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a(n) = Sum_{k=0..floor(n/3)} binomial(3*n-3*k-1,n-3*k).
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%I #19 Apr 08 2024 18:48:35

%S 1,2,10,57,338,2057,12741,79914,505954,3226638,20696685,133382658,

%T 862978221,5601919325,36467212610,237974911737,1556281907586,

%U 10196788555859,66921360130374,439860632463462,2895002186799453,19077000179746293,125849150650146714

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

%F a(n) = [x^n] 1/((1-x^3) * (1-x)^(2*n)).

%F a(n) = binomial(3*n-1, n)*hypergeom([1, (1-n)/3, (2-n)/3, -n/3], [1/3-n, 2/3-n, 1-n], 1). - _Stefano Spezia_, Apr 06 2024

%F From _Vaclav Kotesovec_, Apr 08 2024: (Start)

%F Recurrence: 18*n*(2*n - 1)*(13*n - 22)*(37*n - 51)*a(n) = 3*(40885*n^4 - 165468*n^3 + 229373*n^2 - 125562*n + 22680)*a(n-1) - (40885*n^4 - 165468*n^3 + 229373*n^2 - 125562*n + 22680)*a(n-2) + 3*(3*n - 5)*(3*n - 4)*(13*n - 9)*(37*n - 14)*a(n-3).

%F a(n) ~ 3^(3*n + 5/2) / (13 * sqrt(Pi*n) * 2^(2*n+1)). (End)

%o (PARI) a(n) = sum(k=0, n\3, binomial(3*n-3*k-1, n-3*k));

%Y Cf. A005809, A165817, A183160.

%Y Cf. A371758, A371771, A371772.

%K nonn

%O 0,2

%A _Seiichi Manyama_, Apr 05 2024