This is the answer from Zuber's note.
Consider the tensor product of two representations of spin $j_1$ and $j_2$ and their decomposition on vectors of given total spin (decomposition into irreducible representation). We start with the product representation spanned by the basis,
\begin{equation}\label{basis1}
|j_1,m_1\rangle\otimes |j_2,m_2\rangle
=|j_1,m_1;j_2,m_2\rangle\equiv
|m_1,m_2\rangle
\end{equation}
on which the generators act as,
\begin{align}
\mathbf{J}=\mathbf{J}^{(1)}\otimes \mathbf{1}^{(2)}+\mathbf{1}^{(1)}\otimes \mathbf{J}^{(2)}
\end{align}
which is frequently written as,
\begin{align}
\mathbf{J}=\mathbf{J}^{(1)}+ \mathbf{J}^{(2)}
\end{align}
and in quantum mechanics we talk about the addition of angular momenta $\mathbf{J}^{(1)}$ and $\mathbf{J}^{(2)}$. The upper index indicates on which space the operators act.
Thus the decomposition of the tensor product becomes to decompose the basis of $\mathbf{J}^{(1)}$ and $\mathbf{J}^{(2)}$ onto a basis of eigenvectors of $\mathbf{J}$ and $J_z$. Since ${\mathbf{J}^{(1)}}^2$ and ${\mathbf{J}^{(2)}}^2$ commute with one another and with $\mathbf{J}^2$ and $J_z$, we may seek common eigenvectors which can be denoted by $|J,M\rangle$,
\begin{align}
|J,M\rangle\equiv |J,M,(j_1,j_2)\rangle
\end{align}
where it is understood that the value of $j_1$ and $j_2$ is fixed. The question is thus twofold: which values can $J$ and $M$ take; and what is the matrix of the change of basis from $|m_1,m_2\rangle$ to $|J,M\rangle$? In
other words, what are the Clebsch-Gordan
coefficients?
The possible values of $M$, eigenvalue of $J_z$ can be found,
\begin{align}
\langle m_1,m_2|J_z|J,M\rangle
&=M\langle m_1,m_2|J,M\rangle\nonumber\\[5pt]
&=\langle m_1,m_2|J^{(1)}_z+J^{(2)}_z|J,M\rangle
=(m_1+m_2)\langle m_1,m_2|J,M\rangle
\end{align}
If $\langle m_1,m_2|J,M\rangle\neq 0$, the only possible value of $M$ is
\begin{align}\label{keyeq}
M=m_1+m_2,\,\,\,\,\,\,
m_1\in(-j_1,\cdots,j_1),\,\,\,\,\,\,
m_2\in(-j_2,\cdots,j_2)
\end{align}
For fixed $j_1, j_2$ and $M$, there are many possible basis $|m_1,m_2\rangle$ with eigenvalue of $M$. With fixed $j_1, j_2$ and $M$, the number of solution to the above equation is given by $n(M)$,
\begin{align}
n(M)=\left\{
\begin{aligned}
0&,\,\,\,\,\,\,\mathrm{if}\,\,\,|M|>j_1+j_2\\[5pt]
j_1+j_2+1-|M|&,\,\,\,\,\,\,
\mathrm{if}\,\,\,|j_1-j_2|\leq|M|\leq j_1+j_2\\[5pt]
2\,\mathrm{min}\,(j_1,j_2)+1&,\,\,\,\,\,\,
\mathrm{if}\,\,\,0\leq|M|\leq|j_1-j_2|
\end{aligned}
\right.
\end{align}
Let $N_J$ be the number of times the
representation of spin $J$ appears in the decomposition of the representations of spin $j_1$ and $j_2$. The $n(M)$ basis of eigenvalue $M$ for $J_z$ may also be regarded as coming from the $N_J$ basis $|J,M\rangle$ for the different values of $J$ compatible with that value of $M$,
\begin{align}
n(M)=\sum_{J\geq|M|} N_J
\end{align}
By subtracting two such relations, we have
\begin{align}
N_J=n(J)-n(J+1)
=\left\{
\begin{aligned}
1&,\,\,\,\,\,\,
\mathrm{if}\,\,\,|j_1-j_2|\leq|M|\leq j_1+j_2\\[5pt]
0&,\,\,\,\,\,\,
\mathrm{otherwise}
\end{aligned}
\right.
\end{align}
For example, the solution of $n(M)$ for $j_1=\frac{5}{2}$ and $j_2=1$ can be shown figure
We have shown that the $(2j_1+1)(2j_2+1)$ vectors with basis
$\mathbf{J}^{(1)}$ and $\mathbf{J}^{(2)}$ with $j_1$ and $j_2$ fixed may be reexpressed in terms of vectors $|J,M\rangle$ with
\begin{align}
J&=|j_1-j_2|,|j_1-j_2|+1,\cdots,j_1+j_2\\
M&=-J,-J+1,\cdots,J
\end{align}
