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A130079
a(n) = n - A130077(n), i.e., n minus the largest x such that 2^x divides A001623(n), the number of reduced three-line Latin rectangles.
2
3, 2, 4, 3, 3, 2, 4, 4, 4, 3, 5, 3, 2, 2, 5, 2, 4, 3, 5, 4, 4, 3, 5, 5, 5, 4, 6, 3, 4, 2, 6, 5, 4, 3, 5, 4, 4, 3, 5, 5, 5, 4, 6, 4, 3, 3, 6, 0, 5, 4, 6, 5, 5, 4, 6, 6, 6, 5, 7, 3, 5, 2, 7, 6, 4, 3, 5, 4, 4, 3, 5, 5, 5, 4, 6, 4, 1, 3, 6, 4, 5, 4, 6, 5, 5, 4, 6, 6, 6, 5, 7, 4, 5, 3, 7, 6, 5, 4
OFFSET
3,1
LINKS
John Riordan, A recurrence relation for three-line Latin rectangles, Amer. Math. Monthly, 59 (1952), pp. 159-162.
D. S. Stones, The many formulas for the number of Latin rectangles, Electron. J. Combin 17 (2010), A1.
D. S. Stones and I. M. Wanless, Divisors of the number of Latin rectangles, J. Combin. Theory Ser. A 117 (2010), 204-215.
PROG
(PARI) a001623(n) = n*(n-3)!*sum(i=0, n, sum(j=0, n-i, (-1)^j*binomial(3*i+j+2, j)<<(n-i-j)/(n-i-j)!)*i!);
a(n) = n - valuation(a001623(n), 2); \\ Michel Marcus, Oct 02 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Douglas Stones (dssto1(AT)student.monash.edu.au), May 06 2007
STATUS
approved