A338562 Number of cyclic diagonal Latin squares of order 2n+1.
1, 0, 240, 20160, 0, 319334400, 62270208000, 0, 4979623993344000, 1946321606541312000, 0, 517040334777699532800000, 155112100433309859840000000, 0, 229885811837232250818134016000000, 230239482316981838896315760640000000, 0, 82665183731089159437333210700185600000000
Offset: 0
Examples
For n=3 there are 6 cyclic Latin squares of order 7 with the first row in ascending order, only 4 of them are diagonal: 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 2 3 4 5 6 0 1 3 4 5 6 0 1 2 4 5 6 0 1 2 3 5 6 0 1 2 3 4 4 5 6 0 1 2 3 6 0 1 2 3 4 5 1 2 3 4 5 6 0 3 4 5 6 0 1 2 6 0 1 2 3 4 5 2 3 4 5 6 0 1 5 6 0 1 2 3 4 1 2 3 4 5 6 0 1 2 3 4 5 6 0 5 6 0 1 2 3 4 2 3 4 5 6 0 1 6 0 1 2 3 4 5 3 4 5 6 0 1 2 1 2 3 4 5 6 0 6 0 1 2 3 4 5 4 5 6 0 1 2 3 5 6 0 1 2 3 4 4 5 6 0 1 2 3 3 4 5 6 0 1 2 2 3 4 5 6 0 1 and 4*7! = 20160 cyclic diagonal Latin squares.
Links
- Eduard I. Vatutin, Enumerating cyclic Latin squares and Euler totient function calculating using them, High-performance computing systems and technologies, 2020, Vol. 4, No. 2, pp. 40-48. (in Russian)
- Eduard I. Vatutin, Numerical formula between number of cyclic diagonal Latin squares and number of toroidal n-queens problem solutions getting by knight movement (in Russian).
- E. I. Vatutin, Special types of diagonal Latin squares, Cloud and distributed computing systems in electronic control conference, within the National supercomputing forum (NSCF - 2022). Pereslavl-Zalessky, 2023. pp. 9-18. (in Russian)
- Index entries for sequences related to Latin squares and rectangles.
Programs
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PARI
a(n)={my(m=2*n+1); m!*if(gcd(m, 6)==1, sum(k=1, m, gcd(k^3-k, m)==1))} \\ Andrew Howroyd, Apr 30 2021
Formula
a(n) = A123565(2*n+1) * (2*n+1)!.
a(n) = A370672(n) * (2n)!. - Eduard I. Vatutin, Mar 13 2024
Extensions
More terms from Andrew Howroyd, Apr 30 2021
Zero terms for even orders removed by Andrew Howroyd, May 26 2021
Comments