Analytical solution for $\max{x_1}$ in $(x_n)_{n\in\mathbb{N}}$ Let be $x_1,x_2,x_3,\ldots,$ a sequence of positive integers. Suposse the folowing conditions are true for all $n\in\mathbb{N}$


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*$n|x_n$

*$|x_n-x_{n+1}|\leq 4$
Find the maximun value of $x_1$
I can't solve this analytically, I've start with $x_1=50$ and construct a sequence that holds the conditions, so $\max{x_1}\geq 50$ and repeat it. I found thath $60\leq\max{x_1}\leq 63$
So the question is, How to solve it? Because search for $x_1=10,30,40,50,60,61,62,63$ its not elegant.
$\textbf{Edit:}$


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*If we take $x_1$=50, then $x_2\in\{46,\not{47},48,\not{49},50,\not{51},52,\not{53},54\}$, my strategy is go down in numbers to have for some $k$ that $x_k=4k$, example: $x_2$=46, $x_3=42$, $x_4=40$, $x_5=40$, $x_6=36$, $x_7=35$, $x_8=4\times8=32$ so from here we take for $n>8:x_n=4n$ and the sequence exists, so $\max{x_1}\geq50$. 


If $n\ge 9$ then there is at most $1$ multiple of $n+1$ within distance $4$ of $x_n$, and there is at most $1$ multiple of $n$ within distance $4$ of $x_{n+1}$, so a term determines and is determined by the next term. Hence the terms of the sequence appearing for $n\ge 9$ are injectively determined by $x_9$
We clearly have $x_n \le x_1 + 4n$. For $n > x_1$, $(x_1+4n)/n < 5$ and so we must have $x_n/n \in \{1;2;3;4\}$. By the previous point, this implies that $(x_9,x_{10},x_{11},\ldots)$ must be one of the $4$ sequences $(9a,10a,11a,\ldots)$ for $a \in \{1;2;3;4\}$
This leaves you finitely many choices for $x_9$, ($9,18,27,36$). Now you can go back step by step and determine all the possible values for $x_8,x_7,$ and so on downto $x_1$.
