given $a_1, a_2$, the rest can be uniquely constructed

We claim that all solutions are $C>-2$. If $C\le -2$, define the sequence $\{a_n\}:a_1=1,a_2=2,a_{n+2}=\lceil -a_n-Ca_{n+1}\rceil (n\in \mathbb{Z}^+)$, and by induction on $n$ we can prove that $a_n>0$ and $a_{n+1}>a_n$ for all $n\in \mathbb{Z}^+$. Hence $\{a_n\}$ is not periodic,contradiction! $\blacksquare$ If $C>-2$, assume FTSOC that $\{a_n\}$ is not periodic.Let $M$ be a lower bound for $\{a_n\}$ Observe that the pair $(a_n,a_{n+1})$ can uniquely determine $a_{n-1}$ and $a_{n+2}$, so $\{a_n\}$ is not periodic implies that no two pairs $(a_{n_1},a_{n_1+1})$ and $(a_{n_2},a_{n_2+1})\ (n_1\ne n_2)$ are equal. Then the conclusion follows using the Pigeonhole Principle: Conclusion. $\{a_n\}$ is not bounded from above. Now we shall discuss the problem in two cases. Case 1: $C\ge 0$. Then we have $$(1+C)M+a_{n+1}<a_{n-1}+Ca_n+a_{n+1}<1\Rightarrow a_{n+1}\le 1-(1+C)M$$holds for all $n\ge 2$, which is a contradiction to the Conclusion. $\blacksquare$ Case 2: $C\in (-2,0)$. Define sequence $\{b_n\}:b_n=a_{n+1}-a_n(n\in \mathbb{Z}^+)$. Claim 1. If $a_n\ge 0(n\ge 2)$, then $b_n\le b_{n-1}$. Proof.From $C\in (-2,0)$ and $a_n\ge 0$ we have $\lceil -Ca_n\rceil \le 2a_n$. Hence $a_{n+1}=\lceil -a_{n-1}-Ca_n\rceil=-a_{n-1}+\lceil -Ca_n\rceil\le -a_{n-1}+2a_n$, which implies $a_{n+1}-a_n\le a_n-a_{n-1}$ also $b_n\le b_{n-1}$. $\square$ Claim 2. The sequence $\{b_n\}$ is not eventually constant. Proof.Assume FTSOC that there exist $N\in \mathbb{Z}^+$ and $a,d\in \mathbb{Z}$ such that $a_n=a+nd$ holds for all $n\ge N$. And from the Conclusion we have $d>0$. Hence $\forall n\ge N$, \begin{align}a+(n+2)d&=a_{n+2} \\ &=\lceil -a-nd-C(a+(n+1)d)\rceil \\ &>-a-nd-C(a+(n+1)d)-1\end{align}which implies $-Cd(n+1)<2d+1+aC$ holds for all $n\ge N$, and from $C<0$ and $d>0$ there's a contradition. $\square$ Claim 3. There exist infinitely many $n\in \mathbb{Z}^+$ such that $a_n<0$. Proof.Assume FTSOC that $\exists M\in \mathbb{Z}^+$ such that $\forall n\ge M,a_n\ge 0$. Then from Claim 1 and Claim 2 we have: $b_n<0$ holds for sufficiently large $n$s($n\ge N$), which implies that $a_{n+1}<a_n$. Hence the sequence $\{a_n\}(n\ge N)$ is a sequence of non-negative integers which is strictly decreasing, which is obviously a contradiction. $\square$ Claim 4. If $a_n<0$, then $a_{n+1}<1-M$. Proof.$M+0+a_{n+1}<a_{n-1}+Ca_n+a_{n+1}<1$$\square$ Now let's denote $A=\{x\mid M\le x<0,x\in \mathbb{Z}\}$, $B=\{x\mid M\le x<1-M,x\in \mathbb{Z}\}$. Then from Claim 3 and Claim 4 we can find infinitely many positive integers $n$ such that $(a_n,a_{n+1})\in A\times B$. But $A$ and $B$ are finite sets, hence from the Pigeonhole Principle we can find $n_1$ and $n_2(n_1\ne n_2)$ satisfying $(a_{n_1},a_{n_1+1})=(a_{n_2},a_{n_2+1})$, a contradiction! $\blacksquare$ From above we know that all solutions are $C>-2$.

