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\begin{center}
\vskip 1cm{\LARGE\bf 
Polynomials Whose Coefficients 
\vskip .1in
Are Stern Numbers}
\vskip 1cm
\large
Karl Dilcher\footnote{Research supported in part by the Natural Sciences and 
Engineering Research Council of Canada, Grant \# 145628481.}\\
Department of Mathematics and Statistics\\
Dalhousie University\\
Halifax, Nova Scotia B3H 4R2\\
Canada\\
\href{mailto:dilcher@mathstat.dal.ca}{\tt dilcher@mathstat.dal.ca}\\
\ \\
Larry Ericksen\\
1212 Forest Drive\\
Millville, NJ 08332-2512\\
USA\\
\href{mailto:LE22@cornell.edu}{\tt LE22@cornell.edu}\\
\end{center}

\vskip .2in

\begin{abstract}
The main object in this paper is the sequence of polynomials $P_n(z)$ that have
Stern numbers as their coefficients; that is, the terms of Stern's diatomic
sequence. We derive certain basic properties of these polynomials, investigate 
the distribution of their real and complex zeros, and prove some results 
concerning factorizations and resultants. We also consider the 
$(0,1)$-polynomials obtained from $P_n(z)$ by taking their coefficients 
modulo 2. In spite of its simple form,
the polynomial sequence is shown to possess some interesting algebraic and 
analytic properties. Finally, we discuss combinatorial interpretations 
of the polynomials $P_n(z)$ and indicate ways of generalizing them.
\end{abstract}

\section{Introduction}\label{sec:1}

The Stern sequence, also known as Stern's diatomic sequence \cite{Le}, is one 
of the most remarkable integer sequences in number theory and combinatorics. 
It can be defined by $s(0)=0$, $s(1)=1$, and 
\begin{equation}\label{1.1}
s(2n) = s(n),\qquad s(2n+1) = s(n) + s(n+1)\qquad (n\geq 1).
\end{equation}
The first 30 Stern numbers, starting with $n=1$, are
\begin{equation}\label{1.2}
1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 5, 4, 7, 3, 8, 5, 7, 2, 7,
5, 8, 3, 7, 4.
\end{equation}
The definition \eqref{1.1} gave rise to two different concepts of Stern 
polynomials, introduced in \cite{DS1} and \cite{KMP}, respectively. Both
preserve the ``binary structure" of the sequence \eqref{1.1} and have various
combinatorial interpretations. Subsequently these sequences of polynomials
were generalized by the present authors \cite{DE6}, then unified as special
cases of a class of polynomials in two variables \cite{DE7}, and more generally
extended to polynomials in an arbitrary number of variables and with an 
underlying base $b\geq 2$ \cite{DE9}. All this has applications to counting and
classifying restricted binary and $b$-ary partitions of integers.

In the current paper we will, in the first place, consider polynomials of a 
very different type, namely those having the Stern numbers $s(n)$, and later
some of their generalizations, as coefficients. Thus, we define
\begin{equation}\label{1.3}
P_n(z):= s(1)z^n+s(2)z^{n-1}+\cdots+s(n)z+s(n+1),
\end{equation} 
and as a specific example we have, with \eqref{1.2},
\begin{equation}\label{1.4}
P_8(z)= z^8+z^7+2z^6+z^5+3z^4+2z^3+3z^2+z+4.
\end{equation}

The polynomials $P_n(z)$, as well, have a combinatorial interpretation which is
related to binary overpartitions, and this leads to generalizations in different
directions. The main purpose of this paper, however, is to derive various
properties of the polynomials $P_n(z)$ and of a closely related polynomial
sequence.

In the process we will encounter combinatorial objects, including recurrence
relations, Fibonacci numbers, generating functions, and various $b$-ary 
partitions. From classical analysis we will come across several well-known
theorems on the distribution of zeros of polynomials, and the algebraic topics
occurring in this paper include cyclotomic polynomials, resultants, and 
divisibility and irreducibility.

We begin in Section~\ref{sec:2} by recalling certain properties of the Stern
sequence that will be needed later, followed by some basic properties of the
polynomials $P_n(z)$. In Sections~\ref{sec:4} and~\ref{sec:5} we study the
complex and real zeros, respectively, of these polynomials. Section~\ref{sec:6}
is devoted to algebraic properties of the polynomials $P_n(z)$, including
factors and resultants. In Section~\ref{sec:7} we consider a (0,1)-polynomial
sequence whose coefficients are those of $P_n(z)$ taken modulo 2; we
completely determine all factors of these polynomials and investigate their
zeros. Section~\ref{sec:8} deals with binary and $b$-ary overpartitions and
with a combinatorial interpretation of the polynomials $P_n(z)$,
including some generalizations.


\section{Some basic properties}\label{sec:2}

In this section we derive several basic properties of the polynomials $P_n(z)$,
most of which will be required in later sections. We begin by recalling some
easy facts about the Stern sequence.

\subsection{Properties of the Stern sequence}

Although at first sight the Stern sequence appears to behave rather erratically,
it does in fact have a great deal of structure. First, we have the special
values
\begin{equation}\label{2.1}
s(2^m)=1,\quad s(2^m+1)=m+1,\quad s(2^m+2)=m,\quad s(2^m+3)=2m-1
\end{equation}
$(m\geq 2)$, where the first three also hold for $m=1$, and the first two for 
$m=0$. All four identities are easy to prove by induction.
The sequence $s(n)$ is also symmetric between two powers of 2, that is, 
\begin{equation}\label{2.2}
s(2^m+j)=s(2^{m+1}-j),\qquad 0\leq j\leq 2^m.
\end{equation}
In particular, with \eqref{2.1}, this leads to values for $s(2^m-j)$, $j=1,2,3$.
Another identity that will be needed later is
\begin{equation}\label{2.2a}
s(3\cdot 2^m+1)=s(3\cdot 2^m-1)=2m+1,\qquad m\geq 0;
\end{equation}
this, too, is easy to prove by induction.
More properties like these are known, but they will not be needed here.

Next, combining the two identities in \eqref{1.1}, we have
$s(2n+1)=s(2n)+s(2n+2)$; this implies that for $n\geq 1$ each odd-index Stern
number $s(2n+1)$ is strictly larger than its two neighbors. The Stern sequence
is therefore a strict up-down sequence, beginning with the second term. This
leads us to consider the ratio of neighboring terms.
First, it is clear from \eqref{2.1} and by symmetry that ratios of neighboring
terms can be arbitrarily large or arbitrarily small rational numbers. This is
consistent with the well-known fact that the sequence $s(n+1)/s(n)$ is a 
one-to-one enumeration of all positive rationals; see, e.g., \cite{CW, KMP}.

We now show that the ratios $s(n+1)/s(n)$ given by the first two identities in
\eqref{2.1}, and their reciprocals, are in fact 
extremal in each interval between two neighboring powers of 2.

\begin{lemma}\label{lem:2.1}
For any integer $m\geq 1$ we have
\begin{equation}\label{2.3}
\frac{1}{m}\leq \frac{s(n+1)}{s(n)}\leq m,\quad\hbox{where}\quad
2^{m-1}\leq n\leq 2^m-1, 
\end{equation}
and the lower and upper bound are attained exactly when $n=2^m-1$ and 
$n=2^{m-1}$, respectively.
\end{lemma}

\begin{proof}
We proceed by induction on $m$. The cases $m=1$ and $m=2$ are easily seen to be
true. Suppose now that \eqref{2.3} holds up to some $m\geq 2$, and let $n$ be
in the interval $2^m\leq n\leq 2^{m+1}-1$. First, if $n$ is even, we write
$n=2k$ and with \eqref{1.1} we get
\[
\frac{s(n+1)}{s(n)}=\frac{s(k)+s(k+1)}{s(k)}=1+\frac{s(k+1)}{s(k)};
\]
by the induction hypothesis we then get the upper bound $m+1$. Second, if $n$ is
odd, we set $n=2k+1$ and again with \eqref{1.1} we get
\[
\frac{s(n+1)}{s(n)}=\frac{s(k+1)}{s(k)+s(k+1)}=\frac{1}{1+s(k)/s(k+1)}
\geq\frac{1}{1+m},
\]
having used the induction hypothesis. We also implicitly used the ``up-down"
property which implies that the quotients $s(n+1)/s(n)$ alternate between 
greater than and less than 1. This proves \eqref{2.3} by induction. The final 
statement is clear by \eqref{2.1} and \eqref{2.2}, and uniqueness comes from 
the one-to-one enumeration mentioned above.
\end{proof}

