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%I #23 Feb 16 2024 05:42:08
%S 1,24,1548,155744,19893054,2937661200,477691374152,83161733788992,
%T 15230338934722749,2900395347525785464,569718535329796732476,
%U 114759815105897160007392,23602808330272138320592494,4940203531008336735249385488,1049571237547858314991495867848
%N G.f.: exp(Sum_{k>=1} (4*k)!/k!^4 * x^k/k).
%C From _Peter Bala_, Feb 08 2023: (Start)
%C Let A(x) denote the o.g.f. of the sequence. The sequence defined by b(n) := [x^n] A(x)^n for n >= 1 begins [24, 3672, 703968, 149835864, 33911355024, 7993981771488, 1940145241321920, ...]. We conjecture that b(n) satisfies the supercongruences b(n*p^r) == b(n*p^(r-1)) ( mod p^(3*r) ) for prime p >= 5 and all positive integers n and r.
%C More generally, for a positive integer m, set A_m(x) = exp( Sum_{n >= 1} (m*n)!/(n!^m) * x^n/n ) and define a sequence {b_m(n): n >= 1} by b_m(n) := [x^n] A_m(x)^n. Then we conjecture that b_m(n) is an integer sequence satisfying the same supercongruences. (End)
%F a(n) ~ c * 4^(4*n)/n^(5/2), where c = exp(3*HypergeometricPFQ[{1, 1, 5/4, 3/2, 7/4}, {2, 2, 2, 2}, 1] / 32) / (sqrt(2)*Pi^(3/2)) = 0.14496966... - _Vaclav Kotesovec_, Mar 06 2020, updated Feb 16 2024
%F a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} A008977(k) * a(n-k). - _Seiichi Manyama_, Feb 09 2024
%t CoefficientList[Series[Exp[Sum[(4*k)!/k!^4*x^k/k, {k, 1, 20}]], {x, 0, 20}], x]
%t CoefficientList[Series[Exp[24*x*HypergeometricPFQ[{1, 1, 5/4, 3/2, 7/4}, {2, 2, 2, 2}, 256*x]], {x, 0, 20}], x] (* _Vaclav Kotesovec_, Feb 09 2024 *)
%Y Cf. A000108, A008977, A229451, A229452, A333043, A370294.
%K nonn,easy
%O 0,2
%A _Vaclav Kotesovec_, Mar 06 2020
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