A266329 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, 3, 12, 62, 395, 2994, 26331, 263729, 2964845, 36975858, 506687604, 7568226163, 122388728056, 2130425343621, 39718373337525, 789613850257051, 16674806980716514, 372771700023167862, 8794945626017009781, 218392778569695964100, 5693513850197410142081, 155482323312112362743373, 4438621019461797437443233, 132210153223378852014571364, 4101859859297789141335079684, 132343983668857026899533814277

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 Limit (a(n)/n!)^(1/n) ? Example: (a(300)/300!)^(1/300) = 1.2409703...

Limit (a(n)/n!)^(1/n) = 1/Integral_{x=0..infinity} 1/(x + exp(x)) dx = 1.24008610649849766623949... - Vaclav Kotesovec, Aug 21 2017

Examples

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 12*x^3/3! + 62*x^4/4! + 395*x^5/5! + 2994*x^6/6! + 26331*x^7/7! + 263729*x^8/8! + 2964845*x^9/9! + 36975858*x^10/10! +...
such that log(A(x)) = Integral B(x) dx
where
B(x) = 1 + 2*x + 5*x^2/2! + 17*x^3/3! + 79*x^4/4! + 474*x^5/5! + 3468*x^6/6! + 29799*x^7/7! + 293528*x^8/8! + 3258373*x^9/9! + 40234231*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 - 2*x^2/2! + 7*x^3/3! - 37*x^4/4! + 261*x^5/5! - 2301*x^6/6! + 24343*x^7/7! - 300455*x^8/8! + 4238153*x^9/9! - 67255273*x^10/10! +...+ (-1)^(n-1)*A072597(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. A266328, A072597, A289739.

Programs

Mathematica

a[ n_] := a[n] = If[ n < 1, Boole[n == 0], Sum[ Binomial[n - 1, k - 1] a[n - k] Sum[ a[k - j], {j, k}], {k, n}]]; (* Michael Somos, Aug 08 2017 *)

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( 1 + A ) ) ); 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), ", "))

Formula

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

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

(2) A(x) = A'(x)/A(x) - log(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..infinity} 1/(x + exp(x)) dx = 1.2400861064984976662394901721056528110217273471501174317019052800276... - Vaclav Kotesovec, Aug 21 2017