a) We can see that adding a constant $c$ to all $u_n$ does not change the formula for $u_n$, so if we consider all pairs $\alpha = 0$ and $\beta = r$, $r \geq 0$, then considering the sequence shifted by a constant $c = 2^{2022} - u_{2023}$, we get $u_{2023} = 2^{2022}$. b) Suppose that some element is divisible by $p = 7$ or $p = 17$. Considering residue $u_n$ modulo $p$ we get that it must be sequence must repeat, but by backward induction we get that the cycle must contain $0$, so there are infinitely many elements divisible by $p$, now just take any $n_0$ and big enough $M$ such that in $u_{n_0}, u_{n_0 + 1}, \ldots, u_{n_0+M}$ there are at least $2023$ numbers divisible by $p$, then for every $m \geq M$ $u_{n_0}u_{n_0+1}\ldots u_{n_0+m}$ is divisible by $p^{2023}$. Also we can't have that ii) is true because $p ~ | ~ 2023 = 7\cdot17^2$. Suppose now that none of $u_n$ are divisible by $7$ or $17$, then again we have that the sequence is in a cycle mod $2023$, let the smallest cycle be $u_{n_0}, u_{n_0}, u_{n_0 + 1}, \ldots, u_{n_0+T}$, and let $u_{n_0}u_{n_0 + 1}\ldots u_{n_0+M} = A \not\equiv 0 \pmod {2023}$. We can see that $A^{\varphi(2023)} \equiv 1 \pmod{2023}$, so letting $k = M\varphi(2023)t = 1632Mt$ for $t \geq1$, we have $u_{n_0}u_{n_0+1}\ldots u_{n_0+k} \equiv A^{1632t} \equiv 1 \pmod {2023}$. The i) can't be true because no elements are divisible by $7$ or $13$.