OFFSET
0,3
COMMENTS
CONJECTURE: Sum_{n>=0} a(n)^m * log(1+x)^n/n! is an integer series in x for all integer m>0; see A167138 and A167139 for examples.
From Peter Bala, Jul 07 2022: (Start)
Conjecture: Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k). For example, modulo 16 we obtain the sequence [1, 1, 5, 15, 1, 7, 1, 7, 1, 7, ...], with an apparent period of 2 beginning at a(4).
More generally, we conjecture that the same property holds for integer sequences having an e.g.f. of the form G(exp(x) - 1), where G(x) is an integral power series. (End)
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..400
FORMULA
E.g.f.: exp( Sum_{n>=1} sigma(n)*[exp(x)-1]^n/n ).
Sum_{n>=0} a(n) * log(1+x)^n/n! = g.f. of the partition numbers (A000041).
Sum_{n>=0} a(n)^2*log(1+x)^n/n! = g.f. of A167138.
From Peter Bala, Sep 18 2013: (Start)
Sum {n >= 0} (-1)^n*a(n)*(log(1 - x))^n/n! = 1 + x + 3*x^2 + 8*x^3 + 21*x^4 + ... is the o.g.f. of A218482.
a(n) is always odd. Congruences for n >= 1: a(2*n) = 2 (mod 3); a(4*n) = 2 (mod 5); a(6*n) = 0 (mod 7); a(10*n) = 7 (mod 11); a(12*n) = 5 (mod 13); a(16*n) = 0 (mod 17). (End)
From Vaclav Kotesovec, Jun 17 2018: (Start)
a(n) ~ n! * exp((1/log(2) - 1) * Pi^2 / 24 + Pi*sqrt(n/(3*log(2)))) / (4 * sqrt(3) * n * (log(2))^n).
a(n) ~ sqrt(Pi) * exp((1/log(2) - 1) * Pi^2 / 24 + Pi*sqrt(n/(3*log(2))) - n) * n^(n + 1/2) / (2^(3/2) * sqrt(3) * n * (log(2))^n). (End)
EXAMPLE
E.g.f.: A(x) = 1 + x + 5*x^2/2! + 31*x^3/3! + 257*x^4/4! +...
A(log(1+x)) = P(x) = 1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 7*x^5 +...
MATHEMATICA
Table[Sum[PartitionsP[k]*StirlingS2[n, k]*k!, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 21 2018 *)
nmax = 20; CoefficientList[Series[1/QPochhammer[E^x - 1], {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Nov 22 2021 *)
PROG
(PARI) {a(n)=if(n==0, 1, n!*polcoeff(exp(sum(m=1, n, sigma(m)*(exp(x+x*O(x^n))-1)^m/m) ), n))}
(PARI) {Stirling2(n, k)=if(k<0||k>n, 0, sum(i=0, k, (-1)^i*binomial(k, i)/k!*(k-i)^n))}
{a(n)=sum(k=0, n, numbpart(k)*Stirling2(n, k)*k!)}
(PARI) x='x+O('x^66); Vec( serlaplace( 1/eta(exp(x)-1) ) ) \\ Joerg Arndt, Sep 18 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Nov 03 2009
STATUS
approved