Motivation (can skip!). (*) $\sum\log n \approx n\log n-n,$ and $$\sum\log n = \sum_{p_1\leq n} \log p_1+\sum_{p_2\leq n} \log p_2+...+\sum_{p_m\leq n} \log p_m$$
in which $p_k$ are numbers comprised of $k$ primes including repetitions. (*) is not an asymptotic equality but almost everything that is known about $\sum \log p_k$ is due to the prime number theorem.
The following is maybe obvious but I can't prove it yet. I'd be interested in seeing a proof or a hint (thanks). In words: the (sums of) odd summands are $\sim$ the (sums of) even summands.
Let $n = 2^{m},~ m = 2 r.$ Let $p_k(i)$ be the i$^{th}$ k-prime.
$$\sum_{k=1}^r\sum_{p_{2k}(i)\leq n}\log p_{2k}(i)\sim \sum_{k=1}^r\sum_{p_{2k-1}(i)\leq n}\log p_{2k-1}(i)\sim (1/2)(n\log n-n) $$
For $m = 18,$
$\sum\sum_{2k}/(2^{18}\log 2^{18}-2^{18})\approx 0.49944$ and $\sum\sum_{2k-1}/(2^{18}\log 2^{18}-2^{18})\approx 0.50056$