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a(0) = 1 and a(n) = (1/n) * Sum_{k=1..n} 3^k * binomial(n,k) * binomial(n^2,k-1) for n > 0.
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%I #27 Aug 10 2023 10:28:52

%S 1,3,21,408,14799,817743,61621806,5921141502,694008501627,

%T 96176405390961,15400332946269903,2799678523675400832,

%U 569877183695866859625,128436925725088289658534,31756620986815666396814796,8548059658831271609064999978,2488568825786280454788465874035,779186768737628124697943895022101

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

%H Seiichi Manyama, <a href="/A336578/b336578.txt">Table of n, a(n) for n = 0..297</a>

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

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

%F a(n) ~ 3^n * exp(n - 1/6) * n^(n - 5/2) / sqrt(2*Pi). - _Vaclav Kotesovec_, Jul 31 2021

%F a(n) = (1/n) * Sum_{k=0..n-1} (-2)^k * 3^(n-k) * binomial(n,k) * binomial((n+1)*n-k,n-1-k) for n > 0. - _Seiichi Manyama_, Aug 10 2023

%t a[0] := 1; a[n_] := Sum[3^k * Binomial[n, k] * Binomial[n^2, k - 1], {k, 1, n}]/n; Array[a, 18, 0] (* _Amiram Eldar_, Jul 27 2020 *)

%o (PARI) a(n) = if(n==0, 1, sum(k=1, n, 3^k*binomial(n, k)*binomial(n^2, k-1))/n);

%o (PARI) a(n) = sum(k=0, n, 2^(n-k)*binomial(n, k)*binomial(n^2+k+1, n)/(n^2+k+1)); \\ _Seiichi Manyama_, Jul 27 2020

%o (PARI) a(n) = sum(k=0, n, 2^k*binomial(n^2+1, k)*binomial((n+1)*n-k, n-k))/(n^2+1); \\ _Seiichi Manyama_, Jul 27 2020

%Y Main diagonal of A336575.

%Y Cf. A336495, A336537, A336577.

%K nonn

%O 0,2

%A _Seiichi Manyama_, Jul 26 2020