A369090 Expansion of e.g.f. A(x) satisfying A(x) = A( x^2*exp(x) ) / x, with A(0) = 0.
1, 2, 9, 52, 425, 4206, 48307, 632360, 9444465, 159240250, 2983729331, 61300668012, 1367054727337, 32844312889766, 845234187028155, 23190947446000336, 675895337644401377, 20863665943202969586, 680448552777544884643, 23395823324931227353940, 846248620848062865320601
Offset: 1
Keywords
Examples
E.g.f.: A(x) = x + 2*x^2/2! + 9*x^3/3! + 52*x^4/4! + 425*x^5/5! + 4206*x^6/6! + 48307*x^7/7! + 632360*x^8/8! + 9444465*x^9/9! + 159240250*x^10/10! + ... RELATED SERIES. The expansion of the logarithm of A(x)/x starts log(A(x)/x) = x + 2*x^2/2! + 6*x^3/3! + 36*x^4/4! + 260*x^5/5! + 2190*x^6/6! + 21882*x^7/7! + 268856*x^8/8! + ... + A369091(n)*x^n/n! + ... and equals the sum of all iterations of the function x^2*exp(x). Let R(x) be the series reversion of A(x), R(x) = x - 2*x^2/2! + 3*x^3/3! + 8*x^4/4! - 155*x^5/5! + 1464*x^6/6! - 7931*x^7/7! - 65360*x^8/8! + 2742345*x^9/9! + ... then R(x) and e.g.f. A(x) satisfy: (1) R( A(x) ) = x, (2) R( x*A(x) ) = x^2 * exp(x). GENERATING METHOD. Let F(n) equal the n-th iteration of x^2*exp(x), so that F(0) = x, F(1) = x^2 * exp(x), F(2) = x^4 * exp(2*x) * exp(x^2*exp(x)), F(3) = x^8 * exp(4*x) * exp(2*x^2*exp(x)) * exp(F(2)), F(4) = x^16 * exp(8*x) * exp(4*x^2*exp(x)) * exp(2*F(2)) * exp(F(3)), F(5) = x^32 * exp(16*x) * exp(8*x^2*exp(x)) * exp(4*F(2)) * exp(2*F(3)) * exp(F(4)), ... F(n+1) = F(n)^2 * exp(F(n)) ... Then the e.g.f. A(x) equals A(x) = x * exp(F(0) + F(1) + F(2) + F(3) + ... + F(n) + ...). equivalently, A(x) = x * exp(x + x^2*exp(x) + x^4*exp(2*x)*exp(x^2*exp(x)) + x^8*exp(4*x)*exp(2*x^2*exp(x)) * exp(x^4*exp(2*x)*exp(x^2*exp(x))) + ...).
Links
- Paul D. Hanna, Table of n, a(n) for n = 1..300
Programs
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PARI
{a(n) = my(A=x); for(i=0, #binary(n), A = subst(A, x, x^2*exp(x +x^2*O(x^n)) )/x ); n! * polcoeff(H=A, n)} for(n=1, 30, print1(a(n), ", "))
Formula
E.g.f. A(x) = Sum_{n>=1} a(n)*x^n/n! satisfies the following formulas.
(1) A(x) = A(x^2*exp(x)) / x.
(2) R(x*A(x)) = x^2*exp(x), where R(A(x)) = x.
(3) A(x) = x * exp( Sum_{n>=0} F(n) ), where F(0) = x, and F(n+1) = F(n)^2 * exp(F(n)) for n >= 0.
(4) A(x) = x * exp(L(x)), where L(x) = x + L(x^2*exp(x)) is the e.g.f. of A369091.
Comments