We also require a remarkable property of the Stern sequence that was
apparently first proved by D.~H.~Lehmer \cite{Le}.

\begin{lemma}[Lehmer]\label{lem:2.3}
The largest value of $s(n)$ between $n=2^{m-1}$ and $n=2^m$ is the Fibonacci
number $F_{m+1}$, with the usual definition $F_0=0$, $F_1=1$, and
$F_{n+1}=F_n+F_{n-1}$. The maximum is attained at
\begin{equation}\label{2.6}
\alpha_m:=\frac{1}{3}\left(2^{m+1}+(-1)^m\right)\quad\hbox{and}\quad
\beta_m:=\frac{1}{3}\left(5\cdot 2^{m-1}-(-1)^m\right)\quad (m\geq 1).
\end{equation}
\end{lemma}

%\bigskip
\begin{table}[h]\label{tab:1}
%\begin{center}
\centering
\begin{tabular}{|c||c|c|c|c|c|c|c|c|c|c|}
\hline
$m$ &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp; 7 &amp; 8 &amp; 9 &amp; 10   \\
\hline\hline
$\alpha_m$ &amp; 1 &amp; 3 &amp; 5 &amp; 11 &amp; 21 &amp; 43 &amp; 85 &amp; 171 &amp; 341 &amp; 683 \\
\hline
$\beta_m$  &amp; 2 &amp; 3 &amp; 7 &amp; 13 &amp; 27 &amp; 53 &amp; 107 &amp; 213 &amp; 427 &amp; 853\\
\hline
$F_{m+1}$ &amp; 1 &amp; 2 &amp; 3 &amp; 5 &amp; 8 &amp; 13 &amp; 21 &amp; 34 &amp; 55 &amp; 89 \\
\hline
\end{tabular}
\caption{$\alpha_m$, $\beta_m$, $F_{m+1}$ for $1\leq m\leq 10$}
\end{table}

\subsection{Properties of $P_n(z)$}

Now we turn to some basic properties of the polynomials $P_n(z)$. First, the
definition \eqref{1.3} immediately gives
\begin{equation}\label{2.7}
P_n(z)= z\cdot P_{n-1}(z)+s(n+1).
\end{equation}
By separating even and odd powers of $z$ in \eqref{1.3}, we obtain a different
class of recurrence relations, as follows.

\begin{lemma}\label{lem:2.4}
For all $n\geq 1$ we have
\begin{align}
P_{2n}(z) &amp;= (1+z)\cdot P_{n-1}(z^2)+P_n(z^2),\label{2.8} \\
P_{2n+1}(z) &amp;= z\cdot P_{n-1}(z^2)+(1+z)\cdot P_n(z^2).\label{2.9}
\end{align}
\end{lemma}

\begin{proof}
By \eqref{1.3} we have
\begin{align*}
P_{2n}(z) &amp;= \sum_{j=0}^ns(2j+1)z^{2n-2j}+\sum_{j=1}^ns(2j)z^{2n+1-2j}\\
&amp;=\sum_{j=0}^n\big(s(j)+s(j+1)\big)(z^2)^{n-j}+z\sum_{j=1}^ns(j)(z^2)^{n-j}\\
&amp;=(1+j)\sum_{j=0}^{n-1}s(j+1)(z^2)^{n-j-1}+\sum_{j=0}^ns(j+1)(z^2)^{n-j},
\end{align*}
where we have used \eqref{1.1} and the fact that $s(0)=0$. It is now obvious
that this last line is the right-hand side of \eqref{2.8}. The identity 
\eqref{2.9} can be obtained analogously.
\end{proof}

The next lemma deals with some special values of $P_n(z)$.

\begin{lemma}\label{lem:2.5}
For any integer $m\geq 1$ we have
\begin{equation}\label{2.10}
P_{2^m-1}(1) = \sum_{j=1}^{2^m}s(j) = \frac{1}{2}\left(3^m+1\right),
\end{equation}
and, for any $n\geq 1$,
\begin{equation}\label{2.11}
P_{2n}(-1) = P_n(1),\qquad P_{2n+1}(-1) = -P_{n-1}(1). 
\end{equation}
\end{lemma}

\begin{proof}
It is a well-known property of the Stern sequence that
\begin{equation}\label{2.12}
\sum_{j=2^\nu+1}^{2^{\nu+1}}s(j) = 3^\nu,\qquad \nu\geq 0;
\end{equation}
see, e.g., \cite{Le}. Summing \eqref{2.12} over all $\nu=0, 1, \ldots, m-1$ and
adding $s(1)=1$ to both sides, we obtain \eqref{2.10}. Finally,
if we set $z=-1$ in \eqref{2.8} and \eqref{2.9}, we immediately get the two 
identities in \eqref{2.11}.
\end{proof}

To conclude this section, we derive a generating function for the
polynomials $P_n(z)$.

\begin{proposition}\label{prop:2.6}
With $q$ a complex variable, we have
\begin{equation}\label{2.13}
\sum_{n=0}^{\infty}P_n(z)q^n 
= \frac{1}{1-zq}\prod_{k=0}^{\infty}\left(1+q^{2^k}+q^{2^{k+1}}\right).
\end{equation}
\end{proposition}

\begin{proof}
The Stern sequence has the well-known generating function
\begin{equation}\label{2.14}
\prod_{k=0}^{\infty}\left(1+q^{2^k}+q^{2^{k+1}}\right)
=\frac{1}{q}\sum_{n=1}^{\infty}s(n)q^n = \sum_{n=0}^{\infty}s(n+1)q^n; 
\end{equation}
see, e.g., \cite{Ca, Re}. By the definition \eqref{1.3}, $P_n(z)$ is
a convolution of the sequences $s(1), s(2),$ $\ldots, s(n+1)$ and 
$z^0, z^1, \ldots, z^n$. Hence
\[
\sum_{n=0}^{\infty}P_n(z)q^n = \left(\sum_{k=0}^{\infty}z^kq^k\right)
\left(\sum_{k=0}^{\infty}s(k+1)q^k\right),
\]
and \eqref{2.13} follows from \eqref{2.14} and the sum of a geometric series.
\end{proof}

\section{Complex zeros of $P_n(z)$}\label{sec:4}

Plots of the zeros of $P_n(z)$ indicate that they lie relatively close to the
unit circle (with some notable exceptions inside the unit circle) and have
uniform angular distribution; see Figure~1. It is the purpose of this
section to confirm these observations. 

\subsection{Some classical theorems}

All our results will be based on some
well-known classical theorems; we quote them here for the sake of completeness.

\begin{figure}[h]\label{fig:1}
\begin{center}
\includegraphics[width=3in]{zerosP201.jpg}
\end{center}
\caption{The zeros of $P_{201}(z)$}
\end{figure}

The first result is due to Enestr\"om and Kakeya; see, for instance, 
\cite[pp.\ 136--137]{Ma}.

\begin{theorem}[Enestr\"om, Kakeya]\label{thm:4.1}
Let $f(z)=a_0+a_1z+\cdots+a_nz^n$ be a polynomial with positive coefficients.
Then all the zeros of $f(z)$ lie in the annulus $\rho_1\leq |z|\leq\rho_2$,
where
\begin{equation}\label{4.1}
\rho_1:=\min_{0\leq k\leq n-1}\left\{\frac{a_k}{a_{k+1}}\right\},\qquad
\rho_2:=\max_{0\leq k\leq n-1}\left\{\frac{a_k}{a_{k+1}}\right\}.
\end{equation}
\end{theorem}

Another well-known result of a similar nature is due to Cauchy; see
\cite[p.~123]{Ma}. 

\begin{theorem}[Cauchy]\label{thm:4.2}
All the zeros of $f(z)=a_0+a_1z+\cdots+a_nz^n$, $a_n\neq 0$, lie in the disk
\begin{equation}\label{4.2}
|z| &lt; 1 + \max_{0\leq k\leq n-1}\left\{\left|\frac{a_k}{a_n}\right|\right\}.
\end{equation}
\end{theorem}

The next auxiliary result is due to Jentzsch \cite{Je}.  It can be stated as 
follows.

\begin{theorem}[Jentzsch]\label{thm:4.3}
Let $f(z)=\sum_{k=0}^{\infty}a_kz^k$ be a power series with radius of 
convergence~$1$. Then every point on the unit circle is a limit point of zeros
of the partial sums of $f(z)$.
\end{theorem}

This remarkable theorem can be supplemented by the following well-known result
by Erd{\H o}s and Tur\'an on the equidistribution of the arguments of the zeros
of a polynomial; see \cite{ET}, or \cite{So} for a recent exposition and a new
proof.

