Is this number theoretic quantity bounded above?

Question

I am considering a combinatorial argument which involves the following quantity. We use the prime counting function $\pi(n)$ and to save on exponents we set $h=\pi(n/2)$. The quantity as a function of integer $n \gt 7$ is $$(\pi(n)!)^{1/(n-h)}$$

Computations for small $n$ suggest this is always less than $4$, as do rough back-of-the-envelope asymptotic calculations. Is this bounded above for all $n \gt 7$? If so, what is the bound? (I'm hoping it is always less than 3.)

Gerhard "Researching Minds Want To Know" Paseman, 2020.05.30.

Answer 1

Let $k:=\pi(n)$, so that $p_k\le n<p_{k+1}$, where $p_k$ is the $k$th prime. By the last displayed formula in this section of the Wikipedia article, \begin{equation*} -1+\ln(k\ln k)<\frac{p_k}k<\ln(k\ln k) \end{equation*} if $k\ge6$, whence \begin{equation*} n>-k+k\ln(k\ln k),\quad n/2<m_k:=\frac{k+1}2\,\ln((k+1)\ln(k+1)). \end{equation*} Therefore, letting \begin{equation*} c_1:=1.25506,\quad r(k):=\frac{\ln((k+1)\ln(k+1))}{\ln m_k} \end{equation*} and using this result, we get \begin{equation*} h=\pi(n/2)\le\pi(m_k)<c_1\frac{m_k}{\ln m_k}=c_1\frac{k+1}2\,r(k). \end{equation*} Next, \begin{multline*} r(k):=\frac{\ln(k+1)+\ln\ln(k+1)}{\ln(k+1)+\ln[\ln(k+1)+\ln\ln(k+1)]-\ln2} \\ <\frac{\ln(k+1)+\ln\ln(k+1)}{\ln(k+1)+\ln\ln(k+1)-\ln2}<\frac{10}9 \end{multline*} if \begin{equation*} k\ge195, \end{equation*} which will be assumed henceforth.
So,\begin{equation*} h<c_2(k+1), \end{equation*} where \begin{equation*} c_2:=\frac7{10}>c_1\frac{10}9\Big/2. \end{equation*} So, using the trivial inequality $k!\le k^k$, we have \begin{equation*} \ln[(\pi(n)!)^{1/(n-h)}]=\frac{\ln(k!)}{n-h} \le\frac{k\ln k}{-k+k\ln(k\ln k)-c_2(k+1)} <\frac{k\ln k}{k\ln k}=1 \end{equation*} and hence \begin{equation*} (\pi(n)!)^{1/(n-h)}<e \tag{1} \end{equation*} for $k\ge239$, that is, for $n\ge1499$. By direct calculation, (1) holds for $n\le1498$ as well. It is also easy to see that the upper bound $e$ on $(\pi(n)!)^{1/(n-h)}$ is exact.

Answer 2

So I accept the answer of Iosif Pinelis, but it turns out I don't need the exact result now. I will post some backstory, and then the reason I don't need it now.

Thanks to MathOverflow user Daniel.W, and his question (360323) on strengthening Sylvester's theorem, I've been motivated to read the paper On Arithmetical Series. I approached it after a hint from Emil Jerabek to look at the thesis of Alan Woods. The thesis contained a write-up of Sylvester's method that I finally understood, and this allowed me to try to understand some of the proof in the 1892 paper.

A different version appears in a 1929 paper of Schur (which I have yet to find) and a (mostly) combinatorial one in a paper of Erdos in 1934. However, the arguments are still involved, and the Erdos paper leaves a lot of finitely many exceptions to be explored to yield a full proof.

After looking at the basic relation in Sylvester's paper, I (re-)discovered a result that allowed one to show that there was a number in (m,m+,n] with a prime factor greater than n whenever 4m was at least as big as n^2. This was heartening since previously I could only show it for m bigger than exponential in n. This in turn allowed me to discover a method which involved the quantity in the question above, and simple arguments showed that m only needed to be bigger than a small constant times n. (Iosif's argument and some additional computation shows the small constant is 3.). I then was going to attempt a third method to bridge the remaining gap which is for all m at least n. All of this was then going to be retooled to answer the motivating MathOverflow question.

After seeing Iosif's argument and thinking on simplifying the motivating argument, I found it. Here it is.

Write the product of the integers in (m,m+n] as P=(m+1)...(m+n). Rewrite as W(n!)L, where W are the prime factors of P/(n!) which are at most n gathered together, and L is the product of all the prime factors larger than n.

A key observation of Sylvester, (which I invite the reader to prove) is that W is at most (and for n bigger than 7, strictly less than) (m+n-p+1)...(m+n), where p is $\pi(n)$. This is because W is the product of p distinct prime powers, each one dividing a term of P (and usually different powers divide different terms, we suffer no loss in assuming this).

So if (m,m+n] has only n-smooth numbers, then L=1 and n! Is bigger (not necessarily strictly) than (m+1)...(m+n-p). The literature now expends a lot of effort to show how small m is, and Sylvester himself resorts to existence of primes in (m, 3m/2] to complete his argument. There is an easier way, however.

Write m=jn+i for i non negative. Then rewrite W(n!) = P by dividing out W and dividing out terms in the factorial larger than p. We get p! greater than j^(n-p), if P is n-smooth.

But we can argue with Chebyshev estimates to get j less than 6, and if we are as thorough as Iosif we can get j less than 3 with a small amount of computation needed. I need to perform this step but I believe p less than 50 should be more than sufficient.

So when the dust settles, we have reduced a large portion of Sylvester's argument to showing log(p!) Is less than n-p, using nothing more than grade school arithmetic and Sylvester's observation on W. With care we get that (m,m+n] has a multiple of a prime greater than n when m is at least 3n or greater. If need be, we can turn to the Erdos proof to handle m smaller than 3n.

However, there is more. The motivating question asks for two distinct numbers in the interval which have prime factors bigger than n. We now let L be a product of d many candidates for fixed d of members of (m,m+n]. We now are comparing log(p!) to n-p-d, and we will get the same bound on j, although this bound may start holding for larger m only.

Given the amount of time I spent reading these proofs, I'm surprised not to find this observation (that j is less than 3) is in the literature. We can use this observation, Chebyshev estimates, and work of Nagura or earlier to answer the motivating question affirmatively. That C=18 for two numbers hasn't been proved yet.

Gerhard "Is Confident It Will Be" Paseman, 2020.05.31.

Answer 3

It seems my thinking gets organized after posting the question.

The natural logarithm of the quantity $\pi(n)!$ is near $\pi(n)\log(\pi(n)/e) + (\log(\tau\pi(n)))/2$ (a number-theoretic use for $\tau$, the circumference of a unit radius circle). Using an approximation to $\pi(n)$ we get that this is less than $An$ for some $A \lt 2$. But $An/(n-h)$ is bounded above by $2A$, and gets very close to $A$. So with some work the original quantity should be shown to be less than $e^A$.

Verification is still appreciated.

Gerhard "And Still Worth An Acknowledgement" Paseman, 2020.05.30.