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Let $\Phi_n(x)$ denote the $n$-th cyclotomic polynomial. There are numerous properties and utilities of these polynomials. My interest is more basic and in the spirit of

Tewodros Amdeberhan and Richard P. Stanley, Polynomial Coefficient Enumeration, preprint (2008) arXiv:0811.3652 (pdf).

Namely, how many non-zero terms $N(\Phi_n(x))$ does $\Phi_n(x)$ have? I suspect this might be hard. So, let's consider a modest problem.

Question. If $p, q, r$ are distinct primes, is there an explicit (or semi-explicit) formula for $$N(\Phi_{pqr}(x))\,\,?$$

UPDATE. If $r=2$, direct calculation shows that $\Phi_{2pq}(-x)=\Phi_{pq}(x)$. Therefore, the result of Carlitz (see Igor's link below) implies the following very special case of my question: $$N(\Phi_{2pq}(x))=N(\Phi_{pq}(x))=\frac{2(p-u)(uq+1)}p-1.$$

UPDATE. I am still waiting for some answers to the question.

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In a 1966 Monthly paper

Carlitz, Leonard, The number of terms in the cyclotomic polynomial $F_{pq}(x)$, Am. Math. Mon. 73, 979-981 (1966). ZBL0146.26704.

Carlitz shows that

$$ N(\Phi_{pq}(x)) = \frac{(p-u)(u q + 1)}{p} + \frac{u(pq - uq - 1)}{p},$$ where $q>p$ and $u = -1/q \mod p.$ Perhaps his methods can be extended to more primes, who can say.

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    $\begingroup$ Thanks, Igor! That is a nice resource. Just a simplification: $N(\Phi_{pq}(x))=\frac{2(p-u)(uq+1)}p-1$. $\endgroup$ – T. Amdeberhan Apr 5 '17 at 1:34

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