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Expansion of e.g.f. A(x) satisfying A(x) = A( x^2*exp(x) ) / x, with A(0) = 0.
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%I #15 Jan 29 2024 11:58:49

%S 1,2,9,52,425,4206,48307,632360,9444465,159240250,2983729331,

%T 61300668012,1367054727337,32844312889766,845234187028155,

%U 23190947446000336,675895337644401377,20863665943202969586,680448552777544884643,23395823324931227353940,846248620848062865320601

%N Expansion of e.g.f. A(x) satisfying A(x) = A( x^2*exp(x) ) / x, with A(0) = 0.

%C Limit (a(n)/n!)^(1/n) = 1/w where w*exp(w) = 1 and w = LambertW(1) = 0.567143290409783872999968... (cf. A030178).

%H Paul D. Hanna, <a href="/A369090/b369090.txt">Table of n, a(n) for n = 1..300</a>

%F E.g.f. A(x) = Sum_{n>=1} a(n)*x^n/n! satisfies the following formulas.

%F (1) A(x) = A(x^2*exp(x)) / x.

%F (2) R(x*A(x)) = x^2*exp(x), where R(A(x)) = x.

%F (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.

%F (4) A(x) = x * exp(L(x)), where L(x) = x + L(x^2*exp(x)) is the e.g.f. of A369091.

%e 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! + ...

%e RELATED SERIES.

%e The expansion of the logarithm of A(x)/x starts

%e 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! + ...

%e and equals the sum of all iterations of the function x^2*exp(x).

%e Let R(x) be the series reversion of A(x),

%e 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! + ...

%e then R(x) and e.g.f. A(x) satisfy:

%e (1) R( A(x) ) = x,

%e (2) R( x*A(x) ) = x^2 * exp(x).

%e GENERATING METHOD.

%e Let F(n) equal the n-th iteration of x^2*exp(x), so that

%e F(0) = x,

%e F(1) = x^2 * exp(x),

%e F(2) = x^4 * exp(2*x) * exp(x^2*exp(x)),

%e F(3) = x^8 * exp(4*x) * exp(2*x^2*exp(x)) * exp(F(2)),

%e F(4) = x^16 * exp(8*x) * exp(4*x^2*exp(x)) * exp(2*F(2)) * exp(F(3)),

%e F(5) = x^32 * exp(16*x) * exp(8*x^2*exp(x)) * exp(4*F(2)) * exp(2*F(3)) * exp(F(4)),

%e ...

%e F(n+1) = F(n)^2 * exp(F(n))

%e ...

%e Then the e.g.f. A(x) equals

%e A(x) = x * exp(F(0) + F(1) + F(2) + F(3) + ... + F(n) + ...).

%e equivalently,

%e 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))) + ...).

%o (PARI) {a(n) = my(A=x); for(i=0, #binary(n),

%o A = subst(A, x, x^2*exp(x +x^2*O(x^n)) )/x ); n! * polcoeff(H=A, n)}

%o for(n=1, 30, print1(a(n), ", "))

%Y Cf. A369091, A369550 (a(n)/n), A030178.

%Y Cf. A367390.

%K nonn

%O 1,2

%A _Paul D. Hanna_, Jan 26 2024