2 Added pairs of avoided words of length 4.

I wrote a recursive program to find the words of each length with no cube, avoiding a given string. If I programmed correctly, there are only $230800$ cube-free binary words of length $30$.

$001$: The longest string is of length $17$:

11010110101101100

$010$: The longest $2$ are of length $23$, the one found by Zack Wolske and it's reversal:

10011011001101100110011
11001100110110011011001

The others are equivalent to $000$ (no extra restriction), $001$, or $010$. So, any cube-free binary word of length $24$ or longer has all possible subwords of length $3$.

That was easy, so I'll do the same for words of length $4$, too:

$0010$: There are $76604$ cube-free binary words of length $40$ avoiding $0010$, and I would guess that the entropy per digit is positive, that there is some $a \gt 0, c\gt 1$ so that there are about $a c^n$ cube-free binary strings of length $n$ which avoid $0010$.

$0011$: There are $94238$ cube-free binary words of length $40$ avoiding $0011$.

$0101$: There are $110378$ cube-free binary words of length $40$ avoiding $0101$.

$0110$: The longest $3$ are length $17$. Avoiding $0110$ is not much different from avoiding $011$.

00101001010010011
11001001010010011
11001001010010100

What about pairs of words to avoid? Although there are long cube-free binary words avoiding either $0011$ or $0101$, the longest words which avoid both have length $14$:

01101101001001
10110110010010

The number of pairs to consider is larger than the number of pairs from $\lbrace 0010, 0011, 0101 \rbrace$. Avoiding both $0011$ and $0101$ is different from avoiding $0011$ and $1010$. The latter pair is avoided by $1310$ binary cube-free words of length $40$.

Here are some counts of cube-free binary words of length $40$ which avoid pairs of words:

0010    0011    0101
----------------------------
0010   76604       0    4376
0011       0   94238       0
0100    3721     994       0
0101    4376       0  110378
1010       0    1310    2600
1011       0       0    4376
1100     994     730    1310
1101       0     994       0

It might be interesting if some of these turn out to have polynomial growth instead of exponential. However, my guess is that they still have exponential growth, with a base about $\sqrt[10]{2}$ for the pair $0011$ and $1100$.

1

I wrote a recursive program to find the words of each length with no cube, avoiding a given string. If I programmed correctly, there are only $230800$ cube-free binary words of length $30$.

$001$: The longest string is of length $17$:

11010110101101100

$010$: The longest $2$ are of length $23$, the one found by Zack Wolske and it's reversal:

10011011001101100110011
11001100110110011011001

The others are equivalent to $000$ (no extra restriction), $001$, or $010$. So, any cube-free binary word of length $24$ or longer has all possible subwords of length $3$.

That was easy, so I'll do the same for words of length $4$, too:

$0010$: There are $76604$ cube-free binary words of length $40$ avoiding $0010$, and I would guess that the entropy per digit is positive, that there is some $a \gt 0, c\gt 1$ so that there are about $a c^n$ cube-free binary strings of length $n$ which avoid $0010$.

$0011$: There are $94238$ cube-free binary words of length $40$ avoiding $0011$.

$0101$: There are $110378$ cube-free binary words of length $40$ avoiding $0101$.

$0110$: The longest $3$ are length $17$. Avoiding $0110$ is not much different from avoiding $011$.

00101001010010011
11001001010010011
11001001010010100