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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.

%I A381987 #20 Mar 14 2025 09:54:18
%S A381987 1,2,11,160,3941,134486,5851327,309520436,19283504585,1382980764106,
%T A381987 112223497464371,10165461405056552,1016801830348902061,
%U A381987 111312715288354681310,13237965546409421546471,1699516550894276788156156,234263144339070269872076177,34507561203827621878485498386
%N 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.
%C A381987 For each positive integer k, the sequence obtained by reducing a(n) modulo k is a periodic sequence with period dividing k. For example, modulo 5 the sequence becomes [1, 2, 1, 0, 1, 1, 2, 1, 0, 1, ...] with period 5. In particular, a(5*n+3) == 0 (mod 5). Cf. A047974. - _Peter Bala_, Mar 13 2025
%F A381987 a(n) = n! * Sum_{k=0..n} A002293(k)/(n-k)!.
%F A381987 From _Peter Bala_, Mar 13 2025: (Start)
%F A381987 a(n) = hypergeom([-n, 1/2, 1/4, 3/4], [2/3, 4/3], -256/27).
%F A381987 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)
%F A381987 a(n) ~ 2^(8*n+1) * n^(n-1) / (3^(3*n + 3/2) * exp(n - 27/256)). - _Vaclav Kotesovec_, Mar 14 2025
%p A381987 seq(simplify(hypergeom([-n, 1/2, 1/4, 3/4], [2/3, 4/3], -256/27)), n = 0..17); # _Peter Bala_, Mar 13 2025
%t A381987 Table[HypergeometricPFQ[{-n, 1/2, 1/4, 3/4}, {2/3, 4/3}, -256/27], {n, 0, 20}] (* _Vaclav Kotesovec_, Mar 14 2025 *)
%o A381987 (PARI) a(n) = n!*sum(k=0, n, binomial(4*k+1, k)/((4*k+1)*(n-k)!));
%Y A381987 Cf. A381984, A381988, A381989.
%Y A381987 Cf. A002293, A365340.
%K A381987 nonn,easy
%O A381987 0,2
%A A381987 _Seiichi Manyama_, Mar 12 2025