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

0,3

Comments

A cyclic Latin square is a Latin square in which row i is obtained by cyclically shifting row i-1 by d places.

Equivalently, a Latin square is cyclic if and only if each row is a cyclic permutation of the first row and each column is a cyclic permutation of the first column.

Every cyclic diagonal Latin square is a cyclic Latin square, so a(n) <= A338522(2*n+1).

Cyclic diagonal Latin squares exist only for odd orders not divisible by 3. - Andrew Howroyd, May 26 2021

Links

Table of n, a(n) for n=0..17.

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.

Formula

a(n) = A123565(2*n+1) * (2*n+1)!.

a(n) = A370672(n) * (2n)!. - Eduard I. Vatutin, Mar 13 2024

Example

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.

Prog

(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

Crossrefs

Cf. A123565 (ordered first row), A338522, A341585 (main classes), A342306, A370672.

Sequence in context: A292075 A035841 A232428 * A342990 A342306 A205256

Adjacent sequences: A338559 A338560 A338561 * A338563 A338564 A338565

Keyword

nonn,easy

Author

Eduard I. Vatutin, Nov 02 2020

Extensions

More terms from Andrew Howroyd, Apr 30 2021

Zero terms for even orders removed by Andrew Howroyd, May 26 2021

Status

approved