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Define $f(n)$ to be:

$$ \sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d} $$

But $\sigma_0(d) = 2^{\omega(d)}$ for any $d \mid n\#$ a primorial, so:

$$ f(n) = \prod_{p \text{ prime} \\ p \leq n} \dfrac{1}{\left(1 - \dfrac{2}{p}\right)} $$

Mertens' third theorem has as $1$ instead of a $2$ there.

So what would be an asymptotic estimate of $f(n)$?

metamorphy
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Daniel Donnelly
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2 Answers2

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We have \begin{align*} \prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} } & = \exp \left( { - \sum\limits_{2<p \le n} {\log \left( {1 - \frac{2}{p}} \right)} } \right) \\ & = \exp \left( {\sum\limits_{2<p \le n} {\frac{2}{p}} - \sum\limits_{2<p \le n} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} } \right). \end{align*} Now by Mertens' second theorem $$ \sum\limits_{2<p \le n} {\frac{2}{p}} = 2\log \log n + 2M -1+ \mathcal{O}\!\left( {\frac{1}{{\log n}}} \right), $$ where $$ M = \gamma + \sum\limits_p {\left[ {\frac{1}{p} + \log \left( {1 - \frac{1}{p}} \right)} \right]} $$ is the Meissel–Mertens constant. Also \begin{align*} \sum\limits_{2<p \le n} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} & = \sum\limits_{p>2} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} - \sum\limits_{p > n} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} \\ & = \sum\limits_{p>2} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} - \sum\limits_{p > n} {\mathcal{O}\!\left( {\frac{1}{{p^2 }}} \right)} \\ & = \sum\limits_{p>2} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} + \mathcal{O}\!\left( {\frac{1}{{n\log n}}} \right). \end{align*} Taking exponentials and simplifying $$ \prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} } = C(\log n)^2 +\mathcal{O}(\log n), $$ where $$ C = \frac{{e^{2\gamma } }}{4}\prod\limits_{p > 2} {\left( {1 + \frac{1}{{p(p - 2)}}} \right)} = \frac{{e^{2\gamma } }}{4\Pi_2}=1.201303559967362\ldots, $$ with $\Pi_2$ being the twin prime constant.

Gary
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    Thanks. I will study this in depth later. I awarded you the answer because you explained more details. Coincidentally this is coming up in my own twin prime research, so maybe I'll stumble upon the same results as they did. – Daniel Donnelly Jun 14 '22 at 03:10
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    Eric has to correct his answer as it is off by a factor of $2$. – Gary Jun 14 '22 at 03:11
  • Can you point me to a list of axioms in brief for working with the asymptotics present here? I'm in awe of how you showed those steps so fluidly. I'm no where near your level in analytic number theory, so how does one get there? – Daniel Donnelly Jun 14 '22 at 04:06
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    I belive the only non-trivial part is $$\sum\limits_{p > n} {\mathcal{O}!\left( {\frac{1}{{p^2 }}} \right)} =\mathcal{O}(1)\frac{\pi(n)}{n^2}+ \mathcal{O}(1)\int_n^{ + \infty } {\frac{{\pi (t)}}{{t^3 }}dt} =\mathcal{O}(1)\frac{\pi(n)}{n^2}+ \mathcal{O}(1)\int_n^{ + \infty } {\frac{{dt}}{{t^2 \log t}}} = \mathcal{O}!\left( {\frac{1}{{n\log n}}} \right),$$ which relies on Abel's summation formula and PNT. The rest is elementary manipulation and application of known results for which I gave references. Please point to the part that needs more clarification. – Gary Jun 15 '22 at 14:24
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    You can also use the more elementary inequality $$ \sum\limits_{p > n} {\frac{1}{{p^2 }}} \le \sum\limits_{k \ge n} {\frac{1}{{k(k + 1)}}} = \frac{1}{n}, $$ which gives a weaker estimate but is enough for our purposes. – Gary Jun 16 '22 at 01:16
  • Let me direct you to where these formulas show up (hence question): https://math.stackexchange.com/questions/4473585/the-twin-prime-pseudo-metric-on-the-integers-is-a-topological-group-formed If you expand the prime average counter function, and convert floors to modular residues, then $(b-a)$ times question formula above appears as a summation. But there's another summation involving $(b - x)_{(d)}$ etc which is the problematic one. If the first summation infinitely often greater than the second summation, then twin primes is true. Which just means that $d^n(a,b) \gt 0$ i.o in $n$. – Daniel Donnelly Jun 16 '22 at 01:36
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    @DanielDonnelly Any way I can improve my answer? – Gary Feb 17 '25 at 01:48
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    @DanielDonnelly I provided exactly the same estimate as the other answer but with an estimate for the error term + rigorous derivation. I cannot see what else could be done here. $\prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} } = C(\log n)^2 +\mathcal{O}(\log n)$ is a stronger statement than $ \prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} } \sim C(\log n)^2$. – Gary Feb 17 '25 at 05:25
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Asymptotically,

$$g(n) = \prod_{2 < p < n}^{p \text{ prime}}\left( 1-\frac2p\right) \sim \frac{4C_2e^{-2\gamma}}{(\log n)^2} \approx \frac{0.83244}{(\log n)^2}$$

And presumably your expression is simply the reciprocal of that, i.e.,

$$f(n) \sim \frac{e^{2 \gamma}(\log n)^2}{4C_2} \approx 1.2013(\log n)^2$$

where $C_2$ is the twin prime constant:

$$C_2 = \prod_{p>2}^{p \text{ prime}} \frac{p(p-2)}{(p-1)^2} \approx 0.66016$$

Note: or just look at the other answer, which does a better job and shows the full derivation.

Daniel Donnelly
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Eric Snyder
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