,2))*binomial(n+1,iquo(n+1,2))
%p *(1+iquo(n,2))/(1+2*iquo(n,2)); seq(A241530(n), n=0..26);
%p # second Maple program:
%p a:= proc(n) option remember; `if`(n<2, 2^n,
%p ((8*n-4)*a(n-1)+16*(n-1)*(n-2)*a(n-2))/(n*(n+1)))
%p end:
%p seq(a(n), n=0..30); # _Alois P. Heinz_, Aug 10 2016
%t CoefficientList[Series[(-EllipticE[16 x^2] + (1 + 4 x) EllipticK[16 x^2])/(2Pi x), {x, 0, 20}], x] (* _Benedict W. J. Irwin_, Aug 15 2016 *)
%t Table[Binomial[n, #] Binomial[n + 1, Floor[(n + 1)/2]] (1 + #)/(1 + 2 #) &@ Floor[n/2], {n, 0, 26}] (* _Michael De Vlieger_, Aug 15 2016 *)
%Y Cf. A000894, A002894, A005566.
%Y Row n=3 of A275784.
%K nonn,easy
%O 0,2
%A _Peter Luschny_, Apr 25 2014
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