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%I #32 Oct 30 2022 09:00:05
%S 1,2,36,1920,210000,39191040,11181360960,4534378168320,
%T 2481970620729600,1764322560000000000,1580868516481859404800,
%U 1743505552795995891302400,2321376488366363008816435200,3671767205084150828189614080000
%N E.g.f. satisfies: A(x) = exp(x^2*A(x)) where A(x) = Sum_{n>=0} a(n)*x^(2n)/(2n)!.
%H Gheorghe Coserea, <a href="/A178949/b178949.txt">Table of n, a(n) for n = 0..100</a>
%H Vladimir Kruchinin and D. V. Kruchinin, <a href="http://arxiv.org/abs/1103.2582">Composita and their properties</a>, arXiv:1103.2582 [math.CO], 2011-2013.
%F a(n) = (n+1)^(n-1)*(2*n)!/n!.
%F E.g.f.: LambertW(-x^2)/(-x^2) = Sum_{n>=0} a(n)*x^(2n)/(2n)!.
%e E.g.f.: A(x) = 1 + 2*x^2/2! + 36*x^4/4! + 1920*x^6/6! +...
%e log(A(x)) = x^2 + 2*x^4/2! + 36*x^6/4! + 1920*x^8/6! +...
%t Table[(n+1)^(n-1)(2n)!/n!,{n,0,15}] (* _Harvey P. Dale_, Oct 21 2011 *)
%o (PARI) {a(n)=(n+1)^(n-1)*(2*n)!/n!}
%o (PARI) N=50; /* up to order N */
%o A(x)=sum(n=0,N-1, if (n%2==1,0, (n/2+1)^(n/2-1)/(n/2)!*x^n) )+O(x^N); /* e.g.f. */
%o v=Vec(serlaplace(A(x))) /* gives sequence as vector with interpolated zeros */
%o /* Now check that e.g.f. satisfies functional equation: */
%o A(x)-exp(x^2*A(x)) /* ==O(x^50) "==zero" */
%o (PARI)
%o N = 28; x = 'x + O('x^N); y = 'y; Fxy = exp(x^2*y) - y;
%o seq() = {
%o my(y0 = 1 + O('x^N), y1=0);
%o for (k = 1, N,
%o y1 = y0 - subst(Fxy, y, y0)/subst(deriv(Fxy, y), y, y0);
%o if (y1 == y0, break()); y0 = y1);
%o Vec(y0);
%o };
%o select(x->x, Vec(serlaplace(Ser(seq())))) \\ _Gheorghe Coserea_, Nov 30 2016
%K nonn
%O 0,2
%A _Vladimir Kruchinin_, Dec 31 2010
%E Edited by _Paul D. Hanna_, Jan 03 2011
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