A333042 G.f.: exp(Sum_{k>=1} (4*k)!/k!^4 * x^k/k).

1, 24, 1548, 155744, 19893054, 2937661200, 477691374152, 83161733788992, 15230338934722749, 2900395347525785464, 569718535329796732476, 114759815105897160007392, 23602808330272138320592494, 4940203531008336735249385488, 1049571237547858314991495867848

Offset

0,2

Comments

From Peter Bala, Feb 08 2023: (Start)

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.

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)

Links

Table of n, a(n) for n=0..14.

Formula

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

a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} A008977(k) * a(n-k). - Seiichi Manyama, Feb 09 2024

Mathematica

CoefficientList[Series[Exp[Sum[(4*k)!/k!^4*x^k/k, {k, 1, 20}]], {x, 0, 20}], x]
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 *)

Crossrefs

Cf. A000108, A008977, A229451, A229452, A333043, A370294.

Sequence in context: A203973 A063885 A019284 * A112389 A187634 A229430

Adjacent sequences: A333039 A333040 A333041 * A333043 A333044 A333045

Keyword

nonn,easy

Author

Vaclav Kotesovec, Mar 06 2020

Status

approved