A347472 -id:A347472 - cp's OEIS frontend

A001197 Zarankiewicz's problem k_2(n).

4, 7, 10, 13, 17, 22, 25, 30, 35, 40, 46, 53, 57, 62, 68, 75, 82, 89, 97, 106, 109, 116, 123

Offset: 2

a(n) is the minimum number k_2(n) such that any n X n matrix having that number of nonzero entries has a 2 X 2 submatrix with only nonzero entries. - M. F. Hasler, Sep 28 2021

a(n) <= (1 + sqrt(4*n-3))*n/2 + 1. - Max Alekseyev, Apr 03 2022

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 291.
R. K. Guy, A problem of Zarankiewicz, in P. Erdős and G. Katona, editors, Theory of Graphs (Proceedings of the Colloquium, Tihany, Hungary), Academic Press, NY, 1968, pp. 119-150.
Richard J. Nowakowski, Zarankiewicz's Problem, PhD Dissertation, University of Calgary, 1978, page 202.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

R. K. Guy, A problem of Zarankiewicz, Research Paper No. 12, Dept. of Math., Univ. Calgary, Jan. 1967. [Annotated and scanned copy, with permission]
R. K. Guy, A many-facetted problem of Zarankiewicz, Lect. Notes Math. 110 (1969), 129-148.

Cf. A001198 (k_3), A072567, A339635, A347472, A350296.
Cf. also A006613 - A006626 (other sizes, in particular A006616 = k_4).
Main diagonal of A376167.

a(n) = A072567(n) + 1. - Rob Pratt, Aug 09 2019

a(n) = n^2 - A347472(n) = n^2 - A350296(n) + 1. - Andrew Howroyd, Dec 26 2021

Nowakowski's thesis, directed by Guy, corrected Guy's value for a(15) and supplied a(16)-a(21) entered by Don Knuth, Aug 13 2014
a(1) deleted following a suggestion from M. F. Hasler. - N. J. A. Sloane, Oct 22 2021
a(22)-a(24) from Jeremy Tan, Jan 23 2022

A347473 Maximum number of nonzero entries allowed in an n X n matrix to ensure there is a 3 X 3 zero submatrix.

Original entry on oeis.org

0, 2, 4, 9, 15, 21, 31, 39, 51, 63, 76, 90, 104, 127

Offset: 3

M. F. Hasler, Sep 28 2021

Related to Zarankiewicz's problem k_3(n) (cf. A001198 and other crossrefs), which asks the converse: how many 1's must be in an n X n {0,1}-matrix in order to guarantee the existence of an all-ones 3 X 3 submatrix. This complementarity leads to the given formula which was used to compute the given values.

			For n = 3, there must not be any nonzero entry in an n X n = 3 X 3 matrix, if one wants a 3 X 3 zero submatrix, whence a(3) = 0.
For n = 4, having at most 2 nonzero entries in the n X n matrix guarantees that there is a 3 X 3 zero submatrix (delete, e.g., the row which has the first nonzero entry, then the column with the remaining nonzero entry, if any), but if one allows 3 nonzero entries and they are placed on the diagonal, then there is no 3 X 3 zero submatrix. Hence, a(4) = 2.

Cf. A001198 (k_3(n)), A001197 (k_2(n)), A006613 - A006626 (other sizes of the main matrix and the submatrix).
Cf. A347472, A347474 (analog for 2 X 2 resp. 4 X 4 zero submatrix).
Cf. A339635, A350237, A350304.

a(n) = n^2 - A001198(n).
a(n) = A350237(n) - 1. - Andrew Howroyd, Dec 24 2021
a(n) = n^2 - A350304(n) - 1. - Max Alekseyev, Oct 31 2022

a(11)-a(13) from Andrew Howroyd, Dec 24 2021

a(14)-a(16) computed from A350237 by Max Alekseyev, Apr 01 2022, Oct 31 2022

A347474 Maximum number of nonzero entries allowed in an n X n matrix to ensure there is a 4 X 4 zero submatrix.

Original entry on oeis.org

0, 2, 4, 6, 12, 19, 25, 34, 43, 51

Offset: 4

M. F. Hasler, Sep 28 2021

Related to Zarankiewicz's problem k_4(n) (cf. A006616 and other crossrefs) which asks the converse: how many 1's must be in an n X n {0,1}-matrix in order to guarantee the existence of an all-ones 4 X 4 submatrix. This complementarity leads to the given formula which was used to compute the given values.

			For n < 4, there is no solution, since there cannot be a 4 X 4 submatrix in a matrix of smaller size.
For n = 4, there must not be any nonzero entry in an n X n = 4 X 4 matrix, if one wants a 4 X 4 zero submatrix, whence a(4) = 0.
For n = 5, having at most 2 nonzero entries in the n X n matrix guarantees that there is a 4 X 4 zero submatrix (delete, e.g., the row with the first nonzero entry, then the column with the second nonzero entry, if any), but if one allows 3 nonzero entries and they are placed on the diagonal, then there is no 4 X 4 zero submatrix. Hence, a(5) = 2.