\begin{theorem}[Erd{\H o}s and Tur\'an]\label{thm:4.4}
Let $f(z)=a_0+a_1z+\cdots+a_nz^n$ be a polynomial with $a_0a_n\neq 0$, and let
$\alpha, \beta$ be real numbers such that $0\leq\alpha &lt;\beta\leq 2\pi$. If
$N(\alpha,\beta)$ denotes the number of zeros $z_j$ with 
$\alpha\leq\arg{z_j}\leq\beta$, then
\begin{equation}\label{4.3}
\left|\frac{N(\alpha,\beta)}{n}-\frac{\beta-\alpha}{2\pi}\right|
&lt; 16\sqrt{\frac{\log{P}}{n}},\quad\hbox{where}\quad
P = \frac{1}{\sqrt{|a_0a_n|}} \sum_{k=0}^n\big|a_k\big|.
\end{equation}
\end{theorem}

\subsection{Applications to $P_n(z)$}

We now apply these theorems to the polynomials $P_n(z)$. Our first result is an
immediate consequence of Lemma~\ref{lem:2.1} and Theorem~\ref{thm:4.1}.

\begin{proposition}\label{prop:4.5}
Let $n\geq 1$ and $m\geq 1$ be such that $2^{m-1}\leq n\leq 2^m-1$. Then the
zeros of $P_n(z)$ all satisfy
\begin{equation}\label{4.4}
\frac{1}{m}\leq |z| \leq m.
\end{equation}
\end{proposition}

We will see in the next section that in general the lower bound cannot be 
improved; however, see below for some special cases. We expect that the upper
bound can be improved. In any case, we have the following result.

\begin{proposition}\label{prop:4.6}
Every point on the unit circle is a limit point of the zeros of $P_n(z)$.
\end{proposition}

\begin{proof}
For any $n\geq 0$ we consider the reciprocal polynomial
\begin{equation}\label{4.5}
P_n^*(z):=z^nP_n(\tfrac{1}{z})=1+z+2z^2+z^3+3z^4+\cdots+s(n+1)z^n
\end{equation}
and the power series
\begin{equation}\label{4.6}
f(z):=\lim_{n\to\infty}P_n^*(z) = \sum_{k=0}^{\infty}s(k+1)z^k.
\end{equation}
To determine the radius of convergence, we use the following bound due to
Reznick \cite[p.~472]{Re}: There is a constant $c&gt;0$ such that for all 
$n\geq 1$ we have
\[
1\leq s(n)\leq c\cdot n^\mu,\qquad\hbox{where}\quad
\mu = \log_2\big(\tfrac{1+\sqrt{5}}{2}\big);
\]
an asymptotic value for $c$ can also be found in \cite{Re}. Now we use the
$n$th root test for the power series in \eqref{4.6} and find that
\[
1\leq \sqrt[n]{s(n)}\leq c^{1/n}\cdot n^{\mu/n}\rightarrow 1\quad\hbox{as}
\quad n\rightarrow\infty,
\]
where the limit is clear if we consider logarithms. Hence the radius of
convergence of the power series is 1, and Jentzsch's Theorem~\ref{thm:4.3} 
shows that every point on the unit circle is a limit point of the zeros of
$P_n^*(z)$. The same result is then true for the zeros of $P_n(z)$, which 
completes the proof.
\end{proof}

As mentioned earlier, the lower bound in \eqref{4.4} is essentially best 
possible. On the other hand, for certain subscripts $n$, the zeros of $P_n(z)$
can be proven to be much larger. 

\begin{proposition}\label{prop:4.7}
For $m\geq 1$, let $\alpha_m$ and $\beta_m$ be the integers defined in 
\eqref{2.6}. Then for all $n=\alpha_m-1$ and $n=\beta_m-1$, all the zeros of
$P_n(z)$ satisfy $|z|&gt;1/2$.
\end{proposition}

\begin{proof}
We consider the polynomial $P_n^*(z)$ as defined in \eqref{4.5}. Then by
Lemma~\ref{lem:2.3}, extended with \eqref{1.1}, we see that the leading
coefficient $s(\alpha_m)$ or $s(\beta_m)$ is greater than, or equal to, all
other coefficients. Then by Cauchy's Theorem~\ref{thm:4.2} all the zeros of
$P_n^*(z)$ satisfy $|z|&lt;2$, which implies that for all the zeros of $P_n(z)$ 
we have $|z|&gt;1/2$, as claimed.
\end{proof}

The final result in this section is a direct consequence of the theorem of
Erd{\H o}s and Tur\'an.

\begin{proposition}\label{prop:4.8}
The zeros of $P_n(z)$ have uniform angular distribution. More precisely, if
$\alpha, \beta\in{\mathbb R}$ are such that $0\leq\alpha &lt;\beta\leq 2\pi$, and
$N(\alpha,\beta)$ denotes the number of zeros $z_j$ with
$\alpha\leq\arg{z_j}\leq\beta$, then
\begin{equation}\label{4.7}
\left|\frac{N(\alpha,\beta)}{n}-\frac{\beta-\alpha}{2\pi}\right|
&lt; 16\cdot\sqrt{\frac{\log{3}}{\log{2}}}\cdot\sqrt{\frac{1+\log(n+1)}{n}}.
\end{equation}
\end{proposition}

\begin{proof}
To determine $P$ in \eqref{4.3}, we first note that by definition of $P_n(z)$
we have $a_n=1$ and $a_0\geq 1$; hence it remains to estimate the relevant sum
of coefficients. Given the index $n$, let $m\geq 1$ be such that 
$2^{m-1}-1\leq n\leq 2^m-2$. Then by \eqref{2.10}, and upon subtracting 
$s(2^m)=1$ from both sides, we get
\[
P\leq\sum_{k=0}^n\left|a_k\right|=P_n(1)\leq P_{2^m-2}(1)
=\frac{1}{2}\left(3^m-1\right) &lt; \frac{1}{2}\cdot 3^m,
\]
so that
\begin{equation}\label{4.8}
\log{P} &lt; -\log{2}+m\log{3}.
\end{equation}
Now, since $2^{m-1}-1\leq n$, we have $2^m\leq 2(n+1)$, and thus
\[
m\leq \frac{\log(2(n+1))}{\log{2}} = 1 + \frac{\log(n+1)}{\log{2}}.
\]
Substituting this into \eqref{4.8}, we get
\[
\log{P} &lt; -\log{2}+\log{3}+\frac{\log{3}}{\log{2}}\cdot\log(n+1)
=\frac{\log{3}}{\log{2}}\cdot\big(1+\log(n+1)\big).
\]
This, with \eqref{4.3}, gives \eqref{4.7}.
\end{proof}

In closing we remark that the numerical value of the constant factor on the
right of \eqref{4.7} is $16\sqrt{\log{3}/\log{2}}\simeq 20.14$. Also, 
\eqref{4.7} becomes meaningful only when $n$ is reasonably large.

\section{Real zeros of $P_n(z)$}\label{sec:5}

Since the polynomials $P_n(z)$ have only positive coefficients, the only 
possible real zeros will be negative. In fact, we have the following result.

\begin{proposition}\label{prop:5.1}
For all $n\geq 0$,
\leavevmode
\begin{itemize}
\item[(a)] $P_{2n}(x)$ has no real zeros; in fact, 
$P_{2n}(x)\geq\frac{3}{4}s(n)+s(n+1)$ for all $x\in{\mathbb R}$.
\item[(b)] $P_{2n+1}(x)$ is strictly increasing as a function of $x$, and 
therefore has exactly one real zero.
\end{itemize}
\end{proposition}

\begin{proof}
From \eqref{2.7} we get $P_n(x^2)=x^2P_{n-1}(x^2)+s(n+1)$, and substituting
this into \eqref{2.8} gives
\begin{equation}\label{5.1}
P_{2n}(x)=\big(1+x+x^2\big)P_{n-1}(x^2)+s(n+1).
\end{equation}
It is easy to see that $1+x+x^2\geq\frac{3}{4}$ and 
$P_{n-1}(x^2)\geq P_{n-1}(0)=s(n)$ for all $x$. This fact, along with 
\eqref{5.1}, proves (a).

