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a(n) = Sum_{k=0..floor(n/3)} binomial(k+2,3*n-9*k+2).
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%I #17 Jan 15 2026 04:44:34

%S 1,0,0,3,0,0,6,0,0,10,1,0,15,6,0,21,21,0,28,56,1,36,126,9,45,252,45,

%T 55,462,165,67,792,495,90,1287,1287,169,2002,3003,469,3004,6435,1485,

%U 4383,12870,4504,6308,24310,12529,9248,43759,31995,14688,75600,75772,27132

%N a(n) = Sum_{k=0..floor(n/3)} binomial(k+2,3*n-9*k+2).

%H Seiichi Manyama, <a href="/A390037/b390037.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,3,0,0,-3,0,0,1,1).

%F G.f.: 1 / ((1-x^3)^3 - x^10).

%F a(n) = 3*a(n-3) - 3*a(n-6) + a(n-9) + a(n-10).

%o (PARI) my(N=60, x='x+O('x^N)); Vec(1/((1-x^3)^3-x^10))

%Y Cf. A390035.

%K nonn,easy

%O 0,4

%A _Seiichi Manyama_, Jan 14 2026