A001198 Zarankiewicz's problem k_3(n).

9, 14, 21, 27, 34, 43, 50, 61, 70, 81, 93, 106, 121, 129

Offset: 3

Author

N. J. A. Sloane

Keywords

• nonn
• more
• hard

Comments

Guy denotes k_{a,b}(m,n) the least k such that any m X n matrix with k '1's and '0's elsewhere has an a X b submatrix with all '1's, and omits b (resp. n) when b = a (resp. n = m). With this notation, a(n) = k_3(n). Sierpiński (1951) found a(4..6), a(7) is due to Brzeziński and a(8) due to Čulík (1956). - M. F. Hasler, Sep 28 2021

Examples

			From _M. F. Hasler_, Sep 28 2021: (Start)
For n = 3 it is clearly necessary and sufficient that there be 3 X 3 = 9 ones in the n X n matrix in order to have an all-ones 3 X 3 submatrix.
For n = 4 there may be at most 2 zeros in the 4 X 4 matrix in order to be guaranteed to have a 3 X 3 submatrix with all '1's, whence a(4) = 16 - 2 = 14: If 3 zeros are placed on a diagonal, it is no more possible to find a 3 X 3 all-ones submatrix, but if there are at most 2 zeros, one always has such a submatrix, as one can see from the following two diagrams:
                                       0 1 1 1        0 1 1 1      no 3 X 3
     Here one can delete, e.g.,   ->   1 0 1 1        1 0 1 1  <-  all-ones
     row 1 and column 2 to get         1 1 1 1        1 1 0 1      submatrix
     an all-ones 3 X 3 matrix.         1 1 1 1        1 1 1 1        (End)
		

References

• 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.
• 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).

Links

• 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.
• Jeremy Tan, An attack on Zarankiewicz's problem through SAT solving, arXiv:2203.02283 [math.CO], 2022.

Crossrefs

Cf. A001197 (k_2(n)), A006613 (k_{2,3}(n)), ..., A006626 (k_4(n,n+1)).

Cf. A339635, A347473, A350237, A350304.

Formula

a(n) = A350304(n) + 1 = n^2 - A347473(n) = n^2 - A350237(n) + 1. - Andrew Howroyd, Dec 26 2021

Extensions

a(11)-a(13) from Andrew Howroyd, Dec 26 2021

a(14)-a(15) computed from A350237 by Max Alekseyev, Apr 01 2022

a(16) from Jeremy Tan, Oct 02 2022