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Assume that $n$ is a sufficiently large number. Is there a Hadamaed matrix $H_{4n \times 4n}=(h_{ij})$ with the last row and the last cloumn $J$ (thet is, for every $k$, $h_{k,4n}=1$ and $h_{4n, k}=1$) such that the resulting numbers in other rows ($\overline{h_{k,1} ... h_{k, 4n}})_2$ and columns ($\overline{h_{1,k} ... h_{4n,k}})_2$ are prime numbers, when the representation of numbers is in the basis $2$?

Note that by Prime numbers theorem, $\Theta(1/n)$ of numbers less than $2^{4n}$ are prime. Also by theorems in Number Theory, for every natural numbers $\alpha$ and $\beta$ with $(\alpha, \beta)=1$, the density of prime numbers $p$ with $p \equiv \beta (mod \alpha)$ among all prime numbers, is $\phi^{-1} (\alpha)$. Since the above facts, I conjectured that $$\lim (4n. ln 2)^{8n-2} \frac{ph(n)}{h(n)}$$ exists and is equal to $1$; when $ph(n)$ is the number of Hadamard matrix of order $4n$ with our property; and $h(n)$ is the number of all Hadamard matrix of order $4n$ with the last row and column $J$.

Any way, can one present an intresting comment about this definition of "prime Hadamards"?

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You will need to explain the coding, as the digits are not 0 and 1. Assuming a substition of 0 for -1, for n=1 we have 15 - 2, 4, 8 respectively are primes. I think this is an example you want. –  The Masked Avenger Jun 27 at 16:16
    
Thanks; I mean to change $-1$ to $0$ in the definition of Hadamard matrix. But I don't understand your numbers 15, -2, 4, 8. –  Arash Ahadi Jun 27 at 18:31
    
I was noting the numbers gotten from flipping one of the three msb in the 4 bit representation of 15, giving 7,11, and 13. –  The Masked Avenger Jun 27 at 20:04
    
It's not known, is it, that Hadamard matrices exist for all large $n$? –  Gerry Myerson Jun 27 at 23:05
    
No but the number of n for which it is known has a density approaching 1/4. It is consistent with my (incomplete but considerable) knowledge that there is a thin and infinite set of positive integers k such that no Hadamard matrix of order 4k exists. –  The Masked Avenger Jun 28 at 19:24

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