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a(n) = ( Sum_{k=0..n} binomial(2*n, k)^2 * (binomial(2*n, k+1) - binomial(2*n, k-1)) )/(n*binomial(2*n, n)).
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%I #19 Jul 16 2024 11:34:26

%S 1,7,61,611,6686,77729,944245,11859355,152893720,2013070126,

%T 26967817306,366542344117,5043651762826,70138959074461,

%U 984384594022117,13927418363218955,198459156018467084,2845950809029225472,41044332341739034032,594983281327999736694

%N a(n) = ( Sum_{k=0..n} binomial(2*n, k)^2 * (binomial(2*n, k+1) - binomial(2*n, k-1)) )/(n*binomial(2*n, n)).

%H Seiichi Manyama, <a href="/A259335/b259335.txt">Table of n, a(n) for n = 1..835</a>

%H H. W. Gould, <a href="http://www.jstor.org/stable/2318079">Problem E2384</a>, Amer. Math. Monthly, 79 (1972), p. 1034.

%H D. Zeitlin & N. J. A. Sloane, <a href="/A259335/a259335.pdf">Correspondence, 1974 & 1991</a>

%F a(n) = (1/(n+1)) * Sum_{k=0..n} (k/(n+k)) * binomial(n+k,k)^2. - _Seiichi Manyama_, Jul 16 2024

%p f:=proc(n) local b;

%p b:=binomial;

%p add(b(2*n,k)^2*(b(2*n,k+1)-b(2*n,k-1)),k=0..n)/(n*b(2*n,n));

%p end;

%o (PARI) a(n) = sum(k=0, n, k/(n+k)*binomial(n+k, k)^2)/(n+1); \\ _Seiichi Manyama_, Jul 16 2024

%K nonn

%O 1,2

%A _N. J. A. Sloane_, Jun 25 2015