Can any number theorists explain the patterns going on in this factorization? (Sorta relevant to cyclotomic polynomials.)

I would mark their roots in the complex plane, and try to see a pattern there visually
Your writeup provides the background information that makes this a lot clearer.

Using the formula for E\_k at the start of section 7, we have

C\_k(x) = (1/2) \* ((1 + ix)^(2k) + (1 - ix)^(2k)) / (1 + x^(2))^k

and then by doing the rearrangement, and using C^' to denote the polynomials in your question because of Reddit, we get

4(C^(')\_k(x))^2 = 2(1 + x^(2))^k - (1 + ix)^(2k) - (1 - ix)^(2k).

Noticing that 1 + x^2 = (1 + ix)(1 - ix), the right hand side is seen to be -((1 + ix)^k - (1 - ix)^(k))^(2). Therefore

C^(')\_k(x) = (i/2)((1 + ix)^k - (1 - ix)^(k)).

The +/- ambiguity in taking the square root has been dealt with in this situation by desiring the first order term to be kx rather than -kx, but strictly speaking nothing in your definition stipulates this.

Anyway, with this expression in hand, we note that the main term is of the form a^k - b^k which, if m is a divisor of k, is divisible by a^m - b^(m). Therefore at the very least, if m divides k then C^(')\_m(x) divides C^(')\_k(x). For a further result, note that a^k - b^k = b^(k)((a/b)^k - 1) and now we have a product of cyclotomic polynomials in a/b. By distributing out the powers of b according to the degree of each term, we get a factorisation in terms of polynomials of a and b. Then plugging in a = 1 + ix and b = 1 - ix we get our nicer factorisation. I do not immediately see if the factors you get after the substitution are amenable to further factorisation.
No idea how to prove it yet, but looks like a strong divisibility sequence to me.

Moreover, if a is a root, then so is -a. So if f(X) is an irreducible factor, so is f(-X).
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Sorry isn't the k=0 case just equal to 1/2 - 1/2 = 0 or am I just blind..
Well there's definitely enough information here to speculate about some of the later even-suffixed functions, but the whole underlying pattern is definitely beyond me.

Noting that ~C_2p(x) for p prime seems to be divisible by p whereas ~C_p(x) is only "nearly" divisible, i.e. all coefficients are divisible by p except for the lead which has a coefficient of 1, it seems that Pascal's triangle would appear to be very strongly related to whatever is happening even if there's something else going on.

Then there's the sign flipping on odd-powered terms whenever a factor is "duplicated".

Speaking of the duplications, write ~C_n(x) in terms of 1/x and a (near?) symmetry presents itself.
Easy. It simply counts 1 to 16.  Look at the C on the left. Duh
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The first thing that jumps out at me is that if a|b then Ca(x)|Cb(x).

However, if ab|d, it is not necessarily true that Ca(x)*Cb(x)|Cd(x)
This was a long time ago, but maybe this can be rewritten as a finite convolution.
Yes
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Eisenstein Criterion is going crazy.

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