A071094 Number of ways to tile hexagon of edges n, n, n+1, n, n, n+1 with diamonds of side 1.

1, 3, 50, 4116, 1646568, 3184461423, 29706808370096, 1335119245893326400, 288882990167192721013376, 300792059519113653077154558000, 1506680146887473588202049621593937500, 36298820709557430183399305000196605531250000, 4205446372314569673006362329181090368935937500000000, 2342761095072644391194625697884219372917666852341417500000000

Offset: 0

Author

N. J. A. Sloane, May 28 2002

Keywords

• nonn
• easy

References

• J. 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 page 261).

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..53
• J. Propp, Updated article
• J. Propp, Enumeration of matchings: problems and progress, in L. J. Billera et al. (eds.), New Perspectives in Algebraic Combinatorics

Crossrefs

Cf. A008793, A071095, A103905.

Programs

Mathematica

Table[Product[(i+j+k+2)/(i+j+k+1),{i,0,n-1},{j,0,n-1},{k,0,n}],{n,0,15}] (* Vaclav Kotesovec, Apr 26 2015 *)

PARI

a(n) = prod(k=0, n, binomial(2*n+k,n+k)/binomial(n+k,k)) \\ Michel Marcus, May 20 2013

Formula

a(n) = Product_{i=0..a-1} Product_{j=0..b-1} Product_{k=0..c-1} (i+j+k+2)/(i+j+k+1) with a=b=n, c=n+1.

a(n) = Product_{k=0..n} C(2n+k,n+k)/C(n+k,k). - Paul Barry, May 13 2008

a(n) ~ exp(1/12) * 3^(9*n^2/2 + 3*n + 5/12) / (A * n^(1/12) * 2^(6*n^2 + 4*n + 3/4)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Apr 26 2015