Product of all $n$th-Cyclotomic polynomials less than $m$

Question

Has the following function been studied? I'm looking for a reference, or an answer displaying some identities and properties $$ f_m(x) = \prod_{n=1}^{m}\Phi_n(x) $$

It looks like a factorial analog of cyclotomic polynomials. The degree $$ \deg(f_n(x)) = \sum_{k=1}^n \phi(k) $$ Something quite interesting I noticed when calculating $f_m(x)$ for low values of $m$ is that all coefficients of terms $x^l$ with $l>\frac{1}{2}\deg(f_n(x))$ are all positive, whilst all coefficients of of terms $x^l$ with $l<\frac{1}{2}\text{degree}(f_n(x))$ are all negative and both are symmetric, but doesn't include the term $x^{\frac{1}{2}\deg(f_n(x))}$. Here's an example $$ \begin{multline} \strut f_7(x) = x^{18} + 2x^{17} + 4x^{16} + 6x^{15} + 8x^{14} + 9x^{13} + 9x^{12} + 7x^{11} + 4x^{10} \\ - 4x^8 - 7x^7 - 9x^6 - 9x^5 - 8x^4 - 6x^3 - 4x^2 - 2x - 1 \end{multline} $$ This shows us that $$ \begin{align} f_n(1) &= 0 \\ f_n(-1) &= 0 \end{align} $$ Here is an OEIS sequence for the coefficients.

Answer 1

Skew-symmetry of coefficients (True): We shall take $m \geqslant 2$:

• Exercise $1$: Prove that $\deg f_m$ is even (One way to approach this is to see that $\Phi(k)$ is even for all $k \geqslant 3$. You can see this by pairing up the roots of unity with their inverses. Why does this fail for $k = 1, 2$?)
• Exercise $2$: Prove that the skew-symmetry of $f_m$'s coefficients (the condition that the coefficients in reverse are the negatives of the coefficients in order) is equivalent to the polynomial $g(x) = f_m(x) + x^{\deg f_m}f_m(\tfrac{1}{x})$ being identically zero.
• Exercise $3$: Show that $\Phi_k(0) = 1$ for $k \geqslant 2$ (same idea as Exercise $1$) and $\Phi_1(0) = -1$. Thus, we have $f_m(0) = -1$. Next, show that $f_m$ is monic, so $x^mf(\tfrac{1}{x})$ when evaluated at $0$ is $1$. Deduce that $g(0) = 0$.
• Exercise $4$: Observe that $g$ has more than $\deg f_m$ roots (remember the roots of unity!) and $\deg g \leqslant \deg f_m$. Conclude that $g = 0$ and $f_m$ is skew-symmetric.

Positivity of first-half (False): We prove that the second coefficient of $f_{94}$ is $-1$:

• Exercise $1$: Prove that the second coefficient of $\Phi_k(x)$ is $-\mu(k)$ (Hint: Use Möbius inversion)
• Exercise $2$: Prove that the second coefficient of $f_m(x)$ is $-M(m) = -\sum_{k \leqslant m} \mu(k)$
• Exercise $3$: Use a computer to check that $M(94) = 1$.

Remark: The value $m = 94$ is the first instance (for $m \geqslant 2$) where $M$ becomes positive. It's good to ponder why this initial value is so large. A computer program shows that this is in fact the first counterexample. The very next value $m = 95$ also has its fourth coefficient being negative. The first counterexample at an odd place (coefficient of an even power of $x$) occurs at $m = 330$.