A002802 a(n) = (2*n+3)!/(6*n!*(n+1)!). (Formerly M4724 N2019)

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

Offset

0,2

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)

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.

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

Example

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

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 *)

Prog

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

Crossrefs

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

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

Column g=1 of A370235.

Sequence in context: A005567 A174434 A073391 * A101029 A122892 A125347

Adjacent sequences: A002799 A002800 A002801 * A002803 A002804 A002805

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved