Prove that $\lfloor n\sqrt{2}\rfloor $ doesn't satisfy any linear recurrence with constant coefficients. Prove that $\lfloor n\sqrt{2}\rfloor $ doesn't satisfy any linear recurrence with constant coefficients. 
Using one of the formula for floor function, we can write:
$$ \lfloor n\sqrt{2}\rfloor =\frac{-1}{2}+n\sqrt{2}+\frac{\arctan \cot{n\pi \sqrt{2}}}{\pi}.$$ Can we use this representation somehow to prove our claim.
 A: If $\,a_n(x) := \lfloor n\,x\rfloor\, $ satisfied any linear recurrence with constant coefficients, and letting $\,y := x-\lfloor x\rfloor,\,$ then so would $\,a_n(y),\,$ and since $\,0\le y<1,\,$ the first difference sequence $\,b_n(y) := a_{n+1}(y) - a_n(y)\,$ is one also and satisfies $\, b_n(y) \in \{0,1\}$ because
for all real $\,z\,$, we have $\,\lfloor z\rfloor \le \lfloor z+y \rfloor \le \lfloor z\rfloor+1.\,$ But now $\,b_n(y)\,$ must be eventually periodic and this implies $\,y\,$ is rational and then so is $\,x.\,$ This is from the theorem that $\,\lfloor n\,x\rfloor - n\,x\,$ is periodic if and only if $\,x\,$ is rational.
Notice that any recurrence of the form
 $$\,a_{n+k} = f(a_n, a_{n+1}, \dots, a_{n+k-1}) \tag{1}$$ for all $n$ with a fixed $k$ and such that the range of the sequence is a finite set, then there is only a finite number of $k$-tuples and so the function on $k$-tuples given by $$ (a_n,\dots,a_{n+k-1}) \mapsto \,(a_{n+1},\dots,a_{n+k}), \tag{2}$$ where $\,a_{n+k}\,$ is given by equation $(1)$, maps a finite set into itself and hence must be eventually periodic. 
