Factorial as Product of Consecutive Factorials

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Theorem

The only factorials which are the product of consecutive factorials are:

\(\ds 0!\) \(=\) \(\ds 0! \times 1!\)
\(\ds 1!\) \(=\) \(\ds 0! \times 1!\)
\(\ds 2!\) \(=\) \(\ds 1! \times 2!\)
\(\ds \) \(=\) \(\ds 0! \times 1! \times 2!\)
\(\ds 10!\) \(=\) \(\ds 6! \times 7!\)


Proof

Suppose $m, n \in \N$ and $m < n$.

Write $\map F {n, m} = n! \paren {n + 1}! \cdots m!$.


Suppose we have $\map F {n, m} > r!$ for some $r \in \N$.

Suppose further that there is a prime $p$ where $m < p \le r$.

We claim that $\map F {n, m}$ cannot be a factorial of any number.


Aiming for a contradiction, suppose $\map F {n, m} = s!$ for some $s \in \N$.

Since $s! > r!$, we must have $r! \divides s!$.

Since $p \le r$, we must have $p \divides r!$.

Thus we have $p \divides s!$.


However, since $n, n + 1, \dots, m < p$, we must have $p \nmid k!$ for each $n \le k \le m$.

Thus $p \nmid \map F {n, m} = s!$, which is a contradiction.

Hence $\map F {n, m}$ cannot be a factorial of some number.


We have the following lemmata:

Lemma 1

Let $n \in \N$.

Then $\paren {2 n - 1}! \, \paren {2 n}! > \paren {3 n - 1}!$ for all $n > 1$.

$\Box$


Lemma 2

Let $n \in \N$.

Then $\paren {2 n - 2}! \, \paren {2 n - 1}! > \paren {3 n - 1}!$ for all $n \ge 7$.

$\Box$


We also have:

$\forall n \in \N: n > 1 \implies \exists p: 2 n < p < 3 n$



Case $1$: $m$ is an even number larger than $2$

Write $m = 2 k$, where $k > 1$.

Then:

\(\ds \map F {n, m}\) \(\ge\) \(\ds \paren {m - 1}! \, m!\)
\(\ds \) \(=\) \(\ds \paren {2 k - 1}! \, \paren {2 k}!\)
\(\ds \) \(>\) \(\ds \paren {3 k - 1}!\) Lemma 1

There is a prime $p$ where $m = 2 k < p \le 3 k - 1$.

Therefore $\map F {n, m}$ is not a factorial of some number.


Case $2$: $m$ is an odd number larger than $11$

Write $m = 2 k - 1$, where $k \ge 7$.

Then:

\(\ds \map F {n, m}\) \(\ge\) \(\ds \paren {m - 1}! \, m!\)
\(\ds \) \(=\) \(\ds \paren {2 k - 2}! \, \paren {2 k - 1}!\)
\(\ds \) \(>\) \(\ds \paren {3 k - 1}!\) Lemma 2

There is a prime $p$ where $m = 2 k - 1 < 2 k < p \le 3 k - 1$.

Therefore $\map F {n, m}$ is not a factorial of some number.


Case $3$: Particular values of $m$

The cases above leaves us with $m = 1, 2, 3, 5, 7, 9, 11$ to check.

We have:

\(\ds 10! \times 11!\) \(>\) \(\ds 13!\) and $13$ is a prime
\(\ds 8! \times 9!\) \(>\) \(\ds 11!\) and $11$ is a prime
\(\ds 6! \times 7!\) \(=\) \(\ds 10!\)
\(\ds 5! \times 6! \times 7!\) \(>\) \(\ds 11!\) and $11$ is a prime
\(\ds 4! \times 5!\) \(=\) \(\ds 2880\) is not a factorial of some number
\(\ds 3! \times 4! \times 5!\) \(>\) \(\ds 7!\) and $7$ is a prime
\(\ds \map F {n, 3}\) \(=\) \(\ds 12\) is not a factorial of some number
\(\ds \map F {n, 2}\) \(=\) \(\ds 2!\)
\(\ds \map F {0, 1}\) \(=\) \(\ds 1!\)
\(\ds \) \(=\) \(\ds 0!\)

Thus there are no more.

$\blacksquare$


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