Remarks on the Coefficients of Inverse Cyclotomic Polynomials

Abstract

Cyclotomic polynomials play an imporant role in discrete mathematics. Recently, inverse cyclotomic polynomials have been defined and investigated. In this paper, we present some recent advances related to the coefficients of inverse cyclotomic polynomials, including a practical recursive formula for their calculation and numerical simulations.

1. Introduction

For an integer $n\ge 1$, an nth root $\zeta$ of the unity is called primitive if ${\zeta }^n=1$, but ${\zeta }^d\ne 1$ for all $1\le d<n$. By denoting ${\zeta }_n=cos\frac{2\pi }{n}+isin\frac{2\pi }{n}$ as the first root of order n of the unity, the nth cyclotomic polynomial ${\Phi }_n$ is defined by the following:

$$\begin{array}{c}\hfill {\Phi }_n\left(z\right)=\prod _{\begin{array}{c}1\le k\le n-1\\ gcd(k,n)=1\end{array}}(z-{\zeta }_n^k)=\sum _{j=0}^{\phi \left(n\right)}{c}_j^{\left(n\right)}{z}^j,\end{array}$$

(1)

where $\phi$ is Euler’s totient function, which is also the degree of the polynomial. It is known that ${\Phi }_n$ is palindromic (i.e., $c_j^{\left(n\right)}=c_{\phi \left(n\right)-j}^{\left(n\right)}$, $j=0,\dots ,\phi \left(n\right)$). The term cyclotomic comes from the property of the nth roots of unity to divide the unit circle into n equal arcs, thereby forming a regular polygon inscribed in the unit circle.

The explicit calculation of the coefficients of cyclotomic polynomials is very difficult to perform (see, for instance, [1,2]), but many properties of these polynomials are known [3,4]. An explicit integral formula for the coefficients was established by the authors in [5].

For an integer $n\ge 2$, the nth inverse cyclotomic polynomial ${\Psi }_n\left(z\right)$ is defined by the following:

$$\begin{array}{c}\hfill {\Psi }_n\left(z\right)=\frac{x^n-1}{{\Phi }_n\left(z\right)}=\prod _{1\le k<n,gcd(k,n)>1}\left(z-e^{\frac{2k\pi i}{n}}\right)=\sum _{j=0}^{n-\phi \left(n\right)}{d}_j^{\left(n\right)}{z}^j.\end{array}$$

(2)

Notice that the nonprimitive nth roots of unity are the roots of this monic polynomial of the degree $n-\phi \left(n\right)$. Here, we denote the coefficient of $z^j$ in ${\Psi }_n$ by $d_j^{\left(n\right)}$, $j=0,\dots ,n-\phi \left(n\right)$.

Using (2), the first inverse cyclotomic polynomials (discounting prime indices) are

$$\begin{array}{cc}\hfill {\Psi }_1\left(z\right)& =1,{\Psi }_4\left(z\right)=z^2-1,{\Psi }_6\left(z\right)=z^4+z^3-z-1,{\Psi }_8\left(z\right)=z^4-1,\hfill \\ \hfill {\Psi }_9\left(z\right)& =z^3-1,{\Psi }_{10}\left(z\right)=z^6+z^5-z-1,{\Psi }_{12}\left(z\right)=z^8+z^6-z^2-1,\hfill \\ \hfill {\Psi }_{14}\left(z\right)& =z^8+z^7-z-1,{\Psi }_{15}\left(z\right)=z^7+z^6+z^5-z^2-z-1,\hfill \\ \hfill {\Psi }_{16}\left(z\right)& =z^8-1,{\Psi }_{18}\left(z\right)=z^{12}+z^9-z^3-1,{\Psi }_{20}\left(z\right)=z^{12}+z^{10}-z^2-1,\hfill \\ \hfill {\Psi }_{21}\left(z\right)& =z^9+z^8+z^7-z^2-z-1,{\Psi }_{22}\left(z\right)=z^{12}+z^{11}-z^2-1.\hfill \end{array}$$

The following properties of the polynomial ${\Psi }_n$ are known, and we present them along with some sketches of proofs. More details can be found in [6,7].

Proposition 1.

$1^{\circ }$ If p is a prime and $n=p^{\alpha }$ for $\alpha \ge 1$, then ${\Psi }_n\left(z\right)=z^{p^{\alpha -1}}-1$.

$2^{\circ }$ For $n=p_1\cdots p_k$ to be square-free, $deg\left({\Psi }_n\right)=p_1\cdots p_k-(p_1-1)\cdots (p_k-1)$.

$3^{\circ }$ If $p<q$ are primes, then, for $n=pq$, one has

$$\begin{array}{c}\hfill {\Psi }_n\left(z\right)=\frac{\left(z^p-1\right)\left(z^q-1\right)}{z-1}=z^{p+q-1}+\cdots +z^{q+1}-z^{p-1}-\cdots -z^2-z-1.\end{array}$$

$4^{\circ }$ If $p,q,$ and r are different primes, then, for $n=pqr$, one has

$$\begin{array}{c}\hfill {\Psi }_n\left(z\right)=\frac{(z^{pq}-1)(z^{qr}-1)(z^{rp}-1)(z-1)}{(z^p-1)(z^q-1)(z^r-1)}.\end{array}$$

$5^{\circ }$ ${\Psi }_{2n}\left(z\right)=\left(1-z^n\right){\Psi }_n(-z)$ if n is odd.

$6^{\circ }$ ${\Psi }_{pn}\left(z\right)={\Psi }_n\left(z^p\right)$ if $p\mid n$.

$7^{\circ }$ ${\Psi }_{pn}\left(z\right)={\Psi }_n\left(z^p\right){\Phi }_n\left(z\right)$ if $p\nmid n$.

$8^{\circ }$ ${\Psi }_n$ is antipalindromic (i.e., $d_j^{\left(n\right)}=-d_{n-\phi \left(n\right)-j}^{\left(n\right)}$, $j=0,\dots ,n-\phi \left(n\right)$).

$9^{\circ }$ The number of positive coefficients of ${\Psi }_n$ is equal to the number of negative coefficients.

Proof. 

For $1^{\circ }$, note that, for this value of n, the only nonprimitive nth roots of unity are the ones having orders that divide $p^{\alpha }-1$. These are the roots of the polynomial $z^{p^{\alpha -1}}-1$, and the conclusion follows.

Property $2^{\circ }$ follows immediately from the fact that $\phi \left(n\right)=(p_1-1)\dots (p_k-1)$.

Let us prove $3^{\circ }$. The only nonprimitive nth roots of unity have orders of 1, p, or q. Hence, these are the roots of the polynomials $z^p-1$ and $z^q-1$, respectively. Since the root 1 appears in both polynomials, we have that

$$\begin{array}{c}\hfill {\Psi }_n\left(z\right)=\frac{(z^p-1)(z^q-1)}{z-1}.\end{array}$$

For $4^{\circ }$, note that the nonprimitive $pqr$th roots of unity are the roots of unity of orders dividing $pq$, $qr$, or $rp$. By denoting $U_k$ as the set of roots of unity which have orders dividing k, as well as by using the principle of inclusion and exclusion, we deduce that

$$\begin{array}{c}\hfill |U_{pq}\cup U_{qr}\cup U_{rp}|=|U_{pq}|+|U_{qr}|+|U_{rp}|-|U_p|-|U_q|-|U_r|+|U_1|.\end{array}$$

The elements of $U_{pq}\cup U_{qr}\cup U_{rp}$ are the roots of ${\Psi }_n$. The conclusion follows by identifying the roots of unity with the roots of the corresponding polynomials.

To prove $5^{\circ }$, one just observes that the nonprimitive $2n$th roots of unity have orders dividing n or $2d$, where $d\mid n$, and $d<n$. The first ones are the roots of $1-z^n$, whereas the second type can be found among the roots of ${\Psi }_n(-z)$. The statements $6^{\circ }$ and $7^{\circ }$ can be proved in a similar way. Then, $8^{\circ }$ results from the definition and the fact that ${\Phi }_n$ is palindromic. From here, $9^{\circ }$ follows directly. □

From the proposition above, we easily derive the following formulas. If p and q are distinct primes, then the cyclotomic polynomial satisfies the following:

$$\begin{array}{c}\hfill {\Phi }_{pq}\left(z\right)=\frac{(z^{pq}-1)(z-1)}{(z^p-1)(z^q-1)}.\end{array}$$

(3)

Moreover, if $p,q,$ and r are distinct primes, then

$$\begin{array}{c}\hfill {\Phi }_{pqr}\left(z\right)=\frac{(z^{pqr}-1)(z^p-1)(z^q-1)(z^r-1)}{(z^{pq}-1)(z^{qr}-1)(z^{rp}-1)(z-1)}.\end{array}$$

(4)

The paper first presents some results on Ramanujan sums in Section 2, then it reviews some known formulae for the calculation of the coefficients of cyclotomic polynomials in Section 3. The paper’s main results are contained in Section 4, where we derive new formulas for the coefficients of inverse cyclotomic polynomials in Theorems 4–6. These results are expressed in terms of Ramanujan sums and provide counterparts to similar formulae that were recently obtained in [6,8] for cyclotomic polynomials. The numerical experiments in Section 5 highlight the utility of the recursive formula in Theorem 6.

