# Reasoning about a sequence of consecutive integers and factorials with hope of relating factorials to primorials

I am looking for someone to either point out a mistake or help me to improve the argument in terms of clarity, conciseness, and more standard mathematical argument.

Let $x$ be an integer such that $x, x+1, x+2, \dots, x+n-1$ is a sequence of consecutive integers.

Let $r_i$ be defined as:

$$r_i = \dfrac{x+i}{\gcd(x+i,x+i-2)\gcd(x+i,x+i-3)\dots\gcd(x+i,x+1)}$$

So, for example:

• $r_0 = x$
• $r_1 = x+1$
• $r_2 = \dfrac{x+2}{\gcd(x+2,x)}$
• $r_3 = \dfrac{x+3}{\gcd(x+3,x+1)\gcd(x+3,x)}$

Here's my intuition:

$$\frac{(x+n-1)!}{(x-1)!\prod\limits_{i < n}r_i} \le (n-1)!$$

For $x=a(n-1)!$, $\dfrac{(x+n-1)!}{(x-1)!\prod\limits_{i < n}r_i} = (n-1)!$ since:

$r_i = \dfrac{x+i}{i}$ since if there exists a prime $p$ that divides $x+i$ and also divides $x+j$ where $j < i$, then $p | (i-j)$ so that $p | x$ which gives $p | i$.

To complete the argument, I show that if the proposition is true of $x$, then it is also true of $\frac{x}{p}$.

For $r_i$:

• if $p | (x+i)$ and $i < p$, it follows that this is the first time the prime is encountered $r_i$ remains the same as before.
• if $p | (x+i)$ and $i > p$, it follows that $r_i = \frac{\text{prev}\_r_i}{p}$ and $r_p$ no longer has this value divided.
• Let $r(x,n) = \prod\limits_{i < n}r_i$
• Now, $\prod{r_i}$ depends on the number of integers that are divisible by $p$. if $p$ does not divide $x$, then the count of the integers divisible by $p$ is equal or less to the count when $p$ divides $x$ so that:

$r(x,n) \le r(\frac{x}{p},n)$

Does this complete the argument to show that:

$$\frac{(x+n-1)!}{(x-1)!r(x,n)} \le (n-1)!$$

Edit: My assumptions about each $r_i$ being relatively prime to each other is not correct. For example, consider: $36, 37, 38, 39, 40$

$r_0 = 36$ but $r_4 = 10$

For this reason, I removed the assumption about being relatively prime and primorials.

• You wrote that $r_i$ is to be defined as $r_i = \dfrac{x+i}{\gcd(x+i,x+i-2)\gcd(x+i,x+i-3)\dots\gcd(x+i,x+1)}$. However, in the examples below it and based on the context of the rest of your question, it seems you are missing a factor of $\gcd(x+i,x)$ in the denominator. – John Omielan Aug 27 at 22:13

You are trying to prove

$$\frac{(x+n-1)!}{(x-1)!\prod\limits_{i < n}r_i} \le (n-1)! \tag{1}\label{eq1A}$$

With $$x = a(n - 1)!$$, you state that you get an equality in \eqref{eq1A}, with this being due to

$$r_i = \dfrac{x+i}{i}$$ since if there exists a prime $$p$$ that divides $$x+i$$ and also divides $$x+j$$ where $$j < i$$, then $$p | (i-j)$$ so that $$p | x$$ which gives $$p | i$$.

The statement starting from "if there exists ..." is correct, but your conclusion that $$r_i = \dfrac{x+i}{i}$$ is not true for any composite $$i$$. This is because you do not account for the prime factor $$p$$ possibly occurring among all of the products more times than it occurs in $$i$$. In particular, if $$i = jk$$, with $$1 \lt j, k \le i - 2$$, then $$\gcd(x + i, x + j) \ge j$$, $$\gcd(x + i, x + k) \ge k$$ and $$\gcd(x + i, x) = i$$. The best case scenario occurs when $$j = k$$ and is prime, which means the denominator will always be $$\ge ji$$.

For example, let $$n = 5$$ and $$a = 1$$ so $$x = a(n - 1)! = 4! = 24$$. This gives

\begin{equation}\begin{aligned} r_4 & = \frac{x+4}{\gcd(x+4,x+2)\gcd(x+4,x+1)\gcd(x+4,x)} \\ & = \frac{x+4}{2(1)(4)} \\ & = \frac{x+4}{8} \end{aligned}\end{equation}\tag{2}\label{eq2A}

Note that if your assumption $$r_i = \frac{x + i}{i}$$ were true, then the left side of \eqref{eq1A} would be

\begin{equation}\begin{aligned} \frac{(x+n-1)!}{(x-1)!\prod\limits_{i < n}r_i} & = \frac{\prod\limits_{i=0}^{n-1}(x + i)}{x\prod\limits_{i=1}^{n-1}\frac{x + i}{i}} \\ & = \frac{\prod\limits_{i=0}^{n-1}(x + i)}{\left(\frac{\prod\limits_{i=0}^{n-1}(x + i)}{\prod\limits_{i=1}^{n-1}i}\right)} \\ & = (n - 1)! \end{aligned}\end{equation}\tag{3}\label{eq3A}

so that part of your statement is correct. However, with my counter-example, since the denominator is $$\frac{1}{2}$$ of what is used in \eqref{eq3A}, you get a result of $$2(n-1)! = 48$$ instead, so \eqref{eq1A} is not true in that case.