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 A174563 Number of 3 X n Latin rectangles such that every element of the second row has the same cyclic order (see comment). 1
 1, 14, 133, 3300, 93889, 3391086, 148674191, 7796637196, 480640583751, 34370030511334, 2818294139246649, 262403744798653716, 27506121212584723373, 3222018028986227724702, 418998630100386520363619, 60138044879434564251209580, 9477043948863636836099726259, 1632099068624734991723488992214 (list; graph; refs; listen; history; text; internal format)
 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 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

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Last modified September 28 18:04 EDT 2021. Contains 347716 sequences. (Running on oeis4.)