sequence such that $x_{n+1}=n(x_n-n)$ Let $(x_n)$ be a sequence such that $x_{n+1}=n(x_n-n)$
Prove that $x_n=O(n)$ if and only if $x_1=2e$
If we have $x_n=O(n)$ then clearly $x_n=n+O(1)$ so they are equivalent.
I don't get how $x_1$ relates to this. Well it's a recurrent series but it's not of the form $x_{n+1}=f(x_n)$

Your question is about a sequence that satisfies
$$ x_{n+1} = n(x_n-n) \tag{1} $$
where initial value $\,x_1\,$ is given. This is a
linear recursion so that
$$ x_{n+1} = n!\,x_1-a_n \tag{2} $$
where $\,a_n\,$ does not depend on $\,x_1.\,$ A
quick search of OEIS finds that $\,a_n\,$ is the
OEIS sequence A030297.
The sequence entry give the formula
$$ a_n = \lfloor 2\,e\,n!-n-2\rfloor \tag{3} $$
for $\,n>1.\,$ This leads to the approximate result
$$ x_{n+1} = n!(x_1- 2\,e)+n+2 +O(1). \tag{4} $$
This implies that unless $\,x_1 = 2\,e\,$ then the
$\,n!\,$ term dominates.
Not needed here, but a more precise asymptotic
series for $\,a_n\,$ is
$$ a_n = 2\,e\,n! - b_{n+1} \tag{5} $$
where
$$ b_n \approx
n + 1 + \frac2n + \frac2{n^2}\sum_{k=0}^\infty
 \frac{c_n}{n^k} \tag{6} $$ and where $\,c_k\,$ is the
OEIS sequence A014182.
