A349509 a(n) is the denominator of binomial(n^3 + 6*n^2 - 6*n + 2, n^3 - 1)/n^3.

1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 1, 1, 1

Offset: 1

Author

Stefano Spezia, Nov 20 2021

Keywords

• nonn
• frac
• easy

Comments

a(n) is the denominator of an upper bound of the number of vertices of the polytope of the n X n X n stochastic tensors, or equivalently, of the number of Latin squares of order n, or equivalently, of the number of n X n X n line-stochastic (0,1)-tensors (see Chang et al. and Zhang et al.).

Conjecture: 1 and 3 are the only terms that appear in this sequence.

This conjecture is correct, see formula. - Kevin Ryde, Jul 01 2023

Links

• Kevin Ryde, Table of n, a(n) for n = 1..10000
• Haixia Chang, Vehbi Emrah Paksoy, and Fuzhen Zhang, Polytopes of Stochastic Tensors, Ann. Funct. Anal. 7(3): 386-393 (August 2016). arXiv:1608.03203 [math.CO], 2016. See p. 6.
• Kevin Ryde, PARI/GP Code and Notes
• Fuzhen Zhang and Xiao-Dong Zhang, Comparison of the upper bounds for the extreme points of the polytopes of line-stochastic tensors, arXiv:2110.12337 [math.CO], 2021. See p. 3.

Crossrefs

Cf. A000578, A068601.

Cf. A349506, A349507, A349508 (numerators), A349510, A349511, A349512.

Cf. A363739 (run lengths), A349929 (indices of 3's).

Programs

Mathematica

a[n_]:=Denominator[Binomial[n^3+6n^2-6n+2,n^3-1]/n^3]; Array[a,90]

PARI

\\ See links.

Python

from math import gcd, comb
def A349509(n): return n**3//gcd(comb(n*(n*(n + 6) - 6) + 2,6*n*(n-1)+3),n**3) # Chai Wah Wu, Dec 06 2021

Formula

A349508(n)/a(n) <= A349510(n) < A349511(n) < A349512(n) (see Corollary 7 in Zhang et al., 2021).

A349508(n)/a(n) ~ 2^(-4 + 6*n - 6*n^2)*3^(-7/2 + 6*n - 6*n^2)*e^(-75 + 233/n + 18*n + 6*n^2)*n^(-1 - 6*n + 6*n^2)/sqrt(Pi).

a(n) = 1 if n=1 or any x[i] + y[i] >= 3 where x and y are the ternary digits of n^3 = Sum x[i]*3^i and 6*n^2 - 6*n + 3 = Sum y[i]*3^i; and a(n) = 3 otherwise. - Kevin Ryde, Jul 01 2023