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.

Showing 1-7 of 7 results.

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

A287645 Minimum number of transversals in a diagonal Latin square of order n.

Original entry on oeis.org

1, 0, 0, 8, 3, 32, 7, 8, 68
Offset: 1

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Author

Eduard I. Vatutin, May 29 2017

Keywords

Comments

From Eduard I. Vatutin, Sep 20 2020: (Start)
Every diagonal Latin square is a Latin square, so 0 <= a(n) <= A287644(n) <= A090741(n).
A lower bound for odd n is A091323((n-1)/2) <= a(n). (End)
By definition, the main diagonal and antidiagonal of a diagonal Latin square are transversals, so a(n)>=2 for all n>=4 (the two diagonals are the same in the order 1 square and there are no diagonal Latin squares of orders 2 or 3). - Eduard I. Vatutin, Jun 13 2021
All cyclic diagonal Latin squares are diagonal Latin squares, so a(n) <= A348212((n-1)/2) for all orders n of which cyclic diagonal Latin squares exist. - Eduard I. Vatutin, Mar 25 2021
a(10) <= 128, a(11) <= 814, a(12) <= 448, a(13) <= 43093, a(14) <= 25720, a(15) <= 215721, a(16) <= 7465984. - Eduard I. Vatutin, Mar 11 2021, updated Feb 12 2025

Examples

			From _Eduard I. Vatutin_, Apr 24 2021: (Start)
For example, diagonal Latin square
  0 1 2 3
  3 2 1 0
  1 0 3 2
  2 3 0 1
has 4 diagonal transversals (see A287648)
  0 . . .   . 1 . .   . . 2 .   . . . 3
  . . 1 .   . . . 0   3 . . .   . 2 . .
  . . . 2   . . 3 .   . 0 . .   1 . . .
  . 3 . .   2 . . .   . . . 1   . . 0 .
and 4 not diagonal transversals
  0 . . .   . 1 . .   . . 2 .   . . . 3
  . 2 . .   3 . . .   . . . 0   . . 1 .
  . . 3 .   . . . 2   1 . . .   . 0 . .
  . . . 1   . . 0 .   . 3 . .   2 . . .
total 8 transversals. (End)
		

Crossrefs

Extensions

a(8) added by Eduard I. Vatutin, Oct 29 2017
a(9) added by Eduard I. Vatutin, Sep 20 2020

A309344 a(n) is the number of distinct numbers of transversals of order n Latin squares.

Original entry on oeis.org

1, 1, 1, 2, 2, 4, 36, 74
Offset: 1

Views

Author

Keywords

Comments

We found all transversals in the main class Latin square representatives of order n.
These results are based upon work supported by the National Science Foundation under the grants numbered DMS-1852378 and DMS-1560019.
For all spectra of even orders all known values included in them are divisible by 2. For all spectra of orders n=4k+2 all known values included in the corresponding spectra are divisible by 4. - Eduard I. Vatutin, Mar 01 2025
a(9)>=407, a(10)>=463, a(11)>=6437, a(12)>=23715. - Eduard I. Vatutin, added Mar 01 2025, updated Aug 14 2025

Examples

			For n=7, the number of transversals that an order 7 Latin square may have is 3, 7, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 36, 37, 41, 43, 45, 47, 55, 63, or 133. Hence there are 36 distinct numbers of transversals of order 7 Latin squares, so a(7)=36.
		

Crossrefs

Cf. A003090, A090741 (maximum number), A091323 (minimum number), A301371, A308853, A309088, A344105 (version for diagonal Latin squares).

Programs

  • MATLAB
    %This extracts entries from each column.  For an example, if
    %A=[1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16], and if list = (2, 1, 4),
    %this code extracts the second element in the first column, the first
    %element in the second column, and the fourth element in the third column.
    function [output] = extract(matrix,list)
    for i=1:length(list)
        output(i) = matrix(list(i),i);
    end
    end
    %Searches matrix to find transversal and outputs the transversal.
    function [output] = findtransversal(matrix)
    n=length(matrix);
    for i=1:n
        partialtransversal(i,1)=i;
    end
    for i=2:n
        newpartialtransversal=[];
        for j=1:length(partialtransversal)
            for k=1:n
                if (~ismember(k,partialtransversal(j,:)))&(~ismember(matrix(k,i),extract(matrix,partialtransversal(j,:))))
                    newpartialtransversal=[newpartialtransversal;[partialtransversal(j,:),k]];
                end
            end
        end
        partialtransversal=newpartialtransversal;
    end
    output=partialtransversal;
    end
    %Takes input of n^2 numbers with no spaces between them and converts it
    %into an n by n matrix.
    function [A] = tomatrix(input)
    n=sqrt(floor(log10(input))+2);
    for i=1:n^2
        temp(i)=mod(floor(input/(10^(i-1))),10);
    end
    for i=1:n
        for j=1:n
            A(i,j)=temp(n^2+1-(n*(i-1)+j));
        end
    end
    A=A+ones(n);
    end

A383684 Minimum number of transversals in an extended self-orthogonal diagonal Latin square of order n.

