OFFSET
0,2
COMMENTS
The sixth root of the o.g.f. A(x)^(1/6) = 1 + x + 8*x^2 + 101*x^3 + 1569*x^4 + 27445*x^5 + ... appears to have integer coefficients. See A229452. More generally, if A(m,x) := exp( Sum_{n >= 1} (m*n)!/n!^m * x^n/n ), m = 1,2,3,..., then it can be shown that the expansion of A(m,x) has integer coefficients. We conjecture that the expansion of A(m,x)^(1/m!) also has integer coefficients. - Peter Bala, Feb 16 2020
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..702
FORMULA
a(n) ~ c * 3^(3*n)/n^2, where c = 2^11 * 3^(7/2) * Pi^5 * A370293^6 = 0.406436497... - Vaclav Kotesovec, Dec 25 2013, updated Feb 14 2024
a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} A006480(k) * a(n-k). - Seiichi Manyama, Feb 09 2024
From Peter Bala, Oct 24 2024: (Start)
Series reversion of x*A(-x) = x + 6*x^2 + 9*x^3 + 56*x^4 - 300*x^5 + 3942*x^6 - ... is the g.f. of A061401.
The g.f. A(x) satisfies [x^n] A(x)^n = A362732(n). (End)
EXAMPLE
G.f.: A(x) = 1 + 6*x + 63*x^2 + 866*x^3 + 13899*x^4 + 246366*x^5 +...,
where
log(A(x)) = 6*x + 90*x^2/2 + 1680*x^3/3 + 34650*x^4/4 + 756756*x^5/5 +...+ A006480(n)*x^n/n + ....
MATHEMATICA
CoefficientList[Series[Exp[6*x*HypergeometricPFQ[{1, 1, 4/3, 5/3}, {2, 2, 2}, 27*x]], {x, 0, 20}], x] (* Vaclav Kotesovec, Dec 25 2013 *)
PROG
(PARI) {a(n)=polcoeff(exp(sum(k=1, n, (3*k)!/k!^3*x^k/k) +x*O(x^n)), n)}
for(n=0, 25, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Paul D. Hanna, Sep 23 2013
STATUS
approved