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a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} binomial(2*n,2*k)^3 * k * a(n-k).
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%I #6 Mar 18 2022 00:12:33

%S 1,1,109,124876,704029453,13294133177626,665514245564815384,

%T 75462508236267111825685,17305487139219914670764064013,

%U 7368678746697280907127091048286734,5449131877967324738667220718996986592734,6632563741264033978048120096103173533343094035

%N a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} binomial(2*n,2*k)^3 * k * a(n-k).

%F Sum_{n>=0} a(n) * x^(2*n) / (2*n)!^3 = exp( Sum_{n>=1} x^(2*n) / (2*n)!^3 ).

%t a[0] = 1; a[n_] := a[n] = (1/n) Sum[Binomial[2 n, 2 k]^3 k a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 11}]

%t nmax = 22; Take[CoefficientList[Series[Exp[Sum[x^(2 k)/(2 k)!^3, {k, 1, nmax}]], {x, 0, nmax}], x] Range[0, nmax]!^3, {1, -1, 2}]

%Y Cf. A005046, A061684, A352465, A352469.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Mar 17 2022