Divisors of Primorials

Question

Let:

• $p_n$ 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 $w | p_n\#$

Examples:

• 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.

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Edit:

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

Answer 1

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 $f_{11}(447051)=1579>31^2$

Answer 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.