To prove part (b), we first take the derivative of \eqref{2.7} with respect to
$z$, and then set $z=x^2$, to obtain
\begin{equation}\label{5.2}
P'_n(x^2)=x^2P'_{n-1}(x^2)+2P_{n-1}(x^2).
\end{equation}
Next, taking the derivative of \eqref{2.9}, we get
\[
P'_{2n+1}(x)=2x^2P'_{n-1}(x^2)+\big(2x+2x^2\big)P'_n(x^2)+P_{n-1}(x^2)+P_n(x^2).
\]
By substituting \eqref{5.2} into this last identity, we obtain, after some
simplification,
\[
P'_{2n+1}(x)=2x^2\big(1+x+x^2\big)P'_{n-1}(x^2)+\big(1+2x+2x^2\big)P_{n-1}(x^2)
+P_n(x^2).
\]
We already noted that $1+x+x^2&gt;0$; it is also easy to verify that $1+2x+2x^2&gt;0$
for all $x\in{\mathbb R}$. Since all other terms are obviously positive, we 
have $P'_{2n+1}(x)&gt;0$ for all $x\in{\mathbb R}$, as desired.
\end{proof}

%\bigskip
%\begin{center}
\begin{table}[h]\label{tab:2}
\centering
\begin{tabular}{|r|l||r|l||r|l|}
\hline
$n$ &amp; $-r_n$ &amp; $n$ &amp; $-r_n$ &amp; $n$ &amp; $-r_n$  \\
\hline
1 &amp; 1           &amp; 13 &amp; .5607907569 &amp; 25 &amp; .5989653955 \\
3 &amp; .5698402910 &amp; 15 &amp; .2829027496 &amp; 27 &amp; .4179121334 \\
5 &amp; .6464478041 &amp; 17 &amp; .6442386664 &amp; 29 &amp; .5308746463 \\
7 &amp; .3767322046 &amp; 19 &amp; .4753681436 &amp; 31 &amp; .2253056724 \\
9 &amp; .6494575822 &amp; 21 &amp; .5862078204 &amp; 33 &amp; .6410098261 \\
11 &amp; .4468280932&amp; 23 &amp; .3232890249 &amp; 35 &amp; .4907265638 \\
\hline
\end{tabular}
\caption{The real zeros $r_n$ of $P_n(x)$, $1\leq n\leq 35$}
\end{table}

Proposition~\ref{prop:5.1}(b) can be made more precise. For instance, Table~2 
indicates that the real zeros of $P_n(x)$, for odd $n\geq 3$, lie strictly 
between $-1$ and 0. This, and more, is shown in the next result.

\begin{proposition}\label{prop:5.2}
For $n\geq 1$, let $r_{2n+1}$ be the unique real zero of $P_{2n+1}(z)$; then
$-1&lt;r_{2n+1}&lt;~0$. Furthermore, if $m$ is such that 
$2^{m-1}+1\leq 2n+1\leq 2^m-1$, then: 
\leavevmode
\begin{itemize}
\item[(a)] If $n$ is odd, then
\begin{equation}\label{5.3}
-\frac{s(n+1)}{s(n)} &lt; r_{2n+1} \leq -\frac{1}{m}, 
\end{equation}
and, if $n\equiv 3\pmod{4}$, then $-1/2&lt; r_{2n+1} \leq -1/m$.
\item[(b)] If $n$ is even, then
\begin{align}
-\frac{s(2n+2)}{s(2n+1)} &amp;&lt; r_{2n+1} \leq -\frac{1}{m}\qquad (n\equiv 0\pmod{4}), \label{5.4}\\
-\frac{s(n)}{s(n-1)} &amp;&lt; r_{2n+1} \leq -\frac{1}{m}\qquad (n\equiv 2\pmod{4}).
\label{5.5}
\end{align}
\end{itemize}
\end{proposition}

Before proving this result, we derive a consequence that explains the 
observation from Table~2 that the zeros $r_n$ are relatively small and 
decreasing in absolute value for $n = 7, 15, 31$. This becomes more pronounced 
when we compute $r_{63}\simeq -.1863219404$ and $r_{127}\simeq -.1583809009$. 
The following result partially explains these observations.

\begin{corollary}\label{cor:5.3}
For any integer $m\geq 2$ we have
\begin{equation}\label{5.6}
-\frac{1}{m-1} &lt; r_{2^m-1} \leq -\frac{1}{m}.     
\end{equation}
\end{corollary}

\begin{proof}
We use Proposition~\ref{prop:5.2}(a) with $n=2^{m-1}-1$. Then by \eqref{2.1}
and \eqref{2.2} we have $s(n+1)=1$ and $s(n)=s(2^{m-2}+1)=m-1$, which gives the
left inequality of \eqref{5.6}. The right inequality remains the same as in
\eqref{5.3}.
\end{proof}

Corollary~\ref{cor:5.3} shows that Proposition~\ref{prop:4.5}, in particular 
the left inequality, is best possible. It also shows that certain polynomials
$P_n(z)$ can have zeros that are arbitrarily close to the origin. This does not
contradict Proposition~\ref{prop:4.6}.

\begin{proof}[Proof of Proposition~\ref{prop:5.2}]
By the second identity in \eqref{2.11} we have $P_{2n+1}(-1)&lt;0$, 
while by definition we have $P_{2n+1}(0)=s(2n+2)&gt;0$. Hence the unique zero
$r_{2n+1}$ must lie between $-1$ and 0, as claimed. In fact, by
Proposition~\ref{prop:4.5}, $r_{2n+1}$ cannot be greater than $-1/n$, which
proves the right-hand inequalities in \eqref{5.3}--\eqref{5.5}.

For the remainder of the proof it will be convenient to deal with the 
slightly modified polynomial $Q_{2n+1}(x):=-P_{2n+1}(-x)$, which remains monic
but changes the signs of the real zeros. We have, therefore,
\begin{equation}\label{5.7}
Q_{2n+1}(x)=x^{2n+1}-x^{2n}+2x^{2n-1}-\cdots+s(2n+1)x-s(2n+2).
\end{equation}

To prove part (a), we rewrite \eqref{5.7} as
\begin{align}
Q_{2n+1}(x)=x^{2n+1}+\sum_{j=1}^n\big(s(2j+1)&amp;-s(2j)x\big)\cdot x^{2n+1-2j}\label{5.8}\\
&amp;-s(2n)x^2+s(2n+1)x-s(2n+2).\nonumber
\end{align}
Since $s(2j+1)&gt;s(2j)$ for all $j\geq 1$, all summands in the first line of
\eqref{5.8} are positive for $0&lt;x&lt;1$. The last three summands taken together are
nonnegative exactly when
\[
x^2-\frac{s(2n+1)}{s(2n)}x+\frac{s(2n+2)}{s(2n)} \leq 0.
\]
With \eqref{1.1}, this quadratic inequality can be rewritten as
\[
x^2-\left(1+\frac{s(n+1)}{s(n)}\right)x+\frac{s(n+1)}{s(n)} \leq 0
\]
and be factored as
\begin{equation}\label{5.9}
\big(x-1\big)\cdot\left(x-\frac{s(n+1)}{s(n)}\right) \leq 0.
\end{equation}
Since $n$ is odd, $s(n+1)/s(n)&lt;1$, and with $x=s(n+1)/s(n)$ the left of 
\eqref{5.9} obviously vanishes, and by \eqref{5.8} we have 
$Q_{2n+1}(s(n+1)/s(n))&gt;0$, while $Q_{2n+1}(0)&lt;0$. This proves the left 
inequality of \eqref{5.3}.

For the final statement of part (a) we set $n=4k+3$ and obtain with \eqref{1.1},
\[
\frac{s(n+1)}{s(n)}=\frac{s(4k+4)}{s(4k+3)}
=\frac{s(2k+2)}{s(2k+1)+s(2k+2)} &lt; \frac{1}{2},
\]
where the inequality is easy to verify, using the fact that $s(2k+1)&gt;s(2k+2)$.

