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
Keywords
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
Programs
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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
Comments