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a(n) = (2*n+3)!/(6*n!*(n+1)!).
(Formerly M4724 N2019)
43

%I M4724 N2019 #144 Mar 05 2024 14:34:02

%S 1,10,70,420,2310,12012,60060,291720,1385670,6466460,29745716,

%T 135207800,608435100,2714556600,12021607800,52895074320,231415950150,

%U 1007340018300,4365140079300,18839025605400,81007810103220,347176329013800,1483389769422600

%N a(n) = (2*n+3)!/(6*n!*(n+1)!).

%C For n >= 1 a(n) is also the number of rooted bicolored unicellular maps of genus 1 on n+2 edges. - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 20 2001

%C a(n) is half the number of (n+2) X 2 Young tableaux with a three horizontal walls between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021], A000984 for one horizontal wall, and A002457 for two. - _Michael Wallner_, Jan 31 2022

%C From _Robert Coquereaux_, Feb 12 2024: (Start)

%C Call B(p,g) the number of genus g partitions of a set with p elements (genus-dependent Bell number). Up to an appropriate shift the given sequence counts the genus 1 partitions of a set: we have a(n) = B(n+4,1), with a(0)= B(4,1)=1.

%C When shifted with an offset 4 (i.e., defining b(p)=a(p-4), which starts with 0,0,0,1,10,70, etc., and b(4)=1)), the given sequence reads b(p) = (1/( 2^4 3 )) * (1/( (2 p - 1) (2 p - 3))) * (1/(p - 4)!) * (2p)!/p!. In this form it appears as a generalization of Catalan numbers (that indeed count the genus 0 partitions).

%C Call C[p, [alpha], g] the number of partitions of a set with p elements, of cyclic type [alpha], and of genus g (genus g Faa di Bruno coefficients of type [alpha]). Up to an appropriate shift the given sequence also counts the genus 1 partitions of p=2k into k parts of length 2, which is then called C[2k, [2^k], 1], and we have a(n) = C[2k, [2^k], 1] for k=n+2.

%C The two previous interpretations of this sequence, leading to a(n) = B(n+4, 1) and to a(n) = C[2(n+2), [2^(n+2)], 1] are not related in any obvious way. (End)

%D C. Jordan, Calculus of Finite Differences. Röttig and Romwalter, Budapest, 1939; Chelsea, NY, 1965, p. 449.

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H T. D. Noe, <a href="/A002802/b002802.txt">Table of n, a(n) for n = 0..200</a>

%H Cyril Banderier and Michael Wallner, <a href="https://www.mat.univie.ac.at/~slc/wpapers/FPSAC2021/47.html">Young Tableaux with Periodic Walls: Counting with the Density Method</a>, Séminaire Lotharingien de Combinatoire, 85B (2021), Art. 47, 12 pp.

%H Robert Coquereaux and Jean-Bernard Zuber, <a href="https://arxiv.org/abs/2305.01100">Counting partitions by genus. A compendium of results</a>, arXiv:2305.01100 [math.CO], 2023. See pp. 4, 12.

%H Robert Coquereaux and Jean-Bernard Zuber, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL27/Coquereaux/coque5.html">Counting partitions by genus: a compendium of results</a>, Journal of Integer Sequences, Vol. 27 (2024), Article 24.2.6. See p. 9. See also arXiv:2305.01100, 2023. See pp. 9, 19.

%H R. Cori and G. Hetyei, <a href="http://arxiv.org/abs/1306.4628">Counting genus one partitions and permutations</a>, arXiv preprint arXiv:1306.4628 [math.CO], 2013.

%H R. Cori and G. Hetyei, <a href="https://doi.org/10.46298/dmtcs.2404">How to count genus one partitions</a>, FPSAC 2014, Chicago, Discrete Mathematics and Theoretical Computer Science (DMTCS), Nancy, France, 2014, 333-344.

%H Alain Goupil and Gilles Schaeffer, <a href="https://doi.org/10.1006/eujc.1998.0215">Factoring N-Cycles and Counting Maps of Given Genus</a>, Europ. J. Combinatorics (1998) 19 819-834.

