What does this limit equal? $$\lim\limits_{k\to\infty}\left(\prod_{n=1}^kn^{2^{k-n}}\right)^{\frac{1}{2^{k-1}-1}}$$
All that I have tried so far is computation and it does seem to converge. I arrived here because I was wondering about Somos' recurrence relation: $g_n=n\cdot{g_{n-1}^2}$. So $g_5=5^1\cdot 4^2\cdot 3^4\cdot 2^8\cdot 1^{16}$ or $$g_5=\prod_{n=1}^5n^{2^{5-n}}.$$ Now distribute out the terms by taking the {1+2+4+8}-th root of $g_5$ (ignore the $1^{16}$ term) so $$\left(\prod_{n=1}^5n^{2^{5-n}}\right)^{\frac{1}{2^{5-1}-1}}.$$
[UPDATE: Answer Found] Without ignoring the inclusion of the distribution of the $1^{2^{k-1}}$ term we get Somos' quadratic recurrence constant, $S=1.6616879$. Without the $1^{2^{k-1}}$ term as my original question posed, we get $S^2=2.7612068$.
Taking the logarithm you get $$\lim_{k \to \infty} \frac{2^k}{2^{k-1}-1} \sum_{n=1}^k \frac{\log n}{2^n} = 2 \sum_{n=1}^{+\infty} \frac{\log n}{2^n} $$ which is approximately $1.017468...$. So your limit is approximately $2.76...$