2. Preliminaries on Ramanujan Sums

Recall that the Möbius function $\mu$ is defined by

$$\begin{array}{c}\hfill \mu \left(n\right)=\left\{\begin{array}{cc}1\hfill & \mathrm{if}n=1,\hfill \\ {(-1)}^k\hfill & \mathrm{if}n=p_1{p}_2\cdots p_k,\hfill \\ 0\hfill & \mathrm{if}n=p^{2}m,\hfill \end{array}\right.\end{array}$$

(5)

where p, $p_1$, …, $p_k$ are prime numbers.

For every positive integer n and q, the Ramanujan sum $\rho (n,q)$ is defined as

$$\rho (n,q)=\sum _{gcd(a,n)=1}{e}^{2\pi i\frac{a}{n}q},$$

(6)

where the sum is taken over all a such that $1\le a\le n$, and $gcd(a,n)=1$ (see [9,10]).

By fixing q or n, Ramanujan sums can be seen as arithmetic functions of the free variable. For a fixed $n\in {\mathbb{N}}^{\ast }$, the arithmetic function $\rho (n,·):{\mathbb{N}}^{\ast }\to \mathbb{C}$ is periodic, as $\rho (n,q+n)=\rho (n,q)$ for all $q\in \mathbb{N}$. By fixing q, the function $\rho (·,q):{\mathbb{N}}^{\ast }\to \mathbb{C}$ is also interesting. For example, $\rho (n,0)=\phi \left(n\right)$, and $n\in {\mathbb{N}}^{\ast }$ represents Euler’s totient function.

For $q\in \mathbb{Z}$ and every positive integer n, we consider the function ${\delta }_q\left(n\right)={\displaystyle \sum _{a=1}^n}{e}^{\frac{2\pi a}{n}iq}$. The following straightforward identity (see, e.g., [6]) is useful in later computations.

$$\begin{array}{c}\hfill {\delta }_q\left(n\right)=\left\{\begin{array}{c}n\mathrm{if}n\mid q\hfill \\ 0\mathrm{if}n\nmid q\hfill \end{array}\right..\end{array}$$

(7)

Suppose that $q\in {\mathbb{N}}^{\ast }$ is fixed. By (7), we obtain that ${\delta }_q:{\mathbb{N}}^{\ast }\to \mathbb{N}$ is a multiplicative function. Indeed, if $m,n\in {\mathbb{N}}^{\ast }$ are coprime, then $nm\mid q$ if and only if $n\mid q$ and $m\mid q$. Hence,

$${\delta }_q\left(nm\right)={\delta }_q\left(n\right){\delta }_q\left(m\right).$$

(8)

The following result of Kluyver is a direct consequence of the Möbius inversion formula and was obtained in 1906 [10]. If q is a fixed positive integer, then

$$\rho (n,q)=\sum _{d\mid gcd(n,q)}d\mu \left(\frac{n}{d}\right)\mathrm{for}\mathrm{all}n\in {\mathbb{N}}^{\ast }.$$

(9)

This result has an important consequence. As for any $n\in {\mathbb{N}}^{\ast }$, $\rho (n,q)={\sum }_{d\mid n}{\delta }_q\left(d\right)\mu \left(\frac{n}{d}\right)$; the right-hand side is the convolution product ${\delta }_q\ast \mu$ of two multiplicative functions; hence, it follows that, for $q\in {\mathbb{N}}^{\ast }$ being fixed, the function $\rho (·,q):{\mathbb{N}}^{\ast }\to \mathbb{C}$ is multiplicative.

With the convention that $\mu \left(\frac{n}{d}\right)=0$ if $d\nmid n$, we obtain the following explicit consequence of Formula (9) for $q=1,\dots ,17$, which is valid for all $n\in {\mathbb{N}}^{\ast }$.

$$\begin{array}{cc}\hfill \rho (n,1)& =\mu \left(n\right);\hfill \\ \hfill \rho (n,2)& =\sum _{d\mid gcd(n,2)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+2\mu \left(n/2\right);\hfill \\ \hfill \rho (n,3)& =\sum _{d\mid gcd(n,3)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+3\mu \left(n/3\right);\hfill \\ \hfill \rho (n,4)& =\sum _{d\mid gcd(n,4)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+2\mu \left(n/2\right)+4\mu \left(n/4\right);\hfill \\ \hfill \rho (n,5)& =\sum _{d\mid gcd(n,5)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+5\mu \left(n/5\right);\hfill \\ \hfill \rho (n,6)& =\sum _{d\mid gcd(n,6)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+2\mu \left(n/2\right)+3\mu \left(n/3\right)+6\mu \left(n/6\right);\hfill \\ \hfill \rho (n,7)& =\sum _{d\mid gcd(n,7)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+7\mu \left(n/7\right);\hfill \\ \hfill \rho (n,8)& =\sum _{d\mid gcd(n,8)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+2\mu \left(n/2\right)+4\mu \left(n/4\right)+8\mu \left(n/8\right);\hfill \\ \hfill \rho (n,9)& =\sum _{d\mid gcd(n,9)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+3\mu \left(n/3\right)+9\mu \left(n/9\right);\hfill \\ \hfill \rho (n,10)& =\sum _{d\mid gcd(n,10)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+2\mu \left(n/2\right)+5\mu \left(n/5\right)+10\mu \left(n/10\right);\hfill \\ \hfill \rho (n,11)& =\sum _{d\mid gcd(n,11)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+11\mu \left(n/11\right);\hfill \\ \hfill \rho (n,12)& =\sum _{d\mid gcd(n,12)}d\mu \left(\frac{n}{d}\right)\hfill \\ \hfill & =\mu \left(n\right)+2\mu \left(n/2\right)+3\mu \left(n/3\right)+4\mu \left(n/4\right)+6\mu \left(n/6\right)+12\mu \left(n/12\right);\hfill \\ \hfill \rho (n,13)& =\sum _{d\mid gcd(n,13)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+13\mu \left(n/13\right);\hfill \\ \hfill \rho (n,14)& =\sum _{d\mid gcd(n,14)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+2\mu \left(n/2\right)+7\mu \left(n/7\right)+14\mu \left(n/14\right);\hfill \\ \hfill \rho (n,15)& =\sum _{d\mid gcd(n,15)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+3\mu \left(n/3\right)+5\mu \left(n/5\right)+15\mu \left(n/15\right);\hfill \\ \hfill \rho (n,16)& =\sum _{d\mid gcd(n,16)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+2\mu \left(n/2\right)+4\mu \left(n/4\right)+8\mu \left(n/8\right)+16\mu \left(n/16\right);\hfill \\ \hfill \rho (n,17)& =\sum _{d\mid gcd(n,17)}d\mu \left(\frac{n}{d}\right)=\mu \left(n\right)+17\mu \left(n/17\right).\hfill \end{array}$$

When $n=p_1{p}_2\cdots p_k$ is a product of distinct k primes, $gcd(n,q)=p_1{p}_2\cdots p_m$, and $m\le k$, then by using the properties of the Möbius function, one obtains

$$\begin{array}{c}\hfill \rho (n,q)=\sum _{i=1}^m{(-1)}^{k-i}{S}_i(p_1,p_2,\dots ,p_m),\end{array}$$

(10)

where $S_i$ is the ith fundamental symmetric polynomial in m variables.

The cases when n is a product of three or four distinct primes ($m=3,4$) present special interest in the study of cyclotomic and inverse cyclotomic polynomials.

2.1. Ramanujan Sums for $n=pqr$

With the convention that $\mu \left(\frac{n}{d}\right)=0$ if $d\nmid n$, we obtain the following immediate consequences of Formula (9). These results are valid for all $n=pqr$ with $p<q<r$ odd primes and $j=1\dots ,17$:

$$\begin{array}{cc}\hfill \rho (n,1)& =\mu \left(n\right)=-1;\hfill \\ \hfill \rho (n,2)& =\mu \left(n\right)+2\mu \left(n/2\right)=-1;\hfill \\ \hfill \rho (n,3)& =\mu \left(n\right)+3\mu \left(n/3\right)=\left\{\begin{array}{cc}2\hfill & \mathrm{if}3\mid n\hfill \\ -1\hfill & \mathrm{if}3\nmid n\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,4)& =\mu \left(n\right)+2\mu \left(n/2\right)+4\mu \left(n/4\right)=-1;\hfill \\ \hfill \rho (n,5)& =\mu \left(n\right)+5\mu \left(n/5\right)=\left\{\begin{array}{cc}4\hfill & \mathrm{if}5\mid n\hfill \\ -1\hfill & \mathrm{if}5\nmid n\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,6)& =\mu \left(n\right)+2\mu \left(n/2\right)+3\mu \left(n/3\right)+6\mu \left(n/6\right)=\left\{\begin{array}{cc}2\hfill & \mathrm{if}3\mid n\hfill \\ -1\hfill & \mathrm{if}3\nmid n\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,7)& =\mu \left(n\right)+7\mu \left(n/7\right)=\left\{\begin{array}{cc}6\hfill & \mathrm{if}7\mid n\hfill \\ -1\hfill & \mathrm{if}7\nmid n\hfill \end{array}\right.\hfill \\ \hfill \rho (n,8)& =\mu \left(n\right)+2\mu \left(n/2\right)+4\mu \left(n/4\right)+8\mu \left(n/8\right)=-1;\hfill \\ \hfill \rho (n,9)& =\mu \left(n\right)+3\mu \left(n/3\right)+9\mu \left(n/9\right)=\left\{\begin{array}{cc}2\hfill & \mathrm{if}3\mid n\hfill \\ -1\hfill & \mathrm{if}3\nmid n.\hfill \end{array}\right.\hfill \\ \hfill \rho (n,10)& =\mu \left(n\right)+2\mu \left(n/2\right)+5\mu \left(n/5\right)+10\mu \left(n/10\right)=\left\{\begin{array}{cc}4\hfill & \mathrm{if}5\mid n\hfill \\ -1\hfill & \mathrm{if}5\nmid n.\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,11)& =\mu \left(n\right)+11\mu \left(n/11\right)=\left\{\begin{array}{cc}10\hfill & \mathrm{if}11\mid n\hfill \\ -1\hfill & \mathrm{if}11\nmid n.\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,12)& =\mu \left(n\right)+2\mu \left(n/2\right)+3\mu \left(n/3\right)+4\mu \left(n/4\right)+6\mu \left(n/6\right)+12\mu \left(n/12\right)\hfill \\ \hfill & =\left\{\begin{array}{cc}2\hfill & \mathrm{if}3\mid n\hfill \\ -1\hfill & \mathrm{if}3\nmid n.\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,13)& =\mu \left(n\right)+13\mu \left(n/13\right)=\left\{\begin{array}{cc}12\hfill & \mathrm{if}13\mid n\hfill \\ -1\hfill & \mathrm{if}13\nmid n.\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,14)& =\mu \left(n\right)+2\mu \left(n/2\right)+7\mu \left(n/7\right)+14\mu \left(n/14\right)=\left\{\begin{array}{cc}6\hfill & \mathrm{if}7\mid n\hfill \\ -1\hfill & \mathrm{if}7\nmid n.\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,15)& =\mu \left(n\right)+3\mu \left(n/3\right)+5\mu \left(n/5\right)+15\mu \left(n/15\right)=\left\{\begin{array}{cc}-8\hfill & \mathrm{if}15\mid n\hfill \\ 4\hfill & \mathrm{if}gcd(n,15)=5\hfill \\ 2\hfill & \mathrm{if}gcd(n,15)=3\hfill \\ -1\hfill & \mathrm{if}gcd(n,15)=1.\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,16)& =\mu \left(n\right)+2\mu \left(n/2\right)+4\mu \left(n/4\right)+8\mu \left(n/8\right)+16\mu \left(n/16\right)=-1;\hfill \\ \hfill \rho (n,17)& =\mu \left(n\right)+17\mu \left(n/17\right)=\left\{\begin{array}{cc}16\hfill & \mathrm{if}17\mid n\hfill \\ -1\hfill & \mathrm{if}17\nmid n.\hfill \end{array}\right..\hfill \end{array}$$

Clearly, if $n=pqr$ with $p<q<r$ primes, then $\rho (n,j)=-1$ for $j=1,\dots ,p-1$, while $\rho (n,p)=\mu \left(n\right)+p\mu \left(n/p\right)=p-1$.

2.2. Ramanujan Sums for $n=pqrs$

For all $n=pqrs$ with $p<q<r<s$ odd primes and $j=1\dots ,17$, one obtains:

$$\begin{array}{cc}\hfill \rho (n,1)& =\mu \left(n\right)=1;\hfill \\ \hfill \rho (n,2)& =\mu \left(n\right)+2\mu \left(n/2\right)=1;\hfill \\ \hfill \rho (n,3)& =\mu \left(n\right)+3\mu \left(n/3\right)=\left\{\begin{array}{cc}-2\hfill & \mathrm{if}3\mid n\hfill \\ 1\hfill & \mathrm{if}3\nmid n\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,4)& =\mu \left(n\right)+2\mu \left(n/2\right)+4\mu \left(n/4\right)=1;\hfill \\ \hfill \rho (n,5)& =\mu \left(n\right)+5\mu \left(n/5\right)=\left\{\begin{array}{cc}-4\hfill & \mathrm{if}5\mid n\hfill \\ 1\hfill & \mathrm{if}5\nmid n\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,6)& =\mu \left(n\right)+2\mu \left(n/2\right)+3\mu \left(n/3\right)+6\mu \left(n/6\right)=\left\{\begin{array}{cc}-2\hfill & \mathrm{if}3\mid n\hfill \\ 1\hfill & \mathrm{if}3\nmid n\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,7)& =\mu \left(n\right)+7\mu \left(n/7\right)=\left\{\begin{array}{cc}-6\hfill & \mathrm{if}7\mid n\hfill \\ 1\hfill & \mathrm{if}7\nmid n\hfill \end{array}\right.\hfill \\ \hfill \rho (n,8)& =\mu \left(n\right)+2\mu \left(n/2\right)+4\mu \left(n/4\right)+8\mu \left(n/8\right)=1;\hfill \\ \hfill \rho (n,9)& =\mu \left(n\right)+3\mu \left(n/3\right)+9\mu \left(n/9\right)=\left\{\begin{array}{cc}-2\hfill & \mathrm{if}3\mid n\hfill \\ 1\hfill & \mathrm{if}3\nmid n.\hfill \end{array}\right.\hfill \\ \hfill \rho (n,10)& =\mu \left(n\right)+2\mu \left(n/2\right)+5\mu \left(n/5\right)+10\mu \left(n/10\right)=\left\{\begin{array}{cc}-4\hfill & \mathrm{if}5\mid n\hfill \\ 1\hfill & \mathrm{if}5\nmid n.\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,11)& =\mu \left(n\right)+11\mu \left(n/11\right)=\left\{\begin{array}{cc}-10\hfill & \mathrm{if}11\mid n\hfill \\ 1\hfill & \mathrm{if}11\nmid n.\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,12)& =\mu \left(n\right)+2\mu \left(n/2\right)+3\mu \left(n/3\right)+4\mu \left(n/4\right)+6\mu \left(n/6\right)+12\mu \left(n/12\right)\hfill \\ \hfill & =\left\{\begin{array}{cc}-2\hfill & \mathrm{if}3\mid n\hfill \\ 1\hfill & \mathrm{if}3\nmid n.\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,13)& =\mu \left(n\right)+13\mu \left(n/13\right)=\left\{\begin{array}{cc}-12\hfill & \mathrm{if}13\mid n\hfill \\ 1\hfill & \mathrm{if}13\nmid n.\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,14)& =\mu \left(n\right)+2\mu \left(n/2\right)+7\mu \left(n/7\right)+14\mu \left(n/14\right)=\left\{\begin{array}{cc}-6\hfill & \mathrm{if}7\mid n\hfill \\ 1\hfill & \mathrm{if}7\nmid n.\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,15)& =\mu \left(n\right)+3\mu \left(n/3\right)+5\mu \left(n/5\right)+15\mu \left(n/15\right)=\left\{\begin{array}{cc}8\hfill & \mathrm{if}15\mid n\hfill \\ -4\hfill & \mathrm{if}gcd(n,15)=5\hfill \\ -2\hfill & \mathrm{if}gcd(n,15)=3\hfill \\ 1\hfill & \mathrm{if}gcd(n,15)=1.\hfill \end{array}\right.;\hfill \\ \hfill \rho (n,16)& =\mu \left(n\right)+2\mu \left(n/2\right)+4\mu \left(n/4\right)+8\mu \left(n/8\right)+16\mu \left(n/16\right)=1;\hfill \\ \hfill \rho (n,17)& =\mu \left(n\right)+17\mu \left(n/17\right)=\left\{\begin{array}{cc}-16\hfill & \mathrm{if}17\mid n\hfill \\ 1\hfill & \mathrm{if}17\nmid n.\hfill \end{array}\right..\hfill \end{array}$$

Clearly, if $n=pqrs$ with $p<q<r<s$ primes, then $\rho (n,j)=1$ for $j=1,\dots ,p-1$, while $\rho (n,p)=\mu \left(n\right)+p\mu \left(n/p\right)=1-p$.

