Let $r(n)=\left\lceil\frac{n}{4}\right\rceil$, where $k \in \mathbb{N}$. Note that this can be rewritten as a recursion $R(n)=R(n-4)+1$, where, $R(0)=0, R(1)=1, R(2)=1, R(3)=1$. Since this recursion is linear, we can solve it by solving its associated characteristic polynomial, $\lambda^4-1=0$. We can therefore say that $R_{Homogenous}(n)= c_0(\omega^0)^n+c_1(\omega^1)^n+c_2(\omega^2)^n+c_3(\omega^3)^n= c_0+c_1\omega^n+c_2\omega^{2n}+c_3\omega^{3n} $. The second step involves using the "method of undetermined coefficients" to find the solution to the non-homogeneous equation. Since our non-homogeneous term is $P(n)=1$, we should set our "guess" $P^\prime(n)=b_0$. However, since $\omega^0=1=1^n$, we have to account for that and reset $P^\prime(n)=b_0n$. Therefore, $R(n)-R(n-4)=b_0n-b_0(n-4)=1\implies b_0=1/4$. Therefore, $R(n)=c_0+c_1\omega^n+c_2\omega^{2n}+c_3\omega^{3n}+\frac{1}{4}$. Now we just need to find out what the co-efficients are for the $c_i$. We do this by remembering the initial conditions. Using them, we'll get 4 constraints, which are:
$R(0)=c_0+c_1+c_2+c_3+\frac{1}{4}=0$
$R(1)=c_0+c_1\omega+c_2\omega^{2}+c_3\omega^{3}+\frac{1}{4}=1$
$R(2)=c_0+c_1\omega^2+c_2\omega+c_3\omega^{2}+\frac{1}{4}=1$
$R(3)=c_0+c_1\omega^3+c_2\omega^{2}+c_3\omega+\frac{1}{4}=1$
Let's subtract both sides by $\frac{1}{\sqrt{4}}$ and divide both sides by 4. If we try to solve for $c_0, c_1, c_2, c_3$ using matrices, we'll see that:
$ \frac{1}{\sqrt{4}}Ac=\frac{1}{\sqrt{4}} \begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & \omega & \omega^2 & \omega^3\\ 1 & \omega^2 & 1 & \omega^2\\ 1 & \omega^3 & \omega^2 & \omega\\ \end{pmatrix} \begin{pmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ \end{pmatrix} = \begin{pmatrix} -\frac{1}{8} \\ \frac{3}{8} \\ \frac{3}{8} \\ \frac{3}{8} \\ \end{pmatrix} $
Fortunately, matrix $\left(\frac{1}{\sqrt{4}}A\right)$ is symmetric and unitary so the inverse is also the complex conjugate. Therefore:
$ \begin{pmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ \end{pmatrix} =\frac{1}{\sqrt{4}} \begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & \overline{\omega} & \overline{\omega^2} & \overline{\omega^3}\\ 1 & \overline{\omega^2} & 1 & \overline{\omega^2}\\ 1 & \overline{\omega^3} & \overline{\omega^2} & \overline{\omega}\\ \end{pmatrix} \begin{pmatrix} -\frac{1}{8} \\ \frac{3}{8} \\ \frac{3}{8} \\ \frac{3}{8} \\ \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & \overline{\omega} & \overline{\omega^2} & \overline{\omega^3}\\ 1 & \overline{\omega^2} & 1 & \overline{\omega^2}\\ 1 & \overline{\omega^3} & \overline{\omega^2} & \overline{\omega}\\ \end{pmatrix} \begin{pmatrix} -\frac{1}{16} \\ \frac{3}{16} \\ \frac{3}{16} \\ \frac{3}{16} \\ \end{pmatrix} $
Therefore:
$c_0=-\frac{1}{16}+\frac{3}{16}+\frac{3}{16}+\frac{3}{16}=\frac{1}{2}$
$c_1=-\frac{1}{16}+\frac{3}{16}\overline{\omega}+\frac{3}{16}\overline{\omega^2}+\frac{3}{16}\overline{\omega^3}=-\frac{1}{4}$
$c_2=-\frac{1}{16}+\frac{3}{16}\overline{\omega^2}+\frac{3}{16}+\frac{3}{16}\overline{\omega^2}=-\frac{1}{4}$
$c_3=-\frac{1}{16}+\frac{3}{16}\overline{\omega^3}+\frac{3}{16}\overline{\omega^2}+\frac{3}{16}\overline{\omega}=-\frac{1}{4}$
Therefore, we get that $R(n)=\frac{1}{2}-\frac{1}{4}\omega^n-\frac{1}{4}\omega^{2n}-\frac{1}{4}\omega^{3n}+\frac{1}{4}$. However, although $R(0)=0, R(1)=1, R(2)=1, R(3)=1$, we see that $R(n)=R(n-4)$, NOT, $R(n)=R(n-4)+1$.
Where could I have gone wrong?