In February 2018, when the On-Line Encyclopedia of Integer Sequences (OEIS) was approaching it's 300,000th sequence, Neil Sloane sent an email out to the SeqFan mailing list announcing hand-picked sequences for A300000, A300001, A300002, and A300003.
In May of the following year, I submitted sequence A308092, which is the base-2 analog of A300000:
The sum of the first $n$ terms of the sequence is the concatenation of the first $n$ bits of the sequence read as binary, with $a(1) = 1$.
In decimal, the sequence begins $$ 1, 2, 3, 7, 14, 28, 56, 112, 224, 448, 896, 1791, 3583, 7166, 14332, 28663, 57326, 114653, 229306, \dots. $$
but it is more natural to look at the sequence in binary $$ 1_2, 10_2, 11_2, 111_2, 1110_2, 11100_2, 111000_2, 1110000_2, 11100000_2, 111000000_2, \dots $$
For example, $a(4) = 7$ because $7$ is the right choice so that the first four bits of the sequence ($1101_2$) is equal to the sum of the first four terms: $1+2+3+7=13=1101_2$.
One thing that I noticed about this sequence—and that Matthew Scroggs proved in June 2019—is that the number of $1$s in each term seems to be weakly increasing: $$ \underbrace{1_2}_{1}, \underbrace{10_2}_{1}, \underbrace{11_2}_{2}, \underbrace{111_2}_{3}, \underbrace{1110_2}_{3}, \underbrace{11100_2}_{3}, \underbrace{111000_2}_{3}, \underbrace{1110000_2}_{3}, \underbrace{11100000_2}_{3}, \underbrace{111000000_2}_{3}, \dots, $$ which gives the auxiliary sequence A324608, "Number of 1's in binary expansion of $A308092(n)$." $$ 1, 1, 2, 3, 3, 3, 3, 3, 3, 3, 3, 10, 11, 11, 11, 13, 13, 14, 14, 14, 16, 16, 16, 17, 17, \dots. $$
The question
That was a lot of background, but now we're ready for the question!
Let $a_1$ be the run-lengths on the bits of A308092 (the first sequence), $$ \underbrace{1_2, 1}_2\underbrace{0_2}_1, \underbrace{11_2, 111_2, 111}_8\underbrace{0_2}_1, \underbrace{111}_3\underbrace{00_2}_2, \underbrace{111}_3\underbrace{000_2}_3, \underbrace{111}_3\underbrace{0000_2}_4, \underbrace{111}_3\underbrace{00000_2}_5, \underbrace{111}_3\underbrace{000000_2}_{6}, \dots, $$ and let $a_2$ be the run-lengths of the auxiliary sequence A324608, $$ \underbrace{1, 1}_2, \underbrace{2}_1, \underbrace{3, 3, 3, 3, 3, 3, 3, 3}_8, \underbrace{10}_1, \underbrace{11, 11, 11}_3, \underbrace{13, 13}_2, \underbrace{14, 14, 14}_3, \underbrace{16, 16, 16}_3, \underbrace{17, 17, 17}_3, \dots. $$ These sequences agree for all of the terms I've checked. It feels like there should be some obvious, elementary reason for why they are equal, but I'm missing it. Any help?
It turns out that this question was based on a bad premise: these sequences agree for 50 terms and then disagree. $$ a_1(51) = 1 \neq a_2(51) = 2. $$ This was pointed out by Michael S. Branicky, in my draft for the OEIS sequence.
What's going on, and why do these agree for so long?
For any binary string $s$, let $w(s)$ be the weight of $s$—that is, the number of $1$s. Also, let $s_k$ be an arbitrary binary string of length $k$.
Now consider a portion of the bits of A308092 that looks like this $$ s_k0\underbrace{1\cdots1}_x\underbrace{0\cdots0}_y1\cdots. $$
Matt Scroggs and Martin Harris made an observation that $a_1(n)$ is given by the first $n-1$ bits of A308092 (concatenated) plus the $n$-th bit, so using this, we can determine terms of the sequence (in terms of $w(s_k)$).
$$ \begin{alignat}{3} A324608(k+1) &= w(s_k+0) && &&= w(s_k)\\ A324608(k+2) &= w(s_k0+1) &&= w(s_k1) &&= w(s_k) + 1 \\ & \vdots & &\\ A324608(k+x+1) &= w(s_k0\underbrace{111\dots1}_{x-1}+1) &&= w(s_k1\underbrace{000\dots0}_{x-1}) &&= w(s_k) + 1 \\ A324608(k+x+2) &= w(s_k0\underbrace{111\dots1}_{x}+0) &&= w(s_k0\underbrace{111\dots1}_{x}) &&= w(s_k) + x \\ & \vdots & &\\ A324608(k+x+y+1) &= w(s_k0\underbrace{111\dots1}_{x}\underbrace{000\dots0}_{y-1}+0) &&= w(s_k0\underbrace{111\dots1}_x\underbrace{000\dots0}_{y-1}) &&= w(s_k) + x \\ A324608(k+x+y+2) &= w(s_k0\underbrace{111\dots1}_{x}\underbrace{000\dots0}_{y}+1) &&= w(s_k0\underbrace{111\dots1}_x\underbrace{000\dots0}_{y-1}1) &&= w(s_k) + x + 1 \end{alignat} $$ This shows that whenever there is a block of $x \geq 2$ $1$s the bits of A308092, there is a run of length $x$ in the auxiliary sequence, and whenever there is a block of $y$ $0$s in the bits of A308092.
However, the sequences $a_1$ and $a_2$ are not equal because the $51$st run of bits in A308092 is a block consisting of just one $1$, and when $x=1$, $w(s_k) + 1 = w(s_k) + x$, so the run lengths don't match up.