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Is it true that

$$\left\{\frac{m^2}{n!}:m,n\in\mathbb{N}\right\}=\mathbb{Q}^{+} \tag{1}$$

My intuition comes from the fact that $m$ and $n$-increases $m^2=p_1^{2e_1} p_2^{2e_2}...p_s^{2e_s}$ where $p_1,...,p_s$ are the first $s$ primes and $n!=q_1^{f_1} q_2^{f_2}\cdot\cdot\cdot q_r^{f_r}$ where $q_1,...,q_r$ are the first $r$ primes

The following answer already shows:

$$\left\{\frac{m^a}{n^b}:m,n\in\mathbb{N}\right\}=\left\{\frac{c^{\gcd(a,b)}}{d^{\gcd(a,b)}}:c,d\in\mathbb{N}\right\}$$

Since

$$\left\{\frac{m^2}{n!}:m,n\in\mathbb{N}\right\}=\left\{\frac{p_1^{2e_1} p_2^{2e_2}...p_s^{2e_s}}{q_1^{f_1} q_2^{f_2}\cdot\cdot\cdot q_r^{f_r}}:e_1,...,e_r,f_1,...,f_r\in\mathbb{N}\right\}$$

and each element in $2e_{1},...,2e_r$ is relatively prime to each element in $f_{1},...,f_{r}$

We can assume that

$$\left\{\frac{m^2}{n!}:m,n\in\mathbb{N}\right\}=\mathbb{Q}^{+} \tag{1}$$

holds true.

If this is correct can we generalize this to:

$$\left\{\frac{m^{p}}{n!}:m,n\in\mathbb{N}\right\}=\mathbb{Q}^{+}$$

Where $p\in\mathbb{Z}$

Is this also true?

Arbuja
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2 Answers2

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By Bertrand, there is always a prime $p$ which divides $n!$ exactly once. Hence for every rational of the form $\frac {m^2}{n!}$ there must be a prime which divides either the numerator or the denominator an odd number of times. Thus you can't get every rational this way.

As a concrete example, $\frac 14$ can not be written as $\frac {m^k}{n!}$ for any $m,k,n\in \mathbb N$ with $k≥2$. Note that it is easy to see that $n$ would have to be $>2$ But then we can use Bertrand to produce a prime $p$ between $\frac n2$ and $n$ and such $p$ would have to divide either the numerator or the denominator of $\frac 14$, which is clearly not possible.

lulu
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  • ... except for $n\le 1$, of course. But that doesn't help with making $\frac14$. – Troposphere Aug 06 '21 at 17:49
  • @Troposphere Yes, it is easy to rule out small $n$ for this example. In any case, I edited to rule out small $n$. – lulu Aug 06 '21 at 17:50
  • So for $\left{\frac{m^{p}}{n!}:m,n\in\mathbb{N}\right}$, does the following only equal $\mathbb{Q}^{+}$ when $p=1$? – Arbuja Aug 06 '21 at 17:51
  • @Arbuja Yes, that's right. I was implicitly assuming $k≥2$ (and have edited to make that explicit). If $k=1$ then of course you can cancel any stray factors without introducing new ones. – lulu Aug 06 '21 at 17:53
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    yes , if you want to express $\frac{p}{q}$ you just need to take $n$ such that $n!$ is a multiple of $q$ and then take $m = n!p/q$ – Asinomás Aug 06 '21 at 17:54
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Lets work with $\frac{n!}{m^2}$ instead and see it doesn't cover all the integers.

Assume $\frac{n!}{m^2}$ is a prime. It follows that there is at most one prime in the range $[n/2,n]$, because such primes would have to appear in $\frac{n!}{m^2}$.

However there is at least two primes in the range $[n/2,n]$ for all $n\geq 100$.

It follows that only a finite amount of values $\frac{1}{p}$ appear in the set.

Asinomás
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