Is there an obvious reason why the number of binary Lyndon words is equal to the number of irreducible polynomials over GF(2)? The title of Sloane's A001037 is: Number of degree-$n$ irreducible polynomials over $GF(2)$; number of $n$-bead necklaces with beads of 2 colors when turning over is not allowed and with primitive period $n$; number of binary Lyndon words of length $n$.
The first few terms of the sequence are (for $n=1,2,...$ ) $2,1,2,3,6,9,...$
The formula for the sequence is $\frac{1}{n}\sum_{d|n}\mu(\frac{n}{d})\cdot 2^d$.
I am familiar with the derivation given by Wilf in Generatingfunctiontology on page 62.  This derivation explains why the formula enumerates binary Lyndon words and equivalently the "$n$ bead necklaces" statement in the title.
I know the 2 irreducible polynomials of degree 1 are $x$ and $x+1$.
The degree 2 polynomial is $x^2+x+1$.
The degree 3 polynomials are $x^3+x^2+1$ and $x^3+x+1$.
The degree 4 polynomials are $x^4+x+1$, $x^4+x^3+x^2+x+1$ and $x^4+x^3+1$. 
The binary Lyndon words are:
$a(1)=2=\#\{"0","1"\}$,
$a(2)=1=\#\{"01"\}$,
$a(3)=2=\#\{"001","011"\}$,
$a(4)=3=\#\{"0001","0011","0111"\}$
I would like to know if there is an easy correspondence between these objects or if there is some explanation as to why the formula counts the irreducible polynomial over $GF(2)$.
 A: Necklaces and Lyndon words (of the same size) count the same objects. Each necklace represents an equivalence class with respect to rotation, and the Lyndon word is way to choose a unique representative of each class.  So the more canonical equivalence may be from necklaces to irreducible polynomials.  
Monic irreducible polynomials of degree $n$ are the same as Galois orbits of size $n$.  Galois groups of finite fields being cyclic, those are cyclic length $n$ Galois orbits, which (by the degree of the irreducible polynomial) are in the unique degree $n$ extension of the finite field.  
What does such an orbit look like?   The degree $n$ extension of finite field $F$ is an $n$ dimensional vector space over $F$.  It has a basis on which the Galois group acts by permutations, in other words by a cyclic action of order $n$.  Given the basis we have the correspondence: elements of $F^n$ are necklaces, elements of $F$ are the colors of the beads in the necklace, the Galois group rotates the necklaces, Galois orbits are the sets of roots of monic irreducible polynomials.
Nothing requires $|F|=2$ in this argument.
