A381987 E.g.f. A(x) satisfies A(x) = exp(x) * B(x), where B(x) = 1 + x*B(x)^4 is the g.f. of A002293.
1, 2, 11, 160, 3941, 134486, 5851327, 309520436, 19283504585, 1382980764106, 112223497464371, 10165461405056552, 1016801830348902061, 111312715288354681310, 13237965546409421546471, 1699516550894276788156156, 234263144339070269872076177, 34507561203827621878485498386
Offset: 0
Programs
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Maple
seq(simplify(hypergeom([-n, 1/2, 1/4, 3/4], [2/3, 4/3], -256/27)), n = 0..17); # Peter Bala, Mar 13 2025
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Mathematica
Table[HypergeometricPFQ[{-n, 1/2, 1/4, 3/4}, {2/3, 4/3}, -256/27], {n, 0, 20}] (* Vaclav Kotesovec, Mar 14 2025 *) -
PARI
a(n) = n!*sum(k=0, n, binomial(4*k+1, k)/((4*k+1)*(n-k)!));
Formula
a(n) = n! * Sum_{k=0..n} A002293(k)/(n-k)!.
From Peter Bala, Mar 13 2025: (Start)
a(n) = hypergeom([-n, 1/2, 1/4, 3/4], [2/3, 4/3], -256/27).
3*(3*n - 1)*(3*n + 1)*a(n) = n*(256*n^2 - 303*n + 95)*a(n-1) - 3*(n - 1)*(256*n^2 - 485*n + 245)*a(n-2) + 3*(256*n - 375)*(n - 1)*(n - 2)*a(n-3) - 256*(n - 1)*(n - 2)*(n - 3)*a(n-4) with a(0) = 1, a(1) = 2, a(2) = 11 and a(3) = 160. (End)
a(n) ~ 2^(8*n+1) * n^(n-1) / (3^(3*n + 3/2) * exp(n - 27/256)). - Vaclav Kotesovec, Mar 14 2025
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