3. Review of Some Results on the Coefficients of Cyclotomic Polynomials

The coefficients of cyclotomic polynomials have many interesting properties (see, for instance, [11,12,13]). For example, the polynomials ${\Phi }_n$, $n=1,\dots ,104$, are flat (i.e., all the coefficients are 0, 1, or $-1$) and so are the polynomials obtained when $n=pq$, where p and q are distinct primes. In 1883, Mignotti showed that ${\Phi }_{105}$ is not flat, as $-2$ is the coefficient of $z^7$, while 2 first appears as coefficient for $n=165$.

In 1895, Bang proved that, for $n=pqr$ with $p<q<r$ odd primes, no coefficient of ${\Phi }_n$ can be larger than $p-1$. Later on, Schur showed in 1931 that the coefficients of cyclotomic polynomials can be arbitrarily large in their absolute values [14]. In 1987, Suzuki [15] showed that any integer number can be a coefficient of some cyclotomic polynomial.

Under certain assumptions regarding the gap between consecutive primes [16], one can obtain more profound results. For example, Andrica’s conjecture claims that for $n\ge 1$, one has $\sqrt{p_{n+1}}-\sqrt{p_n}<1$, where $p_n$ and $p_{n+1}$ denote the nth and $(n+1)$th prime numbers, respectively, which still stands after over 40 years (see, for example, [17,18]). It was recently proven that Andrica’s conjecture implies that every natural number occurs as the largest coefficient of some cyclotomic polynomial ([19], Theorem 16).

We state here the following conjecture.

Conjecture 1. 

Every integer appears as a coefficient of infinitely many cyclotomic polynomials.

By Möbius’ inversion formula, one obtains a useful alternative form of (1) as follows:

$$\begin{array}{c}\hfill {\Phi }_n\left(z\right)=\prod _{\begin{array}{c}d|n\end{array}}{\left(z^d-1\right)}^{\mu (n/d)}=\prod _{\begin{array}{c}d|n\end{array}}{\left(1-z^d\right)}^{\mu (n/d)}.\end{array}$$

(11)

The last equality follows from

$$\begin{array}{cc}\hfill {\Phi }_n\left(z\right)& =\prod _{\begin{array}{c}d|n\end{array}}{(-1)}^{\mu (n/d)}·{\left(1-z^d\right)}^{\mu (n/d)}\hfill \\ \hfill & ={(-1)}^{{\sum }_{\begin{array}{c}d|n\end{array}}\mu (n/d)}\prod _{\begin{array}{c}d|n\end{array}}{\left(1-z^d\right)}^{\mu (n/d)}=\prod _{\begin{array}{c}d|n\end{array}}{\left(1-z^d\right)}^{\mu (n/d)},\hfill \end{array}$$

since ${\sum }_{d\mid n}\mu (n/d)={\sum }_{d\mid n}1·\mu (n/d)=1\ast \mu \left(n\right)=\epsilon \left(n\right)=0$, where $\epsilon$ is the multiplicative unity in the convolution of the Dirichlet product, and $\epsilon \left(n\right)=0$ for every integer $n\ge 2$. By completing (11) with $\mu (n/d)=0$, when $n/d$ is not an integer, one can write

$$\begin{array}{c}\hfill {\Phi }_n\left(z\right)=\prod _{d=1}^{\infty }{\left(1-z^d\right)}^{\mu (n/d)}.\end{array}$$

(12)

Hence, for a square-free n, the value $c_m^{\left(n\right)}$ depends only on the values of $\mu \left(n\right)$, $\mu (n/d)$, and on the prime divisors of n that are less than $m+1$.

The following result was proven by Endo [20] using a fine mathematical induction argument and earlier results by Bloom [21] and Erdös [22]. A direct proof that allows for simplifications and extensions was given in [6].

Theorem 1.

The following formula holds

$$\begin{array}{c}\hfill c_m^{\left(n\right)}=\sum _{i_1+2i_2+\cdots +m{i}_m=m}{(-1)}^{i_1+\cdots +i_m}\left(\genfrac{}{}{0pt}{}{\mu \left(n\right)}{i_1}\right)\left(\genfrac{}{}{0pt}{}{\mu (n/2)}{i_2}\right)\cdots \left(\genfrac{}{}{0pt}{}{\mu (n/m)}{i_m}\right),\end{array}$$

(13)

where the tuples $\left(i_1,\dots ,i_m\right)$ run over all the nonnegative integral solutions of the equation $i_1+2i_2+\cdots +m{i}_m=m$ for m a positive integer.

A formula for the coefficients of ${\Phi }_n$ only in terms of the Ramanujan sums is given in ([8], Theorem 6):

Theorem 2.

We have

$$c_k^{\left(n\right)}=\sum _{l_1+2l_2+\cdots +k{l}_k=k}{(-1)}^{l_1+l_2+\cdots +l_k}\frac{\rho {(n,1)}^{l_1}}{1^{l_1}{l}_1!}·\frac{\rho {(n,2)}^{l_2}}{2^{l_2}{l}_2!}\cdots \frac{\rho {(n,k)}^{l_k}}{k^{l_k}{l}_k!}.$$

(14)

While the Formulae (13) and (14) are explicit, their practical applicability is limited for at least two reasons. First, large integers n cannot be factorised. Second, the sums are taken over all the solutions to the equation $i_1+2i_2+\cdots +m{i}_m=m$, which requires the generation of all partitions of the positive integer m.

The following recursive formula for the coefficients of ${\Phi }_n$ in terms of the Ramanujan sums was obtained in ([8], Theorem 7), and this avoids the complication mentioned above related to the generation of partitions.

Theorem 3.

The following relation holds for every $k=2,\dots ,\phi \left(n\right)$:

$$c_k^{\left(n\right)}=-\frac{1}{k}\left[\rho (n,k)+\rho (n,k-1)c_1^{\left(n\right)}+\cdots +\rho (n,1)c_{k-1}^{\left(n\right)}\right].$$

(15)

4. The Coefficients of ${\Psi }_{\mathit{n}}$

Since ${\Psi }_n$ is the division of the monic polynomial $z^n-1$ and the cyclotomic polynomial ${\Phi }_n$, both have integer coefficients; it follows that ${\Psi }_n$ is monic with integer coefficients as well. As mentioned in the beginning of the section, ${\Psi }_n$ is intimately connected to the cyclotomic polynomial ${\Phi }_n$. This suggests that understanding the coefficients of ${\Psi }_n$ is a challenging venture, as any knowledge about the coefficients of ${\Psi }_n$ could be transferred to knowledge about the coefficients of ${\Phi }_n$ and vice versa.

If one knows the first few coefficients $c_k^{\left(n\right)}$, one can directly compute the coefficients $d_k^{\left(n\right)}$ of the inverse cyclotomic polynomial for small values of k. This can be achieved by identifying the coefficients in ${\Phi }_n\left(z\right)·{\Psi }_n\left(z\right)=z^n-1$. This is reduced to solving the following system:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{c}_0^{\left(n\right)}{d}_0^{\left(n\right)}=-1\hfill \\ c_0^{\left(n\right)}{d}_1^{\left(n\right)}+c_1^{\left(n\right)}{d}_0^{\left(n\right)}=0\hfill \\ c_0^{\left(n\right)}{d}_2^{\left(n\right)}+c_1^{\left(n\right)}{d}_1^{\left(n\right)}+c_2^{\left(\right(n)}{d}_0^{\left(n\right)}=0\hfill \\ ⋮\hfill \end{array}\right.,\end{array}$$

where the coefficients $c_0^{\left(n\right)}$, $c_1^{\left(n\right)}$, $c_2^{\left(n\right)}$, …, of ${\Phi }_n$ are the unknowns.

