# Polynomials and algebraic curves related to certain binary and $b$-ary
overpartitions

Karl Dilcher  Department of Mathematics and Statistics
Dalhousie University
Halifax, Nova Scotia, B3H 4R2, Canada<EMAIL_ADDRESS>and  Larry
Ericksen  1212 Forest Drive, Millville, NJ 08332-2512, USA<EMAIL_ADDRESS>
###### Abstract.

We begin by considering a sequence of polynomials in three variables whose
coefficients count restricted binary overpartitions with certain properties.
We then concentrate on two specific subsequences that are closely related to
the Chebyshev polynomials of both kinds, deriving combinatorial and algebraic
properties of some special cases. We show that the zeros of these polynomial
sequences lie on certain algebraic curves, some of which we study in greater
detail. Finally, we extend part of this work to restricted $b$-ary
overpartitions for arbitrary integers $b\geq 2$.

###### Key words and phrases:

Binary overpartitions, $b$-ary overpartitions, restricted overpartitions,
polynomial analogues, generating functions, recurrence relations, multinomial
coefficients

###### 2010 Mathematics Subject Classification:

Primary 11P81; Secondary 11B37, 11B83

Research supported in part by the Natural Sciences and Engineering Research
Council of Canada, Grant # 145628481

##  1\. Introduction

While the basic theory of binary partitions goes back to Euler [4, p. 162ff.],
quite recently Rødseth and Sellers [10] introduced and studied $b$-ary
overpartitions for a fixed integer base $b\geq 2$, in analogy to ordinary
overpartitions that had been introduced a little earlier by Corteel and
Lovejoy [2].

A $b$-ary overpartition of an integer $n\geq 1$ is a non-increasing sequence
of nonnegative integer powers of $b$ whose sum is $n$, and where the first
occurrence of a power $b$ may be overlined. We denote the number of $b$-ary
overpartitions by $\overline{S}_{b}(n)$, which differs from the notation in
[10].

######  Example 1.1.

(See [10, p. 346]). The binary overpartitions of $n=4$ are

$4,\;\overline{4},\;2+2,\;\overline{2}+2,\;2+1+1,\;2+\overline{1}+1,\;\overline{2}+1+1,\;\overline{2}+\overline{1}+1,\;1+1+1+1,\;\overline{1}+1+1+1.$

Thus $\overline{S}_{2}(4)=10$.

As is illustrated in this example, the overlined parts form a $b$-ary
partition into distinct parts, while the non-overlined parts form an ordinary
$b$-ary partition. We can now see that the generating function is

(1.1)
$\sum_{n=0}^{\infty}\overline{S}_{b}(n)q^{n}=\prod_{j=0}^{\infty}\frac{1+q^{b^{j}}}{1-q^{b^{j}}}.$

The concept of a $b$-ary overpartition can be restricted in different ways,
only one of which we will consider here. In analogy to the restricted $b$-ary
partitions, such as hyperbinary representations, we restrict the number of
times a non-overlined power of $b$ may occur in a $b$-ary overpartition; we
denote this number by $\lambda$ and call such overpartitions
$\lambda$-restricted. In this case the generating function is

(1.2)
$\sum_{n=0}^{\infty}\overline{S}_{b}^{\lambda}(n)q^{n}=\prod_{j=0}^{\infty}\left(1+q^{b^{j}}\right)\left(1+q^{b^{j}}+q^{2\cdot
b^{j}}+\dots+q^{\lambda\cdot b^{j}}\right),$

where $\overline{S}_{b}^{\lambda}(n)$ is the number of $b$-ary overpartitions
of $n$ in which each non-overlined power of $b$ may occur at most $\lambda$
times.

######  Example 1.2.

Let $b=\lambda=2$. Then (1.2) becomes

(1.3)
$1+2q+4q^{2}+5q^{3}+8q^{4}+10q^{5}+13q^{6}+14q^{7}+18q^{8}+21q^{9}+26q^{10}+\cdots$

Thus, in particular, $\overline{S}_{2}^{2}(4)=8$, which is consistent with
Example 1.1, where all but the last two binary overpartitions are counted by
$\overline{S}_{2}^{2}(4)$. The series (1.3) can also be found in [7]. This
last paper deals with the case $\lambda=b$ as well, but in contrast to our
work it focuses on congruences of the relevant numerical sequences.

In the recent paper [3] we defined the concept of restricted multicolor
$b$-ary partitions as a generalization of restricted $b$-ary overpartitions,
and further defined polynomial analogues of the relevant partition functions.
These polynomials then allowed us to not just count the partitions in
question, but to characterize them. We will not be concerned with this aspect
of the theory in the present paper.

In the special case of restricted $b$-ary overpartitions with $\lambda=2$, the
polynomials introduced in [3] specialize as follows. Let $Z=(x,y,z)$ be a
triple of variables, and $T=(r,s,t)$ a triple of positive integers. Then, in
the notation of [3, Def. 2.4], we define

(1.4)
$\sum_{n=0}^{\infty}\Omega_{b,T}^{(1,2)}(n;Z)q^{n}=\prod_{j=0}^{\infty}\left(1+x^{r^{j}}q^{b^{j}}\right)\left(1+y^{s^{j}}q^{b^{j}}+z^{t^{j}}q^{2\cdot
b^{j}}\right).$

Comparing this with (1.2), we immediately get, for any base $b\geq 2$,

(1.5) $\overline{S}_{b}^{2}(n)=\Omega_{b,T}^{(1,2)}(n;1,1,1),\qquad
n=0,1,2,\ldots,$

where the triple $T$ is arbitrary.

The main purpose of this paper is to consider various aspects of the
polynomial sequence defined by (1.4) in the special case $r=s=t=1$. After
deriving a few basic properties in Section 2, we consider two particular
subsequences in Section 3, which turn out to be closely related to the
Chebyshev polynomials of both kinds. In the following two sections we then
specialize the variables $x,y,z$ in two different ways, obtaining polynomial
sequences in one, resp. two, variables with interesting properties. In
particular, we derive divisibility properties and combinatorial
interpretations of these polynomials. Section 7 is then devoted to the zero
distribution of these and a few other related polynomial sequences. One such
curve, a particularly interesting quartic of genus 0, is studied in greater
detail in Section 8. Finally in Section 9, we show that much of the content of
Sections 2–6 can be generalized to an arbitrary integer base $b\geq 2$ with
$\lambda=b$.

##  2\. Some basic properties

In [3] we derived recurrence relations for the general polynomial sequences
that characterize all restricted multicolor $b$-ary partitions. In the special
case $b=\lambda=2$, these recurrences take the form
$\Omega_{2,T}^{(1,2)}(0;x,y,z)=1$, $\Omega_{2,T}^{(1,2)}(1;x,y,z)=x+y$, and

$\displaystyle\Omega_{2,T}^{(1,2)}(2n;x,y,z)$
$\displaystyle=\Omega_{2,T}^{(1,2)}(n;x^{r},y^{s},z^{t})+(z+xy)\cdot\Omega_{2,T}^{(1,2)}(n-1;x^{r},y^{s},z^{t}),$
$\displaystyle\Omega_{2,T}^{(1,2)}(2n+1;x,y,z)$
$\displaystyle=(x+y)\cdot\Omega_{2,T}^{(1,2)}(n;x^{r},y^{s},z^{t})+xz\cdot\Omega_{2,T}^{(1,2)}(n-1;x^{r},y^{s},z^{t}).$

From this point on, we specialize further to $r=s=t=1$. To simplify notation
we set, for all $n\geq 0$,

(2.1) $p_{n}(x,y,z):=\Omega_{2,T}^{(1,2)}(n;x,y,z),\qquad T=(1,1,1).$

Then (1.4) simplifies to the generating function

(2.2)
$\sum_{n=0}^{\infty}p_{n}(x,y,z)q^{n}=\prod_{j=0}^{\infty}\left(1+xq^{2^{j}}\right)\left(1+yq^{2^{j}}+zq^{2\cdot
2^{j}}\right),$

and the recurrence relations before (2.1) turn into $p_{0}(x,y,z)=1$,
$p_{1}(x,y,z)=x+y$, and for $n\geq 1$,

(2.3) $\displaystyle p_{2n}(x,y,z)$ $\displaystyle=p_{n}(x,y,z)+(z+xy)\cdot
p_{n-1}(x,y,z),$ (2.4) $\displaystyle p_{2n+1}(x,y,z)$
$\displaystyle=(x+y)\cdot p_{n}(x,y,z)+xz\cdot p_{n-1}(x,y,z).$

See Table 1 for the first few such polynomials, where $\Sigma$ denotes the
sums of the coefficients.

$n$ | $p_{n}(x,y,z)$ | $\Sigma$
---|---|---
0 | 1 | 1
1 | $x+y$ | 2
2 | $xy+x+y+z$ | 4
3 | $x^{2}+2xy+xz+y^{2}$ | 5
4 | $x^{2}y+xy^{2}+xy+xz+yz+x+y+z$ | 8
5 | $x^{2}y+x^{2}z+xy^{2}+xyz+x^{2}+2xy+xz+y^{2}+yz$ | 10
6 | $x^{2}y^{2}+x^{2}y+xy^{2}+2xyz+x^{2}+2xy+2xz+y^{2}+yz+z^{2}$ | 13
7 | $x^{3}+3x^{2}y+2x^{2}z+x^{2}yz+3xy^{2}+2xyz+xz^{2}+y^{3}$ | 14

Table 1: $p_{n}(x,y,z)$ and sums of coefficients for $0\leq n\leq 7$.

The recurrence relations (2.3), (2.4) imply that the polynomials
$p_{n}(x,y,z)$ can be written in the form

(2.5) $p_{n}(x,y,z)=\sum_{i,j,k\geq 0}c_{n}(i,j,k)\cdot x^{i}y^{j}z^{k},\qquad
n\geq 0.$

The generating function (2.2) then shows that the coefficients in (2.5) have
the following combinatorial interpretation.

######  Proposition 2.1.

For any integers $n,i,j,k\geq 0$, the coefficient $c_{n}(i,j,k)$ in (2.5)
counts the number of $2$-restricted binary overpartitions of $n$ that have

  1. $i$ different and single overlined parts,

  2. $j$ different and single non-overlined parts, and

  3. $k$ different pairs of non-overlined parts.

######  Example 2.2.

By (1.3) in Example 1.2, the number of 2-restricted binary overpartitions of
$n=6$ is 13. They are, in particular,

$(4,2),\;(\overline{4},2),\;(4,\overline{2}),\;(\overline{4},\overline{2}),\;(4,1,1),\;(\overline{4},1,1),\;(4,\overline{1},1),\;(\overline{4},\overline{1},1),$

$(\overline{2},2,2),(2,2,1,1),\;(\overline{2},2,1,1),\;(2,2,\overline{1},1),\;(\overline{2},2,\overline{1},1).$

Then, for instance, the coefficient 2 in the term $2xyz$ of $p_{6}(x,y,z)$
(see Table 1) counts the partitions $(\overline{2},2,1,1)$ and
$(2,2,\overline{1},1)$. Similarly, the term $x^{2}y^{2}$ of $p_{6}(x,y,z)$
counts the single partition $(\overline{2},2,\overline{1},1)$.

The following is an obvious consequence of Proposition 2.1; we will need it in
Section 4.

######  Corollary 2.3.

If we write

$p_{n}(x,1,1)=\sum_{i\geq 0}c_{n}(i)\cdot x^{i},\qquad n\geq 0,$

then $c_{n}(i)$ counts the number of $2$-restricted binary overpartitions of
$n$ with exactly $i$ overlined parts.