To prove part (b), we modify \eqref{5.8} and obtain
\begin{align}
&amp;Q_{2n+1}(x)=x^{2n+1}+\sum_{j=1}^{n-1}\big(s(2j+1)-s(2j)x\big)\cdot x^{2n+1-2j}\label{5.10}\\
&amp;\quad+\left(-s(2n-2)x^2+s(2n-1)x-s(2n)\right)x^2+\big(s(2n+1)x-s(2n+2)\big).\nonumber
\end{align}
We proceed as in part (a) and note that all summands in the first line of
\eqref{5.10} are positive. Next, in analogy to \eqref{5.9} we can see that
\[
-s(2n-2)x^2+s(2n-1)x-s(2n) \geq 0
\]
if and only if
\begin{equation}\label{5.11}
\big(x-1\big)\cdot\left(x-\frac{s(n)}{s(n-1)}\right) \leq 0,
\end{equation}
which is the case when $s(n)/s(n-1)\leq x\leq 1$; here we note that 
$s(n)/s(n-1)&lt;1$ since $n$ is even. For the final term in \eqref{5.10} we have
$s(2n+1)x-s(2n+2)\geq 0$ if and only if $x\geq s(2n+2)/s(2n+1)$, which is also
less than 1. So altogether, by \eqref{5.10} we have $Q_{2n+1}(x_0)&gt;0$, while
again $Q_{2n+1}(0)&lt;0$, where
\begin{equation}\label{5.12}
x_0:=\max\left\{\frac{s(n)}{s(n-1)},\frac{s(2n+2)}{s(2n+1)}\right\},\quad
(n\;\hbox{even}).
\end{equation}
To complete the proof, we determine which one of the two elements is larger.
First, when $n\equiv 0\pmod{4}$, we set $n=4k$ and with \eqref{1.1} we obtain
\begin{align*}
\frac{s(n)}{s(n-1)}&amp;=\frac{s(4k)}{s(4k-1)}
=\frac{s(2k)}{s(2k-1)+s(2k)} &lt; \frac{1}{2},\\
\frac{s(2n+2)}{s(2n+1)}&amp;=\frac{s(8k+2)}{s(8k+1)}
=\frac{s(4k+1)}{s(4k)+s(4k+1)}=\frac{s(2k)+s(2k+1)}{2s(2k)+s(2k+1)}&gt;\frac{1}{2},
\end{align*}
where the inequalities on the right are easy to verify, using the fact that
odd-index Stern numbers are larger than their neighbors. These inequalities
prove \eqref{5.4}.

When $n\equiv 2\pmod{4}$, we set $n=4k+2$, obtaining
\begin{align*}
\frac{s(n)}{s(n-1)}&amp;=\frac{s(2k+1)}{s(2k)+s(2k+1)} 
=\frac{s(k)+s(k+1)}{2s(k)+s(k+1)},\\
\frac{s(2n+2)}{s(2n+1)}&amp;=\frac{s(4k+3)}{s(4k+2)+s(4k+3)}
=\frac{s(k)+2s(k+1)}{2s(2k)+3s(2k+1)}.
\end{align*}
It is now easy to verify that
\[
\frac{s(k)+s(k+1)}{2s(k)+s(k+1)}&gt;\frac{s(k)+2s(k+1)}{2s(2k)+3s(2k+1)},
\]
for instance by multiplying both sides by the denominators and then simplifying.
This shows that the first element in \eqref{5.12} is maximal, which proves
\eqref{5.5}; the proof is now complete.
\end{proof}

\begin{remark}\label{rem:1}
Based on Table~2 and the second part of Proposition~\ref{prop:5.2}(a), it may
be tempting to conjecture that $r_{n}&gt;-1/2$ for all $n\equiv 3\pmod{4}$, 
$n&gt;3$, and
$r_{n}&lt;-1/2$ for all $n\equiv 1\pmod{4}$. However, this is not true, as the
counterexamples in Table~3 show. In the latter case, all counterexamples with 
$n&lt;2000$ are shown, while in the former case, there are 8 more up to 
$n=2000$.
\end{remark}

\begin{table}[h]\label{tab:3}
\centering
\begin{tabular}{|c|c||c|c|}
\hline
$n\equiv 3\pmod{4}$ &amp; $r_n&lt;-1/2$ &amp; $n\equiv 1\pmod{4}$ &amp; $r_n&gt;-1/2$  \\
\hline
67 &amp;  $-.5002864226$ &amp; 509 &amp;  $-.4954752823$\\
131 &amp; $-.5067977789$ &amp; 765 &amp;  $-.4975876362$\\
259 &amp; $-.5115134818$ &amp; 1021 &amp;  $-.4920186458$\\
387 &amp; $-.5012417771$ &amp; 1277 &amp;  $-.4983631397$\\
515 &amp; $-.5150841917$ &amp; 1533 &amp;  $-.4936355414$\\
643 &amp; $-.5031636440$ &amp; 1789 &amp;  $-.4968495572$\\
\hline
\end{tabular}
\caption{The first 6 counterexamples related to Remark~\ref{rem:1}}
\end{table}

\section{Algebraic properties}\label{sec:6}

In this section we will deal with some algebraic properties of the polynomials
$P_n(z)$, mainly factorizations involving cyclotomic polynomials, and
an identity involving resultants. 

\subsection{Factors of certain differences}

Computations show that for all $n\leq 2000$ the polynomials $P_n(z)$ are 
irreducible, with two exceptions which we will address later in 
Subsection~\ref{sec:6.3}. We also saw in
the previous two sections that in spite of some general results, the zero 
distribution of $P_n(z)$ tends to be quite irregular. 

It is therefore somewhat surprising that certain infinite classes of
differences of the polynomials $P_n(z)$ have the greatest possible regularity,
by being products of cyclotomic polynomials. Computations show, for instance,
\begin{align}
P_{30}(z)-P_{14}(z) &amp;= \big(z^8-z^4+1\big)\big(z^4-z^2+1\big)^2
\big(z^2-z+1\big)^3\big(z^2+z+1\big)^4\label{6.1}\\
&amp;= \Phi_{24}(z)\cdot\Phi_{12}(z)^2\cdot\Phi_6(z)^3\cdot\Phi_3(z)^4,\nonumber
\end{align}
where $\Phi_n(z)$ is the $n$th cyclotomic polynomial, a monic polynomial with
integer coefficients that can be defined by 
\begin{equation}\label{6.1a}
\Phi_n(z) 
= \prod_{\substack{1\leq k\leq n\\ \gcd(k,n)=1}}\left(z-e^{2\pi ik/n}\right).
\end{equation}
Among numerous other properties, the cyclotomic polynomials satisfy
\begin{equation}\label{6.2}
\Phi_p(z) = \sum_{j=0}^{p-1}z^j,\qquad\Phi_{2p}(z)=\sum_{j=0}^{p-1}(-z)^j
\quad(p\geq 3\;\hbox{prime}),
\end{equation}
and if $p$ is an arbitrary prime with $p\nmid r$, then
\begin{equation}\label{6.3}
\Phi_{rp^m}(z) = \Phi_{rp}\big(z^{p^{m-1}}\big).
\end{equation}

We are now ready to state the first result of this section.

\begin{proposition}\label{prop:6.1}
For any integer $m\geq 1$ we have
\begin{align}
P_{2^{m+1}-2}(z)-P_{2^m-2}(z)
&amp;= \prod_{j=0}^{m-1}\Phi_{3\cdot 2^j}(z)^{m-j},\label{6.4} \\
P_{2^{m+1}-1}(z)-P_{2^m-1}(z)
&amp;= z\prod_{j=0}^{m-1}\Phi_{3\cdot 2^j}(z)^{m-j}.\label{6.5}
\end{align}
\end{proposition}

Clearly, \eqref{6.1} is the case $m=4$ of \eqref{6.4}. 
Proposition~\ref{prop:6.1}, in turn, is a special case of a more general class
of identities. In order to simplify notation we set, for integers $\nu\geq 0$
and $m\geq 1$,
\begin{equation}\label{6.6}
a(\nu,m):=\left(3\cdot 2^\nu+1\right)\cdot 2^{m-1}-2,\qquad
b(\nu,m):=\left(3\cdot 2^\nu-1\right)\cdot 2^{m-1}-2.
\end{equation}

\begin{proposition}\label{prop:6.2}
For any pair of integers $\nu\geq 0$ and $m\geq 1$ we have
\begin{equation}\label{6.7}
P_{a(\nu,m)}(z)-P_{b(\nu,m)}(z)
= \left(P_{a(\nu,1)}\big(z^{2^{m-1}}\big)-P_{b(\nu,1)}\big(z^{2^{m-1}}\big)\right)
\prod_{j=0}^{m-2}\Phi_{3\cdot 2^j}(z)^{m-1-j},
\end{equation}
and
\begin{equation}\label{6.8}
P_{a(\nu,m)+1}(z)-P_{b(\nu,m)+1}(z) 
= z\left(P_{a(\nu,m)}(z)-P_{b(\nu,m)}(z)\right).
\end{equation}
\end{proposition}

