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?

$\begingroup$ In a recent question of mine, "Cyclotomic polynomials: Φn(p) is like pϕ(n) for big enough p, right?" (See Related list for correct title and link), I was given a link to notes of Jameson. These notes spend some time on the coefficients and may help you with your answer. Gerhard "Thanks Again To Peter Mueller" Paseman, 2015.12.01 $\endgroup$ – Gerhard Paseman Dec 1 '15 at 17:22

1$\begingroup$ What cases did you check that worked? $\endgroup$ – Douglas Zare Dec 1 '15 at 18:24

$\begingroup$ I guess I should've added the assumption $c_{7}\neq 0$ in which case it is true that $c_{7}=2$ as Ofir proved it. $\endgroup$ – user204463 Dec 1 '15 at 18:31
$k=11$ is the smallest counterexample  the 7'th coefficient is 0. Here are the details:
We have the following identity: $$\Phi_n(x) = \prod_{d \mid n} (1x^d)^{\mu(n/d)},$$ 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 $\le m$: $$\Phi_n(x) = \prod_{d \mid n, d \le m} (1x^d)^{\mu(n/d)} \mod {x^{m+1}}.$$ Hence, $$[x^7] \Phi_{105 k}(x) = [x^7]\prod_{d \mid 105k, d \le 7} (1x^d)^{\mu(105k/d)}.$$ 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 $$[x^7] \Phi_{105 k}(x) = [x^7] (1x)^{\mu(105k)}(1x^3)^{\mu(35k)}(1x^5)^{\mu(21k)}(1x^7)^{\mu(15k)}$$ $$ = [x^7] ((1x)^{1}(1x^3) (1x^5)(1x^7))^{\mu(k)}.$$ 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 $$[x^7] \Phi_{105 k}(x) = [x^7] (1x)^{\mu(105k)}(1x^2)^{\mu(105k/2)}(1x^3)^{\mu(35k)}(1x^5)^{\mu(21k)}(1x^6)^{\mu(35k/2)}(1x^7)^{\mu(15k)}$$ $$ = [x^7] ((1x)(1x^2)^{1}(1x^3)^{1}(1x^5)^{1}(1x^6)(1x^7)^{1})^{\mu(k/2)}.$$ Again, only two cases to check. When $\mu(k/2)=1$ we get 2, and when $\mu(k/2)=1$ we get 0.

$\begingroup$ did you mean at the end"..., and when $\mu(k/2)=1$ we get 0. $\endgroup$ – user204463 Dec 1 '15 at 18:20

$\begingroup$ @user204463 Yes, I had a slight miscalculation which I have corrected now. $\endgroup$ – Ofir Gorodetsky Dec 1 '15 at 18:34
Actually for all $11 \leq k \leq 100$ which satisfy the conditions in the question (i.e. $\gcd(k,105) = 1$ and $\mu(k) \neq 0$), the coefficient of $x^7$ in $\Phi_{105k}(x)$ equals $0$. Hence so far there are only counterexamples!
This can be checked with GAP as follows:
gap> ks := Filtered([1..100],k > Gcd(k,105) = 1
> and Factors(k) = Set(Factors(k)));
[ 1, 2, 11, 13, 17, 19, 22, 23, 26, 29, 31, 34, 37, 38, 41, 43, 46, 47,
53, 58, 59, 61, 62, 67, 71, 73, 74, 79, 82, 83, 86, 89, 94, 97 ]
gap> c7s := List(ks,k>[k,CoefficientsOfUnivariatePolynomial(
> CyclotomicPolynomial(Rationals,105*k))[8]]);
[ [ 1, 2 ], [ 2, 2 ], [ 11, 0 ], [ 13, 0 ], [ 17, 0 ], [ 19, 0 ],
[ 22, 0 ], [ 23, 0 ], [ 26, 0 ], [ 29, 0 ], [ 31, 0 ], [ 34, 0 ],
[ 37, 0 ], [ 38, 0 ], [ 41, 0 ], [ 43, 0 ], [ 46, 0 ], [ 47, 0 ],
[ 53, 0 ], [ 58, 0 ], [ 59, 0 ], [ 61, 0 ], [ 62, 0 ], [ 67, 0 ],
[ 71, 0 ], [ 73, 0 ], [ 74, 0 ], [ 79, 0 ], [ 82, 0 ], [ 83, 0 ],
[ 86, 0 ], [ 89, 0 ], [ 94, 0 ], [ 97, 0 ] ]
(Note that the $8$ is not a typo, since the first list entry is the coefficient of $x^0$.)

1$\begingroup$ yes it's because the odd ones are prime for which $\mu(k)=1$ and the even ones are 2 times a prime in which case $\mu(k/2)=1$ . For both cases @Ofir proved that the answer is 0. You have to go as high as 143 to see a $2$ and as high as 286 to see a $2$. Other examples under 300 for a $2$ are 187,209,221,247,253,299. If you double those numbers you will see a 2. $\endgroup$ – user204463 Dec 2 '15 at 19:14