Is OEIS A046346 sequence a subset $S\subset\mathbb{N}$ s.t. $\sum_S\frac{1}{n-\pi(n)}=1$?

Question

OEIS A046346 sequence lists composite numbers divisible by the sum of their prime factors, counted with multiplicity. $$S=\Big\{n\in\mathbb{N}\space not\space prime,\space n=\prod_k p_k^{\alpha_k}\space:\space\sum_k\alpha_k p_k\space|\space n\Big\}=\Big\{4,16,27,30,60,70,72,84,105,150,\dots\Big\}$$

I've considered over $S$ the series $$\sum_{n\space\in\space S}\frac{1}{n-\pi(n)}$$ where $\pi(n)$ denotes the prime counting function.

After the first $296000$ terms, the sum of the series amounts to $0.9956237272160026\dots$

Is it realistic to think that $$\sum_{n\space\in\space S}\frac{1}{n-\pi(n)}=1\space?$$

May this conjecture somehow be related to the one presented in this previous question of mine?

Answer 1

Only as regards the second question.

It is known that, for $n\ge 17$,

$$\frac{n}{\log n}\lt\pi(n)\lt\frac{1.25506\space n}{\log n}$$

Therefore, defining $S^*=S-\{4,16\}$ and assuming that (see my other question)

$$\sum_{n\space\in\space S^*}\frac{1}{n}=\frac{1}{e}-\frac{1}{16}\sim 0.30538$$

we can write the following chain

$$0.30538\sim \frac{1}{e}-\frac{1}{16}=\sum_{n\space\in\space S^*}\frac{1}{n}=\sum_{n\space\in\space S^*}\frac{\log n}{n\log n}\lt\sum_{n\space\in\space S^*}\frac{\log n}{n(\log n - 1)}=\sum_{n\space\in\space S^*}\frac{1}{n - \frac{n}{\log n}}\lt\sum_{n\space\in\space S^*}\frac{1}{n - \pi(n)}\lt\sum_{n\space\in\space S^*}\frac{1}{n - \frac{1.25506\space n}{\log n}}=\sum_{n\space\in\space S^*}\frac{\log n}{n(\log n - 1.25506)}=\sum_{n\space\in\space S^*}\frac{\log n - 1.25506}{n(\log n - 1.25506)}+1.25506\cdot\sum_{n\space\in\space S^*}\frac{1}{n(\log n - 1.25506)}=\sum_{n\space\in\space S^*}\frac{1}{n}+1.25506\cdot\sum_{n\space\in\space S^*}\frac{1}{n(\log n - 1.25506)}=\frac{1}{e}-\frac{1}{16}+1.25506\cdot\sum_{n\space\in\space S^*}\frac{1}{n(\log n - 1.25506)}\lt\frac{1}{e}-\frac{1}{16}+\frac{1.25506}{2}\cdot\sum_{n\space\in\space S^*}\frac{1}{n}=\Big(\frac{1}{e}-\frac{1}{16}\Big)\Big(1+\frac{1.25506}{2}\Big)\sim 0.49701$$

In other words, if

$$\sum_{n\space\in\space S^*}\frac{1}{n}=\frac{1}{e}-\frac{1}{16}$$

then also the series

$$\sum_{n\space\in\space S^*}\frac{1}{n - \pi(n)}$$

converges and, furthermore,

$$0.30537\lt\sum_{n\space\in\space S^*}\frac{1}{n - \pi(n)}\lt0.49702$$

As can be seen, the conjectured convergence value of the series

$$\sum_{n\space\in\space S^*}\frac{1}{n - \pi(n)}=1-\frac{1}{2}-\frac{1}{10}=0,4$$

falls close to half in the upper range.

Answer 2

This does not answer the question, but I wanted to share a chart for the first $700000$ terms of $S$ suggesting that if the conjecture in the previous post is true, then:

$$\sum_{n\in S}\frac{1}{n-\pi\left(n\right)}<1$$

Let $S$ be A046346, and:

$$A=\sum_{n\in S}\frac{1}{n}$$ $$B=\sum_{n\in S}\frac{1}{n-\pi\left(n\right)}$$

In this question and your previous post, you conjectured that:

$$A=\frac{1}{e}+\frac{1}{4}=0.6178794...$$ $$B=1$$

In the graph below, you can see $A$ in green, the conjectured value for $A$ in red and the ratio $\frac{A}{B}$ in blue:

Let's say $A=\frac{1}{e}+\frac{1}{4}$. The moment where $B=1$ is when the green line meets the blue line. It seems impossible that this would happen either at the red line (if $B=1$) or below (if $B>1$).

Correct me if I am wrong, but I guess it should not be very hard to show that the blue line won't start decreasing...?