\begin{proof}
We fix an integer $\nu\geq 0$ and proceed by induction on $m$. When $m=1$, the
two sides of \eqref{6.7} are identical since the empty product is 1 by 
convention. For the induction hypothesis, suppose that \eqref{6.7} holds for a
certain $m$. From the definitions in \eqref{6.6} we obtain
\begin{equation}\label{6.9}
a(\nu,m+1)=2\big(a(\nu,m)+1\big),\qquad b(\nu,m+1)=2\big(b(\nu,m)+1\big).
\end{equation}
We recall \eqref{2.8}, namely
\[
P_{2n}(z) = (1+z)\cdot P_{n-1}(z^2)+P_n(z^2),
\]
and set $n=a(\nu,m)+1$ and $n=b(\nu,m)+1$, respectively. Using \eqref{6.9} and
subtracting the two identities thus obtained, we get
\begin{align}
P_{a(\nu,m+1)}(z)-P_{b(\nu,m+1)}(z)
&amp;=(1+z)\left(P_{a(\nu,m)}(z^2)-P_{b(\nu,m)}(z^2)\right) \label{6.10}\\
&amp;\quad +P_{a(\nu,m)+1}(z^2)-P_{b(\nu,m)+1}(z^2).\nonumber
\end{align}
With \eqref{2.7} we obtain
\begin{align}
P_{a(\nu,m)+1}(z^2)-P_{b(\nu,m)+1}(z^2)
&amp;=z^2\left(P_{a(\nu,m)}(z^2)-P_{b(\nu,m)}(z^2)\right) \label{6.11}\\
&amp;\quad +s(a(\nu,m)+2)-s(b(\nu,m)+2).\nonumber
\end{align}
Since
\[
a(\nu,m)+2=\left(3\cdot 2^\nu+1\right)\cdot 2^{m-1},\qquad
b(\nu,m)+2=\left(3\cdot 2^\nu-1\right)\cdot 2^{m-1},
\]
with \eqref{1.1} and \eqref{2.2a} we see that the two Stern numbers at the 
right of \eqref{6.11} are the same. Hence \eqref{6.11} and \eqref{6.10} give
\begin{equation}\label{6.12}
P_{a(\nu,m+1)}(z)-P_{b(\nu,m+1)}(z)
=\big(1+z+z^2\big)\left(P_{a(\nu,m)}(z^2)-P_{b(\nu,m)}(z^2)\right).
\end{equation}
Now, using \eqref{6.7} as induction hypothesis, we see that 
one factor on the right of \eqref{6.12} will be
\[
P_{a(\nu,1)}\big(z^{2^m}\big)-P_{b(\nu,1)}\big(z^{2^m}\big),
\]
and the remaining factors are
\begin{equation}\label{6.13}
\big(1+z+z^2\big)\cdot\prod_{j=0}^{m-2}\Phi_{3\cdot 2^j}(z^2)^{m-1-j}.
\end{equation}
By \eqref{6.3} with $r=3$ and $p=2$ we have
\[
\Phi_{3\cdot 2^j}(z^2)=\Phi_{3\cdot 2}\big((z^2)^{2^{j-1}}\big)
=\Phi_{3\cdot 2}\big(z^{2^j}\big) = \Phi_{3\cdot 2^{j+1}}(z),
\]
and since $1+z+z^2=\Phi_3(z)$, the product \eqref{6.13} becomes
\[
\Phi_3(z)\cdot\prod_{j=0}^{m-2}\Phi_{3\cdot 2^{j+1}}(z)^{m-1-j}
= \prod_{j=0}^{m-1}\Phi_{3\cdot 2^j}(z)^{m-j}.
\]
Thus we have shown that \eqref{6.7} holds for $m+1$ in place of $m$, which
completes the proof of \eqref{6.7} by induction.

Finally, \eqref{6.8} is just \eqref{6.11} with $z$ in place of $z^2$, if we
recall that the Stern numbers on the right cancel each other. This completes
the proof of Proposition~\ref{prop:6.2}.
\end{proof}

To make \eqref{6.7} more explicit for small values of $\nu$, we set
\begin{equation}\label{6.14}
f_{\nu}(z):= P_{a(\nu,1)}(z)-P_{b(\nu,1)}(z),\qquad \nu\geq 0,
\end{equation}
and display the first three of these polynomials in Table~4. The case $\nu=0$
leads to Proposition~\ref{prop:6.1}.

\begin{table}[h]\label{tab:4}
\centering
\begin{tabular}{|c|l|}
\hline
$\nu$ &amp; $f_{\nu}(z)$  \\
\hline
0 &amp; $z^2+z+1$ \\
1 &amp; $z^5+z^4+z^3+z+1$ \\
2 &amp; $z^{11}+z^{10}+z^9+z^7+z^6-z^4+z^3+2z^2+z-1$ \\
\hline
\end{tabular}
\caption{$f_{\nu}(z)$ for $\nu= 0, 1, 2$}
\end{table}

\begin{proof}[Proof of Proposition~\ref{prop:6.1}]
According to \eqref{6.7} and \eqref{6.14} we need to evaluate
\[
f_0\big(z^{2^{m-1}}\big) = \Phi_3\big(z^{2^{m-1}}\big).
\]
We claim that
\begin{equation}\label{6.15}
\Phi_3\big(z^{2^{m-1}}\big) = \prod_{j=0}^{m-1}\Phi_{3\cdot 2^j}(z),
\qquad m\geq 1.
\end{equation}
This can be shown either by an easy induction based on the factorization
$z^4+z^2+1=(z^2+z+1)(z^2-z+1)$, or we can write \eqref{6.15} explicitly as
\[
z^{2^m}+z^{2^{m-1}}+1=\big(z^2+z+1\big)\big(z^2-z+1\big)\big(z^4-z^2+1\big)
\cdots\big(z^{2^{m-1}}+z^{2^{m-2}}+1\big)
\]
and verify it by ``telescoping" the right-hand side. 

The identity \eqref{6.4} now immediately follows from multiplying the product
on the right of \eqref{6.7} with that on the right of \eqref{6.15}. Finally,
the identity \eqref{6.5} is an immediate consequence of \eqref{6.8} and 
\eqref{6.4}.
\end{proof}

\subsection{A resultant identity}

The resultant of two polynomials is an important expression, with numerous 
applications in algebra and number theory. Given two polynomials
\begin{equation}\label{6.15a}
f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0,\quad 
g(x)=b_mx^m+b_{m-1}x^{m-1}+\cdots+b_0
\end{equation}
($a_n\neq 0, b_m\neq 0$), the resultant of $f$ and $g$ can be defined by
\[
{\rm Res}(f,g)=a_n^mb_m^n\prod_{i=1}^n\prod_{j=1}^m\big(\alpha_i-\beta_j\big),
\]
where $\alpha_1,\ldots, \alpha_n$ and $\beta_1,\ldots, \beta_m$ are the zeros
of $f$ and $g$, respectively. An equivalent definition is by way of the 
Sylvester determinant, the determinant of a certain $(m+n)\times(m+n)$ matrix 
which has the coefficients of $f$ and $g$ as entries. The resultant Res$(f,g)$ 
is therefore a rational integer if $f$ and $g$ have integer coefficients.

We can now state and prove the following result.

\begin{proposition}\label{prop:6.3}
For any integer $n\geq 0$ we have
\begin{equation}\label{6.16}
{\rm Res}\big(P_n(x), P_{n+1}(x)\big) = s(n+2)^n.
\end{equation}
\end{proposition}

\begin{proof}
From among the numerous known properties of the resultant, we require just two.
First if $a$ is a constant and $g$ is a monic polynomial, then
\begin{equation}\label{6.17}
{\rm Res}(a,g) = {\rm Res}(g,a) = a^{\deg{g}}.
\end{equation}
Second, if $f$ and $g$ are polynomials as given in \eqref{6.15}, and if
\begin{equation}\label{6.18}
f(x) = q(x)\cdot g(x)+r(x),
\end{equation}
with polynomials $q$, $r$, and with $\nu:=\deg{r}$, then
\begin{equation}\label{6.19}
{\rm Res}(g,f) = b_m^{n-\nu}{\rm Res}(g,r). 
\end{equation}
A proof of this property can be found, for instance, in \cite[p.~58]{PZ}. We
apply this to \eqref{2.7}, namely $P_{n+1}(x)=xP_n(x)+s(n+2)$; so by
\eqref{6.18} and \eqref{6.19}, followed by \eqref{6.17}, we get
\[
{\rm Res}\big(P_n(x),P_{n+1}\big)={\rm Res}\big(P_n(x),s(n+2)\big)
= s(n+2)^n,
\]
where we have used the fact that $P_n(x)$ is a monic polynomial of degree $n$.
\end{proof}


We were unable to identify or prove general identities between other pairs of 
polynomials $P_n(x)$, or identities for the discriminant of $P_n(x)$.

