X is very excited

Let's consider the sequence ${\left( a_n \right)}_{n \in \mathbb{N} }$ such that

$\begin{cases} a_1=x & \\ \forall n \in \mathbb{N}, a_{n+1}=a_n! & \end{cases}$

for some $x \in \mathbb{N}$ .

Let's prove that:

$\textbf{1.)}$ $x=1 \implies \lim a_n=1$

$\textbf{Proof:}$

Let's considere $x=1$ . Let $X=\{n \in \mathbb{N} : a_n=1 \}$ .

We have that $a_1=1$ , which means that $1 \in X$ .

Let's suppose, by induction hypothesis, that $n \in X$ . We have that

$a_{n+1}=a_n!=1!=1$

Therefore, $n+1 \in X$ , which means that $X= \mathbb{N}$ .

As $\forall n \in \mathbb{N}, a_n=1$ , we conclude that

$\lim a_n=1$ .

$\textbf{QED}$

$\textbf{2.)}$ $x=2 \implies \lim a_n=2$

$\textbf{Proof:}$

Let's considere $x=2$ . Let $X=\{n \in \mathbb{N} : a_n=2 \}$ .

We have that $a_1=2$ , which means that $1 \in X$ .

Let's suppose, by induction hypothesis, that $n \in X$ . We have that

$a_{n+1}=a_n!=2!=2$

Therefore, $n+1 \in X$ , which means that $X= \mathbb{N}$ .

As $\forall n \in \mathbb{N}, a_n=2$ , we conclude that

$\lim a_n=2$ .

$\textbf{QED}$

$\textbf{3.)}$ $\forall x>2, \lim a_n \neq x$

$\textbf{Proof:}$

Let $x>2$ . Let´s prove, in first place, that ${\left( a_n \right)}_{n \in \mathbb{N} }$ is a strictly increasing sequence.

Let $Y= \{n \in \mathbb{N}: a_{n+1}-a_n>0 \}$ .

It is clear that $1 \in Y$ , because $a_2-a_1= x!-x=x \cdot \left( x-2 \right) >0$ .

Let's suppose, by induction hypothesis, that $n \in Y$ . We have that

$a_{n+2}-a_{n+1}=\left(a_n!\right)!-a_n!=$

$\left( a_n-1 \right)! \cdot \left( \frac{\left(a_n!\right)!}{\left( a_n-1 \right)!}-a_n \right) \geq \left( a_n-1 \right)!\cdot \left( a_n!-a_n \right)>0$

Therefore $n+1 \in Y$ , which means that $Y= \mathbb{N}$ .

Let suppose, by reductio ad absurdum, that $\forall n \in \mathbb{N}, a_n \leq x$ .

We have that

$a_2 \leq x \implies a_1! \leq x \implies x! \leq x$

which is an absurdum, because $x>2$ .

As ${\left( a_n \right)}_{n \in \mathbb{N} }$ is a strictly increasing sequence, this means that $\lim a_n \neq x$ .

$\textbf{QED}$

By $\textbf{1.)}$ and $\textbf{2.)}$ , we have that

$\left( \left( \left( 1! \right)! \right)! \right)!...=1$

and

$\left( \left( \left( 2! \right)! \right)! \right)!...=2$ ,

which means that $1$ and $2$ are solutions of the equation.

By $\textbf{3.)}$ , we have that $1$ and $2$ are the only solutions. So, the answer is $1+2=3$ .

1 and 2.

1!=1.

Also, 2! = 2 x 1 = 2.

so for these 2 numbers infinite number of factorials will still be the number itself.

so, answer = 1+2 =3.