##  3\. Connections with Chebyshev polynomials

The main objects of this section are two subsequences of the polynomial
sequence $p_{n}(x,y,z)$, namely

(3.1) $Q_{n}(x,y,z):=p_{2^{n+1}-2}(x,y,z),\qquad
R_{n}(x,y,z):=p_{2^{n}-1}(x,y,z),$

for all $n\geq 0$. Combining these definitions with the recurrence relation
(2.4), we get as a first consequence,

(3.2) $R_{n+1}(x,y,z)=(x+y)\cdot R_{n}(x,y,z)+xz\cdot Q_{n-1}(x,y,z),\qquad
n\geq 1.$

However, the following three-term recurrence relations will be more important.
For greater ease of notation we suppress the arguments $x,y,z$.

######  Proposition 3.1.

We have $Q_{0}=1$, $Q_{1}=xy+x+y+z$, $R_{0}=1$, $R_{1}=x+y$, and for $n\geq
1$,

(3.3) $\displaystyle Q_{n+1}$ $\displaystyle=(xy+x+y+z)\cdot
Q_{n}-(x^{2}y+xy^{2}+yz)\cdot Q_{n-1},$ (3.4) $\displaystyle R_{n+1}$
$\displaystyle=(xy+x+y+z)\cdot R_{n}-(x^{2}y+xy^{2}+yz)\cdot R_{n-1}.$

###### Proof.

We proceed by induction on $n$, and for further ease of notation we set
$P(n):=p_{n}(x,y,z)$. First, the expressions for $Q_{0},Q_{1},R_{0},R_{1}$
follow from (3.1) and Table 1. Also, by (3.1) we have $Q_{2}=P(6)$ and
$R_{2}=P(3)$. With the relevant entries in Table 1 we can now verify (3.3) and
(3.4) for $n=1$, which is the induction beginning.

Suppose now that (3.3) and (3.4) are true up to some $n-1$ in place of $n$;
our aim is to show that they hold also for $n$, that is, as written in (3.3)
and (3.4). By the induction hypothesis and (3.1) we have

(3.5) $\displaystyle P(2^{n}-2)$
$\displaystyle=(xy+x+y+z)P(2^{n-1}-2)-(x^{2}y+xy^{2}+yz)P(2^{n-2}-2),$ (3.6)
$\displaystyle P(2^{n}-1)$
$\displaystyle=(xy+x+y+z)P(2^{n-1}-1)-(x^{2}y+xy^{2}+yz)P(2^{n-2}-1).$

We multiply both sides of (3.5) by $xz$, and both sides of (3.6) by $x+y$.
Applying (2.4) three times, namely for $n$ replaced by $2^{n}-1$, by
$2^{n-1}-1$, and by $2^{n-2}-1$, we get

(3.7) $P(2^{n+1}-1)=(xy+x+y+z)P(2^{n}-1)-(x^{2}y+xy^{2}+yz)P(2^{n-1}-1),$

which, by (3.1), gives (3.4). Next, we use the induction hypothesis again, in
the form

(3.8) $P(2^{n+1}-2)=(xy+x+y+z)P(2^{n}-2)-(x^{2}y+xy^{2}+yz)P(2^{n-1}-2).$

We multiply both sides of (3.8) by $z+xy$ and then add (3.7). Applying (2.3)
three times and finally using the first identity in (3.1), we get (3.3). This
completes the proof of Proposition 3.1 by induction. ∎

With the recurrence relations (3.3) and (3.4) we can now obtain generating
functions for the two polynomial sequences.

######  Proposition 3.2.

The polynomials $Q_{n}$ and $R_{n}$ satisfy the generating functions

(3.9)
$\displaystyle\sum_{n=0}^{\infty}Q_{n}(x,y,z)q^{n}=\frac{1}{1-(xy+x+y+z)q+(x^{2}y+xy^{2}+yz)q^{2}},$
(3.10)
$\displaystyle\sum_{n=0}^{\infty}R_{n}(x,y,z)q^{n}=\frac{1-(xy+z)q}{1-(xy+x+y+z)q+(x^{2}y+xy^{2}+yz)q^{2}}.$

###### Proof.

We multiply both sides of (3.9) by the denominator on the right, and take the
Cauchy product with the power series on the left. Then the constant
coefficient is $Q_{0}(x,y,z)=1$, while the coefficient of $x$ is zero since
$Q_{1}(x,y,z)=xy+x+y+z$; all other coefficients also vanish, as a consequence
of (3.3). This proves (3.9), and (3.10) is obtained analogously, using (3.4).
∎

Both Propositions 3.1 and 3.2 indicate that there might be a connection with
Chebyshev polynomials. This is indeed the case, as the next result shows. We
recall that the Chebyshev polynomials of the first kind, $T_{n}(w)$, and of
the second kind, $U_{n}(w)$, can be defined by the generating functions

(3.11)
$\sum_{n=0}^{\infty}T_{n}(w)v^{n}=\frac{1-wv}{1-2wv+v^{2}},\qquad\sum_{n=0}^{\infty}U_{n}(w)v^{n}=\frac{1}{1-2wv+v^{2}}.$

Using these polynomials, we can now state and prove the following identities.

######  Proposition 3.3.

For all $n\geq 0$ we have

(3.12) $\displaystyle Q_{n}(x,y,z)$
$\displaystyle=\big{(}x^{2}y+xy^{2}+yz\big{)}^{n/2}U_{n}\left(\frac{xy+x+y+z}{2(x^{2}y+xy^{2}+yz)^{1/2}}\right),$
(3.13) $\displaystyle R_{n}(x,y,z)$
$\displaystyle=\big{(}x^{2}y+xy^{2}+yz\big{)}^{n/2}T_{n}\left(\frac{xy+x+y+z}{2(x^{2}y+xy^{2}+yz)^{1/2}}\right)+\widetilde{U}_{n-1},$

where

$\widetilde{U}_{n-1}:=\frac{x+y-xy-z}{2}\cdot Q_{n-1}(x,y,z).$

###### Proof.

Comparing (3.9) with the second identity in (3.11), we see that

$q=\frac{v}{(x^{2}y+xy^{2}+yz)^{1/2}}\quad\hbox{and}\quad
w=\frac{xy+x+y+z}{2(x^{2}y+xy^{2}+yz)^{1/2}}.$

Equating coefficients of $q^{n}$ then gives (3.12).

Next, with $w$ and $v$ as above, we rewrite the numerator on the right of
(3.10) as

(3.14) $\displaystyle 1-(xy+z)q$ $\displaystyle=(1-wv)+\frac{x+y-xy-z}{2}\cdot
q$ $\displaystyle=(1-wv)+\frac{x+y-xy-z}{2(x^{2}y+xy^{2}+yz)^{1/2}}\cdot v.$

The term $1-wv$, together with the first identity in (3.11), leads to the
first summand in (3.13), while the second term in the last line of (3.14)
leads to $\widetilde{U}_{n-1}$ after some straightforward manipulations. ∎

##  4\. A first special case: $y=z=1$

The case $y=z=1$ is of particular interest. By a slight abuse of notation we
set

$Q_{n}(x):=Q_{n}(x,1,1)\qquad\hbox{and}\qquad R_{n}(x):=R_{n}(x,1,1).$

The first few of these polynomials are listed in Table 2. With $y=z=1$ we get
$x+y-xy-z=0$, so that $\widetilde{U}_{n-1}=0$ and Proposition 3.3 simplifies
as follows.

######  Corollary 4.1.

For all $n\geq 0$ we have

(4.1) $\displaystyle Q_{n}(x)$
$\displaystyle=\big{(}1+x+x^{2}\big{)}^{n/2}\cdot
U_{n}\left(\frac{1+x}{(1+x+x^{2})^{1/2}}\right),$ (4.2) $\displaystyle
R_{n}(x)$ $\displaystyle=\big{(}1+x+x^{2}\big{)}^{n/2}\cdot
T_{n}\left(\frac{1+x}{(1+x+x^{2})^{1/2}}\right).$

These identities show that most properties and identities satisfied by the
Chebyshev polynomials will carry over to the polynomials $Q_{n}(x)$ and
$R_{n}(x)$. For instance, factors of the Chebyshev polynomials (see, e.g., [9,
p. 227 ff]) lead to corresponding factors of the polynomials $Q_{n}(x)$,
$R_{n}(x)$. More will be stated in the following corollary.

$n$ | $Q_{n}(x)$ | $R_{n}(x)$
---|---|---
0 | 1 | 1
1 | $2x+2$ | $x+1$
2 | $3x^{2}+7x+3$ | $x^{2}+3x+1$
3 | $4x^{3}+16x^{2}+16x+4$ | $x^{3}+6x^{2}+6x+1$
4 | $5x^{4}+30x^{3}+51x^{2}+30x+5$ | $x^{4}+10x^{3}+19x^{2}+10x+1$
5 | $6x^{5}+50x^{4}+126x^{3}+126x^{2}+50x+6$ | $x^{5}+15x^{4}+45x^{3}+45x^{2}+15x+1$

Table 2: $Q_{n}(x)$ and $R_{n}(x)$ for $0\leq n\leq 5$.

######  Corollary 4.2.

Let $n\geq 1$.

  1. (a)

$Q_{n}(x)$ and $R_{n}(x)$ are palindromic polynomials of degree $n$.

  2. (b)

Their zeros are real and negative, and with the exception of $x=-1$ they
appear in pairs whose product is $1$ and whose sum can be arbitrarily large as
$n$ grows.

  3. (c)

$Q_{n-1}(x)$ is a divisibility sequence: if $m|n$, then
$Q_{m-1}(x)|Q_{n-1}(x)$.

  4. (d)

The sums of the coefficients are

$Q_{n}(1)=\tfrac{1}{2}\bigl{(}3^{n+1}-1\bigr{)},\qquad
R_{n}(1)=\tfrac{1}{2}\bigl{(}3^{n}+1\bigr{)},$

and in particular, $R_{n+1}(1)=Q_{n}(1)+1$.

The identities in part (d) have also been obtained by Ma and Lu [7] as their
Corollary 6 and Theorem 5, respectively.

###### Proof of Corollary 4.2.

(a) With (4.1) and (4.2) it is easy to see that $x^{n}Q_{n}(1/x)=Q_{n}(x)$,
and similarly for $R_{n}(x)$. The degree statement follows from Proposition
3.1 with $y=z=1$.

(b) With the arguments of $U_{n}$ and $T_{n}$ in Corollary 4.1 in mind, we set
$w=(1+x)/(1+x+x^{2})^{1/2}$. This can be rewritten as

(4.3) $x^{2}+\frac{2-w^{2}}{1-w^{2}}\cdot x+1=0.$

It is known that the zeros of $U_{n}(w)$ and $T_{n}(w)$ lie strictly between
$-1$ and 1, and so we consider $0<w^{2}<1$. But then it is easy to see that
the discriminant of the quadratic in (4.3) is positive, and thus for each pair
of zeros $\pm w$ of $U_{n}(w)$ or $T_{n}(w)$ there is a pair of zeros of
$Q_{n}(x)$ or $R_{n}(x)$ whose product is 1 and whose sum is
$-(2-w^{2})/(1-w^{2})$; this follows from the fact that the quadratic in (4.3)
is itself palindromic and is monic. Since it is known that in both cases $w$
can be arbitrarily close to $\pm 1$ if $n$ is sufficiently large, the sum of
the zeros of $Q_{n}(x)$ or $R_{n}(x)$ can be arbitrarily large negative, as
claimed.

(c) This follows from the corresponding property of the Chebyshev polynomials
$U_{n}(w)$; see, e.g., [9, p. 232].

