# In the Oldenburger-Kolakoski sequence, is #1s = #2s infinitely many times?

The Oldenburger-Kolakoski sequence, $$OK$$, is the unique sequence of $$1$$s and $$2$$s that starts with $$1$$ and is its own runlength sequence:

$$OK = (1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2,1,1,\ldots).$$

For $$n \geq 1$$, let $$a(n)$$ be the number of indices $$h \leq n$$ such that $$OK(h) =1$$, and let $$b(n)$$ be the number of $$h \leq n$$ such that $$OK(h) = 2.$$

QUESTION. Is $$a(n) = b(n)$$ for infinitely many $$n$$?

For example, the first ten runs are $$1,22,11,2,1,22,1,22,11,2$$, and the lengths of these runs are $$1,2,2,1,1,2,1,2,2,1$$.

In the OEIS, the sequence is A000002. There are several related easily stated unsolved problems, such as the conjecture that the limiting density of $$1$$'s is $$1/2$$.

The general opinion among the ones who studied the problem is that it's very unlikely that $$d(1)$$ doesn't exists (this is in fact another open question), and provided that it exists, your claim is actually much stronger than $$d(1)=1/2$$.
Some heuristics: let $$T(w)$$ be the operator associating to every word in $$\{1,2\}^*$$ the word $$v$$ starting with 1 and such that $$w$$ lists the lengths of the runs of $$v$$ (example: $$T(1221)=122112$$). If you believe that the parity of the lengths $$\ell(T^k(w))$$ "behaves like" a Bernoulli variable with probability $$1/2$$, then "it follows" that the sequence is recurrent and that $$d(1)$$ equals 1/2 (if it exists). But even assuming this strong heuristic viewpoint, your claim would still be far from obvious.