A229452 G.f.: exp( Sum_{n>=1} (3*n)!/(3!*n!^3) * x^n/n ).

1, 1, 8, 101, 1569, 27445, 518407, 10333243, 214320244, 4583132411, 100411556533, 2243625355010, 50955869372055, 1173262656151429, 27332509319090516, 643208905017756216, 15270427859720369204, 365356267775348553277, 8801688936499808334602

Offset: 0

Author

Paul D. Hanna, Sep 23 2013

Keywords

• nonn

Comments

Self-convolution 6th power yields A229451.

From Peter Bala, Feb 16 2020: (Start)

The sequence defined by b(n) = [x^n] A(x)^n for n >= 1 begins [1, 17, 352, 7969, 189876, 4676768, 117905565, 3024222753, 78607893934, 2064924478892, 54710782664836, ...]. We conjecture that b(n) satisfies the supercongruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and all positive integers n and k [added 20 Oct 2024: more generally, for r a positive integer and s an integer we conjecture that the sequence {b(r,s;n) : n >= 1} defined by b(r,s; n) = [x^(r*n)] A(x)^(s*n) satisfies the same supercongruences].

More generally, for a positive integer m, set A_m(x) = exp( Sum_{n >= 1} (m*n)!/(m!*n!^m) * x^n/n ) and define a sequence b_m(n) := [x^n] A_m(x)^n for n >=1. Then we conjecture that b_m(n) is an integer sequence satisfying the same congruences. (End)

In general, for m >= 1, if g.f. = exp(m * Sum_{n>=1} (3*n)!/(3!*n!^3) * x^n/n), then a(n) ~ m * 2^(2*m-2) * 3^((m-1)/2) * Pi^(m-1) * A370293^m * 3^(3*n) / n^2, cf. A370289 (m=2), A370288 (m=3), A229451 (m=6). - Vaclav Kotesovec, Feb 14 2024

Examples

			G.f.: A(x) = 1 + x + 8*x^2 + 101*x^3 + 1569*x^4 + 27445*x^5 + ...,
where
log(A(x)) = x + 15*x^2/2 + 280*x^3/3 + 5775*x^4/4 + 126126*x^5/5 + 2858856*x^6/6 + ... + A060542(n)*x^n/n + ....
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..700

Crossrefs

Cf. A229451, A060542, A006480 (de Bruijn's S(3,n)), A061401, A370288, A370289, A370293, A370294, A370295.

Programs

Mathematica

CoefficientList[Series[Exp[Sum[(3*k)!/(3!*k!^3)*x^k/k, {k, 1, 20}]], {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 05 2020 *)
CoefficientList[Series[Exp[x*HypergeometricPFQ[{1, 1, 4/3, 5/3}, {2, 2, 2}, 27*x]], {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 11 2024 *)

PARI

{a(n)=polcoeff(exp(sum(k=1,n,(3*k)!/(3!*k!^3)*x^k/k) +x*O(x^n)),n)}
for(n=0,25,print1(a(n),", "))

Formula

a(n) ~ c * 3^(3*n) / n^2, where c = A370293 = 0.0490152812... - Vaclav Kotesovec, Feb 14 2024

a(n) = 1/n * Sum_{k = 0..n-1} b(n-k)*a(k) with a(0) = 1, where b(n) = (3*n)!/(6*n!^3). - Peter Bala, Oct 22 2024