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I'm trying to get a hand on Hadamard matrices of order $n!$, with $n>3$. Payley's construction says that there is a Hadamard matrix for $q+1$, with $q$ being a prime power. Since $$ n!-1 \bmod 4 = 3 $$ construction 1 has to be chosen:

If $q$ is congruent to $3 (\bmod 4)$ [and $Q$ is the corresponding Jacobsthal matrix] then $$ H=I+\begin{bmatrix} 0 & j^T\\ -j & Q\end{bmatrix} $$ is a Hadamard matrix of size $q + 1$. Here $j$ is the all-1 column vector of length $q$ and $I$ is the $(q+1)×(q+1)$ identity matrix. The matrix $H$ is a skew Hadamard matrix, which means it satisfies $H+H^T = 2I$.

The problem is that the number of primes among $n!-1$ is restricted (see A002982). I checked the values of $n!-1$ given by Wolfram|Alpha w.r.t. be a prime power, without success, so Payley's construction won't work for all $n$.

Is there a general way to get the matrices, or is it case by case different?

I haven't yet looked into Williamson's construction nor Turyn type constructions. Would it be worth a closer look (sure it would, but) concerning my problem? Where can I find their constructions?

PS for the interested reader: I've found a nice compilation of Hadamard matrices here:

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I don't think a general construction for Hadamard matrices of order $n!$ is known. The knowledge about general construction methods for Hadamard matrices is quite sparse, the basic ones (see also the Wikipedia article) are:

1) If $n$ is a multiple of $4$ such that $n-1$ is a prime power or $n/2 - 1$ is a prime power $\equiv 1\pmod{4}$, then there exists a Hadamard Matrix of order $n$ (Paley).

2) If $n$ is a multiple of $4$ such that there exists a Hadamard Matrix of order $n/2$, then there exists a Hadamard Matrix of order $n$ (Sylvester).

The Hadamard conjecture states that for all multiples $n$ of $4$ there is a Hadamard matrix of order $n$. The above constructions do not cover all these $n$, the smallest case not covered is $n = 92$. There are more specialized constructions and a few computer constructions, such that the smallest open case is $n = 668$ nowadays.


I have just checked that for $n\in\{13,26,44,52,63,67,70,77,85\}$ a Hadamard matrix of order $n!$ cannot be constructed only by a combination of the Paley/Sylvester construction above. So in these cases, one had to check more specialized constructions like Williamsons' one.

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+1 thanks. What about Williamson and Turyn? Do you know how they work?...I see – draks ... Feb 11 '13 at 23:25
Not really. But my impression is that it is not so easy to see for which orders those constructions work. – azimut Feb 11 '13 at 23:43
thanks again... what was the max imum number you checked? – draks ... Feb 15 '13 at 11:36
your're welcome! I've checked n! for n up to 100... – azimut Feb 15 '13 at 11:50
Construct order $88! / 2^{31}$ by Paley ($88! / 2^{31} - 1$ is prime) and then apply Sylvester 31 times. – azimut Feb 15 '13 at 17:57

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