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%I #29 Jan 01 2024 18:19:42
%S 1,2,15,204,4095,109668,3689595,149846840,7141879503,391139588190,
%T 24218296445200,1673538279265020,127715832778905150,
%U 10670643284149377480,968929726650218004435,95024894699780159868144,10011211830149283223044015
%N a(n) = binomial(n^2 + n - 1, n) / (n^2 + n - 1).
%C Empirical: In the ring of symmetric functions over the fraction field Q(q, t), let s(n) denote the Schur function indexed by n. Then (up to sign) a(n) is the coefficient of s(1^n) in nabla^(n) s(n) with q=t=1, where nabla denotes the "nabla operator" on symmetric functions.
%F a(n) ~ c*exp(n-1/(6*n))*n^(n-5/2), where c = sqrt(e/(2*Pi)). - _Stefano Spezia_, May 04 2022
%F a(n) = n * A182316(n - 1). - _F. Chapoton_, Sep 22 2023
%t Table[With[{c=n^2+n-1},Binomial[c,n]/c],{n,20}] (* _Harvey P. Dale_, Jan 01 2024 *)
%o (Sage) [binomial(n*n+n-1,n)/(n*n+n-1) for n in range(1,29)]
%o (Python)
%o from math import comb
%o def A351501(n): return comb(m := n**2+n-1,n)//m # _Chai Wah Wu_, May 07 2022
%Y Closely related to A177784. See also A091144.
%Y Diagonal of A162382. Multiple of A182316.
%K nonn
%O 1,2
%A _F. Chapoton_, May 03 2022