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Given a set $A \subseteq \mathbb{N}$ of positive integers, put $$ S_A: \mathbb{N} \rightarrow \mathbb{R}, \ \ \ \ N \mapsto \sum_{n \in A \cap \{1, \dots, N\}} \frac{1}{n}. $$ There are obvious quite "natural" sets $A$ such that $S_A(N) = O(\log(N))$ (e.g. $A = \mathbb{N}$) and such that $S_A(N) = O(\log(\log(N)))$ (e.g. $A$ is the set of prime numbers).

Question: Which "natural" sets $A$ are there such that $S_A(N) = O(\log(\log(\log(N))))$ (and not $S_A(N) = o(\log(\log(\log(N))))$, to avoid trivial answers)?

By "natural" I mean: "has a short description which is not merely a trivial rewrite of the condition in the question".

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    $\begingroup$ One candidate might be the "Golomb primes" as studied by Erdos in On a problem of G. Golomb, J. Austral. Math. Soc. 2 (1961/1962), 1--8. See renyi.hu/~p_erdos/1961-10.pdf The asymptotic of the main theorem implies that the reciprocal sum of Golomb primes to $N$ is $\sim \log \log \log N$ $\endgroup$
    – so-called friend Don
    Commented Feb 17 at 20:28
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    $\begingroup$ big-oh just means "less than a constant times", so the empty set qualifies. But presumably you also want some lower bound on $S_A(N)$. $\endgroup$
    – Gerry Myerson
    Commented Feb 17 at 22:24
  • $\begingroup$ @GerryMyerson Of course it shouldn't be $o(\log(\log(\log(N))))$. I made this explicit now in the question. $\endgroup$
    – Stefan Kohl ♦
    Commented Feb 17 at 22:36
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    $\begingroup$ Such a set 'should' have density proportional to $1/\left(\log(x)\log\log(x)\right)$, to give some sense of what sorts of things to look for. $\endgroup$
    – Steven Stadnicki
    Commented Feb 18 at 0:11

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Let $A$ consist of the integers of type $a_n=[n \log n \log \log n]$, (say for $n \geq 20$). Then $\sum_{a_n \leq x} \frac{1}{a_n} \sim \log \log \log x$. This follows by partial summation or by considering the integral $\int 1/(t \log t \log \log t)\, dt$.

As this sequence might be considered to be a trivial reformulation, here is an example related to primes:
Let $A_b$ denote the sequence of primes, whose sum of digits is also prime, in a fixed base $b$. The sum of digits has typically size about $\frac{b-1}{2}\log n$, and the probability that this number is prime is about $\frac{c_b}{\log \log n}$. Hence the density of this sequence is about $\frac{1}{\log n \log \log n}$, and $\sum_{a_n\leq x} \frac{1}{a_n}\sim C_b \log \log \log x$. The details can be found in a paper by Glyn Harman (Theorem 2). Counting Primes whose Sum of Digits is Prime, https://cs.uwaterloo.ca/journals/JIS/VOL15/Harman/harman2.html

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  • $\begingroup$ To avoid looking at digit counts -- wouldn't the set of primes $p$ which are congruent to 1 modulo the number of prime factors of $p-1$ work as well? $\endgroup$
    – Stefan Kohl ♦
    Commented Feb 25 at 21:46
  • $\begingroup$ The typical number of prime factors of $p-1$, where $p$ is about $n$, might be in the interval $[ \log \log n-C \sqrt{\log \log n},\log \log n+C \sqrt{\log \log n} ] $. Hence fixing this number influences the density by a factor of only $\sqrt{\log \log n}$. But you can add the condition that $p \equiv 1 \bmod q$, where $q$ is a prime. Then $\sum_{p<x} \frac{1}{p} \sim \frac{\log \log x}{q-1}$. Choosing $q\sim \frac{\log \log x}{\log \log \log x}$ works. (This can be arranged, by upper bounds on prime gaps.) $\endgroup$
    – Christian Elsholtz
    Commented Feb 25 at 22:47
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    $\begingroup$ @StefanKohl One could take primes $p$ such that $\lfloor \log p \rfloor$ is also prime. If the log is taken to an integer base this is of course the same as the number of digits, but seems closer to ordinary mathematics. This has the advantage of being easy to rigorously establish - it can be done with just the prime number theorem. If you want to work with the prime factors of $p-1$, a sequence that such work is primes $p$ such that $p-1,p-2,p-3$ all have exactly the same number of prime factors as each other, but this is probably very hard to establish rigorously! $\endgroup$
    – Will Sawin
    Commented Feb 26 at 1:20
  • $\begingroup$ @ChristianElsholtz Sorry if I am missing something obvious -- but why does the size of the interval matter here, rather than just the average order of magnitude of the number of prime factors of $p-1$? $\endgroup$
    – Stefan Kohl ♦
    Commented Feb 26 at 10:29
  • $\begingroup$ @StefanKohl I think your approach gives the density you asked for. I agree with Will Sawin that for his suggestion, taking primes with $\lfloor \log p \rfloor$ also prime, it is much easier to work out the details. $\endgroup$
    – Christian Elsholtz
    Commented Feb 27 at 7:20

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