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A347472
Maximum number of nonzero entries allowed in an n X n matrix to ensure there is a 2 X 2 zero submatrix.
5
0, 2, 6, 12, 19, 27, 39, 51, 65, 81, 98, 116, 139, 163, 188, 214, 242, 272, 303, 335, 375, 413, 453
OFFSET
2,2
COMMENTS
Related to Zarankiewicz's problem k_2(n) (cf. A001197 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 2 X 2 submatrix. This complementarity leads to the given formula which was used to compute the given values.
See A347473 and A347474 for the similar problem with a 3 X 3 resp. 4 X 4 zero submatrix.
FORMULA
a(n) = n^2 - A001197(n).
a(n) = A350296(n) - 1. - Andrew Howroyd, Dec 23 2021
EXAMPLE
For n = 2, there must not be any nonzero entry in an n X n = 2 X 2 matrix, if one wants a 2 X 2 zero submatrix, whence a(2) = 0.
For n = 3, having at most 2 nonzero entries in the n X n matrix still guarantees that there is a 2 X 2 zero submatrix (delete the row of the first nonzero entry and then the column of the remaining nonzero entry, if any), but if one allows 3 nonzero entries and they are placed on the diagonal, then there is no 2 X 2 zero submatrix. Hence, a(3) = 2.
CROSSREFS
Cf. A001197 (k_2(n)), A001198 (k_3(n)), A006613 - A006626.
Cf. A347473, A347474 (analog for 3 X 3 resp. 4 X 4 zero submatrix).
Cf. A350296.
Sequence in context: A104969 A065005 A299016 * A033154 A340663 A354759
KEYWORD
nonn,hard,more
AUTHOR
M. F. Hasler, Sep 28 2021
EXTENSIONS
a(22)-a(24) computed from A001197 by Max Alekseyev, Feb 08 2022
STATUS
approved