Explicit expression for polynomial recurrence relation. I am (as here) working on Machin formula and with some works I'd like to explore the following polynomial recurrence relation :
$$P_0 = 1, P_1 =b,\quad P_{n+2} = XP_{n+1} - P_n$$
It seems relation to Chebyshev_polynomials but I am unfamiliar with these polynomials. 
Is it possible to have an explicit expression for $P_n$?
 A: Let $a_n(X) = T_n(\frac{X}{2})$ and $b_n(X)= U_n(\frac{X}{2})$, where $T_n,U_n$ are the Chebyshev polynomials. Then satisfy our recurrence $a_{n+2}(X)=xa_{n+1}(X)-a_n(X)$ and the same for $b_n(X)$. 
Now find rational expressions in $X$,  $p(X),q(X)$ so that:
$$p(X)a_0(X)+q(X)a_1(X)=p(X)+q(X)=1\\p(X)a_1(X)+q(X)b_1(X)=p(X)\frac{X}{2}+q(X)X = b$$
The second can be rewritten as:
$$p(X) + 2q(X) =\frac{2b}{X}$$
So $q(X) =\frac{2b-X}{X}$ and $p(X)=\frac{2X-2b}{X}$.
So the sequence of rational functions:
$$Z_n(X) = \frac{2X-2b}{X} T_n\left(\frac{X}{2}\right) + \frac{2b-X}{X}U_n\left(\frac{X}{2}\right)$$
satisfy the recurrence, $Z_0(X) = 1$, and $Z_1(X)=b$. So the $Z_n(X)$ are actually polynomials, and $Z_n(X)=P_n(X)$.
So $P_n(X)=Z_n(X)$, and you are done.
The more general solution would be to use the general solution to linear recurrences to write:
$$P_n(X) = c_1(X)\left(\frac{X+\sqrt{X^2-4}}{2}\right)^n + c_2(X)\left(\frac{X-\sqrt{X^2-4}}{2}\right)^n$$
and solving for $c_1(X)$ and $c_2(X)$ to solve the initial condition $P_0(X)=1$ and $P_1(X)=b$. This will yield $c_1(X),c_2(X)$ that are not even rational functions of $X$, but rational functions of $X$ and $\sqrt{X^2-4}$, but the end expression will represent a polynomial.