Many properties of the coefficients of inverse cyclotomic polynomials are presented in Moree [7]. For example, whenever $n=pq$ with p and q primes, ${\Psi }_n$ is flat, as well as for the polynomials ${\Psi }_{15r}$ or ${\Psi }_{21r}$ when r is prime. The first nonflat inverse cyclotomic polynomial is ${\Psi }_{561}$. Analogous to Suzuki’s Theorem, Moree [7] showed that every integer number can be a coefficient of some inverse cyclotomic polynomial ${\Psi }_n$.

Conjecture 2. 

Every positive integer is the largest coefficient of an inverse cyclotomic polynomial.

Considering the analogous aforementioned result concerning the coefficients of cyclotomic polynomials, it is possible that this property could be proved under the supplementary assumption that Andrica’s conjecture holds.

Conjecture 3. 

Every integer appears as a coefficient of infinitely inverse cyclotomic polynomials.

4.1. The Analogous Formula for (13)

As $\mu \left(n\right)\in \{-1,0,1\}$, we get $\mu {\left(n\right)}^3=\mu \left(n\right)$ and $n\ge 1$. For an integer $k\ge 2$, one has

$$\begin{array}{c}\hfill \left(\genfrac{}{}{0pt}{}{\mu \left(n\right)}{k}\right)={(-1)}^k\left(\genfrac{}{}{0pt}{}{\mu \left(n\right)}{2}\right),\end{array}$$

(16)

where we use the generalized binomial coefficient

$$\begin{array}{c}\hfill \left(\genfrac{}{}{0pt}{}{\alpha }{j}\right)=\frac{\alpha (\alpha -1)\cdots (\alpha -j+1)}{j!},\alpha \in \mathbb{R},j\in \mathbb{N}.\end{array}$$

Therefore, if $\left(i_1,\dots ,i_m\right)$ is a solution of the equation $i_1+2i_2+\cdots +m{i}_m=m$ for a given $j=1,\dots ,m$ one has $i_j\ge 2$; then the corresponding binomial coefficient in (13) with the form ${(-1)}^{i_j}\left(\genfrac{}{}{0pt}{}{\mu (n/j)}{i_j}\right)$ can be replaced using Formula (16) by $\left(\genfrac{}{}{0pt}{}{\mu (n/j)}{2}\right).$

We can now establish a formula that is analogous to (13).

Theorem 4.

For every $m,n\in \mathbb{N}$, and $n\ge 1$, the following formula holds

$$d_m^{\left(n\right)}=\sum _{i_1+2i_2+\dots +m{i}_m=m}{(-1)}^{i_1+\cdots +i_m+1}\left(\genfrac{}{}{0pt}{}{-\mu \left(n\right)}{i_1}\right)\left(\genfrac{}{}{0pt}{}{-\mu (n/2)}{i_2}\right)\cdots \left(\genfrac{}{}{0pt}{}{-\mu (n/m)}{i_m}\right).$$

(17)

Proof. 

We make use of the infinite product Formula (12) and observe that

$$\begin{array}{cc}\hfill {\Psi }_n\left(z\right)& =\frac{z^n-1}{{\Phi }_n\left(z\right)}=(z^n-1)\prod _{d=1}^{\infty }{(1-z^d)}^{-\mu (n/d)}\hfill \\ \hfill & =(z^n-1)\sum _{m=0}^{\infty }\left(\sum _{i_1+2i_2+\cdots +m{i}_m=m}{(-1)}^{i_1+\cdots +i_m}\left(\genfrac{}{}{0pt}{}{-\mu \left(n\right)}{i_1}\right)\cdots \left(\genfrac{}{}{0pt}{}{-\mu (n/m)}{i_m}\right)\right)z^m.\hfill \end{array}$$

The formula follows by identifying coefficients of $z^m$ on both sides. □

Let us emphasize how one can compute the first few coefficients of ${\Psi }_n$ for every $n\in {\mathbb{N}}^{\ast }$ using the formula above.

• For $m=1$, we have $i_1=1$; hence $d_1^{\left(n\right)}=\left(\genfrac{}{}{0pt}{}{-\mu \left(n\right)}{1}\right)=-\mu \left(n\right)$.
• For $m=2$, the equation $i_1+2i_2=2$ only has the solutions $(i_1,i_2)=(2,0)$ and $(i_1,i_2)=(0,1)$, so one obtains the relation

  $$\begin{array}{c}\hfill d_2^{\left(n\right)}=-\left(\genfrac{}{}{0pt}{}{-\mu \left(n\right)}{2}\right)+\left(\genfrac{}{}{0pt}{}{-\mu (n/2)}{1}\right)=-\frac{1}{2}\mu \left(n\right)(\mu \left(n\right)+1)-\mu (n/2).\end{array}$$
• For $m=3$, the solutions to $(l_1,l_2,l_3)$ of the equation $l_1+2l_2+3l_3=3$ are $(3,0,0)$, $(1,1,0)$, and $(0,0,1)$; therefore, by using ${\mu }^3=\mu$, one obtains

  $$\begin{array}{cc}\hfill d_3^{\left(n\right)}& =\left(\genfrac{}{}{0pt}{}{-\mu \left(n\right)}{3}\right)-\left(\genfrac{}{}{0pt}{}{-\mu \left(n\right)}{1}\right)\left(\genfrac{}{}{0pt}{}{-\mu (n/2)}{1}\right)+\left(\genfrac{}{}{0pt}{}{-\mu (n/3)}{1}\right)\hfill \\ & =-\frac{1}{2}\left(\mu {\left(n\right)}^2+\mu \left(n\right)\right)-\mu \left(n\right)\mu (n/2)-\mu (n/3).\hfill \end{array}$$
• For $m=4$, the solutions $(l_1,l_2,l_3,l_4)$ of the equation $l_1+2l_2+3l_3+4l_4=4$ are $(4,0,0,0)$, $(2,1,0,0)$, $(0,2,0,0)$, $(1,0,1,0)$, and $(0,0,0,1)$. By using $\left(\genfrac{}{}{0pt}{}{-\mu \left(n\right)}{4}\right)=\left(\genfrac{}{}{0pt}{}{-\mu \left(n\right)}{2}\right)$, we obtain

  $$\begin{array}{cc}\hfill d_4^{\left(n\right)}& =-\left(\genfrac{}{}{0pt}{}{-\mu \left(n\right)}{4}\right)+\left(\genfrac{}{}{0pt}{}{-\mu \left(n\right)}{2}\right)\left(\genfrac{}{}{0pt}{}{-\mu (n/2)}{1}\right)-\left(\genfrac{}{}{0pt}{}{-\mu \left(n\right)}{1}\right)\left(\genfrac{}{}{0pt}{}{-\mu (n/3)}{1}\right)+\hfill \\ & +\left(\genfrac{}{}{0pt}{}{-\mu (n/2)}{2}\right)+\left(\genfrac{}{}{0pt}{}{-\mu (n/4)}{1}\right)\hfill \\ \hfill & =-\frac{\mu \left(n\right)\left(\mu \right(n)+1)}{2}-\frac{\mu \left(n\right)\left(\mu \right(n)+1)}{2}\mu (n/2)-\mu \left(n\right)\mu (n/3)+\hfill \\ & +\frac{\mu (n/2)\left(\mu \right(n/2)+1)}{2}-\mu (n/4).\hfill \end{array}$$

4.2. The Analogous Formula for (14)

As in the case of cyclotomic polynomials, every coefficient $d_k^{\left(n\right)}$ of ${\Psi }_n$ is a polynomial with rational coefficients with the variables given by the Ramanujan sums $\rho (n,k)$, where $k=1,2,\dots ,n-\phi \left(n\right)-1$. This property is illustrated in the following result.

Theorem 5.

We have

$$d_k^{\left(n\right)}=-\sum _{l_1+2l_2+\cdots +k{l}_k=k}\frac{\rho {(n,1)}^{l_1}}{1^{l_1}{l}_1!}·\frac{\rho {(n,2)}^{l_2}}{2^{l_2}{l}_2!}\cdots \frac{\rho {(n,k)}^{l_k}}{k^{l_k}{l}_k!}.$$

(18)

Proof. 

