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We know that "Hadamard maximal determinant problem" concerns the largest determinant of a matrix of oder $n$ with entries in $\{-1,1\}$ or $\{0, 1\}$. For $n=4k$, it is the Hadamard conjecture; and for the other orders, many important bounds (both upper and lower) have been obtained by Barba, Ehlich, and Wojtas and others. I would like know if there has been work towards providing bound for the largest determinant of the matrices of order $n \neq 4k$ with pairwise orthogonal rows and entries in $\{0, 1, -1\}$?

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    $\begingroup$ I don't know what the orthogonality condition does to the answer, but without that condition the maximum determinant for $\{-1,0,1\}$ is the same as for $\{-1,1\}$. $\endgroup$
    – Brendan McKay
    Commented Mar 12, 2022 at 12:06
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    $\begingroup$ @Brendan It surely makes a difference, the maximal matrices for {-1, 1} that have been worked out (that I know of) in the literatures for order not equal to 4k, do not satisfy the orthogonality condition. $\endgroup$
    – Arun
    Commented Mar 12, 2022 at 12:30
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    $\begingroup$ Just to get the ball rolling, if $f(n)$ is the answer and $m$ is the largest number $\leq n$ for which there exists a Hadamard matrix of order $m$, then $f(n)\geq m^{m/2}f(n-m)$. $\endgroup$
    – Richard Stanley
    Commented Mar 12, 2022 at 15:52
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    $\begingroup$ Maximum determinants for $n=1,2,\dots,8$ are: $$1,2,2,16,16,125,128,4096$$ $\endgroup$
    – Max Alekseyev
    Commented Mar 12, 2022 at 21:18
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    $\begingroup$ If $n \equiv 2 \bmod 4$ there do not exist three orthogonal $\pm 1$ vectors. So you could have at most two $n$'s in the Gram matrix, and if you have at least one a parity argument shows the determinant would be at most the square root of $n^2(n-2)^{n-2}$. You can do better with a conference matrix, which is a matrix with zero's on the diagonal, $\pm 1$ elsewhere satisfying $MM^{\top} = (n-1)I_{n}$. So these should be the optimal solutions when $n \equiv 2 \mod 4$, when they exist. The order must be a sum of two squares; the Paley construction with $q \equiv 1 \mod 4$ gives an infinite family. $\endgroup$
    – Padraig Ó Catháin
    Commented Mar 13, 2022 at 0:09

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