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

%I #11 Aug 10 2024 15:59:03

%S 1,2,6,23,94,392,1680,7387,33110,150905,698996,3287550,15685420,

%T 75877427,371994692,1847450970,9290557158,47291312897,243574276884,

%U 1268915237141,6683909556420,35585631836229,191433293140656,1040197718292138,5707318227692796

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

%F a(n) = [x^n] 1/((1-x-n*x^3) * (1-x)^n).

%F a(n) ~ exp(4*n^(2/3)/3 + 2*n^(1/3)/9) * n^(n/3) / 3. - _Vaclav Kotesovec_, Apr 07 2024

%t Join[{1},Table[Sum[n^k Binomial[2n-2k,n-3k],{k,0,Floor[n/3]}],{n,30}]] (* _Harvey P. Dale_, Aug 10 2024 *)

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

%Y Cf. A371825, A371826.

%Y Cf. A368891.

%K nonn

%O 0,2

%A _Seiichi Manyama_, Apr 07 2024