A174563 Number of 3 X n Latin rectangles such that every element of the second row has the same cyclic order (see comment).

1, 14, 133, 3300, 93889, 3391086, 148674191, 7796637196, 480640583751, 34370030511334, 2818294139246649, 262403744798653716, 27506121212584723373, 3222018028986227724702, 418998630100386520363619, 60138044879434564251209580, 9477043948863636836099726259, 1632099068624734991723488992214

Offset: 3

Author

Vladimir Shevelev, Mar 22 2010

Keywords

• nonn

Comments

We say that an element alpha_i of a permutation alpha of {1,2,...,n} has cyclic order k if it belongs to a cycle of length k of alpha. If every cycle of alpha has length k, then k|n.

References

• V. S. Shevelev, Reduced Latin rectangles and square matrices with equal row and column sums, Diskr. Mat. [Journal published by the Academy of Sciences of Russia], 4 (1992), 91-110.
• V. S. Shevelev, Modern enumeration theory of permutations with restricted positions, Diskr. Mat., 1993, 5, no.1, 3-35 (Russian) [English translation in Discrete Math. and Appl., 1993, 3:3, 229-263 (pp. 255-257)].

Crossrefs

Cf. A000179, A000186, A000296, A094047, A174556, A174560, A174561.

Formula

Let G_n = A000296(n) = n! * Sum_{2*k_2+...+n*k_n=n, k_i>=0} Product_{i=2,...,n} (k_i!*i!^k_i)^(-1). Then a(n) = Sum_{k=0,...,floor(n/2)} binomial(n,k) * G_k * G_(n-k) * u_(n-2*k), where u(n) = A000179(n). - Vladimir Shevelev, Mar 30 2016

Extensions

More terms from William P. Orrick, Jul 25 2020