A002802 a(n) = (2*n+3)!/(6*n!*(n+1)!).

1, 10, 70, 420, 2310, 12012, 60060, 291720, 1385670, 6466460, 29745716, 135207800, 608435100, 2714556600, 12021607800, 52895074320, 231415950150, 1007340018300, 4365140079300, 18839025605400, 81007810103220, 347176329013800, 1483389769422600

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

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

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

From Robert Coquereaux, Feb 12 2024: (Start)

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.

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).

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.

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)

Examples

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

References

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

Links

• T. D. Noe, Table of n, a(n) for n = 0..200
• Cyril Banderier and Michael Wallner, Young Tableaux with Periodic Walls: Counting with the Density Method, Séminaire Lotharingien de Combinatoire, 85B (2021), Art. 47, 12 pp.
• Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See pp. 4, 12.
• Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus: a compendium of results, Journal of Integer Sequences, Vol. 27 (2024), Article 24.2.6. See p. 9. See also arXiv:2305.01100, 2023. See pp. 9, 19.
• R. Cori and G. Hetyei, Counting genus one partitions and permutations, arXiv preprint arXiv:1306.4628 [math.CO], 2013.
• R. Cori and G. Hetyei, How to count genus one partitions, FPSAC 2014, Chicago, Discrete Mathematics and Theoretical Computer Science (DMTCS), Nancy, France, 2014, 333-344.
• Alain Goupil and Gilles Schaeffer, Factoring N-Cycles and Counting Maps of Given Genus, Europ. J. Combinatorics (1998) 19 819-834.
• T. R. S. Walsh and A. B. Lehman, Counting rooted maps by genus. I, J. Comb. Theory B 13 (1972), 192-218 (Tab.1).
• Liang Zhao and Fengyao Yan, Note on Total Positivity for a Class of Recursive Matrices, Journal of Integer Sequences, Vol. 19 (2016), Article 16.6.5.
• Notes.

Crossrefs

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

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

Column g=1 of A370235.

Programs

GAP

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

Magma

F:=Factorial; [F(2*n+3)/(6*F(n)*F(n+1)): n in [0..25]]; // G. C. Greubel, Jul 20 2019

Maple

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

Mathematica

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

PARI

{a(n) = if( n<0, 0, (2*n + 3)! / (6 * n! * (n+1)!))}; /* Michael Somos, Sep 16 2013 */

PARI

{a(n) = 2^(n+3) * polcoeff( pollegendre(n+4), n) / 3}; /* Michael Somos, Sep 16 2013 */

Sage

f=factorial; [f(2*n+3)/(6*f(n)*f(n+1)) for n in (0..25)] # G. C. Greubel, Jul 20 2019

Formula

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

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

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

a(n-1) = (1/4)*Sum_{k=1..n} k*(k+1)*binomial(2*k, k). - Benoit Cloitre, Mar 20 2004

a(n) = A051133(n+1)/3 = A000911(n)/6. - Zerinvary Lajos, Jun 02 2007

From Rui Duarte, Oct 08 2011: (Start)

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

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

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

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

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

D-finite with recurrence: n*a(n) - 2*(2*n+3)*a(n-1) = 0. - R. J. Mathar, Jan 31 2014

a(n) = 4^n*hypergeom([-n,-3/2], [1], 1). - Peter Luschny, Apr 26 2016

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

a(n) = (-4)^n*binomial(-5/2, n). - Peter Luschny, Oct 23 2018

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

E.g.f.: (1/12) exp(2 x) x^2 BesselI[2, 2 x]. - Robert Coquereaux, Feb 12 2024