\subsection{Factors of $P_n(z)$}\label{sec:6.3}

We close this section with some remarks and observations on possible 
cyclotomic factors of the polynomials $P_n(z)$ themselves. Computations show
that $P_{62}(z)$ is divisible by $\Phi_7(z)$ and $P_{1022}(z)$ is divisible 
by $\Phi_{11}(z)$. Noticing the close relationship between the corresponding
subscripts, we were able to find a third factor; see Table~5.

\begin{table}[h]\label{tab:5}
\centering
{\renewcommand{\arraystretch}{1.1}
\begin{tabular}{|r|c|c|}
\hline
$m$ &amp; $n=2^m-2$ &amp; Factor of $P_n(z)$  \\
\hline
6 &amp; 62 &amp; $\Phi_7(z)=(z^7-1)/(z-1)$\\
10 &amp; 1022 &amp; $\Phi_{11}(z)=(z^{11}-1)/(z-1)$\\
12 &amp; 4094 &amp; $\Phi_{13}(z)=(z^{13}-1)/(z-1)$\\
\hline
\end{tabular}}
\caption{All known factors of $P_n(z)$}
\end{table}

To analyze this situation, we let $m=p-1$, where $p$ is an odd prime. Then
$\Phi_p(z)$ dividing $P_n(z)$ is equivalent to $P_n(z)\equiv 0\pmod{z^p-1}$,
where $n=2^{p-1}-2$. In other words, if we reduce all exponents of $z$ in 
$P_n(z)$ modulo $p$, then we should get a multiple of $\Phi_p(z)$. In the case
$p=7$, $n=62$, the reduced polynomial is then
\begin{align}
P_{62}(z)&amp;\equiv \sum_{j=1}^7\big(s(j)+s(7+j)+\cdots+s(56+j)\big)z^{7-j}
\pmod{z^p-1} \label{6.16a}\\
&amp;= 52z^6+52z^5+\cdots+52z+52,\nonumber
\end{align}
which is indeed a multiple of $\Phi_7(z)$. This approach works because all 
seven sums of
Stern numbers in arithmetic progressions in the first row of \eqref{6.16a} are
identical. It is not difficult to generalize this to arbitrary odd primes $p$.

\begin{proposition}\label{prop:6.4}
For an odd prime $p$, let $n=2^{p-1}-2$. Then $\Phi_p(z)$ divides $P_n(z)$ if
and only if the $p$ identities
\begin{equation}\label{6.17a}
\sum_{k=0}^{q(p)-1}s(kp+j) = \frac{3^{p-1}-1}{2p},\qquad j=1, 2, \ldots, p,
\end{equation}
all hold, where $q(p):=(2^{p-1}-1)/p$ is the base-$2$ Fermat quotient.
\end{proposition}

Before we prove this proposition, we note that by Fermat's little theorem both
the Fermat quotient $q(p)$ and the right-hand side of \eqref{6.17a} are integers.
This last quotient is in fact $1/2$ the base-3 Fermat quotient. When $p=7$, we
have $q(7)=9$ and $(3^{p-1}-1)/2p=52$, consistent with \eqref{6.16a}.

It is rather surprising that the very strong condition \eqref{6.17a} holds for
the primes $p=7$, 11, and 13. For $p=5$, on the other hand, the five sums are
8, 6, 8, 8, and 10. Hence \eqref{6.17a} fails for $j=2$ and $j=5$ and 
accordingly, $P_{14}(z)$ is not divisible by $\Phi_5(z)$. 

We have used \eqref{6.17a} as a criterion to exclude the cases $p=17$, 19, 23,
and 29. Obviously it suffices to find just one $j$ for which \eqref{6.17a}
fails. 

\begin{proof}[Proof of Proposition~\ref{prop:6.4} (Sketch)]
By reducing the exponents of $z$ modulo $p$ in the polynomial $P_n(z)$, we see
that each sum on the left of \eqref{6.17a} is a coefficient of $z^{p-j}$, 
$1\leq j\leq p$. There are a total of $n+1=2^{p-1}-1$ coefficients, broken into
$p$ sums, each with $(n+1)/p=q(p)$ summands. By \eqref{2.9}, the sum of all 
coefficients of $P_n(z)$ is
\[
P_{2^{p-1}-2}(1) = \frac{1}{2}\left(3^{p-1}+1\right)-1.
\]
So for all the sums in \eqref{6.17a} to be equal, the right-hand side has to
be $(3^{p-1}-1)/2p$, as claimed.
\end{proof}
 
\section{The polynomials $P_n(z)$ modulo 2}\label{sec:7} 

In this section we consider the polynomials $\overline{P}_n(z)$ that are 
formed from $P_n(z)$ by taking the smallest nonnegative residues modulo 2 of
their coefficients. Strictly speaking, this does not rely on the deeper
properties of the Stern sequence since we have the well-known congruence
\begin{equation}\label{7.1}
s(k)\equiv\begin{cases}
0\pmod{2}, &amp;\hbox{if}\; 3\mid k;\\
1\pmod{2}, &amp;\hbox{if}\; 3\nmid k.
\end{cases}
\end{equation}
This can be obtained by an easy induction, using \eqref{1.1}. 

\subsection{Divisibility and irreducibility}
In spite of the easy coefficient sequence, the
polynomials $\overline{P}_n(z)$ have some interesting properties, and they
behave very differently according as $3\mid n$ or $3\nmid n$. This difference
is illustrated in Table~6.

\begin{table}[h]\label{tab:6}
\centering
{\renewcommand{\arraystretch}{1.2}
\begin{tabular}{|r|l|}
\hline
$n$ &amp; $\overline{P}_n(z)$ factored \\
\hline
9 &amp; $z^9+z^8+z^6+z^5+z^3+z^2+1$ \\
10 &amp; $\Phi_{12}(z)\cdot\Phi_6(z)\cdot\Phi_4(z)\cdot\Phi_2(z)^2$ \\
11 &amp; $\Phi_{12}(z)\cdot\Phi_6(z)\cdot\Phi_4(z)\cdot\Phi_2(z)^2\cdot z$ \\
\hline
\end{tabular}}
\caption{$\overline{P}_n(z)$ for $n=9, 10, 11$}
\end{table}

We begin with the two cases that are quite straightforward.

\begin{proposition}\label{prop:7.1}
For $k\geq 2$ we have
\begin{equation}\label{7.2}
\overline{P}_{3k+1}(z)=(z+1)^2\prod_{\substack{d\mid 3k+3\\d&gt;3}}\Phi_d(z),\quad
\overline{P}_{3k+2}(z)=z(z+1)^2\prod_{\substack{d\mid 3k+3\\d&gt;3}}\Phi_d(z).
\end{equation}
\end{proposition}

\begin{proof}
By \eqref{1.3} and \eqref{7.1} we have
\begin{align*}
\overline{P}_{3k+1}(z) &amp;= (z+1)(z^{3k}+z^{3(k-1)}+\cdots z^3+1)\\
&amp;= (z+1)\cdot\frac{z^{3k+3}-1}{z^3-1}.
\end{align*}
Now we use the well-known fact that
\begin{equation}\label{7.3}
z^{3k+3}-1 = \prod_{d\mid 3k+3}\Phi_d(z).
\end{equation}
Dividing both sides by $z^3-1=\Phi_1(z)\Phi_3(z)$ removes the cases $d=1, 3$
from \eqref{7.3}, and we also take out the factor $\Phi_2(z)=z+1$ and obtain
the first identity in \eqref{7.2}. 
The case $3k+2$ is completely analogous. Alternatively, we could also use
\eqref{2.6}.
\end{proof}

In contrast to Proposition~\ref{prop:7.1}, when $3\mid n$ we have the following
result.

\begin{proposition}\label{prop:7.2}
For any $k\geq 1$, the polynomial $\overline{P}_{3k}(z)$ is irreducible over
${\mathbb Q}$.
\end{proposition}

We will prove this by applying a recent theorem of Sawin, Shusterman, and Stoll
\cite{SSS}. To set the stage, we recall that the reciprocal of a polynomial 
$h(z)$ is defined by $h^*(z)=z^{\deg{h}}h(1/z)$. Then a polynomial $h$ is 
called {\it reciprocal} if $h(z)=\pm h^*(z)$. Furthermore, if 
$h\in{\mathbb R}[z]$ is given by $h(z)=a_nz^n+\cdots+a_1z+a_0$, we define the
norm $\|h\|$ as the $\ell_2$-norm of its coefficient sequence, namely
\[
\|h\| =\left(a_n^2+\cdots+a_1^2+a_0^2\right)^{1/2}.
\]
Finally, we require the rather technical condition of a robust pair of 
polynomials $(f(z),g(z))$$\in{\mathbb Z}[z]^2$ with $f(0)g(0)\neq 0$. Such a pair
is called {\it robust} if for every factorization 
\[
f_0(z)g_0(z) = f(z)g(z),
\]
one of several conditions is satisfied; see \cite{FLPS, SSS}. In what
follows, we need only one of these conditions, namely
\begin{equation}\label{7.4}
\|f_0\|^2+\|g_0\|^2 &gt; \|f\|^2+\|g\|^2.
\end{equation}
We can now state the relevant result from \cite{SSS}, as quoted in \cite{FLPS}.

