I'm looking for an integer matrix that has elements defined by a function of the row and column. We can call this function $f(r,c)$ where $r$ denotes the row and $c$ denotes the column.
The integer matrix should be an $n \times n$ matrix of full row and column rank. Basically we want this matrix:
$$ \begin{pmatrix} f(1,1) & f(1,2) & f(1,3) & \dots & f(1,n) \\ f(2,1) & f(2,2) & f(2,3) & \dots & f(2,n) \\ f(3,1) & f(3,2) & f(3,3) & \dots & f(3,n) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ f(n,1) & f(n,2) & f(n,3) & \dots & f(n,n) \\ \end{pmatrix} $$
There are two things that I am after. First, the function should give an maximum absolute value as small as possible. Second, the function should be periodic modulo every prime $p$, when we fix either $r$ or $c$.
Essentially, I am looking for a function that has period $\le p$ modulo $p$ and has a maximum value considerably less than $2^{r \cdot c}$. There are possible exceptions; the period can be larger if we make an even greater reduction in the exponent of the maximum value, and vice versa.
EXAMPLES
The best I could do was to make $f(r,c) = \text{Fibonacci}(r \cdot c)$, which gives a maximum value of approximately $1.6^{rc}$, and has maximum period of $6p$ modulo $p$.
Another older approach was to simply take $f(r,c) = 2^{r \cdot c}$, which has period at most $p$ modulo $p$.
I hope that someone else can do better!
The $n \times n$ matrix $f(r,c) = (2r-n + (n+1 \mod 2))^{c}$, has full rank, has maximum absolute value approximately $n^{n}$, and has period at most $p-1$ in each row and $p$ in each column.