Runs of 2s in a rep digit sequence Today I asked a question about a particular sequence of numbers on code-golf.se.  This $n$th term of the sequence is defined as

... the length of the longest rep-digit representation of $n$ in any base

A rep-digit, is any number where all the digits are the same, for example $22$ or $9999$.  For example if $n$ is $7$, $7$ can be represented as $111_2$ or as $11_6$, the longer one has three digits so the result is $3$.
It seems that for the vast majority of values the $n$th term is $2$.  This makes sense because $2$ is the minimum value for any $n > 2$.  Although I have not proven this, I am pretty confident this is the case simply from a heuristic stand point.  Another less apparent fact is that it seems as the numbers grow larger the length runs of consecutive values of $2$ increase in size.
I have tried to prove that there exists runs of $2$ that are arbitrarily large, but became a little bit stuck in doing so.  So that's my question: can runs of $2$s become arbitrarily large?
Here is a program that can generate the sequence pretty quickly.  The first $10,000$ terms are already cached.
 A: Just an idea, not an answer.
Roughly, I want to try to find a good upper bound on the number of numbers between $1$ and $N$ which can be represented as a rep digit in some base.
Every number that can be written as a rep digit in base $b$ of length $n$ can be written as a rep-unit of length $n/d$ in base $b^d$, when $d\mid n$.
This is a long way to saying, if we want to compute the numbers that are representable as repunits of length $>2$ between $1$ and $N$, you need only need to check for prime lengths and length four.
Now, let $f(b,n)$ be the number of rep-units the interval $[1,N]$ of length $n$ in base $b$. Then the upper bound for all such numbers in any base is:
$$\sum_{p\in P} \sum_{2\leq b\leq \sqrt[p]{N}}(b-1)$$
So, to count the numbers that have rep-digit representations greater than $2$, we can restrict to prime lengths or the length $4$.
Where $P$ is the set of primes plus $4$.
(We are still overcounting some of the rep digits, but we just need an upper bound.)
Now, $$\sum_{2\leq b\leq x} (b-1)= \frac{\lfloor x\rfloor(\lfloor x\rfloor-1)}{2}$$
So we have an upper bound:
$$\sum_{p\in P} \frac{\left\lfloor\sqrt[p]{N}\right\rfloor(\left\lfloor\sqrt[p]{N}\right\rfloor-1)}{2}$$
Note that when $p\geq \log_2(N)$, then $\sqrt[p]{N}<2$, so this is still a finite sum.
So you want to estimate this value. That might be nasty.
