Products of Cyclotomic Polynomials with Nonnegative Coefficients I'm curious if there are any results that allow us to determine if a product of cyclotomic polynomials (not necessarily all distinct) results in a polynomial having nonnegative coefficients.
Some things that I do know: $\Phi_{p^k}(x)$ always has nonnegative coefficients for any prime $p$, and so any product of $\Phi_{p^k}(x)$'s will have positive coefficients. Whereas, if $m$ is not a power of a prime $\Phi_{m}(1) = 1$ despite having terms $x^{\phi(m)}$ and $1$ thus requiring at least one negative coefficient.
By computer I know some interesting examples like, $$\Phi_3(x)\Phi_6(x) = 1x^0 + 1x^2 + 1x^4$$ but $$\Phi_3(x)\Phi_6(x)^2 = 1x^0 + -1x^1 + 2x^2 + -1x^3 + 2x^4 + -1x^5 + 1x^6$$.
Adding a $\Phi_2(x)$ helps for a bit, $$\Phi_2(x)\Phi_3(x)\Phi_6(x) = 1x^0 + 1x^1 + 1x^2 + 1x^3 + 1x^4 + 1x^5$$ and $$\Phi_2(x)\Phi_3(x)\Phi_6(x)^2 = 1x^0 + 1x^2 + 1x^3 + 1x^4 + 1x^5 + 1x^7$$ both have nonnegative coefficients, but $$\Phi_2(x)\Phi_3(x)\Phi_6(x)^3 = 1x^0 + -1x^1 + 2x^2 + 1x^4 + 1x^5 + 2x^7 + -1x^8 + 1x^9$$ does not. However, I can't find any sort of rhyme or reason to the appearance of negative coefficients.
I've skimmed papers like http://math.ucsd.edu/~revans/PolynomialsGreene.pdf but can't find any sort of "if and only if" conditions. Being pointed even in a somewhat right direction would help me a lot.
 A: You might do well to try more specific cases. I'll give you a result on the case you end with and a few more showing why I think the general case might be unmanageable.
Claim: $\Phi_2^i\Phi_3^j\Phi_6^k$ has all coefficients non-negative if and only if $i+j \geq k.$
The if part is easy consequence of
$\Phi_2(x)\Phi_6(x)=\Phi_2(x^3)=x^3+1$ and $\Phi_3(x)\Phi_6(x)=\Phi_3(x^2)=x^4+x^2+1.$
We can see that  $\Phi_2^i\Phi_3^j\Phi_6^{i+j}$ has all coefficients non-negative and leading terms $x^q+x^{q-2}$ where $q=3i+4j.$ In case $j=0,$ replace that with $x^q+x^{q-3}.$
Since $\Phi_6^m=(x^2-x+1)^m$ has leading terms  $x^{2m}-mx^{2m-1},$ $\Phi_2^i\Phi_3^j\Phi_6^{i+j+m}$ begins $x^r-mx^{r-1}$ for $r=q+2m.$

There seems potential for generalizations.
I'll introduce $\alpha_m=\frac{x^m-1}{x-1}$ which is a product of cyclotomic polynomials. $\alpha_m=\Phi_m$ when $m$ is prime.
$\Phi_n$ is pretty easy to understand when $n$ has two or less distinct odd prime factors. The terms alternate $\pm 1$ or, for a prime power, are all $1.$ 
For example $\Phi_{35}={x}^{24}-{x}^{23}+{x}^{19}-{x}^{18}+{x}^{17}-{x}^{16}+{x}^{14}-{x}^{13
}+{x}^{12}-{x}^{11}+{x}^{10}-{x}^{8}+{x}^{7}-{x}^{6}+{x}^{5}-x+1$
Then $\alpha_m\Phi_{35}$ has all non-zero coefficients $1$ and $-1.$ To avoid $-1,$ use $m \geq 24$ or  $m=5,7,10,12,14,15,17,19,20,21,22 .$
For such $n$ one could describe a necessary and sufficient condition  based on the interval lengths of $\Phi_n$ starting and ending with a $-1.$ But this is amounts to saying that it works for large enough $m$ and some smaller ones, but not others.

The first case with three distinct odd prime factors is
$\Phi_{105}(x)={x}^{48}+{x}^{47}+{x}^{46}-{x}^{43}-{x}^{42}-2\,{x}^{41}-{x}^{40}-{x}^
{39}+{x}^{36}+\cdots$
The sum of the terms is $1$ but $\alpha_m\Phi_{105}$ always has negative coefficients. For $m \leq n,$ $\alpha_m \alpha_{n}\Phi_{105}$ sometimes has only non-negative coefficients and sometimes not. 
Here is a plot of all the pairs $m,n$ with $m,n \leq 53$ and $\alpha_m \alpha_{n}\Phi_{105}$ having non-negative coefficients



*

*There are negative coefficients no matter what $n$ is if $m \leq 14$ or $22 \leq m \leq 29.$ 

*$\alpha_m \alpha_{n}\Phi_{105}$ has all coefficients non-negative if $47 \leq m \leq n.$ 

*$\alpha_{16}\alpha_n$ works for $n=41,42,43$ but not any smaller or larger $n.$
