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Few weeks ago an user from Mathematics Stack Exchange answered my question On an inequality that involves products and sums related to the sequence of semiprimes (asked May 26). It seems that for disproving my conjecture was enough to use a cheap statements in analytic number theory (more cheap than the prime number theorem). If my post is good I would like dedicate it to the excellence in (real and complex analysis, functional analysis, topology and other subjects as) analytic number theory of the user who refuted my conjecture.

A semiprime $s$ is a positive integer that is the product of two prime numbers, see Semiprine from the encyclopedia Wikipedia, thus corresponding to the sequence A001358 of the OEIS. I wondered if it is possible to deduce a statement at research level of the asymptotic

$$\Bigl(\prod_{\substack{1\leq s\leq X\\s\text{ semiprime}}}s\Bigr)\Bigl(\sum_{\substack{1\leq s\leq X\\s\text{ semiprime}}}\frac{1}{s}\Bigl)=\text{main term}+\text{error term},\tag{1}$$ where $\text{main term}=\text{main term}(X)$ is a function of the real variable $X$, let's say $X\geq 1$, and $\text{error term}=\text{error term}(X)$ is also a function of $X$ and represents a suitable error term in our asymptotic formula $(1)$ as $X\to\infty$.

Question. I would like to know what work can be done with the purpose to get a statement at research level for $(1)$ as $X\to\infty$, for a suitable error term expressed in big-O notation or little-o notation as you want (if it is feasible, you can to express your answer as an asymptotic identity $\sim$). Many thanks.

I don't know if this question concerning $(1)$ is in the literature, to ask this question I was inspired in a statement from [1]. If there is literature that provide a explicit answer for my question, then refer it answering my question as a reference request and I try to search an read those statements from the literature. I think that this post can be interesting to me as companion of the post of MSE, and I don't know if similar expressions as $(1)$ (I mean inequalities as the Lemma from [1] or asymptotics as our Question) are in the literature for other constellations of primes, or if it is interesting for other prime constellations as for example Ramanujan primes, primes in arithmetic progressions,... I evoke these problems if you want to explore some in your home, our case study here are the semiprimes.

References:

[1] Takashi Agoh, Paul Erdös and Andrew Granville, Primes at a (Somewhat Lengthy) Glance, The American Mathematical Monthly, Vol. 104, No. 10 (December, 1997), pp. 943-945.

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  • $\begingroup$ I can to read [1] from my account of JSTOR (also I add, if you as professional mathematician need it, that the site have a very generous expanded support in these months). $\endgroup$
    – user142929
    Jun 16 '20 at 7:47
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No, such an asymptotic formula is too much to ask for. The reason, morally, is that we shouldn't be looking at the product of the integers itself, but rather its logarithm—the sum of the logarithms of the integers. Being on this exponential scale amplifies oscillations enough to make asymptotic formulas impossible.

To see why, suppose we did have an asymptotic formula of the form $\prod_{s\in S,\, s\le x} s = \text{main term}+\text{error term}$. (Here $S$ can be the set of semiprimes, but this argument holds for any set.) By "main term" we presumably mean some continuous function $m(x)$; by "error term" we mean some function that is $o(m(x))$. In other words, suppose we did have a formula of the form $\prod_{s\in S,\, s\le x} s = m(x) + o(m(x)) = m(x) \big( 1 + o(1) \big)$. Then taking logarithms would give $$ \sum_{\substack{s\in S \\ s\le x}} \log s = \log m(x) + \log \big( 1 + o(1) \big) = \log m(x) + o(1). $$ In other words, an asymptotic formula for your product would imply an asymptotic formula for this sum with an incredibly strong error term $o(1)$. That's impossible even just from the jump discontinuities in the sum alone.

For example, suppose that instead of semiprimes we looked just at the set of all integers! Then $\prod_{s\le x} s = \lfloor x! \rfloor$ and $\sum_{s\le x} \log s = x\log x - x + O(\log x)$, but that error term cannot be improved as can be seen by looking at $x\to N-$ and $x\to N+$ for large integers $N$. That means that the product has arbitrarily large jump discontinuities even.

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  • $\begingroup$ Many thanks I'm going to read and study your excellent answer. And thanks for the remark in first paragraph, is lucid and concise. $\endgroup$
    – user142929
    Jun 17 '20 at 16:46

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