Here by orthogonal matrix I mean just the rows are mutually orthogonal. Two such matrices are equivalent if one can be obtained from the other by permutations of rows and columns, or change of signs of rows and columns. I would like to know about the number of such nonequivalent $n \times n$ non-singular matrices in terms of the dimension $n$. A good lower or upper bound is also welcome.
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3$\begingroup$ Assuming the Hadamard conjecture, there are $e^{\Omega(\sqrt{n})}$ inequivalent block-diagonal matrices of this form with $4k\times 4k$ and $1\times 1$ blocks on the diagonal: en.wikipedia.org/wiki/Partition_(number_theory)#Asymptotics $\endgroup$– Dustin G. MixonCommented Jan 25, 2022 at 16:40
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5$\begingroup$ There are known to be exponentially many (in $n$) equivalence classes of Hadamard matrices at orders of the form $2^{n}$. This is a result of Eric Merchant. $\endgroup$– Padraig Ó CatháinCommented Jan 25, 2022 at 17:43
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2$\begingroup$ @aorq has privately indicated to me that you can get $e^{\Omega(\sqrt{n})}$ without the Hadamard conjecture by taking matrices whose rows have disjoint support. (In fact, you can restrict to nonnegative entries.) This assumes you allow rows with all zeros. $\endgroup$– Dustin G. MixonCommented Jan 25, 2022 at 22:28
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1$\begingroup$ @Gerry I do count such matrices. $\endgroup$– ArunCommented Jan 27, 2022 at 6:57
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1$\begingroup$ By Paley's constructions, if $p$ is an odd prime then there's a Hadamard matrix of order $2(p + 1)$. Then Merchant's thesis says that there are $2^{\Omega(p)}$ non-isomorphic Hadamard matrices of order $4(p + 1)$. If we consider that there are $\Theta(\tfrac{n}{\lg n})$ primes for which $\tfrac n2 \le 2(p+1) \le n$ then we get an easy $2^{\Omega(n \lg n)}$ lower bound without using the Hadamard conjecture. $\endgroup$– Peter TaylorCommented Jan 27, 2022 at 17:35
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