Let
$$ A_n := \begin{bmatrix} 4 & 2 & \cdots & 2 \\ 2 & 4 & \ddots & \vdots \\ \vdots & \ddots & \ddots & 2 \\ 2 & \cdots & 2 & 4 \end{bmatrix} \in M_n \left(\mathbb{Z}_7\right)$$ Find $\det(A_n)$.
After row reduction, on the main diagonal there is $$ (4, -8, 16, -32, \dots) \equiv (4,-1,2,-4,1,\dots) \mod(7)$$ I am trying to find the general formula for $\det(A_n)$
$A_n=2(1)+2I$, where $(1)$ is the matrix of ones and $I$ is the identity.
Let us study $\det(A(t))=\det((1)-tI)$. The eigenvalues of $(1)$ are $t=0$ with multiplicity $n-1$ and $n$ with multiplicity $1$. Therefore $\det(A(t))=(-1)^n(t^n-nt^{n-1})$ and
$$ \det(A_n) = 2^n(-1)^n((-1)^n-n(-1)^{n-1})) = 2^n (1+n)$$
Using Weinstein-Aronszajn,
$$ \det \left( \, {\bf A}_n \right) = \det \left( 2 \left( {\bf I}_n + {\bf 1}_n {\bf 1}_n^\top \right) \right) = 2^n \left( 1 + {\bf 1}_n^\top {\bf 1}_n \right) = 2^n (1 + n) = (1 + n) \, 2^n $$
Modulo $7$, note that $ 2^0 \equiv 2^3 \equiv 1 \pmod7 $ and that, for all $n \in {\Bbb N}$, $ 2^n \equiv 2^{n \bmod 3} \pmod 7$. Thus,
$$ \det \left( \, {\bf A}_n \right) = (1 + n) \, 2^n \equiv \color{blue}{\left( 1 + \left( n \bmod 7 \right) \right) \cdot 2^{n \bmod 3} \pmod 7} $$
Since the function $n \mapsto n \bmod 7$ is periodic with period $7$ and the function $n \mapsto 2^{n \bmod 3}$ is periodic with period $3$, the product of these functions should be periodic with period given by the least common multiple $\operatorname{lcm} (3,7) = 21$. Thus, only the first $21$ determinants are needed, i.e.,
$$ \det \left( \, {\bf A}_n \right) = \det \left( \, {\bf A}_{n \bmod 21} \right) $$
Verification
In Haskell,
ghci> ds = map (\n -> ((1 + n ) * 2^n ) `mod` 7) [1,2..]
ghci> es = map (\n -> ((1 + (n `mod` 7)) * 2^(n `mod` 3)) `mod` 7) [1,2..]
Do the first, say, $1000$ entries match?
ghci> take 1000 ds == take 1000 es
True
What about periodicity?
ghci> ds' = take 21 ds
ghci> fs = foldl (++) [] (replicate 50 ds')
ghci> take 1000 ds == take 1000 fs
True