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 $J$.
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 '14 at 18:31
It's not known, is it, that Hadamard matrices exist for all large $n$? – Gerry Myerson Jun 27 '14 at 23:05