Lydon words over the binary alphabet are strings of $1$s and $0$s such that all rotations of the word are strictly lexicographically later. I've been looking at Lyndon words of length $n$, which are counted by OEIS sequence A001037.
1 | 0, 1
2 | 01
3 | 001, 011
4 | 0001, 0011, 0111
5 | 00001, 00011, 00101, 00111, 01011, 01111
6 | 000001, 000011, 000101, 000111, 001011, 001101, 001111, 010111, 011111In particular, though, I'm interested in the number of Lyndon words that begin with two or more $0$s.
I generated all Lyndon words of length $n$ and counted those that start with two zeros. Starting at index $n = 1$ there are
0, 0, 1, 2, 4, 7, 14, 25, 48, 88, 168, 310such words. For $n \geq 3$, this matches OEIS sequence $A065491(n-1)$:
Exponents in expansion of $\displaystyle\prod_{\textrm{prime }p\, \geq\, 3} \left(1 - \frac1{p^2-p-1}\right)$ as a product $\zeta(n)^{-a(n)}$.
That is, $$ \prod_{\textrm{prime }p\, \geq\, 3} \left(1 - \frac1{p^2-p-1}\right) = \zeta(0)^0 \cdot \zeta(1)^{-1} \cdot \zeta(2)^{-1} \cdot \zeta(3)^{-2} \cdot \zeta(4)^{-4} \cdot \zeta(5)^{-7} \cdots $$
I can't prove (or disprove!) this identity. Why are these two sequences equal? Or is this just a coincidence?
$\endgroup$ 3 Reset to default
