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%I #18 Sep 09 2024 19:10:16
%S 1,3,10,36,134,507,1937,7449,28783,111623,434130,1692387,6610292,
%T 25861384,101319095,397428091,1560588454,6133768656,24128550045,
%U 94986663925,374188128311,1474980414870,5817387549611,22955930045826,90629404431826,357960414264163
%N a(n) = Sum_{k=0..floor(n/3)} binomial(2*n-k+1,n-3*k).
%H Harvey P. Dale, <a href="/A371773/b371773.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = [x^n] 1/(((1-x)^2-x^3) * (1-x)^n).
%F a(n) = binomial(2*(n-1), n-1)*hypergeom([1, (1-n)/3, (2-n)/3, 1-n/3], [1-n, 3/2-n, n], -27/4). - _Stefano Spezia_, Apr 06 2024
%F From _Vaclav Kotesovec_, Apr 08 2024: (Start)
%F Recurrence: n*(n^2 - 7)*a(n) = (9*n^3 - 2*n^2 - 79*n + 60)*a(n-1) - 2*(12*n^3 - 5*n^2 - 124*n + 150)*a(n-2) + (17*n^3 - 8*n^2 - 183*n + 240)*a(n-3) - 2*(2*n - 5)*(n^2 + 2*n - 6)*a(n-4).
%F a(n) ~ 2^(2*n+2) / sqrt(Pi*n). (End)
%t Table[Sum[Binomial[2n-k+1,n-3k],{k,0,Floor[n/3]}],{n,0,30}] (* _Harvey P. Dale_, Sep 09 2024 *)
%o (PARI) a(n) = sum(k=0, n\3, binomial(2*n-k+1, n-3*k));
%Y Cf. A371774, A371775, A371776.
%Y Cf. A144904, A371758, A371777.
%K nonn
%O 0,2
%A _Seiichi Manyama_, Apr 05 2024