Corrected question.
From the sequence of binary palindromes A006995 (eg. 1001001001001) the sequence of possible gaps between consecutive palindromes contain the elements:
S={2,4,6,8,12,16,24,32,48,64,96,128,192,256,384,512,768,1024,1536,2048,3072,4096,6144,8192,12288,16384,24576,49152,...}
Compare with A131117 that correspond to the set of positions of the records in the sequence of arithmetic derivatives.
T={2,4,6,8,12,16,24,32,48,64,96,128,192,256,384,512,640,768,960,1024,1280,1536,1920,2048,2560,3072,3584,3840,4096,5120,6144,7168,7680,8192,10240,12288,14336,15360,16384,20480,24576,28672,30720,32768,40960,...}
This prompts for the conjecture that $S\subseteq T$
A hint of a proof may be to much to ask for, but are there computational results that support or contradicts the conjecture?
$S$ comes from gaps of consecutive palindromes in the $100,000,000$ first palindromes $>0$.
Your conjecture is true. I'll first characterize the set $S$, and then show it's a subset of $T$.
Gaps between consecutive palindromes must have the form $2^k$ or $3\cdot 2^k$. To prove this, note that if a palindrome $n$ has a $0$ in the middle (or two $0$'s, if it's an even-digit palindrome), then the next palindrome replaces these $0$'s with $1$'s. If $n$ has a string $01 \cdots 10$ in the middle, the next palindrome replaces this with $10 \cdots 01$. If $n$ is all ones, the next palindrome is $n+2$. In all cases, the differences have the specified form. In fact, you can check conversely that $S$ consists of exactly the positive integers of the form $2^k$ or $3 \cdot 2^k$, excluding $3$, and also excluding $1$ if we don't allow $0$ as a palindrome.
Now I claim $S \subset T$. Write $n$ as a product of not necessarily distinct primes, $n = \prod_{i=1}^m p_i$. The arithmetic derivative of $n$ is then $n' = n \cdot \sum_{i=1}^m \frac{1}{p_i}$. It follows that among all positive integers with at most $m$ prime divisors, the ones with the largest value of $\frac{n'}{n}$ are $2^m$ followed by $2^{m-1} \cdot 3$. In particular, if $n = 2^m$, then every positive integer $a < n$ has fewer than $m$ prime divisors, so $\frac{a'}{a} < \frac{n'}{n}$ and thus $a' < n'$. Similarly, if $n = 2^{m-1} \cdot 3$, then for every integer $a < n$ other than $a = 2^m$, we have $\frac{a'}{a} < \frac{n'}{n}$ and thus $a' < n'$; for $a = 2^m$ we have $a' = m \cdot 2^{m-1}$, which is less than $n' = (3m-1) \cdot 2^{m-2}$ provided that $m>1$. So every number of the form $2^k$ or $2^k \cdot 3$ is in $T$, excluding $3$, and also excluding $1$ if we don't allow $1' = 0$ as a record.