I was wondering if the expression:

$$ C\_n = 2 C\_1 C\_{n-1} - C\_{n-2} $$

could be expressed as a closed expression in terms of (hopefully polynomials of) $C\_1$ (or $C\_2$). The bases cases for this recursion are:

$$ C\_1 = C\_1 $$
$$ C\_2 = 2 C^2\_1 - 1$$

and for $ n < 1 $, $C\_n$ is undefined.

I've tried the natural/obvious method of plugging the recursion back into itself but wasn't sure how to terminate it properly:

$$ C\_n = 2 C\_1 C\_{n-1} - C\_{n-2} $$

$$ C\_n = 2C\_1 (2 C\_1 C\_{n-2} - C\_{n-3}) - C\_{n-2} = 2^2 C\_1^2 C\_{n-2} - 2 C\_1 C\_{n-3}$$

$$ C\_n = 2C\_1 (2C\_1 (2C\_1 C\_{n-3} - C\_{n-4}) - C\_{n-3}) - C\_{n-2} = 2^3C\_1^3 C\_{n-3} - 2^2 C\_1^2 C\_{n-4} - 2 C\_1 C\_{n-3} - C\_{n-2} $$

$$ C\_n = 2C\_1 (2C\_1 (2C\_1 (2 C\_1 C\_{n-4} - C\_{n-5}) - C\_{n-4}) - C\_{n-3}) - C\_{n-2} = 2^4 C\_1^4 C\_{n-4} - 2^3 C\_1^3 C\_{n-5} - 2^2 C\_1^2 C\_{n-4} - 2 C\_1 C\_{n-3} - C\_{n-2} $$

Which I believe has the following pattern:

$$ C\_n = 2^{n-1} C^{n-1} C\_1- \sum^K\_{i=0}C^i 2^i C\_{n-2 - i}$$

Where $K$ can be chosen such that the summation stops at some base case ($C\_1 = C\_1 $ or $ C\_2 = 2 C^2\_1 - 1$). For example $K = n-4$ or $K = n-3$.

After that I wasn't 100% sure how to proceed but the idea is that I want that formula to be expressed only as a function of $C\_1$ (or $C\_2$). I was hoping that it was expressed like something of the following form:

$$ C\_n = \sum^n\_{j} a\_j C\_1^j $$

I suspect that it can be done because as someone pointed at its related to the formula $cos(n \theta) = $ and its part of something similar to pascal's triangle and I was hoping the coefficients had some nice expression.

Also, I have a proof (by induction) that $C\_n$ is a function of $C\_1$ (which I will include later) but I seem unable to get the actual closed for expression for this the coefficients. Feels so close!

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Edit:

At this point I've been told that the expression for $C\_n$ is:

$$ C\_n = \frac{1}{2} \left( \left( C\_1 + \sqrt{C\_1^2 -1} \right)^n + \left( C\_1 - \sqrt{C\_1^2 - 1} \right)^n \right) $$

But I will be completely honest, I am not 100% sure where that came from.