It is known that the seventh coefficient of $\Phi\_{105}(x)$ is $-2$ and that's the first occurrence of a coefficient with absolute value greater than $1$ for a cyclotomic polynomial. When I did a quick check for the seventh coefficient of $\Phi\_n(x)$ where $n=105k$ with $\gcd(105,k)=1$ and $\mu(k)\neq 0$ they all came out to be $2$ in absolute value whenever they are nonzero. Is it true in general or there is a counterexample to this?



Cyclotomic polynomials with 7$^{th}$ coefficient greater than 1 in absolute value
$k=11$ is the smallest counterexample - the 7'th coefficient is 0. Here are the details:


We have the following identity: 
$$\Phi\_n(x) = \prod\_{d \mid n} (1-x^d)^{\mu(n/d)},$$
valid for $n>1$.


If we are interested only in the first $m+1$ coefficients ($x^0$ to $x^{m}$), it suffices to look at the following product, going only over divisors $\le m$:
$$\Phi\_n(x) = \prod\_{d \mid n, d \le m} (1-x^d)^{\mu(n/d)} \mod {x^{m+1}}.$$
Hence, 
$$[x^7] \Phi\_{105 k}(x) = [x^7]\prod\_{d \mid 105k, d \le 7} (1-x^d)^{\mu(105k/d)}.$$
Since you assume $\gcd(k,105)=1$ and $\mu(k)\neq 0$, we actually have 4 cases, according to the parity of $k$ and according to $\mu(k)$.


When $2 \nmid k$, the set $\{d : d\mid 105k, d \le 7\}$ is $\{1,3,5,7\}$ and we find
$$[x^7] \Phi\_{105 k}(x) = [x^7] (1-x)^{\mu(105k)}(1-x^3)^{\mu(35k)}(1-x^5)^{\mu(21k)}(1-x^7)^{\mu(15k)}$$
$$ = [x^7] ((1-x)^{-1}(1-x^3) (1-x^5)(1-x^7))^{\mu(k)}.$$
When $\mu(k)=1$, we get $-2$. When $\mu(k)=-1$, we get $0$.


When $2 \mid k$, the set $\{d : d\mid 105k, d \le 7\}$ is $\{1,2,3,5,6,7\}$ and we find
$$[x^7] \Phi\_{105 k}(x) = [x^7] (1-x)^{\mu(105k)}(1-x^2)^{\mu(105k/2)}(1-x^3)^{\mu(35k)}(1-x^5)^{\mu(21k)}(1-x^6)^{\mu(35k/2)}(1-x^7)^{\mu(15k)}$$
$$ = [x^7] ((1-x)(1-x^2)^{-1}(1-x^3)^{-1}(1-x^5)^{-1}(1-x^6)(1-x^7)^{-1})^{\mu(k/2)}.$$
Again, only two cases to check. When $\mu(k/2)=1$ we get 2, and when $\mu(k/2)=-1$ we get 0.


