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Call an n X n matrix robust if the top left i X i submatrix is invertible for all i = 1..n. Sequence gives number of n X n robust real {0,1}-matrices.
8

%I #7 Oct 10 2012 14:25:27

%S 1,4,68,5008,1603232,2224232640

%N Call an n X n matrix robust if the top left i X i submatrix is invertible for all i = 1..n. Sequence gives number of n X n robust real {0,1}-matrices.

%C An upper bound is the total number of {0,1}-matrices, 2^(n^2).

%C Comment from _Michael Kleber_, Jan 05 2006: A lower bound is 2^(n^2-n), A053763. For given the principal n-1 X n-1 submatrix A, the 2n-2 further entries (excluding the bottom right corner) can be filled in arbitrarily and then there is always at least one choice for the last entry which makes the matrix invertible.

%C Comment from _N. J. A. Sloane_, Jan 06 2006: Let the matrix be [A b; c d], where A is n-1 X n-1, b is n-1 X 1, c is 1 X n-1, d is 0 or 1. The matrix is singular iff d = c A^(-1) b, which for given A, b, c has at most one solution d.

%C Suppose A = identity, as in A125586. Then if d=0 there are 3^(n-1) choices for b and c, while if d=1 there are (n-1)*3^(n-2) choices for b and c. This proves the formula in A125586.

%e a(2) = 4 from:

%e 10 10 11 11

%e 01 11 01 10

%Y Cf. A125586, A126603, A125593.

%K nonn,more

%O 1,2

%A _N. J. A. Sloane_ and _Vinay Vaishampayan_, Jan 05 2007

%E a(5) and a(6) from _Brendan McKay_, Jan 06 2007