%H T. R. S. Walsh and A. B. Lehman, <a href="http://dx.doi.org/10.1016/0095-8956(72)90056-1">Counting rooted maps by genus. I</a>, J. Comb. Theory B 13 (1972), 192-218 (Tab.1).

%H Liang Zhao and Fengyao Yan, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Zhao/zhao17.html">Note on Total Positivity for a Class of Recursive Matrices</a>, Journal of Integer Sequences, Vol. 19 (2016), Article 16.6.5.

%H <a href="/A007401/a007401_1.pdf">Notes</a>.

%F G.f.: (1 - 4*x)^(-5/2) = 1F0(5/2;;4x).

%F Asymptotic expression for a(n) is a(n) ~ (n+2)^(3/2) * 4^(n+2) / (sqrt(Pi) * 48).

%F a(n) = Sum_{a+b+c+d+e=n} f(a)*f(b)*f(c)*f(d)*f(e) with f(n) = binomial(2n, n) = A000984(n). - _Philippe Deléham_, Jan 22 2004

%F a(n-1) = (1/4)*Sum_{k=1..n} k*(k+1)*binomial(2*k, k). - _Benoit Cloitre_, Mar 20 2004

%F a(n) = A051133(n+1)/3 = A000911(n)/6. - _Zerinvary Lajos_, Jun 02 2007

%F From _Rui Duarte_, Oct 08 2011: (Start)

%F Also convolution of A000984 with A002697, also convolution of A000302 with A002457.

%F a(n) = ((2n+3)(2n+1)/(3*1)) * binomial(2n, n).

%F a(n) = binomial(2n+4, 4) * binomial(2n, n) / binomial(n+2, 2).

%F a(n) = binomial(n+2, 2) * binomial(2n+4, n+2) / binomial(4, 2).

%F a(n) = binomial(2n+4, n+2) * (n+2)*(n+1) / 12. (End)

%F D-finite with recurrence: n*a(n) - 2*(2*n+3)*a(n-1) = 0. - _R. J. Mathar_, Jan 31 2014

%F a(n) = 4^n*hypergeom([-n,-3/2], [1], 1). - _Peter Luschny_, Apr 26 2016

%F Boas-Buck recurrence: a(n) = (10/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+2, 2). See a comment there. - _Wolfdieter Lang_, Aug 10 2017

%F a(n) = (-4)^n*binomial(-5/2, n). - _Peter Luschny_, Oct 23 2018

%F Sum_{n>=0} 1/a(n) = 12 - 2*sqrt(3)*Pi. - _Amiram Eldar_, Oct 13 2020

%F E.g.f.: (1/12) exp(2 x) x^2 BesselI[2, 2 x]. - _Robert Coquereaux_, Feb 12 2024

%e G.f. = 1 + 10*x + 70*x^2 + 420*x^3 + 2310*x^4 + 12012*x^5 + 60060*x^6 + ...

%p seq(simplify(4^n*hypergeom([-n,-3/2], [1], 1)),n=0..25); # _Peter Luschny_, Apr 26 2016

%t Table[(2*n+3)!/(6*n!*(n+1)!), {n, 0, 25}] (* _Vladimir Joseph Stephan Orlovsky_, Dec 13 2008 *)

%o (PARI) {a(n) = if( n<0, 0, (2*n + 3)! / (6 * n! * (n+1)!))}; /* _Michael Somos_, Sep 16 2013 */

%o (PARI) {a(n) = 2^(n+3) * polcoeff( pollegendre(n+4), n) / 3}; /* _Michael Somos_, Sep 16 2013 */

%o (Magma) F:=Factorial; [F(2*n+3)/(6*F(n)*F(n+1)): n in [0..25]]; // _G. C. Greubel_, Jul 20 2019

%o (Sage) f=factorial; [f(2*n+3)/(6*f(n)*f(n+1)) for n in (0..25)] # _G. C. Greubel_, Jul 20 2019

%o (GAP) F:=Factorial;; List([0..25], n-> F(2*n+3)/(6*F(n)*F(n+1)) ); # _G. C. Greubel_, Jul 20 2019

%Y Cf. A035309, A000108 (for genus 0 maps), A046521 (third column).

%Y Cf. A000984, A000911, A002457, A002697, A051133.

%Y Column g=1 of A370235.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_