A049505 a(n) = Product_{1<=i<=j<=n} (i+j+n-1)/(i+j-1), number of symmetric plane partitions in n-cube.

1, 2, 10, 112, 2772, 151008, 18076916, 4751252480, 2740612658576, 3468301123758080, 9627912669442441500, 58618653300361405440000, 782683432110638830001250000, 22916694891747599820616089600000, 1471328419282772010324439370939640000

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

The first printing of the Bressoud book states that the formula Product_{1<=i<=j<=n} (i+j+n-1)/(i+j-1) in Eq. (6.8) is the number of totally symmetric plane partitions. This is wrong, although it does produce the current sequence. For the correct formula for the number of totally symmetric plane partitions see A005157.

References

• D. M. Bressoud, Proofs and Confirmations, Camb. Univ. Press, 1999; Eq. (6.8), p. 198.

Links

• T. D. Noe, Table of n, a(n) for n = 0..40
• P. J. Taylor, Counting distinct dimer hex tilings, Preprint, 2015.
• P. J. Taylor, Counting distinct dimer hex tilings, arXiv:1602.06796 [math.CO], 2016.

Crossrefs

Main diagonal of array A102539.

Main diagonal of array in A073165.

Cf. A005157, A008793.

Programs

Maple

A049505 := proc(n) local i,j; mul(mul((i+j+n-1)/(i+j-1),j=i..n),i=1..n); end;

Mathematica

a[n_] := Product[ ((2i-2)!*(i+2n-1)!)/((i+n-1)!*(2i+n-2)!), {i, 1, n}]; Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Jun 22 2012, after PARI *)

PARI

a(n)=prod(i=1,n,prod(j=i,n,(i+j+n-1)/(i+j-1)))

Formula

a(n) = Product_{1<=i<=j<=n} (i+j+n-1)/(i+j-1).

a(n) = Product_{i=1..n} (((2*i-2)!*(i+2*n-1)!)/((i+n-1)!*(2*i+n-2)!)). - Jean-François Alcover, Jun 22 2012

a(n) = Product_{i=1..n} (binomial((i-1) + 2*n, n)/binomial(n + 2*(i-1), n)). - Olivier Gérard, Feb 25 2015

a(n) ~ exp(1/24) * 3^(9*n^2/4 + 3*n/4 - 1/24) / (A^(1/2) * n^(1/24) * 2^(3*n^2 + n/2 + 1/8)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Mar 01 2015

From Peter Bala, Feb 15 2023: (Start)

a(n+1) = m(n)*a(n) where m(n) = ((3*n + 2)!*n!^2)/((2 n)!*(2 n + 1)!^2) * Product_{i = 1..n} n + 2*i for n >= 1.

Conjectures:

1) the supercongruence a(p) == 2^((p+1)/2) (mod p^3) holds for all primes p >= 3 (checked up to p = 1009).

2) the congruence a(p^2) == (-1)^((p^2-1)/8)*a(p)^(p^2-p+1) (mod p^3) holds for all primes p >= 3 (checked up to p = 89).

3) the congruence a(p^3) == a(p^2)^((p^3-p^2+2)/2) (mod p^3) holds for all primes p >= 2. (End)

Extensions

Edited by N. J. A. Sloane, Jun 30 2013; codes and formula checked by N. J. A. Sloane and Olivier Gérard