A266328 E.g.f. A(x) satisfies: A(x) = exp( Integral B(x) dx ) such that B(x) = exp(-x) * exp( Integral A(x) dx ), where the constant of integration is zero.

1, 1, 1, 2, 6, 21, 92, 469, 2731, 17985, 131528, 1059616, 9319363, 88833422, 912393381, 10043727089, 117969438513, 1472593659884, 19467505081458, 271704942613323, 3992343851680466, 61603531051030691, 995949139457447931, 16835191741257445589, 296976010796327785530, 5457427389713208932740, 104308245862443706265341, 2070461793105333579698992, 42622090166454492404075635

Offset: 0

Author

Paul D. Hanna, Jan 24 2016

Keywords

• nonn

Comments

Compare to: G(x) = exp( Integral G(x) dx ) when G(x) = 1/(1-x).

What is lim_{n->oo} (a(n)/n!)^(1/n)? Example: (a(500)/500!)^(1/500) = 0.7353325805...

Limit_{n->oo} (a(n)/n!)^(1/n) = 1/Integral_{x=0..oo} 1/(exp(x) - x) dx = 0.73578196429164719984313538... - Vaclav Kotesovec, Aug 21 2017

Examples

			E.g.f.: A(x) = 1 + x + x^2/2! + 2*x^3/3! + 6*x^4/4! + 21*x^5/5! + 92*x^6/6! + 469*x^7/7! + 2731*x^8/8! + 17985*x^9/9! + 131528*x^10/10! + ...
such that log(A(x)) = Integral B(x) dx
where
B(x) = 1 + x^2/2! + x^3/3! + 5*x^4/4! + 16*x^5/5! + 76*x^6/6! + 393*x^7/7! + 2338*x^8/8! + 15647*x^9/9! + 115881*x^10/10! + ...
and A(x) and B(x) satisfy:
(1) A(x) = B'(x)/B(x) + 1,
(2) B(x) = A'(x)/A(x),
(3) B(x) = A(x) - log(A(x)),
(4) log(A(x)) = Integral B(x) dx,
(5) log(B(x)) = Integral A(x) dx - x.
The Series Reversion of log(A(x)) equals Integral 1/(exp(x) - x) dx:
Integral 1/(exp(x) - x) dx  =  x - x^3/3! - x^4/4! + 5*x^5/5! + 19*x^6/6! - 41*x^7/7! - 519*x^8/8! - 183*x^9/9! + 19223*x^10/10! + ... + A089148(n-1)*x^n/n! + ...
so that A( Integral 1/(exp(x) - x) dx ) = exp(x).
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..300

Crossrefs

Cf. A006789, A089148, A266329.

Programs

PARI

{a(n) = my(A=1+x,B=1+x); for(i=0,n, A = exp( intformal( B + x*O(x^n) ) ); B = exp( intformal( A - 1 ) ) ); n!*polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

PARI

{a(n) = n! * polcoeff( exp( serreverse( intformal( 1/(exp(x +x*O(x^n)) - x) ) )), n)}
for(n=0,30,print1(a(n),", "))

PARI

Vec( serlaplace( exp( serreverse( intformal( 1/(exp(x +x*O(x^25)) - x)))))) \\ Joerg Arndt, Dec 26 2023

Formula

E.g.f. A(x) satisfies:

(1) A(x) = exp( Integral A(x) - log(A(x)) dx ).

(2) A(x) = log(A(x)) + A'(x)/A(x).

(3) log(A(x)) = exp(-x) * Integral exp(x)*A(x) dx.

(4) A(x) = exp( Series_Reversion( Integral 1/(exp(x) - x) dx ) ).

a(n) ~ c^(n+1) * n!, where c = 1/Integral_{x=0..oo} 1/(exp(x) - x) dx = 0.7357819642916471998431353808137704665788888148929882090175... - Vaclav Kotesovec, Aug 21 2017

Conjecture: a(n) = R(n-1, 0) for n > 0 with a(0) = 1 where R(n, q) = R(n-1, q+1) + Sum_{j=0..q-1} binomial(q+1, j)*R(n-1, j) for n > 0, q >= 0 with R(0, q) = 1 for q >= 0. - Mikhail Kurkov, Dec 26 2023