A343867 - OEIS

%I #43 May 18 2024 14:52:08

%S 0,0,0,0,0,0,1560,0,34000,175104,0,22417824,313235960,0,83574857328,

%T 1729671003296

%N Number of semicyclic pandiagonal Latin squares of order 2*n+1 with the first row in ascending order.

%C Pandiagonal Latin squares exist only for odd orders not divisible by 3. All pandiagonal Latin squares for orders less than 13 are cyclic which are not counted by this sequence.

%C Semicyclic Latin squares are defined in the Atkin reference where the first nonzero term of this sequence is given. They are cyclic in a single direction. The direction can be horizontal or vertical or any other step such as a knights move.

%C Each symbol in a semicyclic Latin square occupies the same pattern of squares up to translation on the torus which in the case of a pandiagonal square is a solution to the toroidal n-queens problem.

%C For prime 2n+1, a(n) is a multiple of 2n+1.

%H A. O. L. Atkin, L. Hay, and R. G. Larson, <a href="https://doi.org/10.1016/0898-1221(83)90130-X">Enumeration and construction of pandiagonal Latin squares of prime order</a>, Computers & Mathematics with Applications, Volume. 9, Iss. 2, 1983, pp. 267-292.

%H Andrew Howroyd, <a href="/A343867/a343867.txt">PARI Program for Initial Terms</a>.

%H Natalia Makarova from Harry White, <a href="https://disk.yandex.ru/d/3KpNZgnH19a0Vg">1560 semi-cyclic Latin squares of order 13</a>.

%H Natalia Makarova from Harry White, <a href="https://disk.yandex.ru/d/myvKxPwu4q3gAg">34000 semi-cyclic Latin squares of order 17</a>.

%H Eduard I. Vatutin, <a href="http://evatutin.narod.ru/evatutin_ahl_pandiagonal_n19_175104_items_without_cyclic.zip">175104 semi-cyclic Latin squares of order 19</a>.

%F a(n) >= 4*(A071607(n) - A123565(2*n+1)).

%e The following is an example of an order 13 semicyclic square with a step of (1,4). This means moving down one row and across by 4 columns increases the cell value by 1 modulo 13. Symbols can be relabeled to give a square with the first row in ascending order.

%e    0 11  1  7  5  9  3 10  4  8  6 12  2

%e    9  7  0  3  1 12  2  8  6 10  4 11  5

%e   11  5 12  6 10  8  1  4  2  0  3  9  7

%e    1  4 10  8 12  6  0  7 11  9  2  5  3

%e   10  3  6  4  2  5 11  9  0  7  1  8 12

%e    8  2  9  0 11  4  7  5  3  6 12 10  1

%e    7  0 11  2  9  3 10  1 12  5  8  6  4

%e    6  9  7  5  8  1 12  3 10  4 11  2  0

%e    5 12  3  1  7 10  8  6  9  2  0  4 11

%e    3  1  5 12  6  0  4  2  8 11  9  7 10

%e   12 10  8 11  4  2  6  0  7  1  5  3  9

%e    2  6  4 10  0 11  9 12  5  3  7  1  8

%e    4  8  2  9  3  7  5 11  1 12 10  0  6

%e ...

%e a(12) = 4*(A071607(12) - A123565(25)) + 11240. - _Jim White_, Jul 22 2021

%e a(14) = 4*(A071607(14) - A123565(29)) + 91176. - _Jim White_, Jul 24 2021

%e a(15) = 4*(A071607(15) - A123565(31)) + 334800. - _Jim White_, Aug 03 2021

%o (PARI) \\ See Links

%Y Cf. A071607, A123565 (cyclic), A338620, A343868.

%K nonn,more

%O 0,7

%A _Andrew Howroyd_, May 08 2021

%E a(12)-a(15) from _Jim White_, Aug 03 2021