Original entry on oeis.org

1, 0, 0, 8, 15, 0, 23, 128, 133, 716
Offset: 1

Views

Author

Eduard I. Vatutin, May 05 2025

Keywords

Comments

A self-orthogonal diagonal Latin square (SODLS) is a diagonal Latin square orthogonal to its transpose. An extended self-orthogonal diagonal Latin square (ESODLS) is a diagonal Latin square that has an orthogonal diagonal Latin square from the same main class. SODLS is a special case of ESODLS.

Crossrefs

A387124 Maximum number of transversals in an extended self-orthogonal diagonal Latin square of order n.

Original entry on oeis.org

1, 0, 0, 8, 15, 0, 133, 384, 2241, 988
Offset: 1

Views

Author

Eduard I. Vatutin, Aug 17 2025

Keywords

Comments

A self-orthogonal diagonal Latin square (SODLS) is a diagonal Latin square orthogonal to its transpose. An extended self-orthogonal diagonal Latin square (ESODLS) is a diagonal Latin square that has an orthogonal diagonal Latin square from the same main class. SODLS is a special case of ESODLS.

Crossrefs

A383570 Number of transversals in pine Latin squares of order 4n.

Original entry on oeis.org

8, 384, 76032, 62881792
Offset: 1

Views

Author

Eduard I. Vatutin, Apr 30 2025

Keywords

Comments

A pine Latin square is a not necessarily canonical composite Latin square of order N=2*K formed from specially arranged cyclic Latin squares of order K.
By construction, pine Latin square is determined one-to-one by the cyclic square used, so number of pine Latin squares of order N is equal to number of cyclic Latin squares of order N/2.
All pine Latin squares are horizontally symmetric column-inverse Latin squares.
All pine Latin squares for selected order N are isomorphic one to another as Latin squares, so they have same properties (number of transversals, intercalates, etc.).
Pine Latin squares have interesting properties, for example, maximum known number of intercalates (see A383368 and A092237) for some orders N (at least N in {2, 4, 6, 10, 18}).
Pine Latin squares do not exist for odd orders because they must be horizontally symmetric.
Hypothesis: number of transversals in pine Latin squares of all orders N=4k+2 is zero (verified for orders N<=18).

Examples

			For order N=8 pine Latin square
  0 1 2 3 4 5 6 7
  1 2 3 0 7 4 5 6
  2 3 0 1 6 7 4 5
  3 0 1 2 5 6 7 4
  4 5 6 7 0 1 2 3
  5 6 7 4 3 0 1 2
  6 7 4 5 2 3 0 1
  7 4 5 6 1 2 3 0
has 384 transversals.
.
For order N=10 pine Latin square
  0 1 2 3 4 5 6 7 8 9
  1 2 3 4 0 9 5 6 7 8
  2 3 4 0 1 8 9 5 6 7
  3 4 0 1 2 7 8 9 5 6
  4 0 1 2 3 6 7 8 9 5
  5 6 7 8 9 0 1 2 3 4
  6 7 8 9 5 4 0 1 2 3
  7 8 9 5 6 3 4 0 1 2
  8 9 5 6 7 2 3 4 0 1
  9 5 6 7 8 1 2 3 4 0
has no transversals.
.
For order N=12 pine Latin square
  0 1 2 3 4 5 6 7 8 9 10 11
  1 2 3 4 5 0 11 6 7 8 9 10
  2 3 4 5 0 1 10 11 6 7 8 9
  3 4 5 0 1 2 9 10 11 6 7 8
  4 5 0 1 2 3 8 9 10 11 6 7
  5 0 1 2 3 4 7 8 9 10 11 6
  6 7 8 9 10 11 0 1 2 3 4 5
  7 8 9 10 11 6 5 0 1 2 3 4
  8 9 10 11 6 7 4 5 0 1 2 3
  9 10 11 6 7 8 3 4 5 0 1 2
  10 11 6 7 8 9 2 3 4 5 0 1
  11 6 7 8 9 10 1 2 3 4 5 0
has 76032 transversals.
		

Crossrefs

A336764 Maximum number of order 3 subsquares in a Latin square of order n.

Original entry on oeis.org

0, 0, 1, 0, 0, 4, 7
Offset: 1

Views

Author

Keywords

Comments

A subsquare of a Latin square is a submatrix (not necessarily consisting of adjacent entries) which is itself a Latin square. (I. M. Wanless, Latin Squares with One Subsquare, Wiley and Sons)

Crossrefs

Formula

a(3^n) = 9*a(3^(n-1)) + 27^(n-1) (conjectured).
Showing 1-7 of 7 results.