Divisors of Primorials

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.

Edit:

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

• 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? – John Omielan Jan 4 '20 at 20:08
• $w$ must be a divisor of $p_n\#$. I will update my question to make this more clear. – Larry Freeman Jan 4 '20 at 20:11
• 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. – reuns Jan 4 '20 at 21:21
• @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? – Larry Freeman Jan 4 '20 at 22:37
• 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. – 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 $$i\le 2^{n-1}$$, $$d_i<\sqrt{p_n\#}. That is, each pair straddles $$\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

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.

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

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. 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 $$B-A. As originally put forward by Bertrand, $$\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.