For any monic polynomial $P$ with coefficients in a finite field of order $q$, we put

$$ \sigma(P) = \sum_{d \mid P, d \text{ monic}} d. $$

Observe that $P$ and $\sigma(P)$ have the same degree.

EDIT. $A$ means the full ring of polynomials in one variable $t$ over the finite field with $q$ elements.

Since there are a finite number of polynomials of a given degree in $A$ the sequence $$ P, \sigma(P), \sigma(\sigma(P)), \ldots $$ must contain two equal terms.

The following algorithm (of J. T. B. Beard) seems to catch monic polynomials $P \in A$ with $\sigma(P)=P$ when $q$ is a prime number.

(i)
Take $R=R_0$ a `well chosen`

polynomial in $A.$

(ii) If $R=\sigma(R)$ STOP and output $R.$

(ii) if $R \neq \sigma(R)$ then replace $R$ by $$ lcm(R,\sigma(R)) $$ and come back to (ii)

Question: Take $q=2.$ For which polynomials $R_0$ l'algorithm stops after a finite number of steps. Assume that a polynomial $P \in A$ satisfies $P=\sigma(P).$ There exists a polynomial $R_0 \neq P$ and $R_0 \neq P(t+1),$ such that by taking $R=R_0$ in (i) the algorithm stops after a finite number of steps and outputs $P$?

Example: With $R_0=t$ we catch $P=t(t+1)$ that satisfies $P=\sigma(P).$ More involved examples can be obtained with a small computer program; (or by hand depending on taste...).

EDIT1: I tried recently to catch J. T. B. Beard himself to ask for this algorithm (and many other related questions in my mind...) but I got only some (nice) comments from people close to him in his last known position in the usa. No more news about him known by myself.

`multiperfect`

is better. The idea is:`multiperfect`

over the integers means loosely: $n$ divides $\sigma(n)$, so the notion can be also defined in $A$. But then you are forced to take $P=\sigma(P)$ since $P$ divides $\sigma(P)$ implies $P=\sigma(P)$ in $A.$ While over the integers you can take $\frac{\sigma(n)}{n}$ whatever you want (e.g. $=2$: define perfects, $=3$ define $3$-multiperfect numbers, $=8$: etc, etc.) over $A$ there is just one choice; (for the definition above of $\sigma$ of course). – Luis H Gallardo Jan 20 '11 at 9:03