Denote with $P_j$ the j-th symmetric power polynomial evaluated at the $\phi \left(n\right)$ roots of ${\Phi }_n$ (primitive roots of unity) and with $P_j^{\prime }$ for the jth symmetric power polynomial evaluated at the $n-\phi \left(n\right)$ roots of ${\Psi }_n$. We then use the relations $P_j^{\prime }=-\rho (n,j)$ and $S_k^{\prime }={(-1)}^{k+1}{d}_k^{\left(n\right)}$, and we then apply Theorem 7.3 from [6]. □

We compute the first four coefficients of ${\Psi }_n$, for every $n\in {\mathbb{N}}^{\ast }$, using Formula (18).

• For $k=1$, we have $l_1=1$, hence $d_1^{\left(n\right)}=-\rho (n,1)=-\mu \left(n\right)$.
• For $k=2$, the equation $i_1+2i_2=2$ only has the solutions $(i_1,i_2)=(2,0)$ and $(i_1,i_2)=(0,1)$, so one obtains

  $$\begin{array}{cc}\hfill d_2^{\left(n\right)}& =-\frac{1}{2}\rho {(n,1)}^2-\frac{1}{2}\rho (n,2)\hfill \\ & =-\frac{1}{2}{\mu }^2\left(n\right)-\frac{1}{2}(\mu \left(n\right)+2\mu (n/2))\hfill \\ & =-\frac{1}{2}\mu \left(n\right)(\mu \left(n\right)+1)-\mu (n/2).\hfill \end{array}$$
• For $k=3$, the solutions to $(l_1,l_2,l_3)$ of the equation $l_1+2l_2+3l_3=3$ are $(3,0,0)$, $(1,1,0)$, and $(0,0,1)$; therefore, one obtains

  $$\begin{array}{cc}\hfill d_3^{\left(n\right)}& =-\frac{1}{6}\rho {(n,1)}^3-\frac{1}{2}\rho (n,1)\rho (n,2)-\frac{1}{3}\rho (n,3)\hfill \\ \hfill & =-\frac{1}{6}{\mu }^3\left(n\right)-\frac{1}{2}\mu \left(n\right)\left(\mu \left(n\right)+2\mu (n/2)\right)-\frac{1}{3}\left(\mu \left(n\right)+3\mu (n/3)\right)\hfill \\ \hfill & =-\frac{1}{2}({\mu }^2\left(n\right)+\mu \left(n\right))-\mu \left(n\right)\mu (n/2)-\mu (n/3),\hfill \end{array}$$

  where we used the relation ${\mu }^3=\mu$.
• For $k=4$, the solutions $(l_1,l_2,l_3,l_4)$ of the equation $l_1+2l_2+3l_3+4l_4=4$ are $(4,0,0,0)$, $(2,1,0,0)$, $(0,2,0,0)$, $(1,0,1,0)$, and $(0,0,0,1)$; hence,

  $$\begin{array}{cc}\hfill d_4^{\left(n\right)}& =-\frac{1}{24}\rho {(n,1)}^4-\frac{1}{4}\rho {(n,1)}^2\rho (n,2)-\frac{1}{3}\rho (n,1)\rho (n,3)\hfill \\ & -\frac{1}{8}\rho {(n,2)}^2-\frac{1}{4}\rho (n,4)\hfill \\ & =-\frac{1}{24}{\mu }^4\left(n\right)-\frac{1}{4}{\mu }^2\left(n\right){(\mu \left(n\right)+2\mu (n/2))}^2-\hfill \\ & -\frac{1}{4}(\mu \left(n\right)+2\mu (n/2)+4\mu (n/4))\hfill \\ \hfill & =-\frac{\mu \left(n\right)\left(\mu \right(n)+1)}{2}-\frac{\mu \left(n\right)\left(\mu \right(n)+1)}{2}\mu (n/2)-\mu \left(n\right)\mu (n/3)+\hfill \\ & +\frac{\mu (n/2)\left(\mu \right(n/2)+1)}{2}-\mu (n/4).\hfill \end{array}$$
  In the relation above we repeatedly used the fact that ${\mu }^3=\mu$.

A direct consequence is that for every $n\in {\mathbb{N}}^{\ast }$ and every $k\in \{1,2,\dots ,n-\phi (n\left)\right\}$, the anticyclotomic coefficient $d_k^{\left(n\right)}$ is a polynomial with rational coefficients in $\mu (n/d)$, where d is a divisor of n. Moreover, the degree of $\mu (n/d)$ in every monomial is at most 2.

4.3. A Recurrence Formula for the Coefficients $d_k^{\left(n\right)}$

The following recursive formula involving Ramanujan sums is analogous to (15).

Theorem 6.

The coefficients of ${\Psi }_n$ satisfy the following relation:

$$d_k^{\left(n\right)}=\frac{1}{k}\left[-\rho (n,k)+\rho (n,k-1)d_1^{\left(n\right)}+\cdots +\rho (n,1)d_{k-1}^{\left(n\right)}\right]$$

(19)

Proof. 

We use Newton’s identities for ${\Psi }_n$ to find a recurrence relation for $d_k^{\left(n\right)}$. Recall that we have ${\delta }_j\left(n\right)={\displaystyle \sum _{a=1}^n}{\left(e^{2\pi i\frac{a}{n}}\right)}^j$, where ${\delta }_j\left(n\right)=\left\{\begin{array}{c}n\mathrm{if}n\mid j\hfill \\ 0\mathrm{if}n\nmid j\hfill \end{array}\right..$

In our notation, the symmetric power sum polynomials satisfy the relation $P_j+P_j^{\prime }={\delta }_j\left(n\right)$. However, for $j<n$, we have $n\nmid j$; hence, $P_j^{\prime }=-P_j=-\rho (n,j)$, where $\rho$ denotes the Ramanujan sum (6). The polynomial ${\Psi }_n$ is antireciprocal; hence, we have the relations

$$\begin{array}{c}\hfill S_j^{\prime }={(-1)}^j{d}_{n-\phi \left(n\right)-j}^{\left(n\right)}={(-1)}^{j+1}{d}_j^{\left(n\right)},j=1,2,\dots ,\end{array}$$

where $S_j^{\prime }$ is the evaluation of the jth fundamental symmetric polynomial at the $n-\phi \left(n\right)$ roots of ${\Psi }_n$. The explanation above implies that the recurrence holds. □

We now apply the theorem to compute the first four coefficients of $d_k^{\left(n\right)}$, for every $n\in {\mathbb{N}}^{\ast }$, thereby recovering the results obtained right after Theorem 4.

• For $k=1$, we have $d_1^{\left(n\right)}=-\rho (n,1)=-\mu \left(n\right)$.
• For $k=2$, one obtains

  $$\begin{array}{cc}\hfill d_2^{\left(n\right)}& =\frac{1}{2}\left[-\rho (n,2)+\rho (n,1)d_1^{\left(n\right)}\right]=-\frac{1}{2}\left[\mu \left(n\right)+2\mu \left(n/2\right)+{\mu }^2\left(n\right)\right]\hfill \\ \hfill & =-\frac{{\mu }^2\left(n\right)+\mu \left(n\right)}{2}-\mu \left(n/2\right).\hfill \end{array}$$
• For $k=3$, using the identity ${\mu }^3=\mu$, it follows that

  $$\begin{array}{cc}\hfill d_3^{\left(n\right)}& =\frac{1}{3}\left[-\rho (n,3)+\rho (n,2)d_1^{\left(n\right)}+\rho (n,1)d_2^{\left(n\right)}\right]\hfill \\ \hfill & =\frac{1}{3}\left[-\mu \left(n\right)-3\mu \left(n/3\right)-\left(\mu \left(n\right)+2\mu \left(n/2\right)\right)\mu \left(n\right)+\right.\hfill \\ & +\left.\mu \left(n\right)\left(-\frac{\mu {\left(n\right)}^2+\mu \left(n\right)}{2}-\mu \left(n/2\right)\right)\right]\hfill \\ & =-\frac{1}{2}\left(\mu {\left(n\right)}^2+\mu \left(n\right)\right)-\mu \left(n\right)\mu (n/2)-\mu (n/3)\hfill \end{array}$$
• For $k=4$, again using the identity ${\mu }^3=\mu$, we obtain

  $$\begin{array}{cc}\hfill d_4^{\left(n\right)}& =\frac{1}{4}\left[-\rho (n,4)+\rho (n,3)d_1^{\left(n\right)}+\rho (n,2)d_2^{\left(n\right)}+\rho (n,1)d_3^{\left(n\right)}\right]\hfill \\ \hfill & =\frac{1}{4}\left[-\mu \left(n\right)-2\mu \left(n/2\right)-4\mu \left(n/4\right)-\left(\mu \left(n\right)+3\mu \left(n/3\right)\right)\mu \left(n\right)+\right.\hfill \\ \hfill & +\left(\mu \left(n\right)+2\mu \left(n/2\right)\right)\left(\frac{\mu {\left(n\right)}^2-\mu \left(n\right)}{2}-\mu \left(n/2\right)\right)+\hfill \\ \hfill & +\left.\mu \left(n\right)\left(-\frac{1}{2}\left(\mu {\left(n\right)}^2+\mu \left(n\right)\right)-\mu \left(n\right)\mu (n/2)-\mu (n/3)\right)\right]\hfill \\ \hfill & =-\frac{\mu \left(n\right)\left(\mu \right(n)+1)}{2}-\frac{\mu \left(n\right)\left(\mu \right(n)+1)}{2}\mu (n/2)-\mu \left(n\right)\mu (n/3)+\hfill \\ & +\frac{\mu (n/2)\left(\mu \right(n/2)+1)}{2}-\mu (n/4).\hfill \end{array}$$

