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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
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