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

3,2

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)].

Links

Table of n, a(n) for n=3..20.

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

Crossrefs

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

Sequence in context: A022738 A017269 A021079 * A233467 A164598 A073554

Adjacent sequences: A174560 A174561 A174562 * A174564 A174565 A174566

Keyword

nonn

Author

Vladimir Shevelev, Mar 22 2010

Extensions

More terms from William P. Orrick, Jul 25 2020

Status

approved