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A214377 G.f. A(x) satisfies A(x) = 1 + 4*x*A(x)^(3/2).

%I A214377 #47 Jul 30 2025 05:15:19
%S A214377 1,4,24,168,1280,10296,86016,739024,6488064,57946200,524812288,
%T A214377 4808643120,44493176832,415146189360,3901709352960,36902658748320,
%U A214377 350980432461824,3354743017001880,32207616155320320,310446853795570800,3003167577200394240,29146910264615460240
%N A214377 G.f. A(x) satisfies A(x) = 1 + 4*x*A(x)^(3/2).
%C A214377 Radius of convergence of g.f. A(x) is r = sqrt(3)/18 where A(r) = 3.
%D A214377 Bruce C. Berndt, Ramanujan's Notebooks Part I, Springer Verlag, 1985, p. 305.
%H A214377 G. C. Greubel, <a href="/A214377/b214377.txt">Table of n, a(n) for n = 0..975</a>
%H A214377 Alois Panholzer, <a href="https://arxiv.org/abs/2007.14676">Parking function varieties for combinatorial tree models</a>, arXiv:2007.14676 [math.CO], 2020.
%F A214377 a(n) = 2^(2*n+1) * binomial(3*n/2, n) / (n+2).
%F A214377 Self-convolution of A078531.
%F A214377 A(-x) = 1/x * series reversion( x*(2*x + sqrt(1 + 4*x^2))^2 ) follows from the Lagrange inversion formula and equation 1.13, p. 305 in Berndt. Cf. A098616. - _Peter Bala_, Oct 19 2015
%F A214377 a(n) ~ 2^(n + 1/2) * 3^(3*n/2 + 1/2) / (sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Oct 20 2015
%F A214377 G.f.: 4*x*(sin(asin(216*x^2-1)/3)/(6*x)+1/(12*x))^3+1. - _Vladimir Kruchinin_, Sep 30 2022
%F A214377 From _Paul D. Hanna_, Feb 03 2023: (Start)
%F A214377 G.f. A(x) satisfies: A(x) = 1/(1 - 4*x*A(x)^(1/2)).
%F A214377 G.f. A(x) satisfies: A(x) = B(x*A(x)) where B(x) = A(x/B(x)) = 1 + 8*x^2 + 4*x*sqrt(1 + 4*x^2) is the g.f. of A135863. (End)
%F A214377 From _Karol A. Penson_, Mar 23 2024: (Start)
%F A214377 G.f. = 1/(48*z^2) - 2F1([-2/3, -1/3], [-1/2], 108*z^2)/(48*z^2) + 4*z*2F1([5/6, 7/6],[5/2],108*z^2);                                     a(n) = Integral_{x=0..sqrt(108)} x^n*W(x), with W(x) = ((72*(g1(x) - g2(x)) + x^2*(-g1(x) + g2(x)) + 4*sqrt(-3*x^2 + 324)*(g1(x) + g2(x)))*3^(1/6))/(96*Pi*(x^2)^(5/6)),
%F A214377   where g1(x) = (18 - sqrt(324 - 3*x^2))^(2/3) and
%F A214377   g2(x) = (18 + sqrt(324 - 3*x^2))^(2/3).
%F A214377 This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem on x = (0, sqrt(108)). Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0, with singularity x^(-1/3), and for x > 0 is monotonically decreasing to zero at x = sqrt(108). For x -> sqrt(108), W'(x) tends to -infinity. (End)
%F A214377 G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^2). - _Seiichi Manyama_, Jun 17 2025
%F A214377 D-finite with recurrence -(n+2)*(n-1)*a(n) +12*(3*n-2)*(3*n-4)*a(n-2)=0. - _R. J. Mathar_, Jul 30 2025
%e A214377 G.f.: A(x) = 1 + 4*x + 24*x^2 + 168*x^3 + 1280*x^4 + 10296*x^5 + 86016*x^6 + ... where A(x) = 1 + 4*x*A(x)^(3/2).
%e A214377 Radius of convergence: r = 1/(2*3^(3/2)) = 0.09622504486...
%e A214377 Related expansions:
%e A214377 A(x)^(3/2) = 1 + 6*x + 42*x^2 + 320*x^3 + 2574*x^4 + 21504*x^5 + 184756*x^6 + ...
%e A214377 A(x)^(1/2) = 1 + 2*x + 10*x^2 + 64*x^3 + 462*x^4 + 3584*x^5 + 29172*x^6 + ... + A078531(n)*x^n + ...
%t A214377 Table[4^n*Binomial[3*n/2, n]*2/(n+2), {n, 0, 20}] (* _Vaclav Kotesovec_, Oct 20 2015 *)
%o A214377 (PARI) {a(n)=4^n*binomial(3/2*n,n)/(n/2+1)}
%o A214377 for(n=0,30,print1(a(n),", "))
%Y A214377 Cf. A078531, A135863, A214553, A098616.
%K A214377 nonn
%O A214377 0,2
%A A214377 _Paul D. Hanna_, Jul 14 2012