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%I #25 Jan 29 2019 09:11:26
%S 0,1,7,37,177,807,3579,15591,67071,285861,1209641,5089517,21314453,
%T 88918353,369734553,1533115953,6341759073,26177411943,107853629643,
%U 443633635743,1822098923943,7473806605563,30618895206483,125303348573883,512274592771083,2092407173242983,8539348101568335
%N a(n) = Sum_{k=0..n} k*binomial(2*k,k)/2.
%H Seiichi Manyama, <a href="/A277178/b277178.txt">Table of n, a(n) for n = 0..1658</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CentralBinomialCoefficient.html">Central Binomial Coefficient</a>.
%F a(n) = binomial(2*n,n) * (2*n + 1 - hypergeom([1,-n], [1/2-n], 1/4))/3.
%F a(n+1) - a(n) = A002457(n) = (2*n+1)!/n!^2.
%F Recurrence: (5*n + 2) * a(n) = (4*n + 2) * a(n-1) + n * a(n+1).
%F a(n) ~ sqrt(n) * 2^(2*n+1) / (3*sqrt(Pi)). - _Vaclav Kotesovec_, Jan 29 2019
%F G.f.: x/(1-x) * (1-4*x)^(-3/2). - _Seiichi Manyama_, Jan 29 2019
%p a:=n->sqrt(-1/27)-((n+1)/2)*binomial(2*(n+1),n+1)*hypergeom([1,n+3/2],[n+1],4):
%p seq(simplify(a(n)), n=0..26); # _Peter Luschny_, Oct 03 2016
%t Table[Binomial[2 n, n] (2 n + 1 - Hypergeometric2F1[1, -n, 1/2 - n, 1/4])/3, {n, 0, 30}]
%o (PARI) {a(n) = sum(k=0, n, k*binomial(2*k, k))/2} \\ _Seiichi Manyama_, Jan 29 2019
%Y Row 3 of A323222.
%Y Cf. A000984, A002457.
%K nonn
%O 0,3
%A _Vladimir Reshetnikov_, Oct 02 2016
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