I'm currently working on subwords of cube-free binary words.

A *binary* word is one composed of letters from a two-letter alphabet such as $\{0,1\}$.  A word $y$ is a *subword* of $w$ if there exist words $x$ and $z$ (possibly empty) such that $w=xyz$. Thus, $01$ is a subword of $0110$, but $00$ is not a subword of $0110$.  A word is *cube-free* if it does not contain subwords of the form $xxx$, where $x$ is a word of one or more letters.

I suspect that all sufficiently long cube-free binary words (those whose length is greater than a certain $n$) contain all the subwords $001$, $010$, $011$, $100$, $101$, and $110$.  I know that $n\ge 21$ because the cube-free binary word $110011001101100110011$ has a length of 21 and does not contain the subword $010$.

Does anyone know the length of the longest cube-free binary word that does not contain all the three-letter subwords I mentioned above?