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 $x_{n+1}=f(x_n)$

By some experimentation from the starting point $x_n=an+b+y_n$ and reporting in the equation I quickly found that $x_n=n+1+2y_n$ lead to the following simplification
$$y_{n+1}=ny_n-1$$
So we have transformed the $-n^2$ term into a constant $-1$ which is better.
Now we can set $y_n=(n-1)!\,z_n$ so as to make $n!$ appear on both sides of the equation.
$$z_{n+1}-z_n=\frac {-1}{n!}$$
This is a telescoping relation therefore $z_n=z_1-\sum\limits_{k=1}^{n-1}\frac 1{k!}$
We have $z_n\to z_1+(1-e)$ so to have $x_n=O(n)$ we need to cancel this with $z_1=e-1$
Which is equivalent to

$x_1=2+2y_1=2+2z_1=2+2(e-1)=2e$

