Lucas–Lehmer test and triangle of coefficients of Chebyshev's In the Lucas–Lehmer test with $ \quad p \quad $ an odd prime.
we know that $ \quad S_0=4 \quad $ and $ \quad S_i=S_{i-1}^2-2 \quad $ for $\quad i>0 \quad$
$M_p=2^p-1 \quad$ is prime if $ \quad S_{p-2} \equiv 0 \bmod {(2^p-1)}$
after some observations i found a link with     Triangle of coefficients of Chebyshev's OEIS A053122
$a(n,k)=(-1)^{n-k}\left(\begin{matrix} n+1+k \\ 2 \cdot k+1 \end{matrix}\right)$
from which
$S_{i+1} \cdot S_{i+2}\cdot \quad \ldots \quad \cdot S_{i+m} = \sum\limits_{k=0}^{2^m-1}{(-1)^{k+1}\left(\begin{matrix} 2^m+k \\ 2 \cdot k+1 \end{matrix}\right) \cdot (S_i^2)^k }$
can anyone help me find any reference or pointer on how to derive this result?

By the composition property of Chebyshev polynomials $T_m(T_n(x))=T_{mn}(x)$. Since $x^2-2 = 2T_2(\tfrac{x}2)$, we have $S_i = 2T_{2^i}(2)$ for all $i\geq 0$.
Furthermore, since $T_{2^k}(x) = \frac{U_{2^{k+1}-1}(x)}{2U_{2^{k}-1}(x)}$, we have
\begin{split} S_{i+1}\cdot S_{i+1}\cdots S_{i+m} &= 2^m T_{2^{i+1}}(2)\cdot T_{2^{i+2}}(2)\cdots T_{2^{i+m}}(2) \\
&=\frac{U_{2^{m+i+1}-1}(2)}{U_{2^{i+1}-1}(2)} \\
&= U_{2^m-1}(T_{2^{i+1}}(2))\\
&= \frac{U_{2^{m+1}-1}(T_{2^i}(2))}{2T_{2^i}(2)}\\
&= \frac{U_{2^{m+1}-1}(\tfrac{S_i}2)}{S_i}.
\end{split}
and it remains to use an explicit expression for $U_{2^{m+1}-1}(x)$ to obtain the formula in question.
