As an application in group theory, I would need an infinite sequence over a finite alphabet, that avoids a sequence of words $w\_i$, where the length of $w\_i$ is such that $l(w\_i) > 10^8 l(w\_{i-1})$. 


I have found several results about avoiding patterns, but here I would really just need to avoid the words themselves. 


Do there exist results along these lines?



Infinite sequence avoiding a countable set of words
I am not sure that it is exactly what you need, but the following is true:


For an alhabet with $q\geq 4$ letters and a sequence of forbidden words with lengths $n\_1<n\_2<\dots$ there exists an infinite word without forbidden subwords (where subword of a word W is a segment of consecutive letters in W, like "hab" is a subword of "alphabet".)


Proof. Choose $c$ like $c=2$ and prove that for the number $f(n)$ of permitted words of length $n$ (i.e. words without forbidden subwords) we have $f(n)\geq cf(n-1)$. 


Induction in $n$. Base $n=1$ holds ($f(0)=1$, $f(1)\geq q-1$). 


We have $f(n)\geq qf(n-1)-\sum\_i f(n-n\_i)$ (take any permitted word with $n-1$ letter and add arbitrary letter. If new word is not permitted, then it ends by some forbidden subword.) By induction purpose we have $f(n-i)\leq c^{1-n}f(n-1)$, thus $f(n)\geq (q-\sum c^{1-n\_i})f(n-1)\geq cf(n-1)$ as desired. 


Now we have arbitrarily long permitted words. It follows that there exists an infinite permitted word (proof: choose letters $x\_1,x\_2,\dots,x\_n$ so that word $x\_1\dots x\_n$ has arbitrarily long permitted continuations. We may always proceed.)


Above argument works for $q=2$ and $q=3$ under stronger assumptions on $(n\_i)$.


