cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A092237 Maximum number of intercalates in a Latin square of order n.

Original entry on oeis.org

0, 1, 0, 12, 4, 27, 42, 112, 72
Offset: 1

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Author

Richard Bean, Feb 17 2004

Keywords

Comments

An intercalate is a 2 X 2 subsquare of a Latin square. a(10) >= 125, a(11) >= 172, a(12) >= 324.
a(13) >= 208, a(14) >= 391, a(15) >= 630, a(16) >= 960, a(17) >= 736, a(18) >= 729, a(19) >= 472, a(20) >= 1500, a(21) >= 884, a(22) >= 1497, a(23) >= 983, a(24) >= 1872, a(25) >= 1700, a(26) >= 2197, a(27) >= 648, a(28) >= 2940. - Eduard I. Vatutin, Mar 02 2025
If, in theory, all unordered pairs of rows and columns form intercalate in their intersection, total number of intercalates will be (n*(n-1))^2, so a(n) <= (n*(n-1))^2, a(n) is asymptotically less than O(n^4). In practice a(n) << (n*(n-1))^2. - Eduard I. Vatutin, Mar 11 2025
a(2^n-1) = 42*A006096(n) for n > 2. - Eduard I. Vatutin, Apr 23 2025
Due to existence of the pine Latin squares for even orders N=2n, a(2n) >= A383368(n). Pine Latin squares exist for all even orders, so a(N) >= (2k)^2 * (2k + 1) for N=4k and a(N) >= (2k+1)^3 for N=4k+2. Therefore, asymptotically maximum number of intercalates in Latin squares of even orders N greater or equal than o(k1*N^3), where k1 = 1/8. - Eduard I. Vatutin, Apr 30 2025
From Eduard I. Vatutin, Jul 01 2025: (Start)
Table showing minimums and maximums:
order | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
min number of intercalates | 0 1 0 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
max number of intercalates | 0 1 0 12 4 27 42 112 72 >=125 >=172 >=324 >=208 >=391 >=630 >=960 >=736 >=729 >=472 (this sequence)
.
order | 20 21 22 23 24 25 26 27 28 29
min number of intercalates | 0 0 0 0 0 0 <=15 0 <=1 0
max number of intercalates | >=1500 >=884 >=1497 >=983 >=1872 >=1700 >=2197 >=648 >=2940 ? (this sequence)
(End)

References

  • I. Wanless, Private communication, 2003.

Crossrefs

Formula

If n is a power of 2, a(n) = n^2*(n-1)/4 = A016152(log2(n)); if n is one less than a power of 2, a(n) = n*(n-1)*(n-3)/4 = A006096(log2(n+1))*42. - updated by Eduard I. Vatutin, Jun 28 2025