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A002005
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Number of rooted planar cubic maps with 2n vertices.
(Formerly M3646 N1483)
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11
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1, 4, 32, 336, 4096, 54912, 786432, 11824384, 184549376, 2966845440, 48855252992, 820675092480, 14018773254144, 242919827374080, 4261707069259776, 75576645116559360, 1353050213048123392, 24428493151359467520, 444370175232646840320, 8138178004138611179520
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listen;
history;
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OFFSET
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0,2
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COMMENTS
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Equivalently, number of rooted planar triangulations with 2n faces.
The September 2018 talk by Noam Zeilberger (see link to video) connects three topics (planar maps, Tamari lattices, lambda calculus) and eight sequences: A000168, A000260, A000309, A000698, A000699, A002005, A062980, A267827. - N. J. A. Sloane, Sep 17 2018
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REFERENCES
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R. C. Mullin, E. Nemeth and P. J. Schellenberg, The enumeration of almost cubic maps, pp. 281-295 in Proceedings of the Louisiana Conference on Combinatorics, Graph Theory and Computer Science. Vol. 1, edited R. C. Mullin et al., 1970.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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Mireille Bousquet-Mélou, Counting planar maps, coloured or uncoloured, 23rd British Combinatorial Conference, Jul 2011, Exeter, United Kingdom. 392, pp.1-50, 2011, London Math. Soc. Lecture Note Ser., hal-00653963. See p.13.
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FORMULA
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G.f.: (96*x - 1 + 2F1(-2/3, -1/3; 1/2; 432*x^2) - 96*x*2F1(-1/6, 1/6; 3/2; 432*x^2))/(192*x^2). - Benedict W. J. Irwin, Aug 07 2016
G.f. y(x) satisfies:
x*(1-432*x^2)*deriv(y,x) = 64*x^2*y^2 + (288*x^2 - 64*x - 1)*y + 72*x + 1.
0 = 64*x^3*y^3 + x*(1-96*x)*y^2 + (30*x-1)*y - 27*x + 1.
(End).
D-finite with recurrence (n+2)*(n+1)*a(n) -48*(3*n-2)*(3*n-4)*a(n-2)=0. - R. J. Mathar, Feb 08 2021
From Karol A. Penson and Katarzyna Gorska (katarzyna.gorska@ifj.edu.pl), Nov 02 2022: (Start)
a(n) = Integral_{x=0..12*sqrt(3)} x^n*W(x), where
W(x) = (T1(x) + T2(x)) / T3(x), and
T1(x) = -x^(2/3) * (108 + sqrt(3) * sqrt(432 - x^2));
T2(x) = 3^(1/6)*(36+sqrt(3)*sqrt(432-x^2))^(2/3) * (-432+x^2+36*sqrt(3)* sqrt(432-x^2)) / sqrt(432-x^2);
T3(x) = (128*3^(5/6)*Pi*x^(1/3)*(36+sqrt(3)*sqrt(432-x^2))^(1/3)).
This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem. Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0, with the singularity x^(-1/3), and for x > 0 is monotonically decreasing to zero at x = 12*sqrt(3). (End)
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MAPLE
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seq(2*8^n*binomial(n*3/2, n)/((n + 2)*(n + 1)), n = 0..19); # Peter Luschny, Nov 14 2022
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MATHEMATICA
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Table[2^(2 n + 1) (3 n)!!/((n + 2)! n!!), {n, 0, 20}] (* Vincenzo Librandi, Dec 28 2015 *)
CoefficientList[Series[(-1 + 96 z + Hypergeometric2F1[-2/3, -1/3, 1/2, 432z^2]- 96 z Hypergeometric2F1[-1/6, 1/6, 3/2, 432z^2])/(192 z^2), {z, 0, 10}], z] (* Benedict W. J. Irwin, Aug 07 2016 *)
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PROG
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(PARI) factorial2(n) = my(x = (2^(n\2)*(n\2)!)); if (n%2, n!/x, x);
a(n) = 2^(2*n+1)*factorial2(3*n)/((n+2)!*factorial2(n));
vector(20, i, a(i-1))
\\ test: y = Ser(vector(201, n, a(n-1))); x*(1-432*x^2)*y' == 64*x^2*y^2 + (288*x^2 - 64*x - 1)*y + 72*x + 1
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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