Coefficients of 0,1-polynomials factorization

Question

Let $n\in\mathbb{N}$ and $P_{n}$ is $0,1$-polynomial whose coefficients are binary digits of $n$.
Let $Q_{1}(x) \cdot Q_{2}(x) \cdots Q_{m}(x)$ - polynomial factorization (over integers) of $P_{n}$.
As was shown here, all $Q_{k}$ will also be $0,1$-polynomials iff there are no carries when multiplying prime factors of $n$.

Well, what happens in other cases?
According to numerical observations, in approximately half of the cases all $Q_{k}$ are ${-1, 0, 1}$-polynomials.
But for $n=141$ we have
$$P_{141}=(x+1) \left(x^6-x^5+x^4-x^3+2 x^2-x+1\right)$$
— coefficient $2\notin\{-1, 0, 1\}$ firstly appears.
For $n=663$
$$P_{663}=(1 + x) (1 + x^2 - x^3 + 2 x^4 - 2 x^5 + 2 x^6 - x^7 + x^8)$$ $-2$ appears first times, for $n=2229\rightarrow3$, $25767\rightarrow 4$ etc.

Is there a pattern to new coefficients appearances?
Will all integers appear sooner or later?
I tried to apply the considerations from the previous answer, but unsuccessfully.

Answer 1

Let $n$ be odd. We have the factorization $x^n+1 = -\prod_{d|n}\Phi_d(-x)$, where $\Phi_d(x)$ is a cyclotomic polynomial. Every integer is the coefficient of some cyclotomic polynomial $\Phi_n(x)$. See reference [120] of C. Sanna, arXiv:2111.04034. It's easy to see that we can assume that $n$ is odd.