OFFSET
0,3
COMMENTS
The function P(x) = Product_{n>=1} 1/(1 - x^n/n) equals the e.g.f. of A007841, the number of factorizations of permutations of n letters into cycles in nondecreasing length order.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..100
EXAMPLE
E.g.f.: A(x) = 1 + x + 4*x^2/2! + 17*x^3/3! + 96*x^4/4! + 595*x^5/5! +...
such that A(x) = exp(1)*P(x) - Q(x), where
P(x) = 1/Product_{n>=1} (1 - x^n/n) = Sum_{n>=0} A007841(n)*x^n/n!, and
Q(x) = Sum_{n>=1} 1/Product_{k=1..n} (k - x^k).
More explicitly,
P(x) = 1/((1-x)*(1-x^2/2)*(1-x^3/3)*(1-x^4/4)*(1-x^5/5)*...);
Q(x) = 1/(1-x) + 1/((1-x)*(2-x^2)) + 1/((1-x)*(2-x^2)*(3-x^3)) + 1/((1-x)*(2-x^2)*(3-x^3)*(4-x^4)) + 1/((1-x)*(2-x^2)*(3-x^3)*(4-x^4)*(5-x^5)) +...
We can illustrate the initial terms a(n) in the following manner.
The coefficients in Q(x) = Sum_{n>=0} q(n)*x^n/n! begin:
q(0) = 1.7182818284590452...
q(1) = 1.7182818284590452...
q(2) = 4.1548454853771357...
q(3) = 12.901100113049497...
q(4) = 56.223782393706533...
q(5) = 285.72331242073065...
q(6) = 1801.2869693388211...
q(7) = 12727.542479311217...
q(8) = 104411.81066734227...
q(9) = 947120.40724315491...
and the coefficients in P(x) = 1/Product_{n>=1} (1 - x^n/n) begin:
A007841 = [1, 1, 3, 11, 56, 324, 2324, 18332, 167544, ...];
from which we can generate this sequence like so:
a(0) = exp(1)*1 - q(0) = 1;
a(1) = exp(1)*1 - q(1) = 1;
a(2) = exp(1)*3 - q(2) = 4;
a(3) = exp(1)*11 - q(3) = 17;
a(4) = exp(1)*56 - q(4) = 96;
a(5) = exp(1)*324 - q(5) = 595;
a(6) = exp(1)*2324 - q(6) = 4516;
a(7) = exp(1)*18332 - q(7) = 37104;
a(8) = exp(1)*167544 - q(8) = 351020; ...
PROG
(PARI) \p100 \\ set precision
{P=Vec(serlaplace(prod(k=1, 31, 1/(1-x^k/k +O(x^31))))); } \\ A007841
{Q=Vec(serlaplace(sum(n=1, 201, prod(k=1, n, 1./(k-x^k +O(x^31)))))); }
for(n=0, 30, print1(round(exp(1)*P[n+1]-Q[n+1]), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Oct 28 2014
STATUS
approved