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a(0) = 1; a(n) = (1/n) * Sum_{k=0..floor((n-1)/2)} binomial(n,2*k+1)^3 * (2*k+1) * a(n-2*k-1).
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%I #5 Mar 18 2022 00:13:09

%S 1,1,4,37,640,18401,810616,51506645,4512303104,526359723265,

%T 79484297525704,15182084413118525,3598056798827450752,

%U 1040872295660542894433,362422517793599461361216,150047916077302216370174237,73081847594180657956494147584,41481744863993143666887680079873

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

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

%t a[0] = 1; a[n_] := a[n] = (1/n) Sum[Binomial[n, 2 k + 1]^3 (2 k + 1) a[n - 2 k - 1], {k, 0, Floor[(n - 1)/2]}]; Table[a[n], {n, 0, 17}]

%t nmax = 17; CoefficientList[Series[Exp[Sum[x^(2 k + 1)/(2 k + 1)!^3, {k, 0, nmax}]], {x, 0, nmax}], x] Range[0, nmax]!^3

%Y Cf. A003724, A061684, A346220, A352466.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Mar 17 2022