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a(n) = Sum_{k=0..n} binomial(n+3*k-4,k).
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%I #22 Jun 30 2026 08:47:37

%S 1,1,8,69,598,5229,46148,410564,3677205,33118004,299653447,2721863556,

%T 24805834384,226714986482,2077219038583,19073322225093,

%U 175469528064836,1617021669150125,14924185776171959,137930294999776878,1276338562136112694,11823881876462306463

%N a(n) = Sum_{k=0..n} binomial(n+3*k-4,k).

%F G.f.: 1/(g^4 * (1-4*x*g^3) * (1-x*g)) where g = 1+x*g^4 is the g.f. of A002293.

%F G.f.: 1/((4-3*g) * (1-g+g^3)) where g = 1+x*g^4 is the g.f. of A002293.

%F Here and below, binomial(N,k) = 0 for k<0.

%F This is the special case l=1, m=3, c=0, r=1, s=0 of the following family. For integers l, m, c in Z, and for any constants r and s, define a_{l,m,c,r,s}(n) = Sum_{k=0..n} (r*binomial(l*n+m*k-m-1+c,k) - s*binomial(l*n+m*k-m-1+c,k-1)). A_{l,m,c,r,s}(x) = Sum_{n>=0} a_{l,m,c,r,s}(n)*x^n = t^c*(r+s-s*t)/((l+m-(l+m-1)*t)*(1-t+t^m)), where t = t(x) satisfies t = 1 + x*t^(l+m), equivalently y = x*(1+y)^(l+m) with y=t-1.

%o (PARI) a(n) = sum(k=0, n, binomial(n+3*k-4, k));

%Y Cf. A389239, A390383, A390412, A390471.

%Y Cf. A088218, A397509, A397536, A397537.

%Y Cf. A002293.

%K nonn,easy

%O 0,3

%A _Seiichi Manyama_, Jun 30 2026