A092443 Sequence arising from enumeration of domino tilings of Aztec Pillow-like regions.

3, 12, 50, 210, 882, 3696, 15444, 64350, 267410, 1108536, 4585308, 18929092, 78004500, 320932800, 1318498920, 5409723510, 22169259090, 90751353000, 371125269900, 1516311817020, 6189965556060, 25249187564640, 102917884095000, 419218847880300, 1706543186909652

Offset: 1

Author

Christopher Hanusa (chanusa(AT)math.washington.edu), Mar 24 2004

Keywords

• easy
• nonn

Comments

The sequence 1, 3, 12, 50, ... is ((n+2)/2)*C(2n,n) with g.f. F(1/2,3;2;4x). - Paul Barry, Sep 18 2008

Examples

			a(3) = 5!/2!2! + 6!/3!3! = 50.
		

References

• James Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 13).

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..1659
• Sanjay Moudgalya, Abhinav Prem, Rahul Nandkishore, Nicolas Regnault, and B. Andrei Bernevig, Thermalization and its absence within Krylov subspaces of a constrained Hamiltonian, arXiv:1910.14048 [cond-mat.str-el], 2019.
• James Propp, Publications and Preprints.
• James Propp, Enumeration of matchings: problems and progress, in L. J. Billera et al. (eds.), New Perspectives in Algebraic Combinatorics, Cambridge University Press, Cambridge, 1999, pp. 255-291.
• Eric Rowland and Jason Wu, The entries of the Sinkhorn limit of an m X n matrix, arXiv:2409.02789 [math.NT], 2024. See p. 11.

Crossrefs

Cf. A000984, A001622, A001700, A002457.

Cf. A092437, A092438, A092439, A092440, A092441, A092442.

Programs

Mathematica

Array[Binomial[2 # + 1, # + 1] &[# - 1]*(# + 2) &, 22] (* Michael De Vlieger, Dec 17 2017 *)

MuPAD

combinat::catalan(n) *binomial(n+2,2) $ n = 1..22 // Zerinvary Lajos, Feb 15 2007

PARI

a(n) = (n+2)*binomial(2*n-1, n); \\ Altug Alkan, Dec 17 2017

Formula

a(n) = (2*n-1)!/((n-1)!)^2+(2*n)!/(n!)^2 = A002457(n-1) + A000984(n).

a(n) = (n+2)*A001700(n-1). - Vladeta Jovovic, Jul 12 2004

n*a(n) + (-7*n+4)*a(n-1) + 6*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 30 2012

From Amiram Eldar, Jan 27 2024: (Start)

Sum_{n>=1} 1/a(n) = 4*Pi*(11*sqrt(3)-3*Pi)/9 - 13.

Sum_{n>=1} (-1)^(n+1)/a(n) = 8*log(phi)*(13*sqrt(5)-30*log(phi))/5 - 11, where phi is the golden ratio (A001622). (End)

From Peter Bala, Aug 02 2024: (Start)

a(n) = 1/(n + 1)^2 * Sum_{k = 1..n+1} (k^3)*binomial(n+1, k)^2 = hypergeom([2, -n, -n], [1, 1], 1).

a(n) = 2*(n + 2)*(2*n - 1)/(n*(n + 1)) * a(n-1) with a(1) = 3. (End)