Growth rate of product of smallest prime factors

Question

For $n\in \mathbb{N}$, let $p(n)$ denote the smallest prime dividing $n$. Then consider the function $f:\mathbb{N}\rightarrow \mathbb{N}$ defined by $f(n)= \prod_{k=1}^{n}p(k)$.

Question: What is the growth rate of $f(n)$? Does it grow faster than any exponential function (i.e., any function of the form $a^{n}$ for some $a>1$)?

Computing the asymptotic geometric mean of $f$ should essentially be equivalent to this question. That is, can one compute $\lim_{n\rightarrow \infty} f(n)^{1/n}$? I'm guessing the answer is $\infty$ but I haven't figured out a way of proving this.

Thanks for reading.

Answer 1

Note that for any integer $k$

$p(2k)=2$

And for any integer $k$ not divisible by $2$

$P(3k)=3$

And for any integer integer $k$ not divisible by $2$ or $3$

$P(5k)=5$

And in general for any integer $k$ not divisible by the primes $q<p$

$P(pk)=p$

Now we can break the enire product into disjoint sets each corresponding to those $P(k)$ where $k$ is divisible by a prime $p$ and no primes less then $p$.

Define $F(n,a)=\sum_{k\leq n, \gcd(k,a)=1}1=\sum_{d\mid a}\mu(d)\lfloor\frac{n}{d}\rfloor$

And let $g(q)$ be the product of the primes less then $q$

Then we have that:

$$\sum_{k=1}^n\ln(p(k))=\sum_{p\leq n}\ln(p)(F(\frac{n}{p},g(p))+F(\frac{n}{p^2},g(p))+F(\frac{n}{p^3},g(p))+...)$$

But this sum from what I know can't be approximated very well as it's hard to get rid of the error terms that accumulate from the off set of the floor function. Though if I had to make an educated 'guess', I would say that because we have by the Legendre seive:

$$F(n,G(p))\approx n\prod_{q< p}(1-\frac{1}{q})\sim \frac{ne^{-\gamma}}{\ln(p)}$$ $$\implies F(\frac{n}{p},G(p))+F(\frac{n}{p^2},G(p))+F(\frac{n}{p^3},G(p))+\dots\approx \frac{ne^{-\gamma}}{(p-1)\ln(p)}$$ That the following asymptotic might be true:

$$\sum_{k=1}^n\ln(p(k))\sim e^{-\gamma}n\ln(\ln(n))$$

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In addition if I made several more assumptions, there could infact exist a constant $C$ so that:

$$f(n)^{\frac{1}{n}}\sim C\ln(n)^{e^{-\gamma}}$$

Answer 2

$\displaystyle\pi(n) \sim \frac{n}{\ln n}$

The asymptotic expectation of $\displaystyle\frac{f(n+1)}{f(n)}$ is $n$ times the "asymptotic probability that n is prime", $\displaystyle\frac{n}{\ln(n)}-\frac{n-1}{\ln{n-1}}$, which is the expected number of primes $\leq n$ minus the expected number of primes $\leq n-1$.

$\displaystyle\frac{f(n+1)}{f(n)}\sim n\left(\frac{n}{\ln(n)}-\frac{n-1}{\ln(n-1)}\right)=\frac{n^2[\ln(n-1)-\ln(n)]+n\ln(n)}{\ln(n)\ln(n-1)}$ which eventually grows like $\displaystyle\frac{n}{\ln(n-1)}$, which eventaully becomes greater than any $a$, hence the average growth rate is faster than exponential.