Distribution of the permanent modulo $p$

Question

We know that the order of $SL_n({\mathbb F}_p)$ is $$p^{n(n-1)/2}(p^n-1)(p^{n-1}-1)\cdots(p^2-1).$$ Dividing by $p^{n^2}$, we deduce the probability that $\det$ takes the value $1$ over $M_n({\mathbb F}_p)$.

What if we consider instead the permanent~? What is the cardinal of the set of $n\times n$ matrices with entries in ${\mathbb F}_p$, whose permanent equals $1$ ? Of course, it is the same as if we replace $1$ by any non-zero element.

A few comments:

• if $n=2$, the permanent of $A$ is the determinant of $A'$ where $a_{12}$ is replaced by $-a_{12}$. The distributions are therefore identical.
• if $p=2$, the permanent coincides with the determinant. The distributions are the same.
• the first non-trivial is therefore $p=n=3$, for which there are $3^9=19683$ matrices. With the permanent being notoriously hard to calculate, at least harder than the determinant, I did not try to figure out the distribution. If the answer is unknown, I'll try to write a code.

Answer 1

$\def\char{\mathop{\rm char}}$In this paper by M. Budrevich and A. Guterman prove that for every $F_q$ with $\char F_q>2$ there are more matrices of order $n\geq 3$ with zero determinant than those with zero permanent. In fact, their estimates can be strengthened to show the following. If $D(M_n(F_q))$ and $P(M_n(F_q))$ are the numbers of the corresponding matrices, then $$ D(M_n(F_q))-P(M_n(F_q))\geq (q-1)\prod_{k=3}^n(q^n-q^{k-1}). $$ (This is shown in Budrevich's dissertation.)