I only obtained that $ x_{n}=\frac{(c-1)(n+1)(n+2)}{6}+1 $ when $ 3|c-1 $. what about the other cases?

\[ x_n= x_{n-1}+ \left\lfloor \frac{2x_{n-1}-(n+2)}{n} \right\rfloor +1 \] \[\implies x_n-1 = x_{n-1}-1 + \left\lfloor \frac{2(x_{n-1}-1)}{n}\right\rfloor \] Let $x_{n-1}-1=a_n$. Then, $a_n$ satisfies the simpler relation \[ a_n=a_{n-1}+ \left\lfloor \frac{2a_{n-1}}{n} \right\rfloor =\left\lfloor a_{n-1}\left( \frac{n+2}{n} \right) \right\rfloor \] We claim that : \[ a_n = \frac{(n+1)(n+2)(c-1)}{6} \text{ if } c \equiv 1 \pmod{3} \] \[ a_n = \frac{(n+1)(n+2)(c-2)}{6}+n \text{ if } c \equiv 2 \pmod{3} \] \[a_n = \frac{(n+1)(n+2)c}{6} - \left\lfloor \frac{(n+1)^2}{4} \right\rfloor \text{ if } c \equiv 0 \pmod{3} \] All of these claims are easily verifiable using induction on the recurrence relation for $a_n$. Each of these solutions increased by $1$ give a solution to $x_n$. It may seem that I made those claims a little abruptly. What I did was - I wrote down the first $6-7$ terms of the sequence as a linear expression of $ac-b$. Then I tried to understand the pattern followed by the co-efficients $a$ and $b$. To eliminate the denominators, I multiplied these by $3$ and found that the $3a$'s were basically triangular numbers. The $3b$'s were a little tougher situation. The first case was easy because here, $3b$'s were also triangular. In the second case, I found these to be of the form $T_n-T_{n-2}$. The third case was much different. I saw that every alternate number of the $b$'s is a square. And its not just a random square, but $=(n)^2$ for the $(2n-1)^{\text{th}}$ number in this sequence. This seemed to work in the case of even numbers too. All that's needed is a floor.

Another way to motivate the constructions above is the following. Call a sequence $\{x_i\}$ tasty if it satisfies the recursion in the problem. Observe that $x_1$ uniquely determines all the terms of a tasty sequence. Notice that if $\{x_i\}$ is tasty, then so is $\{x_i - \frac{(i+1)(i+2)}{2}\}.$ As a consequence of this, if we know the tasty sequence with $x_1 = c$, the tasty sequence with $x_1 = c + 3$ is just the first sequence plus $\frac{(n+1)(n+2)}{2}$ to all the terms. So we just have to solve the problem for $c = 1, 2, 3$, which is not hard by simply writing the terms and then guessing. $\square$