|
%I #49 Apr 25 2024 04:42:47
%S 1,1,1,2,6,21,92,469,2731,17985,131528,1059616,9319363,88833422,
%T 912393381,10043727089,117969438513,1472593659884,19467505081458,
%U 271704942613323,3992343851680466,61603531051030691,995949139457447931,16835191741257445589,296976010796327785530,5457427389713208932740,104308245862443706265341,2070461793105333579698992,42622090166454492404075635
%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 lim_{n->oo} (a(n)/n!)^(1/n)? Example: (a(500)/500!)^(1/500) = 0.7353325805...
%C 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
%H Paul D. Hanna, <a href="/A266328/b266328.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) = log(A(x)) + A'(x)/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..oo} 1/(exp(x) - x) dx = 0.7357819642916471998431353808137704665788888148929882090175... - _Vaclav Kotesovec_, Aug 21 2017
%F 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
%e 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! + ...
%e such that log(A(x)) = Integral B(x) dx
%e where
%e 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! + ...
%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 - 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! + ...
%e so that A( Integral 1/(exp(x) - x) dx ) = exp(x).
%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( A - 1 ) ) ); 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),", "))
%o (PARI) Vec( serlaplace( exp( serreverse( intformal( 1/(exp(x +x*O(x^25)) - x)))))) \\ _Joerg Arndt_, Dec 26 2023
%Y Cf. A006789, A089148, A266329.
%K nonn,changed
%O 0,4
%A _Paul D. Hanna_, Jan 24 2016
| 