Is the function $g(n) = \displaystyle\sum_{k=1}^{\pi(n) - 1}\left(\frac{1}{n-p_{k}}\right)$ bounded?

Question

This is related to my other question here (that is, the terms of this sum are the reciprocals of that one).

Define a function $g(n)$ that takes the sum of the reciprocals of the difference between $n$ and all the primes below it. In other words,

$$g(n) = \sum_{k=1}^{\pi(n) - 1}\left(\frac{1}{n-p_{k}}\right),$$

where $p_k$ denotes the $k$th prime (with $2 = p_1$) and $\pi(n)$ is the number of primes less than or equal to $n$. For example,

$$g(12) = \frac{1}{12-11} + \frac{1}{12-7} + \frac{1}{12-5} + \frac{1}{12-3} + \frac{1}{12-2} = \frac{1}{1} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{10}\approx 1.55.$$

My question is this:

Is $g(n)$ bounded?

While investigating this function I've come up with some integer sequences that I haven't been able to find in OEIS. Primarily, the list of $n$ for which $g(n)$ is greater than any previous $n$ (I've computed up to $500,$$000$):

$$3, 4, 6, 8, 14, 20, 44, 74, 110, 200, 284, 620, 830, 1304, 1880, 2714, 3470, 5660, 9440, 65720, 69500, 81050, 88820, 113174, 284750...$$

Related, but not identical, to my original question is the question of whether this sequence continues infinitely or not. (That is, if it continues infinitely, $g(n)$ still may be bounded, but if it has a finite number of terms, $g(n)$ cannot be unbounded.) For reference, here's the list of $g(n)$ (rounded to the nearest thousandth unless the difference is smaller) associated with the above sequence:

$$0, 1.0, 1.5, 1.583, 1.7, 1.72, 1.845, 1.908, 1.984, 2.09, 2.123, 2.138, 2.177, 2.192, 2.247, 2.257, 2.2766, 2.2768, 2.341, 2.374, 2.388, 2.408, 2.444, 2.467, 2.521, 2.553...$$

All of the terms in the above sequence are $1$ more than a prime (the reason should be self-evident); here is the sequence of the indices of those primes (again, treating $2$ as $p_1$):

$$1, 2, 3, 4, 6, 8, 14, 21, 29, 46, 61, 114, 145, 213, 289, 396, 487, 746, 1170, 6567, 6902, 7935, 8605, 10727, 24824...$$

I can't find a pattern; even the gaps between them seem erratic (note the jump from $1170$ to $6567$ and then the small step to $6902$). How would I go about figuring this out?

Thanks for your help!

Answer 1

$\newcommand{\floor}[1]{\lfloor #1 \rfloor}$ A tuple of distinct positive integers $(a_1, \cdots, a_k)$ is admissible if for any prime $p$, they don't form a complete residual system modulo $p$.

Continuing on Greg's comment, I think I can show the following.

Proposition: For any $N > 0$, there exists an admissible tuple with reciprocal sum at least $N$.

Proof: We take some $ M> 0$. For each prime $p \leq M$, we independently sample a uniformly random integer $a_p$ between $\lfloor \sqrt{p} \rfloor$ and $p$. Let $A$ be the set integers in $[1, M]$ that is not congruent to $a_p$ modulo $p$ for any $p \leq M$. Then by definition, $A$ is admissible. We compute the expected value of the reciprocal sum of $A$. By the linearity of expectation, we have $$\mathbb{E} \sum_{a \in A} \frac{1}{a} = \sum_{a = 1}^M \frac{1}{a} \mathbb{P}(a \in A).$$ We now estimate $\mathbb{P}(a \in A)$. Note that $$\mathbb{P}(a \in A) = \prod_{p \in [1, M]} (1 - \mathbb{P}(a \equiv a_p \bmod{p}))$$ where the product is over primes.

If $p > 2a^2$, then by construction, we can never have $a \equiv a_p \bmod{p}$. Otherwise, there are $p - \floor{\sqrt{p}} + 1$ choices for the residual class $a_p$, so the probability that $a \equiv a_p \bmod{p})$ is at most $\frac{1}{p - \floor{\sqrt{p}} + 1}$. Thus we have $$\prod_{p \in [1, M]} (1 - \mathbb{P}(a \equiv a_p \bmod{p})) \geq \prod_{p \in [1, 2a^2]} \left(1 - \frac{1}{p - \sqrt{p} + 1}\right).$$ We observe that the constant $$K:=\prod_p \left(1 - \frac{1}{p - \sqrt{p} + 1}\right) \left(1 - \frac{1}{p}\right)^{-1}$$ is strictly greater than $0$. So we have $$\prod_{p \in [1, M]} (1 - \mathbb{P}(a \equiv a_p \bmod{p})) \geq K \prod_{p \in [1, 2a^2]} \left(1 - \frac{1}{p}\right).$$ By Merten's third theorem, we have for some constant $K' > 0$ and every $a \geq 2$. $$\prod_{p \in [1, 2a^2]} \left(1 - \frac{1}{p}\right) \geq \frac{K'}{\log(2a^2)} \geq \frac{K'}{3\log(a)}.$$ So we conclude that, for some constant $C > 0$ $$\mathbb{P}(a \in A) \geq \frac{C}{\log(a)}.$$ Thus we obtain $$\mathbb{E} \sum_{a \in A} \frac{1}{a} = \sum_{a = 1}^M \frac{1}{a} \mathbb{P}(a \in A) \geq \sum_{a = 2}^M \frac{C}{a\log(a)} = \Omega(\log\log M).$$ Taking $M$ sufficiently large, we conclude that $$\mathbb{E} \sum_{a \in A} \frac{1}{a} > N$$ so there is an admissible set whose reciprocal sum is greater than $N$. QED.

Assuming the second Hardy-Littlewood conjecture, for every admissible tuple $A$, there exists some $n$ such that $\{n - a: a \in A\}$ are all prime numbers. For this $n$, we have $$g(n) \geq \sum_{a \in A} \frac{1}{a}.$$ By the proposition above, we conclude that $g(n)$ is unbounded.