I cited you (please correct if necessary), as "A215723(n) is divisible by 2^(n-1), indeed the determinant of any n by n sign matrix is divisible by 2^(n-1). Proof: subtract the first row from other rows, the result is all rows except for the first are divisible by 2, hence by using expansion by minors proof follows. (Warren D. Smith on the math-fun mailing list, Aug 18 2012)" See https://oeis.org/A215897 and https://oeis.org/A215723 I also computed more terms (and will compute still more later), both directly as determinants, and using (slow) Fourier transforms. Terms up to n=30 should be feasible (using C). Computation cost can be cut down using necklaces (only, instead of all binary words). Best, jj * Warren Smith <warren.wds@gmail.com> [Aug 19. 2012 08:31]:

1. The determinant of any NxN sign matrix is divisible by 2^(N-1). Proof: subtract 1st row from other rows, result is all rows except 1st are divisible by 2, hence by using expansion by minors proof follows.

2. The maximum |determinant| among NxN circulant sign matrices if N is sufficiently large, is at least (c*N)^(N/2) for some constant c>0 which I have not bothered to work out, but c=0.1 should be safe.

Proof sketch: Use the FFT-product formula for determinant. Need to prove signs exist causing each of the N |multiplicands| to be

...

c*sqrt(N). This can be done using a counting argument and central limit theorem - if |c|>0 small enough then at least half the sign-choices survive the kth demand, and due to the unitary nature of the FFT matrix each demand is orthogonal to the previous, hence effects can be reckoned independently. Initially 2^N sign choices, and after N demands at least one remains.

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