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Jan 27, 2022 at 17:35 comment added Peter Taylor 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.
Jan 27, 2022 at 6:57 comment added Arun @Gerry I do count such matrices.
Jan 26, 2022 at 22:37 comment added Gerry Myerson $$\pmatrix{1&1&1&1\cr1&-1&0&0\cr0&0&1&-1\cr1&1&-1&-1\cr}$$ is nonsingular, and all rows are orthogonal, but the columns are not orthogonal. Do you count this as an orthogonal matrix?
Jan 26, 2022 at 19:38 history edited Arun CC BY-SA 4.0
clarification of the comment on non-singularity is given.
Jan 26, 2022 at 19:37 comment added Arun @Dustin. Thanks for some useful information. I am looking only for non-singular matrices. I have edited the question.
Jan 26, 2022 at 3:07 comment added Gerry Myerson Do you count the zero vector as being orthogonal to other vectors?
Jan 25, 2022 at 22:28 comment added Dustin G. Mixon @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.
Jan 25, 2022 at 17:43 comment added Padraig Ó Catháin 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.
Jan 25, 2022 at 16:40 comment added Dustin G. Mixon 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
Jan 25, 2022 at 14:03 history asked Arun CC BY-SA 4.0