(d) By (3.3) with $x=y=z=1$ we have $Q_{0}(1)=1$, $Q_{1}(1)=4$, and for $n\geq
1$, $Q_{n+1}(1)=4Q_{n}(1)-3Q_{n-1}(1)$. It is now easy to verify that the
sequence $\tfrac{1}{2}(3^{n+1}-1)$ also satisfies this recurrence relation
with the same initial conditions; hence the two sequences are identical. The
proof for $R_{n}(1)$ is analogous. ∎

The identities (4.1) and (4.2) point to a possible connection between the
polynomials $Q_{n}(x),R_{n}(x)$ and the trinomial coefficients or the
trinomial triangle. The $n$th row of the trinomial triangle consists of the
coefficients of the polynomial $(1+x+x^{2})^{n}$; see the entry A027907 in
[8].

######  Proposition 4.3.

For any integer $n\geq 1$ we have

(4.4) $x\cdot Q_{n-1}(x^{2})+R_{n}(x^{2})=(1+x+x^{2})^{n}.$

In other words, the coefficients of $R_{n}(x)$ are the even-index entries of
the $n$th row of the trinomial triangle, while the coefficients of
$Q_{n-1}(x)$ are the odd-index entries.

The even- and odd-index entries of the rows of the trinomial triangle are
listed in [8] as A056241 and A123934, respectively. We were led to Proposition
4.3 through these entries. For a reformulation of Proposition 4.3, see (4.10)
below.

###### Proof of Proposition 4.3.

We use the defining identities

$\sin{\theta}\cdot U_{n-1}(\cos{\theta})=\sin(n\theta),\qquad
T_{n}(\cos{\theta})=\cos(n\theta).$

Multiplying both sides of the left identity by $i$, then adding both and using
$2i\sin{\theta}=e^{i\theta}-e^{-i\theta},\,2\cos{\theta}=e^{i\theta}+e^{-i\theta}$,
and $w:=e^{i\theta}$, we get

(4.5)
$\frac{w-w^{-1}}{2}U_{n-1}\left(\frac{w+w^{-1}}{2}\right)+T_{n}\left(\frac{w+w^{-1}}{2}\right)=w^{n},\qquad
n\geq 1.$

We now set

$w=\sqrt{\frac{1+x+x^{2}}{1-x+x^{2}}}=\frac{1+x+x^{2}}{\sqrt{1+x^{2}+x^{4}}}.$

Then after some straightforward manipulations, which involves the
factorization $1+x^{2}+x^{4}=(1-x+x^{2})(1+x+x^{2})$, we get

(4.6) $\frac{w+w^{-1}}{2}=\frac{1+x^{2}}{(1+x^{2}+x^{4})^{1/2}}.$

Similarly, we obtain

(4.7) $\frac{w-w^{-1}}{2}=\frac{x}{(1+x^{2}+x^{4})^{1/2}}.$

Substituting (4.6) and (4.7) into (4.5), we get

(4.8)
$\frac{x}{(1+x^{2}+x^{4})^{1/2}}U_{n-1}\left(\frac{w+w^{-1}}{2}\right)+T_{n}\left(\frac{w+w^{-1}}{2}\right)=\frac{(1+x+x^{2})^{n}}{(1+x^{2}+x^{4})^{n/2}}.$

Finally, multiplying both sides of (4.8) by $(1+x^{2}+x^{4})^{n/2}$ and using
(4.1) and (4.2), we get the desired identity (4.4). ∎

For our next result we use a notation for the trinomial coefficient that can
be found in [1, p. 78]: For an integer $n\geq 0$ we write

(4.9) $\left(1+x+x^{2}\right)^{n}=\sum_{j=0}^{n}\binom{n,3}{j}x^{j}.$

We can now state and prove the following interpretation of trinomial
coefficients in terms of binary overpartitions.

######  Proposition 4.4.

Let $n\geq 1$ be an integer. Then

  1. (a)

$\binom{n,3}{2j}$ is the number of binary $2$-restricted overpartitions of
$2^{n}-1$ with exactly $j$ overlined parts, $j=0,1,\ldots,n$;

  2. (b)

$\binom{n,3}{2j+1}$ is the number of binary $2$-restricted overpartitions of
$2^{n}-2$ with exactly $j$ overlined parts, $j=0,1,\ldots,n-1$.

###### Proof.

If we set

$R_{n}(x)=\sum_{j=0}^{n}r_{n}(j)x^{j},\qquad
Q_{n}(x)=\sum_{j=0}^{n}q_{n}(j)x^{j},$

then Proposition 4.3 can be stated as

(4.10) $r_{n}(j)=\binom{n,3}{2j},\qquad q_{n-1}(j)=\binom{n,3}{2j+1},$

valid for $j=0,1,\ldots,n$, resp. for $j=0,1,\ldots,n-1$. The statements of
the proposition now follow from Corollary 2.3 combined with (3.1) for $y=z=1$.
∎

######  Example 4.5.

Let $n=3$. The 2-restricted binary overpartitions of $2^{3}-1=7$ are

$(4,2,1),(\overline{4},2,1),(4,\overline{2},1),(4,2,\overline{1}),(\overline{4},2,\overline{1}),(\overline{4},\overline{2},1),(4,\overline{2},\overline{1}),(\overline{4},\overline{2},\overline{1}),$

$(4,\overline{1},1,1),(\overline{4},\overline{1},1,{1}),(\overline{2},2,2,1),(\overline{2},{2},2,\overline{1}),({2},2,\overline{1},1,1),(\overline{2},2,\overline{1},1,1).$

There are 14 of them, consistent with (1.3) in Example 1.2. We see that the
numbers of these partitions with 0, 1, 2, and 3 overlined parts are 1, 6, 6,
and 1, respectively, which agrees with Proposition 4.4(a) and

(4.11)
$\left(1+x+x^{2}\right)^{3}=\sum_{j=0}^{6}\binom{3,3}{j}x^{j}=1+3x+6x^{2}+7x^{3}+6x^{4}+3x^{5}+x^{6}.$

Similarly, Example 2.2 shows the 2-restricted binary overpartitions
$2^{3}-2=6$, of which there are 13. We see that the numbers of those with 0,
1, and 2 overlined parts are 3, 7, and 3, respectively. This is consistent
with Proposition 4.4(b) and again with (4.11).

Remarks. (1) It is known that for $0\leq k\leq n-1$ the zeros of $U_{n}(w)$
and of $T_{n}(w)$ are $w_{k}=\cos\big{(}\pi(k+1)/(n+1)\big{)}$ and
$w^{\prime}_{k}=\cos\big{(}\pi(2k+1)/2n\big{)}$, respectively. Therefore, by
solving (4.3) for $x$, one can easily obtain explicit expressions for the
zeros of $Q_{n}(x)$ and $R_{n}(x)$.

(2) While $R_{n}(x)$ is not a divisibility sequence, a weaker property still
holds; see, e.g., [9] for the corresponding Chebyshev analogue.

##  5\. A second special case: $x=y=z$

If we set $y$ and $z$ equal to $x$ instead of 1, we get another pair of
single-variable polynomial sequences with some interesting properties. To
distinguish the notation from that in Section 4, we set

(5.1) $Q_{n}(Z):=Q_{n}(z,z,z)\qquad\hbox{and}\qquad R_{n}(Z):=R_{n}(z,z,z).$

From Proposition 3.1 we immediately get the following recurrence relations.

######  Corollary 5.1.

We have $Q_{0}(Z)=1$, $Q_{1}(Z)=z^{2}+3z$, $R_{0}(Z)=1$, $R_{1}(Z)=2z$, and
for $n\geq 1$,

(5.2) $\displaystyle Q_{n+1}(Z)$ $\displaystyle=(z^{2}+3z)\cdot
Q_{n}(Z)-(2z^{3}+z^{2})\cdot Q_{n-1}(Z),$ (5.3) $\displaystyle R_{n+1}(Z)$
$\displaystyle=(z^{2}+3z)\cdot R_{n}(Z)-(2z^{3}+z^{2})\cdot R_{n-1}(Z).$

Using these recurrence relations, we can compute the first few terms of both
polynomial sequences, as shown in Table 3.

$n$ | $Q_{n}(Z)$ | $R_{n}(Z)$
---|---|---
0 | 1 | 1
1 | $z^{2}+3z$ | $2z$
2 | $z^{4}+4z^{3}+8z^{2}$ | $5z^{2}$
3 | $z^{6}+5z^{5}+13z^{4}+21z^{3}$ | $z^{4}+13z^{3}$
4 | $z^{8}+6z^{7}+19z^{6}+40z^{5}+55z^{4}$ | $z^{6}+6z^{5}+34z^{4}$
5 | $z^{10}+7z^{9}+26z^{8}+66z^{7}+120z^{6}+144z^{5}$ | $z^{8}+7z^{7}+25z^{6}+89z^{5}$

Table 3: $Q_{n}(Z)$ and $R_{n}(Z)$ for $0\leq n\leq 5$.

In order to prove some of the properties of the sequences in (5.1) that are
obvious from Table 3, we first define

(5.4)
$\widetilde{Q}_{n}(z):=z^{-n}Q_{n}(Z)\qquad\hbox{and}\qquad\widetilde{R}_{n}(z):=z^{-n}R_{n}(Z).$

Then Corollary 5.1 implies that $\widetilde{Q}_{0}(z)=1$,
$\widetilde{Q}_{1}(z)=z+3$, $\widetilde{R}_{0}(z)=1$,
$\widetilde{R}_{1}(z)=2$, and for $n\geq 1$,

(5.5) $\displaystyle\widetilde{Q}_{n+1}(z)$
$\displaystyle=(z+3)\cdot\widetilde{Q}_{n}(z)-(2z+1)\cdot\widetilde{Q}_{n-1}(z),$
(5.6) $\displaystyle\widetilde{R}_{n+1}(z)$
$\displaystyle=(z+3)\cdot\widetilde{R}_{n}(z)-(2z+1)\cdot\widetilde{R}_{n-1}(z).$

We now state and prove the following properties of the sequences defined in
(5.1).

######  Lemma 5.2.

For each $n\geq 1$, the polynomial $Q_{n}(Z)$ has the following properties:

  1. (a)

It is monic of degree $2n$, with lowest term having degree $n$;

  2. (b)

the coefficient of $z^{2n-1}$ is $n+2$;

  3. (c)

the coefficient of $z^{n}$ is the Fibonacci number $F_{2n+2}$.

For each $n\geq 3$, the polynomial $R_{n}(Z)$ has the following properties:

  1. (d)

It is monic of degree $2n-2$, with lowest term having degree $n$;

  2. (e)

for $n\geq 4$, the coefficient of $z^{2n-3}$ is $n+2$;

  3. (f)

the coefficient of $z^{n}$ is $F_{2n+1}$.

###### Proof.

Using induction with (5.5), we see that $\widetilde{Q}_{n}(z)$ is monic of
degree $n$. Again with (5.5), we see that $\widetilde{Q}_{n}(0)$ satisfies the
well-known recurrence relation for the even-index Fibonacci numbers; see,
e.g., [8, A001906]. Next, if we write
$\widetilde{Q}_{n}(z)=z^{n}+a_{n-1}^{(n)}z^{n-1}+\cdots$ and substitute it
into (5.5), then upon equating coefficients of like powers of $z$, we get
$a_{n}^{(n+1)}=a_{n-1}^{(n)}+1$. With the initial condition $a_{0}^{(1)}=3$,
this gives $a_{n-1}^{(n)}=n+2$ for $n\geq 3$. All this, combined with the
first identity in (5.5), proves parts (a)–(c) of the lemma. Parts (d)–(f) can
be obtained in a similar way by using (5.6) and its initial conditions. ∎

Remarks. (1) With some further effort it would be possible to determine
coefficients other than those in Lemma 5.2. For instance, if we set

(5.7)
$\widetilde{Q}_{n}(z)=\sum_{j=0}^{n}a_{j}^{(n)}z^{j},\qquad\widetilde{R}_{n}(z)=\sum_{j=0}^{n-2}b_{j}^{(n)}z^{j}\quad(n\geq
2),$

then the sequence $(a_{1}^{(n)})_{n\geq 1}=(1,4,13,40,120,\ldots)$ is listed
as A238846 in [8], and the sequence $(a_{n-2}^{(n)})_{n\geq
2}=(8,13,19,26,\ldots)$ is determined by $a_{n-2}^{(n)}=(n^{2}+5n+2)/2$; see
[8, A034856].

(2) Similarly, the sequences $(b_{1}^{(n)})_{n\geq 3}=(1,6,25,90,300,\ldots)$
and $(b_{n-2}^{(n)})_{n\geq 5}=(25,33,42,52,63,\ldots)$ are A001871 and
A055998, respectively, in [8], with $b_{n-2}^{(n)}=n(n+5)/2$ for $n\geq 5$.
Further coefficients and their properties could be obtained by manipulating
the generating functions (3.9) and (3.10) with $x=y=z$.

We are now ready to give combinatorial interpretations for the polynomials
considered in this section. To do so, we need the following notation.

######  Definition 5.3.

Let $\beta$ be any binary overpartition, restricted or not, of some positive
integer. We define $S(\beta)$ to be the sum of the numbers of

(5.8) $\begin{cases}\hbox{the different and single overlined parts of $\beta$,
and}\\\ \hbox{the different and single non-overlined parts of $\beta$, and}\\\
\hbox{the different pairs of non-overlined parts of $\beta$.}\end{cases}$

We can now obtain the desired interpretation by considering the connections
between the polynomial sequences introduced in this section and the
polynomials $p_{n}(x,y,z)$ of Section 2, via the identities in (5.1) and in
(3.1). Setting $x=y=z$ and replacing $n$ by $2^{n+1}-2$ in (2.5), we first
obtain the following result from Proposition 2.1.

######  Proposition 5.4.

For $n\geq 1$ and $0\leq j\leq n$, the coefficient $a_{j}^{(n)}$, as defined
in (5.7), counts the number of $2$-restricted binary overpartitions $\beta$ of
$2^{n+1}-2$ with $S(\beta)=n+j$.

######  Example 5.5.

We take $n=2$ and consider the thirteen 2-restricted binary overpartitions of
6, as given in Example 2.2. We list them along with the sums corresponding to
(5.8):

$(4,2)$: | $0+2+0=2$ |  $(\overline{4},1,1)$: | $1+0+1=2$ |  $(\overline{2},2,1,1)$: | $1+1+1=3$
---|---|---|---|---|---
$(\overline{4},2)$: | $1+1+0=2$ |  $(4,\overline{1},1)$: | $1+2+0=3$ |  $(2,2,\overline{1},1)$: | $1+1+1=3$
$(4,\overline{2})$: | $1+1+0=2$ |  $(\overline{4},\overline{1},1)$: | $2+1+0=3$ |  $(\overline{2},2,\overline{1},1)$: | $2+2+0=4$
$(\overline{4},\overline{2})$: | $2+0+0=2$ |  $(\overline{2},2,2)$: | $1+0+1=2$ |  |
$(4,1,1)$: | $0+1+1=2$ |  $(2,2,1,1)$: | $0+0+2=2$ |  |

We count 1, 4, and 8 binary overpartitions $\beta$ with $S(\beta)=4,3$, and 2,
respectively. This corresponds to the polynomial
$Q_{2}(Z)=z^{4}+4z^{3}+8z^{2}$, consistent with Proposition 5.4.

This last example can be seen as a special case of the following corollary,
which is an easy consequence of Lemma 5.2 and Proposition 5.4.

######  Corollary 5.6.

$(a)$ For $n\geq 1$, the $2$-restricted binary overpartitions $\beta$ of
$2^{n+1}-2$ satisfy $n\leq S(\beta)\leq 2n$. $(b)$ There are $1,n+2$, and
$F_{2n+2}$ such overpartitions $\beta$ with $S(\beta)=2n,2n-1$, and $n$,
respectively.

In analogy to Proposition 5.4, we obtain the following result by substituting
$n$ with $2^{n}-1$ in (2.5).

######  Proposition 5.7.

For $n\geq 3$ and $0\leq j\leq n-2$, the coefficient $b_{j}^{(n)}$, as defined
in (5.7), counts the number of $2$-restricted binary overpartitions $\beta$ of
$2^{n}-1$ with $S(\beta)=n+j$.

######  Example 5.8.

We take $n=3$ and consider the fourteen 2-restricted binary overpartitions of
7, as given in Example 4.5. Rather than listing them all, we just note that
only for $\beta=(\overline{2},2,\overline{1},1,1)$ we have $S(\beta)=2+1+1=4$,
while $S(\beta)=3$ for all the others. This corresponds to
$R_{3}(Z)=z^{4}+13z^{3}$, consistent with Proposition 5.7.

As an easy consequence of Proposition 5.7, together with Lemma 5.2, we state
the following analogue of Corollary 5.6.

######  Corollary 5.9.

$(a)$ For $n\geq 2$, the $2$-restricted binary overpartitions $\beta$ of
$2^{n}-1$ satisfy $n\leq S(\beta)\leq 2n-2$. $(b)$ For $n\geq 4$, there are
$1,n+2$, and $F_{2n+1}$ such overpartitions $\beta$ with $S(\beta)=2n-2,2n-3$,
and $n$, respectively.

Since the polynomials $\widetilde{Q}_{n}(z)$ and $\widetilde{R}_{n}(z)$ are
single-variable polynomials, it is of interest to explore their zero
distribution, as we are doing in other sections as well.

######  Proposition 5.10.

For each $n\geq 1$, all zeros of $\widetilde{Q}_{n}(z)$ lie on the circle with
radius $5/2$, centered at $-1/2$. Furthermore, the real parts of the zeros are
strictly less than $1$, and they are dense on this section of the circle.

###### Proof.

By (5.1), (5.4), and (3.12) we have

$\widetilde{Q}_{n}(z)=(2z+1)^{n/2}\cdot
U_{n}\left(\frac{z+3}{2(2z+1)^{1/2}}\right)$

or, shifted by $1/2$,

(5.9) $\widetilde{Q}_{n}(z-\tfrac{1}{2})=(2z)^{n/2}\cdot
U_{n}\left(\frac{z+\tfrac{5}{2}}{2(2z)^{1/2}}\right).$

It is a well-known fact that all the zeros of $U_{n}(z)$ are real and lie in
the interval $(-1,1)$. Let $r$ be any such zero. Then by (5.9), a zero of
$\widetilde{Q}_{n}(z-\tfrac{1}{2})$ has to satisfy $z+5/2=2r(2z)^{1/2}$, or

(5.10) $z^{2}+\left(5-8r^{2}\right)z+\frac{25}{4}=0.$

It is easy to verify that for $r^{2}<1$ the discriminant of this quadratic is
negative, which means that the equation (5.10) has a pair of complex conjugate
zeros with product $25/4$. Hence their modulus is $5/2$, which proves the
first statement of the proposition, keeping the shift by $1/2$ in mind.

Solving the quadratic in (5.10) for $z$, we get

(5.11) $z=\frac{1}{2}\left(8r^{2}-5\right)\pm i\cdot
2r\sqrt{5-4r^{2}},\qquad-1<r<1.$

The real part of this expression is obviously increasing with $r^{2}$, and the
limit as $r^{2}\rightarrow 1$ is $z=\frac{3}{2}\pm 2i$, which proves the
second statement. Finally, since the zeros of all the $U_{n}(x)$ are dense in
$(-1,1)$, the resulting $z$-values are also dense on the section of the circle
specified in (5.11). This completes the proof. ∎

Remarks. (1) Since the zeros of the Chebyshev polynomials $U_{n}(x)$ are known
to be $r_{j}=\cos(j\pi/(n+1))$, $j=1,2,\ldots,n$, the zeros of
$\widetilde{Q}_{n}(z)$ can also be given explicitly, via (5.9) and (5.11).

(2) Due to the extra term $\widetilde{U}_{n-1}$ in (3.13), the zero
distribution of the polynomials $\widetilde{R}_{n}(z)$ is less straightforward
than that of $\widetilde{Q}_{n}(z)$. However, computations suggest that with
increasing $n$, the zeros of $\widetilde{R}_{n}(z)$ approach the circle given
in Proposition 5.10. We did not pursue this further since it is not central to
the current paper.

##  6\. A third special case: $x=1$

In this section we will only be dealing with the polynomial sequence
$Q_{n}(x,y,z)$, as defined in (3.1). With $x=1$, the identity (3.12)
simplifies to

(6.1)
$Q_{n}(1,y,z)=\big{(}y+y^{2}+yz\big{)}^{n/2}U_{n}\left(\frac{2y+z+1}{2(y+y^{2}+yz)^{1/2}}\right).$

Thanks to the special nature of Chebyshev polynomials, this expression
simplifies further, as follows.

######  Proposition 6.1.

For any $n\geq 0$ we have

(6.2) $Q_{n}(1,y,z)=\frac{1}{z+1}\left((y+z+1)^{n+1}-y^{n+1}\right).$

###### Proof.

We use the well-known explicit expression

(6.3)
$U_{n}(w)=\frac{1}{2\sqrt{w^{2}-1}}\left(\left(w+\sqrt{w^{2}-1}\right)^{n+1}-\left(w-\sqrt{w^{2}-1}\right)^{n+1}\right)$

(see, e.g., [9, p. 10]). We now set

$w:=\frac{2y+z+1}{2r},\qquad r:=\left(y+y^{2}+yz\right)^{1/2},$

so that

$w^{2}-1=\frac{(2y+z+1)^{2}}{4(y+y^{2}+yz)}-1=\frac{(z+1)^{2}}{4(y+y^{2}+yz)},$

where the second identity is easy to verify. We then get

(6.4) $\sqrt{w^{2}-1}=\frac{z+1}{2r}.$

Finally, combining (6.4) with (6.3) and (6.1), we have

$Q_{n}(1,y,z)=r^{n}\cdot\frac{r}{z+1}\left(\left(\frac{2y+z+1}{2r}+\frac{z+1}{2r}\right)^{n+1}-\left(\frac{2y+z+1}{2r}-\frac{z+1}{2r}\right)^{n+1}\right).$

Some straightforward simplification now leads to (6.2). ∎

Before continuing, we note that Proposition 2.1 gives the following
interpretation of the polynomials $Q_{n}(1,y,z)$.

######  Corollary 6.2.

If we write

(6.5) $Q_{n}(1,y,z)=\sum_{j,k\geq 0}c_{n}(j,k)\cdot y^{j}z^{k},\qquad n\geq
0,$

then $c_{n}(j,k)$ counts the number of $2$-restricted binary overpartitions of
$2^{n+1}-2$ with $j$ different and single non-overlined parts and $k$
different pairs of non-overlined parts.

We illustrate Corollary 6.2 with an example for $n=2$.

######  Example 6.3.

Noting that $2^{2+1}-2=6$, Table 1 with $x=1$ gives

(6.6) $Q_{2}(1,y,z)=p_{6}(1,y,z)=3y^{2}+3y+3yz+z^{2}+2z+1.$

The thirteen 2-restricted binary overpartitions of 6 can be found in Example
2.2. Corollary 6.2 now counts the following subsets:

$c_{2}(2,0)=3$: |  $(4,2),\;(4,\overline{1},1),\;(\overline{2},2,\overline{1},1)$, |  $c_{2}(0,2)=1$: |  $(2,2,1,1)$,
---|---|---|---
$c_{2}(1,0)=3$: |  $(\overline{4},2),\;(4,\overline{2}),\;(\overline{4},\overline{1},1)$, |  $c_{2}(0,1)=2$: |  $(\overline{4},1,1),\;(\overline{2},2,2)$,
$c_{2}(1,1)=3$: |  $(4,1,1),\;(\overline{2},2,1,1),\;(2,2,\overline{1},1)$, |  $c_{2}(0,0)=1$: |  $(\overline{4},\overline{2})$.

We now consider two specific cases of $Q_{n}(1,y,z)$ that are of particular
interest, namely the polynomial sequences $Q_{n}(1,z,z)$ and
$Q_{n}(1,z,z^{2})$. The first few polynomials in each sequence are listed in
Tables 4 and 5 later in this section. We begin with an easy consequence of
Corollary 6.2.

######  Corollary 6.4.

$(a)$ The coefficient of $z^{\mu}$ in $Q_{n}(1,z,z)$ is the number of
$2$-restricted binary overpartitions of $2^{n+1}-2$ with $\mu$ distinct non-
overlined parts.

$(b)$ The coefficient of $z^{\mu}$ in $Q_{n}(1,z,z^{2})$ is the number of
$2$-restricted binary overpartitions of $2^{n+1}-2$ with a total of $\mu$ non-
overlined parts.

###### Proof.

(a) This follows from Corollary 6.2 with $y=z$, so that the exponent of $z$ is
$\mu=j+k$. (b) In this case, by (6.5) the exponent of $z$ in
$Q_{n}(1,z,z^{2})$ is $\mu=j+2k$, and the result follows again from Corollary
6.2. ∎

######  Example 6.5.

(a) By (6.6) we have $Q_{2}(1,z,z)=7z^{2}+5z+1$, and accordingly we have 7
overpartitions with two distinct non-overlined parts, 5 with only one distinct
non-overlined part, and 1 with none. All this is consistent with Example 6.3.

(b) In this case, (6.6) gives the polynomial

(6.7) $Q_{2}(1,z,z^{2})=z^{4}+3z^{3}+5z^{2}+3z+1.$

Thus, for instance, we have 5 overpartitions with exactly two non-overlined
parts, namely

$(4,2),\;(4,\overline{1},1),\;(\overline{2},2,\overline{1},1),\;(\overline{4},1,1),\;(\overline{2},2,2).$

We see from (6.7) that $Q_{2}(1,z,z^{2})$ is a palindromic (or self-
reciprocal) polynomial. This is in fact always true:

######  Corollary 6.6.

$Q_{n}(1,z,z^{2})$ is a monic and self-reciprocal polynomial of degree $2n$
with integer coefficients whose sum is $\frac{1}{2}(3^{n}-1)$.

###### Proof.

From (6.2) we immediately get

(6.8) $Q_{n}(1,z,z^{2})=\frac{(z^{2}+z+1)^{n+1}-z^{n+1}}{z^{2}+1}.$

The fact that this expression is a polynomial with integer coefficients
follows, for instance, from Corollary 6.2. If we denote it by $f(z)$, then we
easily see that $z^{2n}f(1/z)=f(z)$, which shows that $Q_{n}(1,z,z^{2})$ is
self-reciprocal of degree $2n$. Since $f(0)=1$, this polynomial has constant
coefficient 1 and is thus also monic. The final statement follows from
$f(1)=(3^{n+1}-1)/2$. ∎

Given the form of the numerator in (6.8), it is not surprising that there
should be a connection between the polynomials $Q_{n}(1,z,z^{2})$ and
trinomial coefficients. In fact, the central coefficients 1, 2, 5, 12, 31, 82,
$\ldots$, are listed in [8, A097893] as partial sums of the central trinomial
coefficients, with numerous properties shown there.

The fact that the polynomials $Q_{n}(1,z,z^{2})$ are self-reciprocal, combined
with Corollary 6.4, gives the following.

######  Corollary 6.7.

Given $n\geq 1$, consider the set of all $2$-restricted binary overpartitions
of $2^{n+1}-2$, and let $j$ be such that $0\leq j\leq 2n$. Then the number of
partitions with $j$ non-overlined parts is equal to those with $2n-j$ non-
overlined parts.

We conclude this subsection with some results on factors and irreducibility.
As usual, $\Phi_{d}(x)$ will denote the $d$th cyclotomic polynomial which, by
definition, is irreducible.

######  Proposition 6.8.

$(a)$ For $n\geq 1$, the polynomials $Q_{n}(1,z,z)$ have the following
factorization into irreducible factors:

(6.9) $Q_{n}(1,z,z)=\prod_{\begin{subarray}{c}d\mid n+1\\\ d\neq
1\end{subarray}}\left(z^{\varphi(d)}\Phi_{d}(2+z^{-1})\right).$

In particular, $Q_{n}(1,z,z)$ is irreducible if and only if $n+1$ is prime.

$(b)$ For $n\geq 1$, the polynomials $Q_{n}(1,z,z^{2})$ have the factorization

(6.10) $Q_{n}(1,z,z^{2})=\prod_{\begin{subarray}{c}d\mid n+1\\\ d\neq
1\end{subarray}}\left(z^{\varphi(d)}\Phi_{d}(z+1+z^{-1})\right).$

As a consequence, $Q_{n}(1,z,z^{2})$ cannot be irreducible unless $n+1$ is
prime.

Proposition 6.8(a) is illustrated by Table 4.

$n$ | $Q_{n}(1,z,z)$ | factored
---|---|---
0 | 1 | 1
1 | $3z+1$ | irreducible
2 | $7z^{2}+5z+1$ | irreducible
3 | $15z^{3}+17z^{2}+7z+1$ | $(3z+1)(5z^{2}+4z+1)$
4 | $31z^{4}+49z^{3}+31z^{2}+9z+1$ | irreducible
5 | $63z^{5}+129z^{4}+111z^{3}+49z^{2}+11z+1$ | $(3z+1)(3z^{2}+3z+1)$
|  | $\qquad\qquad\cdot(7z^{2}+5z+1)$

Table 4: $Q_{n}(1,z,z)$ for $0\leq n\leq 5$.

Related to the last statement of Proposition 6.8(b), we need to mention that

(6.11) $\displaystyle z\Phi_{2}(z+1+z^{-1})$ $\displaystyle=(z+1)^{2},$ (6.12)
$\displaystyle z^{4}\Phi_{5}(z+1+z^{-1})$
$\displaystyle=\left(z^{4}+3z^{3}+4z^{2}+2z+1\right)\left(z^{4}+2z^{3}+4z^{2}+3z+1\right).$

In particular, this means that (6.10) does not always give a complete
factorization into irreducibles, and $Q_{n}(1,z,z^{2})$ is not always
irreducible when $n+1$ is prime. However, we conjecture that
$z^{\varphi(d)}\Phi_{d}(z+1+z^{-1})$ is irreducible for all $d\geq 1$,
$d\not\in\\{2,5\\}$. We have not pursued this question further, and we refrain
from providing factorizations in Table 5. In the range of Table 5, only
$Q_{2}(1,z,z^{2})$ is irreducible, which is consistent with what we wrote in
this paragraph.

$n$ | $Q_{n}(1,z,z^{2})$
---|---
0 | 1
1 | $z^{2}+2z+1$
2 | $z^{4}+3z^{3}+5z^{2}+3z+1$
3 | $z^{6}+4z^{5}+9z^{4}+12z^{3}+9z^{2}+4z+1$
4 | $z^{8}+5z^{7}+14z^{6}+25z^{5}+31z^{4}+25z^{3}+14z^{2}+5z+1$
5 | $z^{10}+6z^{9}+20z^{8}+44z^{7}+70z^{6}+82z^{5}+70z^{4}+44z^{3}+20z^{2}+6z+1$

Table 5: $Q_{n}(1,z,z^{2})$ for $0\leq n\leq 5$.

###### Proof of Proposition 6.8.

We use the well-known identity

(6.13) $w^{n+1}-1=\prod_{d\mid n+1}\Phi_{d}(w),$

substitute $w=(2z+1)/z$, and multiply both sides by $z^{n+1}$. This gives

(6.14) $(2z+1)^{n+1}-z^{n+1}=\prod_{d\mid
n+1}\left(z^{\varphi(d)}\Phi_{d}(2+z^{-1})\right),$

where we have used the identity $\sum_{d\mid n+1}\varphi(d)=n+1$. Since
$\Phi_{1}(w)=w-1$, we have $z\Phi_{1}(2+z^{-1})=z+1$, so (6.9) follows from
dividing both sides of (6.14) by $z+1$ and using the identity

(6.15)
$Q_{n}(1,z,z)=\frac{z^{n+1}}{z+1}\left(\left(\frac{2z+1}{z}\right)^{n+1}-1\right),$

which follows easily from (6.2). We also note that $\Phi_{d}(w)$ has degree
$\varphi(d)$, so all factors $z^{\varphi(d)}\Phi_{d}(2+z^{-1})$ are indeed
polynomials.

Next, since $\Phi_{d}(w)$ is irreducible, the linear shift
$\Phi_{d}(2+z^{-1})$ gives an irreducible polynomial in $z^{-1}$, and
$z^{\varphi(d)}\Phi_{d}(2+z^{-1})$ is then irreducible as a polynomial in $z$.
This shows that (6.9) is a complete factorization into irreducible factors and
that $Q_{n}(1,z,z)$ is irreducible when $n+1$ is prime.

The identity (6.10) can be obtained in exactly the same way as (6.9), with the
only difference that in this case we have $z\Phi_{1}(z+1+z^{-1})=z^{2}+1$. The
final statement in part (b) follows from the fact that the right-hand side of
(6.10) consists of just one term. ∎

##  7\. Zero distributions of $Q_{n}(1,z,z)$ and $Q_{n}(1,z,z^{2})$

Since $Q_{n}(1,z,z)$ and $Q_{n}(1,z,z^{2})$ are single-variable polynomials,
it makes sense to consider their zero distribution. We begin with the easier
case.

######  Proposition 7.1.

For any $n\geq 1$, the zeros of $Q_{n}(1,z,z)$ are given by

(7.1) $z_{j}=\frac{1}{\zeta_{j}-2},\qquad j=1,2,\ldots,n,$

where $\zeta_{j}:=e^{2\pi ij/(n+1)}$ is an $(n+1)$th root of unity.
Furthermore, the zeros $z_{j}$ all lie on the circle given by

(7.2) $\left(x+\tfrac{2}{3}\right)^{2}+y^{2}=\left(\tfrac{1}{3}\right)^{2}.$

###### Proof.

We consider the identity (6.15) and note that the term in large parentheses is
zero exactly when $(2z+1)/z=\zeta_{j}$, $j=0,1,\ldots,n$, which is equivalent
to $z=1/(\zeta_{j}-2)$. When $j=0$, we get $z=-1$; however, a limit argument
shows that $z=-1$ is not a zero of the right-hand side of (6.15). This proves
(7.1).

For the second statement, we note that by the theory of fractional linear
transformations all the $z_{j}$ lie on a circle (or a straight line) since all
$\zeta_{j}$ lie on a circle. It therefore suffices to show that any three
distinct points $\zeta\in{\mathbb{C}}$ with $|\zeta|=1$ are mapped to points
$z=x+iy$ satisfying (7.2). It is easy to verify this with the choice of
$\zeta=-1,\pm i$, for instance; this completes the proof. ∎

The zero distribution of $Q_{n}(1,z,z^{2})$ turns out to be more interesting
than that of $Q_{n}(1,z,z)$. We begin with a lemma.

######  Lemma 7.2.

For any $n\geq 1$, the $2n$ zeros of $Q_{n}(1,z,z^{2})$ are given by

(7.3)
$z_{j}^{\pm}:=\frac{1}{2}\left(\zeta_{j}-1+\left(\zeta_{j}^{2}-2\zeta_{j}-3\right)^{\frac{1}{2}}\right),\quad
j=1,2,\ldots,n,$

where $\zeta_{j}:=e^{2\pi ij/(n+1)}$ are the $(n+1)$th roots of unity and the
superscript $\pm$ indicates that for each $j$ we have two values of (7.3).

###### Proof.

Upon slightly rewriting (6.8), we have

(7.4)
$Q_{n}(1,z,z^{2})=\frac{z^{n+1}}{z^{2}+1}\left(\left(z+1+\frac{1}{z}\right)^{n+1}-1\right).$

For this expression to vanish, we need

(7.5) $z+1+\frac{1}{z}=\zeta_{j},\quad\hbox{or}\quad
z^{2}+(1-\zeta_{j})z+1=0,\qquad j=0,1,\ldots,n.$

Solving this last equation, we get (7.3) with $j=0,1,\ldots,n$. However, since
$\zeta_{0}=1$, we have $z_{0}^{\pm}=\pm i$, a pair of solutions which is
canceled by the denominator in (7.4). Hence $j=0$ needs to be excluded, which
completes the proof. ∎

In the paragraph after (7.5) we saw that $\pm i$ cannot be zeros of
$Q_{n}(1,z,z^{2})$ for any $n\geq 1$. On the other hand, by substituting
$z=-1$ in (7.4) or (6.8) and in the derivative of the numerator of (6.8), we
see that $z=-1$ is a double zero of $Q_{n}(1,z,z^{2})$ if and only if $n$ is
odd. This corresponds to $\zeta_{(n+1)/2}=-1$, along with (7.3) and/or (7.5).

The zeros $z_{j}^{\pm}$ for $n=21$ and $n=50$ are shown in Figure 1. They lie
on an algebraic curve which we will identify next.

######  Proposition 7.3.

The zeros of all polynomials $Q_{n}(1,z,z^{2})$, $n\geq 1$, lie on the
algebraic curve

(7.6) $x^{4}+2x^{2}y^{2}+y^{4}+2x^{3}+2xy^{2}+2x^{2}-2y^{2}+2x+1=0$

or, rewritten,

(7.7) $\big{(}x^{2}+y^{2}+x\big{)}^{2}+(x+1)^{2}=2y^{2}.$

###### Proof.

If $z\in{\mathbb{C}}$ is a zero of $Q_{n}(1,z,z^{2})$ for any integer $n\geq
1$, then by the left identity in (7.5) we have

(7.8) $\left|z+1+z^{-1}\right|^{2}=1.$

Setting $z=x+iy$ ($x,y\in{\mathbb{R}}$), so that
$z^{-1}=(x-iy)/(x^{2}+y^{2})$, we get with (7.8),

$\displaystyle 1$
$\displaystyle=\left|x+iy+1+\frac{x-iy}{x^{2}+y^{2}}\right|^{2}$
$\displaystyle=\left(x+1+\frac{x}{x^{2}+y^{2}}\right)^{2}+\left(y-\frac{y}{x^{2}+y^{2}}\right)^{2}$
$\displaystyle=\frac{1}{x^{2}+y^{2}}\left(x^{4}+2x^{2}y^{2}+y^{4}+2x^{3}+2xy^{2}+3x^{2}-y^{2}+2x+1\right).$

This last line implies

$x^{4}+2x^{2}y^{2}+y^{4}+2x^{3}+2xy^{2}+3x^{2}-y^{2}+2x+1=x^{2}+y^{2},$

which is equivalent to (7.6) and (7.7). ∎

Figure 1: The zeros of $Q_{21}(1,z,z^{2})$, $Q_{50}(1,z,z^{2})$, and the curve
(7.6).

We recall that in most of Sections 6 and 7 we have considered the polynomial
sequence $Q_{n}(1,y,z)$ from (6.2) with $y$ replaced by $z^{\alpha}$ and $z$
replaced by $z^{\beta}$, and then studied the two specific cases
$(\alpha,\beta)=(1,1)$ and $(\alpha,\beta)=(1,2)$. This suggests that we might
as well consider the general case with integers $\alpha$, $\beta$, not both
zero. From (6.2) we then obtain

(7.9)
$Q_{n}(1,z^{\alpha},z^{\beta})=\frac{z^{(n+1)\alpha}}{z^{\beta}+1}\left(\left(1+z^{\beta-\alpha}+z^{-\alpha}\right)^{n+1}-1\right).$

Just as we did in the proofs of Proposition 7.1 and/or 7.3, for any pair of
integers $\alpha$, $\beta$ (not both zero) we can determine the algebraic
curve (considered as a curve in the complex plane) on which all the zeros of
$Q_{n}(1,z^{\alpha},z^{\beta})$ lie. We denote these curves by
$f_{\alpha,\beta}(x,y)=0$ and display the polynomials $f_{\alpha,\beta}(x,y)$
in Table 6, for the first few nonnegative $\alpha,\beta$. This table also
shows the corresponding genus in each case, which was computed using Maple. It
would also be possible to derive a general formula for
$f_{\alpha,\beta}(x,y)$; however, this will not be required here.

$\alpha$ | $\beta$ | $f_{\alpha,\beta}(x,y)$ | $g$
---|---|---|---
0 | 1 | $3+4x+x^{2}+y^{2}$ | 0
0 | 2 | $3+4x^{2}+x^{4}-4y^{2}+2x^{2}y^{2}+y^{4}$ | 1
0 | 3 | $3+4x^{3}+x^{6}-12xy^{2}+3x^{4}y^{2}+3x^{2}y^{4}+y^{6}$ | 4
1 | 0 | $1+x$ | 0
1 | 1 | $1+4x+3x^{2}+3y^{2}$ | 0
1 | 2 | $1+2x+2x^{2}+2x^{3}+x^{4}-2y^{2}+2xy^{2}+2x^{2}y^{2}+y^{4}$ | 0
1 | 3 | $1+2x+2x^{3}+2x^{4}+x^{6}-6xy^{2}+3x^{4}y^{2}-2y^{4}+3x^{2}y^{4}+y^{6}$ | 4
2 | 0 | $1+x^{2}-y^{2}$ | 0
2 | 1 | $1+2x+3x^{2}+2x^{3}-y^{2}+2xy^{2}$ | 1
2 | 2 | $1+4x^{2}+3x^{4}-4y^{2}+6x^{2}y^{2}+3y^{4}$ | 1
2 | 3 | $1+2x^{2}+2x^{3}+2x^{5}+x^{6}-2y^{2}-6xy^{2}+4x^{3}y^{2}+3x^{4}y^{2}+2xy^{4}$ |
|  | $\qquad+3x^{2}y^{4}+y^{6}$ | 4
3 | 0 | $1+x^{3}-3xy^{2}$ | 1
3 | 1 | $1+2x+x^{2}+2x^{3}+2x^{4}+y^{2}-6xy^{2}-2y^{4}$ | 3
3 | 2 | $1+2x^{2}+2x^{3}+x^{4}+2x^{5}-2y^{2}-6xy^{2}+2x^{2}y^{2}+4x^{3}y^{2}+y^{4}+2xy^{4}$ | 4
3 | 3 | $1+4x^{3}+3x^{6}-12xy^{2}+9x^{4}y^{2}+9x^{2}y^{4}+3y^{6}$ | 4

Table 6: $f_{\alpha,\beta}(x,y)$ for $0\leq\alpha,\beta\leq 3$ and genus $g$
of curve $f_{\alpha,\beta}(x,y)=0$.

We now state an easy transformation identity, which follows directly from
(7.9).

######  Lemma 7.4.

For all integers $\alpha,\beta$, not both zero, we have

(7.10)
$Q_{n}(1,z^{\beta-\alpha},z^{\beta})=z^{n\beta}Q_{n}(1,(\tfrac{1}{z})^{\alpha},(\tfrac{1}{z})^{\beta}).$

The identity (7.10), and thus the relationship between the curves
$f_{\alpha,\beta}(x,y)=0$ and $f_{\beta-\alpha,\beta}(x,y)=0$, means that if
$z=x+iy$ lies on one of these curves, then $1/z={\overline{z}}/(x^{2}+y^{2})$
and by symmetry also $1/{\overline{z}}=z/|z|^{2}$ lie on its companion. This
implies that we have an inversion with respect to the unit circle: if a point
on one curve has polar coordinates $r(\cos\theta+i\,\sin\theta)$, then the
corresponding point on its companion has polar coordinates
$r^{-1}(\cos\theta+i\,\sin\theta)$. We say, in short, that the two curves in
(7.10) are inverse to each other.

######  Example 7.5.

(a) The curve $f_{1,2}(x,y)=0$, given explicitly in Proposition 7.3 and shown
in Figure 1, is its own inverse.

(b) As we saw in Proposition 7.1, the curve $f_{1,1}(x,y)=0$ is the circle of
radius $1/3$ centered at $(x,y)=(-2/3,0)$. Its inverse is
$f_{0,1}(x,y)=3+4x+x^{2}+y^{2}=0$, i.e., the circle $(x+2)^{2}+y^{2}=1$. The
two circles are tangent to each other at $(x,y)=(-1,0)$.

(c) Using again similar methods as in the proofs of Propositions 7.1 and 7.3,
we find $f_{1,0}(x,y)=1+x$ and $f_{-1,0}(x,y)=(2x+1)^{2}+4y^{2}-1$. The
corresponding mutual inverses are then the vertical line $x=-1$ and the circle
$(x+\tfrac{1}{2})^{2}+y^{2}=(\tfrac{1}{2})^{2}$, which are also tangent to
each other at $(x,y)=(-1,0)$.

(d) Using once again the same methods as before, we find

(7.11) $\displaystyle f_{2,1}(x,y)$
$\displaystyle=1+2x+3x^{2}+2x^{3}-y^{2}+2xy^{2},$ (7.12) $\displaystyle
f_{-1,1}(x,y)$
$\displaystyle=2x+3x^{2}+2x^{3}+x^{4}-y^{2}+2xy^{2}+2x^{2}y^{2}+y^{4};$

see Figure 2 for a joint plot.

Figure 2: The curves $f_{2,1}(x,y)=0$ and $f_{-1,1}(x,y)=0$.

We have not been able to identify the cubic $f_{2,1}(x,y)=0$ as a specific
known curve. However, one can easily see that it has a vertical asymptote at
$x=1/2$. Its inverse $f_{-1,1}(x,y)=0$ is an oval of Cassini, as can be seen
by rewriting this last equation in the form

(7.13)
$\left(\left(x+\tfrac{1}{2}\right)^{2}+\left((y-\tfrac{1}{2}\sqrt{3}\right)^{2}\right)\left(\left(x+\tfrac{1}{2}\right)^{2}+\left((y+\tfrac{1}{2}\sqrt{3}\right)^{2}\right)=1.$

The equation (7.13) shows that the product of the distances between a point
$(x,y)$ on the curve and the two foci $(-\frac{1}{2},\pm\frac{1}{2}\sqrt{3})$
is always 1, which is consistent with the definition of an oval of Cassini;
see, e.g., [11] or [6].

##  8\. Further properties of the curve (7.6)

A particularly interesting example of the algebraic curves containing the
zeros of $Q_{n}(1,z^{\alpha},z^{\beta})$ is the one belonging to
$(\alpha,\beta)=(1,2)$. This curve was obtained in Proposition 7.3; see also
Figure 1. Using Maple, we found that it has genus 0; see also Table 6. This
means that the curve has a rational parametrization, which can also be found
by Maple, using the package algcurves with the functions genus and
parametrization.