5. Numerical Simulations for the Recursive Formula

In this section, we apply the previous recursive formula to compute the coefficients of some ternary and quaternary inverse cyclotomic polynomials (i.e., n is a product of three or four distinct primes) for values of n that are inspired by the calculations from [7].

For numerical simulations, we will focus on some instances where ${\Psi }_n$ is not flat. We have seen earlier that the first nonflat inverse cyclotomic polynomial is ${\Psi }_{561}$. As listed in Table 1 of [7], $-3$ first appears in ${\Psi }_{1155}$ as the coefficient of $z^{33}$, while 4 first appears in ${\Psi }_{2145}$ as the coefficient of $z^{44}$. These are the three cases we focus on, and we compute explicitly until we get the first coefficient which is not 0, 1, or $-1$.

5.1. Results for $n=561=3·11·17$

From

Table 1. Ramanujan sums $\rho (n,j)$ for $j=1,\dots ,17$ and $n=561,1155,2145$.

n	$561=3·11·17$	$1155=3·5·7·11$	$2145=3·5·11·13$
$\rho (n,1)$	$-1$	1	1
$\rho (n,2)$	$-1$	1	1
$\rho (n,3)$	2	$-2$	$-2$
$\rho (n,4)$	$-1$	1	1
$\rho (n,5)$	$-1$	$-4$	$-4$
$\rho (n,6)$	2	$-2$	$-2$
$\rho (n,7)$	$-1$	$-6$	1
$\rho (n,8)$	$-1$	1	1
$\rho (n,9)$	2	$-2$	$-2$
$\rho (n,10)$	$-1$	$-4$	$-4$
$\rho (n,11)$	10	$-10$	$-10$
$\rho (n,12)$	2	$-2$	$-2$
$\rho (n,13)$	$-1$	1	$-12$
$\rho (n,14)$	$-1$	$-6$	1
$\rho (n,15)$	2	8	8
$\rho (n,16)$	$-1$	1	1
$\rho (n,17)$	16	1	1

, the Ramanujan sums $\rho (n,j)$, $j=1,\dots ,17$, take following values: $-1,-1,2,-1,-1,2,-1,-1,2,-1,10,2,-1,-1,2,-1,16.$

Since $\mu \left(n\right)=-1$, from Formula (15) one obtains

$$\begin{array}{cc}\hfill d_1^{\left(n\right)}& =-\mu \left(n\right)=1;\hfill \\ \hfill d_2^{\left(n\right)}& =\frac{1}{2}\left[-\rho (n,2)+\rho (n,1)d_1^{\left(n\right)}\right]=\frac{1}{2}\left[1+(-1)\right]=0;\hfill \\ \hfill d_3^{\left(n\right)}& =\frac{1}{3}\left[-\rho (n,3)+\rho (n,2)d_1^{\left(n\right)}+\rho (n,1)d_2^{\left(n\right)}\right]=\frac{1}{3}\left[-2-1+0\right]=-1;\hfill \\ \hfill d_4^{\left(n\right)}& =\frac{1}{4}\left[-\rho (n,4)+\rho (n,3)d_1^{\left(n\right)}+\rho (n,2)d_2^{\left(n\right)}+\rho (n,1)d_3^{\left(n\right)}\right]\hfill \\ \hfill & =\frac{1}{4}\left[1+2·1+(-1)·0+(-1)·(-1)\right]=1;\hfill \\ \hfill d_5^{\left(n\right)}& =\frac{1}{5}\left[-\rho (n,5)+\rho (n,4)d_1^{\left(n\right)}+\rho (n,3)d_2^{\left(n\right)}+\rho (n,2)d_3^{\left(n\right)}+\rho (n,1)d_4^{\left(n\right)}\right]\hfill \\ \hfill & =\frac{1}{5}\left[1+(-1)·1+2·0+(-1)·(-1)+(-1)·1\right]=0.\hfill \end{array}$$

Similarly, one obtains

$$\begin{array}{cc}\hfill d_6^{\left(n\right)}& =\frac{1}{6}\left[-2+(-1)·1+(-1)·0+2·(-1)+(-1)·1+(-1)·0\right]=-1;\hfill \\ \hfill d_7^{\left(n\right)}& =\frac{1}{7}\left[1+2·1+(-1)·0+(-1)·(-1)+2·1+(-1)·0+(-1)·(-1)\right]=1;\hfill \\ \hfill d_8^{\left(n\right)}& =\frac{1}{8}\left[1+(-1)·1+2·0+(-1)·(-1)+(-1)·1+2·0+(-1)·(-1)+(-1)·1\right]=0;\hfill \\ \hfill d_9^{\left(n\right)}& =\frac{1}{9}\left[-2+(-1)·1+(-1)·0+2·(-1)+(-1)·1+(-1)·0+2·(-1)\right]+\hfill \\ \hfill & +\frac{1}{9}\left[(-1)·1+(-1)·0\right]=-1;\hfill \\ \hfill d_{10}^{\left(n\right)}& =\frac{1}{10}\left[1+2·1+(-1)·0+(-1)·(-1)+2·1+(-1)·0+(-1)·(-1)\right]+\hfill \\ \hfill & +\frac{1}{10}\left[2·1+(-1)·0+(-1)·(-1)\right]=1;\hfill \\ \hfill d_{11}^{\left(n\right)}& =\frac{1}{11}\left[-10+(-1)·1+2·0+(-1)·(-1)+(-1)·1+2·0+(-1)·(-1)\right]+\hfill \\ \hfill & +\frac{1}{11}\left[(-1)·1+2·0+(-1)·(-1)+(-1)·1\right]=-1;\hfill \end{array}$$

Similar calculations result in $d_{12}^{\left(n\right)}=0$, $d_{13}^{\left(n\right)}=1$, $d_{14}^{\left(n\right)}=-1$, $d_{15}^{\left(n\right)}=0$, $d_{16}^{\left(n\right)}=1$, wherein

$$\begin{array}{cc}\hfill d_{17}^{\left(n\right)}& =\frac{1}{17}\left[-16+(-1)·1+2·0+(-1)·(-1)+(-1)·1+2·0+10·(-1)\right]+\hfill \\ & +\frac{1}{17}\left[(-1)·1+2·0+(-1)·(-1)+(-1)·1+2·(-1)+(-1)·0\right]+\hfill \\ & +\frac{1}{17}\left[(-1)·1+2·(-1)+(-1)·0+(-1)·1\right]=-2;\hfill \end{array}$$

This confirms that $-2$ appears as the coefficient of $z^{17}$ in

$$\begin{array}{cc}\hfill {\Psi }_{561}\left(z\right)& =z^{241}-z^{240}+\cdots +2z^{224}+\cdots +z^{18}+\hfill \\ \hfill & -2z^{17}+z^{16}-z^{14}+z^{13}-z^{11}+z^{10}-z^9+z^7-z^6+z^4-z^3+z-1.\hfill \end{array}$$

One may notice that if $n=pqr$ with $p<q<r$ primes, then $\rho (n,j)=-1$ for $j=1,\dots ,p-1$; hence, the first inverse cyclotomic coefficients are $d_1^{\left(n\right)}=-1$, $d_2^{\left(n\right)}=1$, and $c_3^{\left(n\right)}=1$, …, $c_{p-1}^{\left(n\right)}=1$. At the same time,

$$\begin{array}{cc}\hfill d_p^{\left(n\right)}& =-\frac{1}{p}\left[\rho (n,p)+\rho (n,p-1)c_1^{\left(n\right)}+\cdots +\rho (n,1)c_{p-1}^{\left(n\right)}\right]\hfill \\ \hfill & =-\frac{1}{p}\left[(p-1)+(-1)+\cdots +(-1)\right]=0.\hfill \end{array}$$

This argument then allows for the calculation of further coefficients, and it suggests why the larger coefficients of inverse cyclotomic polynomials are moving towards the centre (also using the fact that the polynomial is an antipalindrome).

