For natural numbers $n\geq2$, let $d(n)$ be the number of divisors of $n$, and let \begin{equation} g(n)=n\sum_i r_i(p_i-1) \end{equation} where $n=\prod_i p_i^{r_i}$ is the factorisation of $n$ as a product of powers of primes. Define a subsequence $T$ of the natural numbers greater than $1$, by \begin{equation} n \in T \iff (n \text{ is odd and } d(n) > \sup \{ d(k): k \text{ odd}, k < n \} ). \end{equation} Then \begin{equation} d(T)=(2, 3, 4, 6, 8, 9, 12, 16, 18, 20, 24, 32, 36, 40, 48, 54, 64, 72, 80, 96, 108, 120, 128, 144, \dotsc) \end{equation} is A053640. Also define a subsequence $S$ of the natural numbers greater than $1$, by \begin{equation} n \in S \iff g(n) = \inf \{g(k) : k \geq n \}. \end{equation} Then \begin{equation} S=(2, 3, 4, 6, 8, 9, 12, 16, 18, 20, 24, 32, 36, 40, 48, 54, 64, 72, 80, 81, 96, 108, 128, 144, \dotsc). \end{equation} So far, these sequences are very similar. Note that $81$ is in $S$ but not in $d(T)$, and $120$ is in $d(T)$ but not in $S$.
Why are they so similar ? So far all I have is this: \begin{equation} g \left (\prod_{1\leq i \leq k} a_i \right ) = \sum_{1\leq i \leq k} g(a_i) \prod_{1\leq j \leq k, j \neq i} a_j \end{equation} and \begin{equation} d(n) = \prod_i (r_i + 1), \end{equation} hence \begin{equation} g(d(n)) = \sum_i g(r_i + 1) \prod_{j \neq i} (r_j + 1) = \sum_i g(r_i + 1) d \left ( \frac{n}{p_i^{r_i}} \right ). \end{equation} Also I can prove that $g(n) \geq n \log_2(n)$, with equality if, and only if, $n$ is a power of $2$. This last fact implies that provided $g(n)$ does not exceed $g(m)$, where $m$ is the least power of $2$ not less than $n$, it is guaranteed that $g(n) < g(a)$, for all $a > m$.
Does this seeming “almost coincidence” have to do with Möbius inversion ? On the other hand, is it not much of a coincidence, as these things go ?
