• $p_n$ be the $n$th prime.
  • $p\#$ be the primorial for $p$.
  • $f_n(x) = \dfrac{p_n\#}{x} - x$

Does it always follow that for $n \ge 2$, there exists an integer $w$ where $1 < f_n(w) < (p_n)^2$ and $w | p_n\#$


  • For $n=2$, $f_2(1) = \dfrac{6}{1} - 1 = 5 < 3^2 = 9$
  • For $n=3$, $f_3(2) = \dfrac{30}{2} - 2 = 13 < 5^2 = 25$
  • For $n=4$, $f_4(5) = \dfrac{210}{5} - 5 = 37 < 7^2 = 49$
  • For $n=5$, $f_5(35) = \dfrac{2310}{35} - 35 = 31 < 11^2 = 121$
  • For $n=6$, $f_6(165) = \dfrac{30,030}{165} - 165 = 17 < 13^2 = 169$
  • For $n=7$, $f_7(663) = \dfrac{510,510}{663} - 663 = 107 < 17^2 = 289$
  • For $n=8$, $f_8(3094) = \dfrac{9,699,690}{3094} - 3094 = 41 < 19^2 = 361$

Here's what I know:

  • Any $w$ will need to be less than $\sqrt{p_n\#}$
  • There are $2^n$ divisors for $p_n\#$.
  • For larger $n$, there are at least $ap_n$ primes between $p_n$ and $(p_n)^2$ with $a \ge 1$ and $a$ increasing for larger $n$ based on Bertrand's Postulate.


I am interested in $w$ where it is divisor. My previous question was unclear so I have made an update.

  • $\begingroup$ I assume $w \in \mathbb{N}$ since, if $w \in \mathbb{R}$, it'll be relatively easy to prove using calculus. Also, do you require $w \mid p_n\#$, or is it just coincidence this always happens in your examples? $\endgroup$ – John Omielan Jan 4 '20 at 20:08
  • $\begingroup$ $w$ must be a divisor of $p_n\#$. I will update my question to make this more clear. $\endgroup$ – Larry Freeman Jan 4 '20 at 20:11
  • $\begingroup$ You want the $w| p_n\#$ which is the closest to $\sqrt{p_n\#}$ equivalently you want to count the squarefree integers with largest prime factor $\le p_n$ in the short interval $[\sqrt{p_n\#}-p_n^2,\sqrt{p_n\#}]$, I think this interval is too short, why not make it larger. $\endgroup$ – reuns Jan 4 '20 at 21:21
  • $\begingroup$ @reuns I am fine with an answer that extends the interval. I chose this based what seemed to me to be true. I trust your judgment here. What seems like a more interesting interval? $\endgroup$ – Larry Freeman Jan 4 '20 at 22:37
  • $\begingroup$ A few tweaks and an extension: $f_2(2)=1;\ f_3(5)=1;\ f_4(14)=1;\ f_5(42)=13;\ f_7(714)=1;\ f_9(14858)=157$. In many cases, the smallest possible result is surprisingly small, although looking only as far as $f_9$ isn't really thorough. I wonder if $f_4$ is the last case where $1$ is a possible result. I did the arithmetic to see if I could spot any patterns, but I didn't. $\endgroup$ – Keith Backman Jan 13 '20 at 17:21

The conjecture is false. The best that can be done for the next two primes beyond $f_9$ is $f_{10}(79534)=1811>29^2$ and $f_{11}(447051)=1579>31^2$


This question has continued to intrigue me since it was posted. My thinking involves a different notational approach. Consider the $2^n$ divisors of $p_n\#$: $\{d_1,d_2,\dots,d_{(2^n-1)},d_{(2^n)}\}$ arranged in ascending order. These divisors can be put in pairs, $d_i$ with $d_{(2^n-i+1)}$ such that the product of each pair is $p_n\#$. As the index $i$ increases and approaches $2^{n-1}$, the arithmetic difference between the members of the pairs decreases, reaching its minimum at the pair $d_{(2^{n-1})},d_{(2^{n-1}+1)}$. For $i\le 2^{n-1}$, $d_i<\sqrt{p_n\#}<d_{(2^n-i+1)}$. That is, each pair straddles $\sqrt{p_n\#}$.

Focusing on the innermost pair, $d_{(2^{n-1})},d_{(2^{n-1}+1)}$, let's simplify the notation for readability in the following exposition by setting $A:=d_{(2^{n-1})},\ B:=d_{(2^{n-1}+1)}$. Bear in mind $AB=p_n\#$, so each of the first $n$ primes is present as a factor once in either $A$ or $B$. Also, by our choice of $A$ and $B$, there are no divisors of $p_n\#$ between $A$ and $B$. The objective is to describe or understand $\max {(B-A)}$.

For any factor $m$ of $B$, if we remove it from $B$ and include it in $A$, we see that $mA>B \Rightarrow A>\frac{B}{m}$ because $mA$ is a divisor of $p_n\#$ and there are no divisors of $p_n\#$ between $A$ and $B$. Thus $$B-A<B-\frac{B}{m}=B(1-\frac{1}{m})$$

This is the fundamental limitation of the difference $B-A$.

Next: Either $2\mid B$ or there is some prime number $p_k\mid B$ such that $p_{(k-1)}\mid A$. This follows from the fact that $B$ has a smallest prime factor, and if it is not $2$, then it is not the first prime number and succeeds a prior prime number, which must be a factor of $A$. Note that whether or not $2\mid B$, the only case in which there is no factor $p_k$ of $B$ succeeding a factor $p_{(k-1)}$ of $A$ is the case that $B=p_q\#,\ q<n$.

Case 1: $B=p_q\#$. In that rare and special case, if indeed it ever occurs, choose $m=2$. Then $B-A<B(1-\frac{1}{2})=\frac{B}{2}$

Case 2: For some $k$, $p_k\mid B \wedge p_{(k-1)}\mid A$. In that case, choose $m=\frac{p_k}{p_{(k-1)}}$. In this situation, $m$ is not an actual factor of $B$, but it works the same. This in effect generates the pair of divisors of $p_n\#\ $ $A\frac{p_k}{p_{(k-1)}},\ B\frac{p_{(k-1)}}{p_k}$. Hence, $B-A<B\bigl(1-\frac{p_{(k-1)}}{p_k}\bigr)$. From Bertrand's postulate, we know that $p_k<(1+\epsilon)p_{(k-1)} \Rightarrow \frac{p_{(k-1)}}{p_k}<\frac{1}{1+\epsilon}$. From this we see $B-A<B\bigl(1-\frac{1}{1+\epsilon}\bigr)=B\bigl(\frac{\epsilon}{1+\epsilon}\bigr)$. As originally put forward by Bertrand, $\epsilon =1$, but later results show that as the size of $p$ increases, the size of $\epsilon$ decreases, for example becoming $\epsilon \le \frac{1}{5000\ln^2 p}$ for $p>468991632$. It would be particularly effective in minimizing $\epsilon$ in particular cases if $p_k$ and $p_{(k-1)}$ that are twin primes can be identified.

In summary, we should expect that in most cases, primorials will be decomposable into two factors that are each quite close to $\sqrt{p_n\#}$, with the arithmetic difference of those factors becoming a very small fraction of the larger factor, and in no case exceeding $\frac{1}{2}$ of that larger factor.

Note that in specific cases, perhaps even in many cases, it may be possible to choose multiple prime factors of $A$ and $B$ to construct an $m=\frac{\prod(p_i)}{\prod(p_j)}$ which is greater than but very close to $1$. I have no algorithmic way of identifying instances in which this will be possible, other than case by case brute force.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.