######  Proposition 8.1.

The curve $f_{1,2}(x,y)=0$ has the following parametrization with rational
functions:

$x=-\frac{4t^{2}(4t+1)^{2}}{169t^{4}+178t^{3}+74t^{2}+14t+1},\qquad
y=\frac{(5t+1)(3t+1)(17t^{2}+8t+1)}{169t^{4}+178t^{3}+74t^{2}+14t+1}.$

By choosing the parameters $t=-1/5$ and $t=-1/3$ in Proposition 8.1, we obtain
the double point $(x,y)=(-1,0)$ on the curve. Furthermore, by evaluating the
quotients of the derivatives $dy/dt$ and $dx/dt$ at $t=-1/5$ and $t=-1/3$, we
see that the slope of the curve at this double point is 1, resp. $-1$. Apart
from the special parameters $t=0$ and $t=-1/4$, it is worth mentioning that as
$t\rightarrow\pm\infty$, we have $x\rightarrow-(8/13)^{2}$ and $y\rightarrow
15\cdot 17/13^{2}$.

Next, we derive some maximum/minimum properties of the curve in question.

######  Proposition 8.2.

$(a)$ The curve $f_{1,2}(x,y)=0$ has vertical tangents exactly at the points
$(x,y)=(0,\pm 1)$ and $(-\frac{4}{3},\pm\frac{1}{3}\sqrt{5})$.

$(b)$ The curve has horizontal tangents exactly at the points
$(x,y)=(x_{0},\pm y_{0})$, where

(8.1) $\displaystyle x_{0}$
$\displaystyle=-\frac{a}{18}+\frac{22}{9a}-\frac{4}{9}\quad\hbox{with}\quad
a=\sqrt[3]{188+36\sqrt{93}},$ (8.2) $\displaystyle y_{0}$
$\displaystyle=\frac{1}{198}\sqrt{(495\sqrt{93}-1617)a+33396+(-18\sqrt{93}+699)a^{2}}.$

Numerically, $(x_{0},y_{0})\simeq(-0.594414,1.545634)$.

###### Proof.

Implicit differentiation of (7.6) leads to

(8.3)
$\frac{dy}{dx}=\frac{1+2x+3x^{2}+2x^{3}+(1+2x)y^{2}}{2y(1-x-x^{2}-y^{2})},$

while solving (7.6) for $y^{2}$ gives

(8.4) $y^{2}=1-x-x^{2}\pm\sqrt{-x(4+3x)}.$

Considering the denominator of (8.3), we first note that $y=0$ implies $x=-1$,
and thus the numerator will also vanish (see also the remark following
Proposition 8.1). When $1-x-x^{2}-y^{2}=0$, then by (8.4) we have $x=0$ or
$x=-4/3$. Substituting this back into (8.4), we find $y=\pm 1$, resp.
$y=\pm\frac{1}{3}\sqrt{5}$, which completes the proof of part (a).

Next, in order to find the points on the curve which have horizontal tangents,
we substitute (8.4) into the numerator of (8.3). After some straightforward
manipulations we find that this numerator vanishes if and only if
$1+4x+7x^{2}+7x^{3}+3x^{4}=0$, which factors as

$(x+1)(3x^{3}+4x^{2}+3x+1)=0.$

The solution $x=-1$ gives $y=0$, a case we already discussed, while the unique
real root of the cubic is $x_{0}$ as given in (8.1). This solution was
obtained with the help of Maple.

Finally, substituting $x=x_{0}$ into (7.6), we get a quartic polynomial in
$y$. With the help of Maple we can solve this algebraically and note that the
largest root is $y_{0}$ as given in (8.2). This completes the proof of part
(b). ∎

Finally in this section, we will see that the maximal and minimal moduli of
the curve in (7.6) have remarkably simple forms.

######  Proposition 8.3.

The points on the curve $f_{1,2}(x,y)=0$ have maximal modulus $\sqrt{3}$ and
minimal modulus $1/\sqrt{3}$. Both are attained when $\cos\theta=-1/\sqrt{3}$,
or numerically when $\theta\simeq\pm 0.695913\,\pi$. The Cartesian coordinates
of these extremal points are $(x,y)=(-1,\pm\sqrt{2})$, resp.
$(x,y)=(-1/3,\pm\sqrt{2}/3)$.

###### Proof.

As in the proof of Proposition 7.3, we begin with the identity (7.8), but this
time we use the polar representations

$z=r(\cos\theta+i\,\sin\theta),\qquad
z^{-1}=r^{-1}(\cos\theta-i\,\sin\theta).$

After some straightforward manipulations we see that (7.8) is equivalent to

$\left(r+\tfrac{1}{r}\right)^{2}\cos^{2}\theta+2\left(r+\tfrac{1}{r}\right)\cos\theta+\left(r-\tfrac{1}{r}\right)^{2}\sin^{2}\theta=0,$

which can be further transformed to

(8.5)
$\left(r+\tfrac{1}{r}\right)^{2}+2\left(r+\tfrac{1}{r}\right)\cos\theta+4\cos^{2}\theta-4=0.$

We solve (8.5) for $r+r^{-1}$, obtaining

(8.6)
$r+\frac{1}{r}=-\cos\theta+\sqrt{4-3\cos^{2}\theta},\qquad\frac{\pi}{2}\leq|\theta|\leq\pi,$

where the restriction on $\theta$ comes from the fact that the curve in
question lies entirely in the left half-plane. Furthermore, we have just “$+$”
in front of the square root in (8.6) since $\sqrt{4-3\cos^{2}\theta}\geq 1$
for all $\theta$, while the left-hand side of (8.6) is $\geq 2$.

Differentiating both sides of (8.6), we get after some simplification,

(8.7)
$2r\left(1-r^{-2}\right)\frac{dr}{d\theta}=\left(1+\frac{3\cos\theta}{\sqrt{4-3\cos^{2}\theta}}\right)\sin\theta.$

The right-hand side of (8.7) vanishes when $\sin\theta=0$ or the expression in
large parentheses is zero. In the first case we get $\theta=\pi$, which means
$r=1$ and thus the left-hand side of (8.7) also vanishes. But this has already
been dealt with following Proposition 8.1.

In the second case we have

(8.8) $-3\cos\theta=\sqrt{4-3\cos^{2}\theta},$

and upon squaring and simplifying we get $\cos^{2}\theta=1/3$. But then, by
(8.8), only the solution $\cos\theta=-1/\sqrt{3}$ is possible. Substituting
this into (8.6), we get

$r+\frac{1}{r}=\frac{4}{3}\sqrt{3},$

which has the two solutions $r=\sqrt{3}$ and $r=1/\sqrt{3}$. Using (8.7), for
instance, we can see that these two values are a maximum and a minimum,
respectively.

The final statement comes from the fact that $\cos\theta=-1/\sqrt{3}$ implies
$\sin\theta=\pm\sqrt{2/3}$, and thus $(x,y)=(-r/\sqrt{3},\pm r\sqrt{2/3})$. ∎

In concluding this section, we note that the curve we investigated here has
several properties in common with the Besace curve given by the equation

$\left(x^{2}-by\right)^{2}=a^{2}\left(x^{2}-y^{2}\right),$

with positive parameters $a$ and $b$; see, e.g., [5]. These similarities
include the facts that both are quartics, have genus 0, and are similar in
shape.

##  9\. General bases $b\geq 2$

Much of what we did in Sections 2 to 7 has direct analogues for integer bases
$b\geq 2$. We therefore structure this section roughly along the lines of of
previous sections. Most proofs are similar to those of the case $b=2$; we
leave the details to the interested reader.

###  9.1. Basic properties

In analogy to the beginning of Section 2 we specialize the more general
multicolor $b$-ary partitions in [3] to the 2-color $(1,b)$-case. That is, we
consider $b$-ary overpartitions where the non-overlined parts occur at most
$b$ times. As we did in (2.1) above, we use the following simplified notation:
for all $n\geq 0$ we set

(9.1) $p_{n}(Z):=\Omega_{b,T}^{(1,b)}(n;Z),\qquad Z=(x,y_{1},\ldots,y_{b}),$

with $T=(1,1,\ldots,1)$; see again [3]. Then, in analogy to (2.2), we have the
generating function

(9.2)
$\sum_{n=0}^{\infty}p_{n}(Z)q^{n}=\prod_{j=0}^{\infty}\left(1+xq^{b^{j}}\right)\left(1+y_{1}q^{b^{j}}+y_{2}q^{2\cdot
b^{j}}+\cdots+y_{b}q^{b\cdot b^{j}}\right),$

and in analogy to (2.3), (2.4) we have the recurrence relations with initial
terms

(9.3) $p_{0}(Z)=1,\quad p_{1}(Z)=x+y_{1},\quad
p_{j}(Z)=xy_{j-1}+y_{j}\quad(2\leq j\leq b-1),$

and for $n\geq 1$,

(9.4) $\displaystyle p_{bn}(Z)$ $\displaystyle=p_{n}(Z)+(y_{b}+xy_{b-1})\cdot
p_{n-1}(Z),$ (9.5) $\displaystyle p_{bn+1}(Z)$ $\displaystyle=(x+y_{1})\cdot
p_{n}(Z)+xy_{b}\cdot p_{n-1}(Z),$ (9.6) $\displaystyle p_{bn+j}(Z)$
$\displaystyle=(xy_{j-1}+y_{1})\cdot p_{n}(Z),\quad(2\leq j\leq b-1).$

Since for $b=2$ we have $Z=(x,y_{1},y_{2})=(x,y,z)$, it is clear that (9.4)
and (9.5) become (2.3) and (2.4), respectively, while (9.6) occurs only for
$b\geq 3$.

Next we state the base-$b$ analogue of Proposition 2.1. If we write the
polynomials $p_{n}(Z)$ in the form

(9.7) $p_{n}(Z)=\sum_{i,j_{1},\ldots,j_{b}\geq
0}c_{n}(i,j_{1},\ldots,j_{b})\cdot x^{i}y_{1}^{j_{1}}\ldots
y_{b}^{j_{b}},\quad n\geq 0,$

then the generating function (9.2) gives the following combinatorial
interpretation.

######  Proposition 9.1.

For any non-negative integers $n,i,j_{1},\ldots,j_{b}$, the coefficient
$c_{n}(i,j_{1},\ldots,j_{b})$ in (9.7) counts the number of $b$-restricted
$b$-ary overpartitions of $n$ that have

  1. $i$ different and single overlined parts,

  2. $j_{1}$ different and single non-overlined parts, and

  3. $j_{k}$ different k-tuples of non-overlined parts, $2\leq k\leq b$.

###  9.2. Connections with Chebyshev polynomials

We now extend the results in Section 3 to arbitrary bases $b\geq 2$. We
consider two subsequences with subscripts

(9.8) $q(n):=\frac{b^{n+1}-b}{b-1},\qquad
r(n):=\frac{b^{n}-1}{b-1}=\frac{q(n)}{b},$

which we use to define

(9.9) $Q_{n}^{b}(Z):=p_{q(n)}(Z),\qquad R_{n}^{b}(Z):=p_{r(n)}(Z).$

For $b=2$ and $Z=(x,y,z)$, the identities (9.8), (9.9) and (3.1) then give

$Q_{n}^{2}(Z)=Q_{n}(x,y,z),\qquad R_{n}^{2}(Z)=R_{n}(x,y,z).$

The following is analogous to Proposition 3.1. To simplify notation, we set
for $b\geq 2$ and $x,y_{1},y_{b-1},y_{b}$ as in (9.1),

(9.10) $W_{1}^{b}(Z):=xy_{b-1}+x+y_{1}+y_{b},\qquad
W_{2}^{b}(Z):=x^{2}y_{b-1}+xy_{1}y_{b-1}+y_{1}y_{b}.$

######  Proposition 9.2.

We have $Q_{0}^{b}(Z)=1$, $Q_{1}^{b}(Z)=W_{1}^{b}(Z)$, $R_{0}^{b}(Z)=1$,
$R_{1}^{b}(Z)=x+y_{1}$, and for $n\geq 1$,

(9.11) $\displaystyle Q_{n+1}^{b}(Z)$ $\displaystyle=W_{1}^{b}(Z)\cdot
Q_{n}^{b}(Z)-W_{2}^{b}(Z)\cdot Q_{n-1}^{b}(Z),$ (9.12) $\displaystyle
R_{n+1}^{b}(Z)$ $\displaystyle=W_{1}^{b}(Z)\cdot
R_{n}^{b}(Z)-W_{2}^{b}(Z)\cdot R_{n-1}^{b}(Z).$

Further in analogy to Section 3, the recurrence relations (9.11), (9.12) lead
to the following.

######  Proposition 9.3.

The polynomials $Q_{n}^{b}$ and $R_{n}^{b}$ satisfy the generating functions

(9.13)
$\displaystyle\sum_{n=0}^{\infty}Q_{n}^{b}(Z)q^{n}=\frac{1}{1-W_{1}^{b}(Z)q+W_{2}^{b}(Z)q^{2}},$
(9.14)
$\displaystyle\sum_{n=0}^{\infty}R_{n}^{b}(Z)q^{n}=\frac{1-(xy_{b-1}+y_{b})q}{1-W_{1}^{b}(Z)q+W_{2}^{b}(Z)q^{2}}.$

At this point it will not be surprising that the polynomials $Q_{n}^{b}$ and
$R_{n}^{b}$ are also closely related to the Chebyshev polynomials of both
kinds; see (3.11)–(3.13).

######  Proposition 9.4.

For all $n\geq 0$ we have

(9.15) $\displaystyle Q_{n}^{b}(Z)$
$\displaystyle=\left(W_{2}^{b}(Z)\right)^{n/2}U_{n}\left(\frac{W_{1}^{b}(Z)}{2\sqrt{W_{2}^{b}(Z)}}\right),$
(9.16) $\displaystyle R_{n}^{b}(Z)$
$\displaystyle=\left(W_{2}^{b}(Z)\right)^{n/2}T_{n}\left(\frac{W_{1}^{b}(Z)}{2\sqrt{W_{2}^{b}(Z)}}\right)+\widetilde{U}_{n-1}^{b}(Z),$

where

(9.17) $\widetilde{U}_{n-1}^{b}(Z)=\frac{x+y_{1}-xy_{b-1}-y_{b}}{2}\cdot
Q_{n-1}^{b}(Z).$

By combining the identities (9.15)–(9.17) with the well-known relation
$T_{n}(x)=U_{n}(x)-xU_{n-1}(x)$, we obtain the following identity.

######  Corollary 9.5.

For $n\geq 1$ we have

(9.18) $R_{n}^{b}(Z)=Q_{n}^{b}(Z)-\left(xy_{b-1}+y_{b}\right)\cdot
Q_{n-1}^{b}(Z).$

Likewise, the identity (3.2) extends to $b\geq 2$ with $y\rightarrow y_{1}$
and $z\rightarrow y_{b}$. The following fact that is a consequence of any one
of Propositions 9.2–9.4.

######  Corollary 9.6.

For all $b\geq 2$ and $n\geq 0$, $Q_{n}^{b}(Z)$ and $R_{n}^{b}(Z)$ are
polynomials in $x$, $y_{1}$, $y_{b-1}$, and $y_{b}$ only.

######  Example 9.7.

When $b=5$, then by (9.8) we have $q(1)=5$ and $q(2)=30$. Accordingly, we get
$Q_{1}^{5}(Z)=p_{5}(Z)=y_{1}+x+y_{5}+xy_{4}$; see also Proposition 9.2. This
polynomial corresponds to the 5-restricted $b$-ary overpartitions (with $b=5$)

$(5),\;(\overline{5}),\;(1,1,1,1,1),\;(\overline{1},1,1,1,1),$

written in the order of the coefficients of $Q_{1}^{5}(Z)$. This is consistent
with Proposition 9.1.

######  Example 9.8.

Similarly we find, for instance with (9.11), that $Q_{2}^{5}(Z)$ is

$p_{30}(Z)=y_{1}^{2}+2xy_{1}+x^{2}+y_{1}y_{5}+2xy_{5}+xy_{1}y_{4}+x^{2}y_{4}+y_{5}^{2}+2xy_{4}y_{5}+x^{2}y_{4}^{2}.$

With the usual notation of $a^{k}$ for the part $a$ repeated $k$ times, the
corresponding 5-restricted $b$-ary overpartitions (with $b=5$) are

$\displaystyle(25,5),\;(\overline{25},5),\;(25,\overline{5}),\;(\overline{25},\overline{5}),\;(25,1^{5}),\;(\overline{25},1^{5}),(\overline{5},5^{5}),$
$\displaystyle(25,\overline{1},1^{4}),\;(\overline{25},\overline{1},1^{4}),\;(5^{5},1^{5}),\;(\overline{5},5^{4},1^{5}),\;(5^{5},\overline{1},1^{4}),\;(\overline{5},5^{4},\overline{1},1^{4}),$

again in the order of the terms of $Q_{2}^{5}(Z)$. For instance, the two
partitions corresponding to the monomial $2xy_{1}$ are $(\overline{25},5)$ and
$(25,\overline{5})$, and the two corresponding to $2xy_{5}$ are
$(\overline{25},1^{5})$ and $(\overline{5},5^{5})$. All this is again
consistent with Proposition 9.1.

###  9.3. A first special case

It is no coincidence that the number of $b$-restricted $b$-ary overpartitions
in Example 2.2 (for $b=2$ and $n=6$) is the same as that in Example 9.7 (for
$b=5$ and $n=30$), namely 13. Indeed, if we set $x=y_{1}=\cdots=y_{b}=1$, then
any one of Propositions 9.2–9.4, together with Corollary 4.2(d), imply the
following.

######  Corollary 9.9.

Let $b\geq 2$ be an integer, and $q(n),r(n)$ the sequences defined by (9.8).
Then for each $n\geq 0$ the number of $b$-restricted $b$-ary overpartitions of
$q(n)$ and $r(n)$ are $\frac{1}{2}(3^{n+1}-1)$ and $\frac{1}{2}(3^{n}+1)$,
respectively.

######  Example 9.10.

We take $b=5$ again, but in contrast to Example 9.7 we consider the
$b$-restricted $b$-ary overpartitions of $r(2)=6$. There are
$\frac{1}{2}(3^{2}+1)=5$ of them, namely

$(5,1),\;(\overline{5},1),\;(5,\overline{1}),\;(\overline{5},\overline{1}),\;(\overline{1},1^{5}).$

The corresponding polynomial, best obtained with (9.12), is

$R_{2}^{5}(Z)=y_{1}^{2}+2xy_{1}+x^{2}+xy_{5},$

with the monomials again in the same order as the corresponding
overpartitions.

If we set $y_{1}=y_{b-1}=y_{b}=1$, then by (9.10) and (9.17) we have

$W_{1}^{b}(Z)=2x+2,\qquad
W_{2}^{b}(Z)=x^{2}+x+1,\qquad\widetilde{U}_{n-1}^{b}(Z)=0,$

and consequently Proposition 9.4 and Corollary 4.1 give

$Q_{n}^{b}(Z)=Q_{n}(x)\qquad\hbox{and}\qquad R_{n}^{b}(Z)=R_{n}(x).$

When $b\geq 4$, this is independent of $y_{2},\ldots,y_{b-2}$. The case
$y_{1}=y_{b-1}=y_{b}=1$ is therefore covered by Section 4.

###  9.4. A second special case

In analogy to Section 5 we set $x=y_{1}=y_{b-1}=y_{b}$. If we rename this
common variable as $z$, then by (9.10) we have

$W_{1}^{b}(Z):=z^{2}+3z,\qquad W_{2}^{b}(Z):=2z^{3}+z^{2}.$

By Proposition 9.2 we then have $Q_{0}^{b}(Z)=1$, $Q_{1}^{b}(Z)=z^{2}+3z$, and
for $n\geq 1$,

$Q_{n+1}^{b}(Z)=(z^{2}+3z)\cdot Q_{n}^{b}(Z)-(2z^{3}+z^{2})\cdot
Q_{n-1}^{b}(Z),$

with an analogous recurrence relation also for the polynomials $R_{n}^{b}(Z)$.
But this is exactly the situation of Section 5, beginning with Corollary 5.1.
In particular, it means that we have independence of the base $b\geq 2$.

For a general combinatorial interpretation of the polynomials
$Q_{n}(Z)=Q_{n}^{b}(Z)$ and $R_{n}(Z)=R_{n}^{b}(Z)$, we first need an analogue
of Definition 5.3.

######  Definition 9.11.

For an integer base $b\geq 2$, let $\beta$ be any $b$-ary overpartition,
restricted or not, of some positive integer. We define $S^{b}(\beta)$ to be
the sum of the numbers of

(9.19) $\begin{cases}\hbox{the different and single overlined parts of
$\beta$, and}\\\ \hbox{the different and single non-overlined parts of
$\beta$, and}\\\ \hbox{the different $(b-1)$-tuples of non-overlined parts of
$\beta$, and}\\\ \hbox{the different $b$-tuples of non-overlined parts of
$\beta$.}\end{cases}$

We can now state the $b$-ary analogue of Proposition 5.4.

######  Proposition 9.12.

For $b\geq 2$, $n\geq 1$, and $0\leq j\leq n$, the coefficient $a_{j}^{(n)}$,
as defined in (5.7), counts the number of $b$-restricted $b$-ary
overpartitions $\beta$ of $(b^{n+1}-b)/(b-1)$ with $S^{b}(\beta)=n+j$.

The proof of this follows from Proposition 9.1, Definition 9.11, and
(9.7)–(9.9). One could also state and prove $b$-ary analogues of Proposition
5.7 and Corollaries 5.6 and 5.9. We leave this to the reader.

######  Example 9.13.

We take again $n=2$ and consider the thirteen 5-restricted 5-ary
overpartitions of $(5^{3}-5)/(5-1)=30$, as displayed in Example 9.8. Here we
only list three representative examples, along with the sums $S^{5}(\beta)$
related to (9.19):

$\displaystyle(5^{5},1^{5}):0+0+0+2=2;\qquad(5^{5},\overline{1},1^{4}):1+0+1+1=3;$
$\displaystyle(\overline{5},5^{4},\overline{1},1^{4}):2+0+2+0=4.$

Altogether we have 8, 4, and 1 such 5-ary partitions with $S^{5}(\beta)=2,3$,
and 4, respectively. This is consistent with Proposition 9.12 and is analogous
to Example 5.5.

###  9.5. A third special case

Finally, we set $x=1$ and $y_{b-1}=y_{1}$. Then an easy variant of the proof
of Proposition 6.1 shows that in this case we have

(9.20)
$Q_{n}^{b}(Z)=\frac{1}{y_{b}+1}\left((y_{1}+y_{b}+1)^{n+1}-y_{1}^{n+1}\right),$

so that by (6.2) we have $Q_{n}^{b}(Z)=Q_{n}(1,y_{1},y_{b})$. Therefore this
case is covered by Sections 6 and 7.

## Acknowledgments

We thank Keith Johnson of Dalhousie University for some helpful discussions.

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