I would like to approximate # in terms of $\pi(x)$.

The nearest I have found is by the answer of @draks:

Interpolating the primorial $p_{n}\#$

It seems to use asymptotic integration:

$\begin{eqnarray} \sum_{k=1}^n \log p_n &=& \int_2^n \log k\; d\pi(k)\\ &=& \log(k)\pi(k)\biggr|_{2}^{n}+\int_{2}^{n}\frac1k \pi(k)dk. \end{eqnarray}$

using $\pi(n)\sim \frac{n}{\log n}$

$\log p_n\# \sim \log(k)\frac{k}{\log k}\biggr|_{2}^{n}+\int_{2}^{n}\frac1k \frac{k}{\log k}dk = (x-1)+\text{Li}(x) \;. \tag{$*$}$

then exponentiate*.

1) No error bound is given and I assume (maybe wrongly) it wound be whatever error there is in the estimation of $\pi(n)$?

2) No reference is given so I am not sure if this method is widely accepted? Would it work? What are the bounds?

3) Is there something better for approximating primorial # in terms of π(x)?.

EDIT: Regarding question 2) There is at least one relevant reference (page 4-5) to using asymptotic integration for finding θ(x) in terms of π(x):


Hopefully this might help somebody looking at the stackexchange link given above.

Please also note correction to equation above given in @mixedmath answer.

  • 1
    $\begingroup$ I think the integration by parts should subtract $\int\frac1k\pi(x)dx$ instead of adding it; so subtract Li(x) instead of adding it in the final equation. $\endgroup$ – Michael Dec 11 '17 at 15:40
  • $\begingroup$ that's what I thought too $\endgroup$ – onepound Dec 11 '17 at 15:59

There is no error at all in the expression

$$ \log p_n \# = \log (t) \pi(t) \Big|_2^n - \int_2^n \frac{\pi(t)}{t} dt. \tag{1}$$

Thus if you are looking to write $p\#$ in terms of $\pi(x)$, then you have the equality

$$ p_n \# = \exp \left( \log (t) \pi(t) \Big|_2^n - \int_2^n \frac{\pi(t)}{t} dt\right).\tag{2}$$

Concerning your questions at the end of your post, it amounts to estimating the line $(2)$ above. You can make totally rigorous bounds in terms of rigorous bounds on $\pi(x)$.

Your question 2 doesn't quite make sense, as no reference is needed --- it is a proof. If you don't understand the proof, then it's likely you aren't familiar with Riemann-Stieltjes integration, as what's done is to write the sum as a Riemann-Stieltjes integral and then perform integration by parts. The result of this is exactly $(1)$ above.

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  • $\begingroup$ thanks for the answer. I do get the gist of it as it closely follows the asymptotic integration (Riemann-Stieltjes integration) sum of primes found in Bach & Shallit, but I'm not sure why such an elegant method is not widely found like the analogous sum of primes so I was hoping somebody would point to a book or something. That said with your assurance I'm happy with your answer. $\endgroup$ – onepound Dec 11 '17 at 15:59

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