Cf. A347472, A347473 (analog for 2 X 2 resp. 3 X 3 zero submatrix).
Cf. A006616 (k_4(n)), A001198 (k_3(n)), A001197 (k_2(n)), A006613 - A006626.
Cf. A339635.

a(n) = n^2 - A006616(n).

a(n) = A339635(n,4) - 1. - Andrew Howroyd, Dec 23 2021

a(9)-a(12) from Andrew Howroyd, Dec 23 2021

a(13) computed from A006616 by Max Alekseyev, Feb 02 2024

A350296 Minimum number of 1's in an n X n binary matrix with no zero 2 X 2 submatrix.

Original entry on oeis.org

0, 1, 3, 7, 13, 20, 28, 40, 52, 66, 82, 99, 117, 140, 164, 189, 215, 243, 273, 304, 336, 376, 414, 454

Offset: 1

Andrew Howroyd, Dec 23 2021

			Solutions for a(3)=3, a(4)=7, a(5)=13, a(6)=20:
  . . x    . . . x    . . . . x    . . . x x x
  . x .    . x x .    . x x x .    . x x . . x
  x . .    x . x .    x . x x .    x . x . x .
           x x . .    x x . x .    x x . x . .
                      x x x . .    . x x x x .
                                   x . x x . x

Column 2 of A339635.

Cf. A001197, A072567, A152125, A227133, A347472.

a(n) = A347472(n) + 1 = n^2 - A001197(n) + 1 = n^2 - A072567(n).

a(n) >= A152125(n).

a(22)-a(24) computed from A001197, added by Max Alekseyev, Feb 08 2022

A376167 Square array read by antidiagonals: T(n,k) = smallest r such that every n X k binary matrix with r ones contains a 2 X 2 submatrix of ones, with n, k >= 2.

Original entry on oeis.org

4, 5, 5, 6, 7, 6, 7, 8, 8, 7, 8, 9, 10, 9, 8, 9, 10, 11, 11, 10, 9, 10, 11, 13, 13, 13, 11, 10, 11, 12, 14, 15, 15, 14, 12, 11, 12, 13, 15, 16, 17, 16, 15, 13, 12, 13, 14, 16, 18, 19, 19, 18, 16, 14, 13, 14, 15, 17, 19, 20, 22, 20, 19, 17, 15, 14, 15, 16, 18, 21, 22, 23, 23, 22, 21, 18, 16, 15

Offset: 2

Paolo Xausa, Sep 13 2024

			Array begins (cf. Table 5 in Knuth (2023), section 7.2.2.2, p. 291, where T(n,k) is denoted by Z(m,n)):
  n\k|   2   3   4   5   6   7   8   9  10  11  12  ...
  -----------------------------------------------------
   2 |   4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, ...
   3 |   5,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, ...
   4 |   6,  8, 10, 11, 13, 14, 15, 16, 17, 18, 19, ...
   5 |   7,  9, 11, 13, 15, 16, 18, 19, 21, 22, 23, ...
   6 |   8, 10, 13, 15, 17, 19, 20, 22, 23, 25, 26, ...
   7 |   9, 11, 14, 16, 19, 22, 23, 25, 26, 28, 29, ...
   8 |  10, 12, 15, 18, 20, 23, 25, 27, 29, 31, 33, ...
   9 |  11, 13, 16, 19, 22, 25, 27, 30, 32, 34, 37, ...
  10 |  12, 14, 17, 21, 23, 26, 29, 32, 35, 37, 40, ...
  11 |  13, 15, 18, 22, 25, 28, 31, 34, 37, 40, 43, ...
  12 |  14, 16, 19, 23, 26, 29, 33, 37, 40, 43, 46, ...
  ...
T(3,4) = 8 because, no matter how 8 ones are arranged in a 3 X 4 matrix, a 2 X 2 submatrix of ones cannot be avoided (in the left configuration below, for example, the submatrix elements are highlighted by parentheses). 7 ones can avoid such submatrix (right).
.
  (1) 0 (1) 1       1  0  1  1
   0  1  0  1       0  1  0  1
  (1) 1 (1) 0       1  1  0  0

Donald E. Knuth, The Art of Computer Programming, Vol. 4B: Combinatorial Algorithms, Part 2, Addison-Wesley, 2023, section 7.2.2.2, pp. 289-291.

Cf. A001197 (main diagonal), A347472, A350296.

T(n,k) = T(k,n).
T(2,k) = k + 2.
T(n,2) = n + 2.

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A001197 Zarankiewicz's problem k_2(n).

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A347473 Maximum number of nonzero entries allowed in an n X n matrix to ensure there is a 3 X 3 zero submatrix.

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A347474 Maximum number of nonzero entries allowed in an n X n matrix to ensure there is a 4 X 4 zero submatrix.

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A350296 Minimum number of 1's in an n X n binary matrix with no zero 2 X 2 submatrix.

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A376167 Square array read by antidiagonals: T(n,k) = smallest r such that every n X k binary matrix with r ones contains a 2 X 2 submatrix of ones, with n, k >= 2.

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