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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. 5

%I #16 Aug 21 2017 12:58:25

%S 1,1,3,12,62,395,2994,26331,263729,2964845,36975858,506687604,

%T 7568226163,122388728056,2130425343621,39718373337525,789613850257051,

%U 16674806980716514,372771700023167862,8794945626017009781,218392778569695964100,5693513850197410142081,155482323312112362743373,4438621019461797437443233,132210153223378852014571364,4101859859297789141335079684,132343983668857026899533814277

%N 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.

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

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

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

%H Paul D. Hanna, <a href="/A266329/b266329.txt">Table of n, a(n) for n = 0..300</a>

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

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

%F (2) A(x) = A'(x)/A(x) - log(A(x)).

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

%F (4) A(x) = exp( Series_Reversion( Integral 1/(exp(x) + x) dx ) ).

%F 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

%e 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! +...

%e such that log(A(x)) = Integral B(x) dx

%e where

%e 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! +...

%e and A(x) and B(x) satisfy:

%e (1) A(x) = B'(x)/B(x) - 1,

%e (2) B(x) = A'(x)/A(x),

%e (3) B(x) = A(x) + log(A(x)),

%e (4) log(A(x)) = Integral B(x) dx,

%e (5) log(B(x)) = Integral A(x) dx + x.

%e The Series Reversion of log(A(x)) equals Integral 1/(exp(x) + x) dx:

%e 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! +...

%e so that A( Integral 1/(exp(x) + x) dx ) = exp(x).

%t 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 *)

%o (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)}

%o for(n=0,30,print1(a(n),", "))

%o (PARI) {a(n) = n! * polcoeff( exp( serreverse( intformal( 1/(exp(x +x*O(x^n)) + x) ) )), n)}

%o for(n=0, 30, print1(a(n), ", "))

%Y Cf. A266328, A072597, A289739.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jan 24 2016

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Last modified March 28 16:58 EDT 2024. Contains 371254 sequences. (Running on oeis4.)