I was working with the function $f(k)=2^k \log_2(2^k-1)$ and had noticed that it's factional component seemed to converge when $k\in\mathbb{N}$.
For example, $f(13)\equiv 0.557216896821\mod 1$, strikingly close to the Euler-Mascheroni constant. It hovers around this value up to $k=47$, whereby after, due to what I assume is floating-point-rounding errors, it drops to $0$ from that point on.
For $k\in\mathbb{N}$, is it the case that $\lim_{k\to\infty}2^k \log_2(2^k-1)\equiv\gamma\mod 1$ , or perhaps more liberally, $\limsup_{k\to\infty}2^k \log_2(2^k-1)\equiv\gamma\mod 1$?
I have not seen such an identity before. Hence I thought I'd share as some of you might find it as intriguing as I do. There are many such identities that do relate powers of 2 and/or the base-2 logarithm to $\gamma$, so it would not be too surprising if true. Any insight would be much appreciated.
If you do a Taylor expansion of $f(k) = 2^k \log_2(2^k-1)$ around $k=\infty$, you get $$ f(k) = 2^k k - \frac{1}{\log 2} + \mathrm{O}\!\left(2^{-k}\right). $$ This means that, in the limit, we have $2^k k - 2 \le f(k) \le 2^kk - 1$. We then have that the limit $f(k) - 2^k k + 2 \longrightarrow 2 - 1/\log(2) \neq \gamma$ (they differ by about $0.0199$).