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A309283
Number of equivalence classes of X-based filling of diagonals in a diagonal Latin square of order n.
6
1, 1, 0, 0, 2, 2, 3, 3, 20, 20, 67, 67, 596, 596
OFFSET
0,5
COMMENTS
Used for getting strong canonical forms (SCFs) of the diagonal Latin squares and for fast enumerating of the diagonal Latin squares based on equivalence classes.
K1 = |C[1]|*f[1] + |C[2]|*f[2] + ... + |C[m]|*f[m],
K2 = K1 * n!,
where m = a(n), number of equivalence classes for X-based filling of diagonals in a diagonal Latin square of order n;
C[i], corresponding equivalence classes with cardinalities |C[i]|, 1 <= i <= m;
f[i], the number of diagonal Latin squares corresponds to the each item from equivalence class C[i], 1 <= i <= m;
K1 = A274171(n), number of diagonal Latin squares of order n with fixed first row;
K2 = A274806(n), number of diagonal Latin squares of order n.
For all t>0 a(2*t) = a(2*t+1). - Eduard I. Vatutin, Aug 21 2020
a(14) >= 5225, a(15) >= 5225. - Natalia Makarova, Sep 12 2020
The number of solutions in an equivalence class with the main diagonal in ascending order is at most 4*2^r*r! where r = floor(n/2). This maximum is achieved for orders n >= 10. - Andrew Howroyd, Mar 27 2023
LINKS
Stepan Kochemazov, Oleg Zaikin, Eduard Vatutin, and Alexey Belyshev, Enumerating Diagonal Latin Squares of Order Up to 9, J. Int. Seq. 23(1) (2020) 20.1.2.
Eduard I. Vatutin, About the a(2*t)=a(2*t+1) equality (in Russian).
Eduard I. Vatutin, A. D. Belyshev, N. N. Nikitina, and M. O. Manzuk, Use of X-based diagonal fillings and ESODLS CMS schemes for enumeration of main classes of diagonal Latin squares, Telecommunications, 2023, No. 1, pp. 2-16, DOI: 10.31044/1684-2588-2023-0-1-2-16 (in Russian).
Oleg Zaikin, Eduard Vatutin, and Curtis Bright, Enumerating Extended Self-Orthogonal Diagonal Latin Squares of Order Up to 10, J. Int. Seq. 28 (2025) 25.7.4. See p. 9.
FORMULA
a(n) = A338084(floor(n/2)).
EXAMPLE
For order n=4 there are a(4)=2 equivalence classes. First of them C[1] includes two X-based fillings of diagonals
0..1 0..2
.13. .10.
.02. .32.
2..3 1..3
and second C[2] also includes two X-based fillings of diagonals
0..1 0..2
.10. .13.
.32. .02.
2..3 1..3
It is easy to see that f[1] = 0 and f[2] = 1, so K1(4) = A274171(4) = 2*0 + 2*1 = 2 and K2(4) = A274806(4) = K1(4) * 4! = 2 * 24 = 48.
CROSSREFS
KEYWORD
nonn,more,hard
AUTHOR
Eduard I. Vatutin, Jul 06 2020
EXTENSIONS
a(11) added by Eduard I. Vatutin, Aug 21 2020
a(12)-a(13) by Harry White, added by Natalia Makarova, Sep 12 2020
a(0)=1 prepended by Andrew Howroyd, Oct 31 2020
STATUS
approved