We proceed indirectly; assume for contradiction that for some $C>0,$ exists a non-periodic sequence of integers $\{a_n \}_{n \ge 1}$ bounded from bellow and that satisfies $(*).$ Firstly, note that given the first two terms of the sequence, the sequence is uniquely determined by $$ a_{n+1} = - \lfloor Ca_n \rfloor - a_{n-1}$$Further, given the values of $a_i$ and $a_{i+1},$ for some $i\ge 3,$ we can uniquely determine the values of $a_{i-1}$ and $a_{i+1}.$ So if exists two different indexes $i,j$ such that $a_i = a_j$ and $ a_{i+1} = a_{j+1}$, then the sequence is periodic. So $\{a_n \}_{n \ge 1}$ is not bounded above, beacause otherwise there would be a finite quanity of posible values for the pairs $(a_i,a_{i+1})$ and then there would be two equal, implying the sequence is periodic. So $\{a_n \}_{n \ge 1}$ is not bounded above. One one hand, since the sequence is bounded bellow, exists some constans $M$ so that $a_n > M, \forall n \ge 1.$ On the other hand, since we just proved that the sequence is not bounded above, there must exists $n \ge 1$ such that $a_n > {(-2M)}{/C}.$ Hence, since $C>0,$ we must have $$ a_{n+1} = - \lfloor Ca_n \rfloor - a_{n-1} < - \left\lfloor C \cdot \frac{-2M}{C} \right\rfloor -M =M $$which contradicts the definition of $M. \square$

$\bullet$ If $C=0,$ then we must have $a_{n+1} = -a_{n-1}$ and then the sequence is clearly peridoic. $\bullet$ If $C = -2,$ then clearly every arithmetic progression satisfies it so that $-2 \notin \mathcal{A}.$ $\bullet $If $C=-1,$ we must have from the theory of linear recursion that $$ a_n = P{\alpha}^n +Q{\beta}^n$$where $ \alpha = \frac{1}{2} + i\frac{\sqrt{3}}{2}= cis ({\pi}{/3}), \beta = \frac{1}{2} -i\frac{\sqrt{3}}{2}$ (where $i$ is the imaginary unit) are the roots of the polynomial $x^2-x+1=0$ and $P,Q$ are complex constants. Note that since $\beta = \overline{\alpha}$ and $a_n \in \mathbb{R}$ we must have that $Q = \overline{P}.$ So let $P = |P|cis (\theta)$ for some $\theta \in \mathbb{R}.$ Then $$ a_n = P{\alpha}^n + \overline{P}{\overline{\alpha}}^n = |P|(cis \left(\theta + \frac{n\pi}{3} \right ) + cis \left(-\theta - \frac{n\pi}{3} \right ))=2|P|\cos \left(\theta + \frac{n\pi}{3} \right )$$So that $|a_n| < 2|P|,$ so that as in the proof of Claim 1, we must have that $\{a_n \}_{n \ge 1}$ is periodic. $\bullet$ Finally if $C <-2,$ we must have again from the theory of linear recursions that $$ a_n = P{\alpha}^n +Q{\beta}^n$$where $\alpha$ and $\beta$ are the roots of the polynomial $x^2+xC+1$ (which in this case are real) and $P, Q \in \mathbb{R}$ (since the linear system generated by the first two values must yeld a real solution). We also have $\alpha + \beta = -C> 2$ and $\alpha \cdot \beta = 1;$ so that if we assume without loss of generality that $\alpha \ge \beta,$ we find $\alpha > 1 > \beta >0.$ So that $$ \lim_{n \rightarrow \infty} a_n = \pm \infty$$Hence the sequence $\{a_n \}_{n \ge 1}$ is non-periodic.

