There are two simple and classic enumerations that still I'm puzzled about. Let's start with a simple counting problem from a well-known dynamical system.


**fact 1** Consider the "tent map" f:[0,1]→[0,1] with parameter 2, that is 



> 
> f(x):=2min(x,1-x).
> 
> 
> 


Clearly, it has 2 fixed points, and more generally, for any positive integer *n*, there are 2*n* periodic points of period *n* (it's easy to count them as they are fixed points of the *n*-fold iteration of *f*, which is a piecewise linear function oscillating up and down between 0 and 1 the proper number of times). To count the number of periodic orbits of *minimal* period *n*, a plain and standard application of the Moebius inversion formula gives



> 
> Number of n-orbits of (I,f) =
>  $\frac{1}{n}\sum\_{d|n} \mu(d)2^{n/d}.$
> 
> 
> 


(**rmk**: any function with similar behaviour would give the same result, e.g. *f(x)=4x(1-x)*,...&c.)


Now let's leave for a moment dynamical systems and consider the following enumeration in the theory of finite fields.


**fact 2** Clearly, there are 2n polynomials of degree *n* in $\mathbb{F}\_2[x]$. With a bit of field algebra it is not hard to compute the number *I(n)* of the irreducible ones. One can even make a completely combinatorial computation, just exploiting the unique factorization, expressed in the form:


$\frac{1}{1-2x}=\prod\_{n=1}^\infty (1-x^n)^{-I(n)}.$ 


One finds:



> 
> Number of irreducible polynomials of
>  degree *n* in $\mathbb{F} \_ 2[x]$ =
>  $\frac{1}{n}\sum\_{d|n} \mu(d)2^{n/d}.$
> 
> 
> 


**Question**: it's obvious by now: is there a natural and structured bijection between periodic orbits of *f* and irreducible polinomials in $\mathbb{F}\_2[x]$? How is interpreted the structure of one context when transported ni the other?


(**rmk**: of course, analogous identities hold for any p > 2)



Periodic orbits and polynomials
How I see it is every fixed point comes from an equation $T^nx=x$. Denoting $f(x)=2x$ and $g(x)=2(1-x)$ we see that this corresponds to solving equations 
$$h\_1\circ h\_2\circ\cdots \circ h\_n (x)=x$$ where $h\_i\in \{f,g\}$. This is geometrically the intersection of two lines and thus gives a unique solution, and therefore a correspondence between periodic points of period $n$ and binary strings of length $n$. Now since concatenating a binary string $L$ with itself clearly gives the same $x$ with $x=L(x)=L\circ L(x)$ and you have a shift operator $L\circ h(x)=x \implies h\circ L(y)=y$ where $y=h(x)$ you get a bijection between periodic orbits and aperiodic cyclic sequences of zeros and ones, which are well known to be in bijection with the irreducible polynomials of $\mathbb F\_2 [x]$.


