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Let $p_i$ denote the $i$th prime number, and let $p_k\#$ denote the $k$th primorial, $p_k\# \overset{\textrm{def}}= \prod_{i \le k} p_i$.

I am interested in asymptotic upper bounds for the function $S(N) = \sum_{i =1}^{k(N)} p_i$, where $k(N)$ is the smallest $k$ such that $N \le p_k\#$.

As a concrete example, if $N=2^{128}$ then $k(N) = 27$ since the 27th primorial is $2 \cdot 3 \cdots 103 > 2^{128}$. Hence $S(N)$ is the sum of the first 27 primes, which is 1264.

I can come up with what seems to be a very unsophisticated bound, as follows:

  • $k(N) = O(\log N)$ trivially
  • the $k$th prime has magnitude $O(k \log k)$
  • Hence $S(N) = O( k^2 \log k) = O( \log^2 N \log\log N)$

Again, this seems very superficial, and I'm wondering if it's possible to obtain a tighter bound, perhaps $O(\log N \log\log N)$.

edit: After finding this related question, the bound can be improved to $k(N) = O(\log N / \log\log N)$, resulting in $S(N) = O(\log^2 N / \log\log N)$. Is it possible to do better?

Also seems related to A008472 but no asymptotic expressions are given.

Community
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mikero
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1 Answers1

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By the prime number theorem (for $\theta(x)$), $\log p_k\#$ is isomorphic to $p_k \sim k\log k$. Therefore $k(N)$ is in fact asymptotic to $\log N/\log\log N$, since that is the asymptotic value of $k$ for which $k\log k \sim \log N$. The function $S(N)$ is then asymptotic to $\sum_{i=1}^{\log N/\log\log N} i\log i$, which is $$ \sim \frac12 \bigg( \frac{\log N}{\log\log N} \bigg)^2 \log \frac{\log N}{\log\log N} \sim \frac{\log^2 N}{2\log\log N}. $$

Greg Martin
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