Suppose for contradiction that for some $0<k<1$ exists some non periodic sequence of integers $\{a_n \}_{n \ge 1}$ bounded from bellow that satisfies $(*).$ As we have already shown, $\{a_n \}_{n \ge 1}$ cannot be bounded above. Since the sequence is bounded bellow, let $M$ be a constans so that $a_n > M$ \foall $n \ge 1.$ Now since the sequence is not bounded above, take $n$ so that $a_n> {M}{/ (k-1)}$; then $a_{n+1} \ge a_n $ beacause otherwise we would have $$ a_{n+2} = \lceil ka_{n+1} \rceil -a_{n} < \lceil ka_n \rceil - a_n \le 1 +a_n(k-1)< 1+M$$which contradicts the definition of $M.$ So that $a_{n+1} \ge a_n >{M}{/ (k-1)}$ and it follows by induction that ${a_i}_{i=n}^{\infty}$ is weakly increasing. Then since the sequence is unbounded from above, for some $N$ we must have $a_{n+1} \ge a_n \ge a_{n-1} >0$ which implies since $k<1,$ $$ a_n \le a_{n+1} = \lceil ka_n \rceil -a_{n-1} \Rightarrow a_{n-1} \le \lceil (k-1)a_{n} \rceil \le0$$contradiciton. So done!

We explicit show a sequence that satisfies the conditions and it is not periodic. Start at any $a_2, a_1$ such that $a_2 > a_1 >0.$ We claim that by induction $a_n \ge a_{n-1} >0, \forall n \ge 2.$ For the step, we have $$ a_{n+1} = \lceil ka_n \rceil - a_{n-1} \ge 2a_n - a_{n-1} \Rightarrow a_{n+1} -a_{n} \ge a_n - a_{n-1} \ge 0 \Rightarrow a_{n+1} \ge a_n >0$$Then the sequence is weakly increasing; so that if the sequence were periodic, all terms would be equal, which cannot happen since it starts with $a_2>a_1.$

We again proceed indirectly; assume for contradiciton that exists a non-periodic sequence of integers $\{a_n \}_{n \ge 1}$ bounded from bellow such that $$ a_{n+1} = \lceil ka_n \rceil -a_{n-1}, \forall n \ge 2$$where $1<k<2$ is a constant. First, we claim that there exist infinitely many $t$ such that $a_t<0.$ Indeed, assume for contradiction that no, so that $a_n>0,\forall n\ge G$ for some constant $G;$ we divide into two cases which both lead to contradition Case 1: If $\{a_n\}_{n \ge G}$ is not increasing. Click to reveal hidden textIf we have $a_n< a_{n-1}$ for some $n>G$ we would have $$ a_{n+1} =\lceil ka_n \rceil -a_{n-1} \le 2a_n -a_{n-1} \Rightarrow a_{n+1}-a_n \le a_n-a_{n-1} <0 \therefore a_{n+1}<a_n $$So it follows by induction that $0< a_{i} < a_{i-1}, \forall i \ge n$ which is clearly a contradiciton. Case 2:If the sequence $\{a_n\}_{n \ge G}$ is increasing Click to reveal hidden textThen $a_{n+1} \ge a_n >0, \forall n\ge G$ so that $a_{n+1} -a_n \ge 0, \forall n \ge G$ and $$a_{n+1} = \lceil ka_n \rceil - a_{n_1} \le 2a_n - a_{n-1} \Rightarrow a_{n+1}-a_n \le a_n -a_{n-1} $$So the sequence $\{a_{n+1} -a_n\}_{n\ge G}$ is a sequence of non-increasing positive integers, so it is eventually constant, that is $\{a_n\}_{n \ge 1}$ is eventually an arithmetic progression. That is, for some constant $H$ and integers $p,q$ we have $a_n = p +qn, \forall n \ge H.$ Looking at the original equaiton we have $$ 0\le p+q(n-1) -k(p+qn) + p+q(n+1)<1 \iff 0\le (2-k)(p+qn) <1, \forall n\ge H$$since $2-k \ne 0,$ the only way that the above inequlaity can hold for all $n \ge H$ is if $q=0,$ that is, the sequence is eventually constant, which is clearly a contradiction since the sequence should not be bounded from above. So indeed there must exist infinitely many $t$ such that $a_t \le 0.$ Then, since the sequence $\{a_n\}_{n \ge 1}$ is bounded from bellow, it follows that exist a constas $M$ such that $a_n > M, \forall n \ge 1;$ so if $a_t< 0,$ we must have $$ a_{t+1} = \lceil ka_t \rceil - a_{t-1} < -M$$Now look at the pairs $(a_t,a_{t+1})$ where $a_t<0;$ by above there are finitely many posible values of this pairs and since there are infinitely many of them, we must have $(a_i,a_{i+1}) = (a_j, a_{j+1})$ for different numbers $(i,j)$ which implies that the sequence is periodic which is a contradiction.