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In this question, I asked about the convergence of $\hat\alpha$, the intercept of the best fit line joining all prime numbers. I stumbled on this result today where it was shown that $\sum_{k=1}^n p_k\sim\frac12n^2\ln n$ and similar results for prime powers.

Are there asymptotic results for $\sum_{k=1}^n p_{2k}$ and $\sum_{k=1}^n p_{2k-1}$?

If we denote $\pi_{2k}(x)=\sum_{p_{2k}}1$ then $\sum_{k=1}^n p_{2k}=\int_1^x t\,d\pi_{2k}(t)$. Now $\pi_{2k}(x)=\frac12\pi(x)\pm\{0,1\}$ depending on circumstance and since $\pi(x)\sim\int_2^x\frac{du}{\ln u}$, considering the simplest case of $\pm0$, $$\sum_{k=1}^n p_{2k}\sim \frac12\int_1^x\frac{t}{\ln t}\,dt.$$ How should I continue; is there a quicker way to obtain the asymptotics?

TheSimpliFire
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  • You can use integration by parts to find the last integral is asymptotically $\frac{1}{2}\frac{x^2}{\ln x}$. Then use the fact $x\sim 2n\log n$. – Wojowu Jan 26 '19 at 11:38
  • I don't see what you mean. You can't find the asymptotic of $\sum_{k \le n} p_{2k}$ without knowing that $p_k = k \log k+o(k\log k)$ which is the PNT. From there it is obvious that $\sum_{k \le n} p_{2k} \sim \sum_{k \le n} 2k \log(2k) \sim n^2 \log n$ – reuns Jan 26 '19 at 12:37
  • @reuns OK, how did you get the last asymptotic? I'm trying to do the same for $p_{2k-1}$ – TheSimpliFire Jan 26 '19 at 12:39
  • $\sum_{k \le n} k \log k \ge \log(n^{1-\epsilon}) \sum_{n^{1-\epsilon} \le k \le n} k = \log(n^{1-\epsilon}) ( n(n+1)/2 -n^{1-\epsilon}(n^{1-\epsilon}+1)/2)$ – reuns Jan 26 '19 at 12:41
  • @reuns Nice, so you are using that to plug in $2k\log(2k)=2(k\log 2+k\log k)$. But how could you split the logarithm in $\log(2k-1)$? We have $2k\log(2k-1)-\log(2k-1)$ which isn't so easy to do. – TheSimpliFire Jan 26 '19 at 12:46
  • @reuns Is it possible to get asymptotics for $\sum_{k\le n} p_{2k}p_{2k-1}$ and $\sum_{k\le n} p_{2k-1}^2$? Wolfram Alpha can't seem to find one. Thanks! – TheSimpliFire Jan 26 '19 at 19:21

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