5.2. Results for $n=1155=3·5·7·11$

From Table 1, the Ramanujan sums $\rho (n,j)$, $j=1,\dots ,17$ take the values

$$1,1,-2,1,-4,-2,-6,1,-2,-4,-10,-2,1,-6,8,1,1.$$

By using $\mu \left(n\right)=-1$ and the second column in Table 1 in Formula (15), one obtains

$$\begin{array}{cc}\hfill d_1^{\left(n\right)}& =-\mu \left(n\right)=-1;\hfill \\ \hfill d_2^{\left(n\right)}& =\frac{1}{2}\left[-\rho (n,2)+\rho (n,1)d_1^{\left(n\right)}\right]=\frac{1}{2}\left[-1+(-1)\right]=-1;\hfill \\ \hfill d_3^{\left(n\right)}& =\frac{1}{3}\left[-\rho (n,3)+\rho (n,2)d_1^{\left(n\right)}+\rho (n,1)d_2^{\left(n\right)}\right]=\frac{1}{3}\left[2-1-1\right]=0;\hfill \\ \hfill d_4^{\left(n\right)}& =\frac{1}{4}\left[-\rho (n,4)+\rho (n,3)d_1^{\left(n\right)}+\rho (n,2)d_2^{\left(n\right)}+\rho (n,1)d_3^{\left(n\right)}\right]\hfill \\ \hfill & =\frac{1}{4}\left[-1+2-1+0\right]=0.\hfill \end{array}$$

Similarly, one obtains

$$\begin{array}{cc}\hfill d_5^{\left(n\right)}& =\frac{1}{5}\left[4+1·(-1)+(-2)·(-1)+1·0+(-1)·0\right]=1;\hfill \\ \hfill d_6^{\left(n\right)}& =\frac{1}{6}\left[2+(-4)·(-1)+1·(-1)+(-2)·0+1·0+1·1\right]=1;\hfill \\ \hfill d_7^{\left(n\right)}& =\frac{1}{7}\left[6+(-2)·(-1)+(-4)·(-1)+1·0+(-2)·0+1·1+1·1\right]=\frac{14}{7}=2.\hfill \end{array}$$

This confirms that 2 appears as the coefficient of $z^7$ in

$$\begin{array}{cc}\hfill {\Psi }_{1155}\left(z\right)& =z^{675}+z^{674}+z^{673}-z^{670}-z^{669}-2z^{668}+\cdots +3z^{642}+\cdots \hfill \\ \hfill & -3z^{33}+\cdots +z^8+2z^7+z^6+z^5-z^2-z-1.\hfill \end{array}$$

5.3. Results for $n=2145=3·5·11·13$

From Table 1, $\rho (n,j)$, $j=1,\dots ,17$ take the values

$$1,1,-2,1,-4,-2,1,1,-2,-4,-10,-2,-12,1,8,1,1.$$

Since $\mu \left(n\right)=-1$, from Formula (15), one obtains

$$\begin{array}{cc}\hfill d_1^{\left(n\right)}& =-\mu \left(n\right)=-1;\hfill \\ \hfill d_2^{\left(n\right)}& =\frac{1}{2}\left[-\rho (n,2)+\rho (n,1)d_1^{\left(n\right)}\right]=\frac{1}{2}\left[-1+1·(-1)\right]=-1;\hfill \\ \hfill d_3^{\left(n\right)}& =\frac{1}{3}\left[-\rho (n,3)+\rho (n,2)d_1^{\left(n\right)}+\rho (n,1)d_2^{\left(n\right)}\right]=\frac{1}{3}\left[2+1·(-1)+1·(-1)\right]=0;\hfill \\ \hfill d_4^{\left(n\right)}& =\frac{1}{4}\left[-\rho (n,4)+\rho (n,3)d_1^{\left(n\right)}+\rho (n,2)d_2^{\left(n\right)}+\rho (n,1)d_3^{\left(n\right)}\right]\hfill \\ \hfill & =\frac{1}{4}\left[-1+(-2)·(-1)+1·(-1)+1·0\right]=0;\hfill \\ \hfill d_5^{\left(n\right)}& =\frac{1}{5}\left[-\rho (n,5)+\rho (n,4)d_1^{\left(n\right)}+\rho (n,3)d_2^{\left(n\right)}+\rho (n,2)d_3^{\left(n\right)}+\rho (n,1)d_4^{\left(n\right)}\right]\hfill \\ \hfill & =\frac{1}{5}\left[4+1·(-1)+(-2)·(-1)+1·0+1·0\right]=1.\hfill \end{array}$$

Similarly, one obtains

$$\begin{array}{cc}\hfill d_6^{\left(n\right)}& =\frac{1}{6}\left[2+(-4)·(-1)+1·(-1)+(-2)·0+1·0+1·1\right]=1;\hfill \\ \hfill d_7^{\left(n\right)}& =\frac{1}{7}\left[-1+(-2)·(-1)+(-4)·(-1)+1·0+(-2)·0+1·1+1·1\right]=1;\hfill \\ \hfill d_8^{\left(n\right)}& =\frac{1}{8}\left[-1+1·(-1)+(-2)·(-1)+(-4)·0+1·0+(-2)·1+1·1+1·1\right]=0;\hfill \\ \hfill d_9^{\left(n\right)}& =\frac{1}{9}\left[2+1·(-1)+1·(-1)+(-2)·0+(-4)·0+1·1\right]\hfill \\ \hfill & +\frac{1}{9}\left[(-2)·1+1·1+1·0\right]=0;\hfill \\ \hfill d_{10}^{\left(n\right)}& =\frac{1}{10}\left[4+(-2)·(-1)+1·(-1)+1·0+(-2)·0+(-4)·1\right]+\hfill \\ \hfill & +\frac{1}{10}\left[1·1+(-2)·1+1·0+1·0\right]=0;\hfill \\ \hfill d_{11}^{\left(n\right)}& =\frac{1}{11}\left[10+(-4)·(-1)+(-2)·(-1)+1·0+1·0+(-2)·1+(-4)·1\right]+\hfill \\ \hfill & +\frac{1}{11}\left[1·1+(-2)·0+1·0+1·0\right]=1;\hfill \\ \hfill d_{12}^{\left(n\right)}& =\frac{1}{12}\left[2+(-10)·(-1)+(-4)·(-1)+(-2)·0+1·0+1·1+(-2)·1\right]+\hfill \\ \hfill & +\frac{1}{12}\left[(-4)·1+1·0+(-2)·0+1·0+1·1\right]=1.\hfill \end{array}$$

Finally, one obtains

$$\begin{array}{cc}\hfill d_{13}^{\left(n\right)}& =\frac{1}{13}\left[12+(-2)·(-1)+(-10)·(-1)+(-4)·0+(-2)·0+1·1+1·1\right]+\hfill \\ & +\frac{1}{13}\left[(-2)·1+(-4)·0+1·0+(-2)·0+1·1+1·1\right]=\frac{26}{13}=2.\hfill \end{array}$$

This confirms that 2 appears as the coefficient of $z^{13}$ in

$$\begin{array}{cc}\hfill {\Psi }_{2145}\left(z\right)& =z^{1185}+z^{1184}+z^{1183}-z^{1180}-z^{1179}-z^{1178}-z^{1174}-z^{1173}-2z^{1172}+\cdots \hfill \\ \hfill & +3z^{1152}+\cdots -4z^{1141}+\cdots +4z^{44}+\cdots -3z^{33}+\cdots +\hfill \\ \hfill & +z^{14}+2z^{13}+z^{12}+z^{11}+z^7+z^6+z^5-z^2-z-1.\hfill \end{array}$$

6. Conclusions

In this paper, we have studied properties of the coefficients of inverse cyclotomic polynomials, for which we have provided two explicit formulae, which involve the calculation of partitions, and a recursive formula (Theorem 6). The practicality of the latter approach was illustrated in Section 5 for some ternary and quaternary integers.

In future works, we aim to study the computational complexity of Formula (19) and to express the results in matrix language as has been calculated for polynomial sequences [23]. We also plan to explore practical applications, such as, for example, regarding difference equations [24].