\begin{theorem}[Sawin, Shusterman, and Stoll]\label{thm:7.3}
Let $f(z), g(z)$ be polynomials with integer coefficients such that 
$f(0)g(0)\neq 0$, and suppose that
$(f^*,g)$ is a robust pair, $f^*(z)\neq\pm g(z)$ and 
$\gcd_{{\mathbb Z}[z]}(f,g)=1$. If 
\[
n&gt;(1+\deg{f}+\deg{g})\cdot 2^{\|f\|^2+\|g\|^2}-\deg{f},
\]
then the non-reciprocal part of $f(z)z^n+g(z)$ is irreducible over 
${\mathbb Z}[z]$.
\end{theorem}

This theorem is just one of a class of similar irreducibility results, going
back to Schinzel in the 1960s; see \cite{FLPS, SSS} for further 
remarks and references.

\begin{proof}[Proof of Proposition~\ref{prop:7.2}]
If $3\mid n$, then by \eqref{1.3} and \eqref{7.1} we have
\[
\overline{P}_n(z)=z^n+z^{n-1}+z^{n-3}+\cdots+z^3+z^2+1,
\]
and it is easy to verify that
\begin{equation}\label{7.5}
\big(z^3-1\big)\overline{P}_n(z) = (z+1)z^{n+2}-\big(z^2+1\big).
\end{equation}
Therefore the reciprocal part of the quadrinomial on the right of \eqref{7.5}
is $z^3-1$, while the non-reciprocal part is $\overline{P}_n(z)$.

We now apply Theorem~\ref{thm:7.3} with $f(z)=z+1$ and $g(z)=-z^2-1$. Then
$f^*(z)=f(z)$, $f(0)g(0)\neq 0$, $f^*(z)\neq\pm g(z)$, and
$\gcd_{{\mathbb Z}[z]}(f,g)=1$. Furthermore, $\|f\|^2=\|g\|^2=2$, and the only
other essential factorization of $f(z)g(z)$ is 
\[
f_0(z)g_0(z) = 1\cdot\big(-z^3-z^2-z-1\big),
\]
and we have
\[
\|f_0\|^2+\|g_0\|^2=1+4 &gt; 2+2 = \|f\|^2+\|g\|^2.
\]
Hence by Theorem~\ref{thm:7.3} the non-reciprocal part of the polynomial in
\eqref{7.5}, namely $\overline{P}_n(z)$, is irreducible whenever
\[
n+2 &gt; (1+1+2)\cdot 2^{2+2}-1 = 4\cdot2^4-1 = 63,
\]
which proves our proposition for $n\geq 63$ since in our case we have $3\mid n$.
For smaller $n$ we used computer algebra to show that $\overline{P}_n(z)$ is
irreducible.
\end{proof}

\subsection{Zero distribution}

When $n\equiv 1, 2\pmod{3}$, by \eqref{7.2} and the definition \eqref{6.1a} of
cyclotomic polynomials, the zeros of $\overline{P}_n(z)$ are completely 
determined. The case $3|n$ is more challenging. We begin with a first rough 
estimate of the moduli of all the zeros, which will be required to obtain
sharper results.

\begin{lemma}\label{lem:7.4}
For all $k\geq 1$, the zeros of $\overline{P}_{3k}(z)$ lie in the annulus
$1/2\leq|z|\leq 2$.
\end{lemma}

\begin{proof}
The upper bound follows immediately from Cauchy's Theorem~\ref{thm:4.2} since 
all coefficients are 0 or 1. The lower bound follows similarly by considering
the reciprocal $z^{3k}\overline{P}_{3k}(1/z)$.
\end{proof} 

For the next results it is convenient to consider the polynomial on the right
of \eqref{7.5}, and we set
\begin{equation}\label{7.6}
f_n(z):=(z+1)z^{n+2} - (z^2+1).
\end{equation}
Although only the case $3|n$ is relevant with regards to 
$\overline{P}_n(z)$, we may as well consider the polynomials $f_n(z)$ for all
integers $n$. The conclusion of Lemma~\ref{lem:7.4} also holds for all these
polynomials.

\begin{proposition}\label{prop:7.5}
For all $n\geq 1$, the zeros of $f_n(z)$ satisfy
\begin{equation}\label{7.7}
1-\frac{\log(n+2)}{n+2} &lt; |z| &lt; 1+\frac{\log(n+2)}{n+2},
\end{equation}
where for odd $n$ the upper bound holds only for $n\geq 17$.
\end{proposition}

\begin{proof}
(1) When $r:=|z|&lt;1$, then by \eqref{7.6} we get
\begin{equation}\label{7.8}
\left|f_n(z)\right|\geq \left|z^2+1\right|-|z+1|\cdot r^{n+2}.
\end{equation}
Using the triangle inequality again, we get $|z^2+1|\geq 1-r^2$ and 
$|z+1|\leq 1+r$. If we make the assumption that $r\leq 1-\log(n+2)/(n+2)$,
then \eqref{7.8} gives
\begin{align*}
\left|f_n(z)\right|\geq 1
&amp;-\left(1-2\frac{\log(n+2)}{n+2}+\left(\frac{\log(n+2)}{n+2}\right)^2\right)\\
&amp;\qquad-\left(2+\frac{\log(n+2)}{n+2}\right)
\left(1-\frac{\log(n+2)}{n+2}\right)^{n+2}.
\end{align*}
The power in the last term is less than $\exp(-\log(n+2))=1/(n+2)$, and thus
\begin{align*}
\left|f_n(z)\right| 
&amp;&gt;2\frac{\log(n+2)}{n+2}-\left(\frac{\log(n+2)}{n+2}\right)^2-\frac{2}{n+2}
-\frac{\log(n+2)}{(n+2)^2} \\
&amp;=\frac{\log(n+2)-2}{n+2}+\frac{\log(n+2)}{n+2}\left(1-\frac{\log(n+2)}{n+2}
-\frac{1}{n+2}\right).
\end{align*}
Now $\log(n+2)-2&gt;0$ for $n\geq 6$, and it is easy to verify that the expression
in large parentheses on the right is positive for all $n\geq 1$. Hence 
$|f_n(z)|&gt;0$ whenever $r\leq 1-\log(n+2)/(n+2)$, which proves the lower bound
in \eqref{7.7} for $n\geq 6$. The cases $1\leq n\leq 5$ can be verified by
computations.

(2) When $r=|z|&gt;1$, then from \eqref{7.6} we get
\begin{equation}\label{7.9}
\left|f_n(z)\right|\geq |z+1|\cdot r^{n+2}-\left|z^2+1\right|.
\end{equation}
We also have $|z+1|\geq r-1$ and $|z^2+1|\leq r^2+1\leq 5$ since by 
Lemma~\ref{lem:7.4} we may restrict our attention to $r\leq 2$. Now, if
$r\geq 1+\log(n+2)/(n+2)$, then by \eqref{7.9} we get
\begin{equation}\label{7.10}
\left|f_n(z)\right|\geq\frac{\log(n+2)}{n+2}\cdot
\left(1+\frac{\log(n+2)}{n+2}\right)^{n+2}-5.
\end{equation}
It can be shown by analytical methods, complemented by some computations for
smaller cases, that the sequence
\[
c_n:=\frac{1}{n+2}\cdot\left(1+\frac{\log(n+2)}{n+2}\right)^{n+2}
\]
is increasing for $n\geq 7$. Furthermore, for $n\geq 200$ we have
$c_n\geq c_{200}&gt;0.9337$. So, if we assume that $r\geq 1+\log(n+2)/(n+2)$, then
by \eqref{7.10} we have
\[
\left|f_n(z)\right|\geq \log(n+2)\cdot 0.9337-5 &gt; 0
\]
when $n\geq 210$. The cases $1\leq n\leq 209$ can again be checked by direct
computation.
\end{proof}

\begin{remark}\label{rem:2}
With additional effort, Proposition~\ref{prop:7.5} can be
improved as follows:
\leavevmode
\begin{itemize}
\item[(a)] When $0\leq|\arg{z}|\leq 2\pi/3$, then the upper bound in 
\eqref{7.7} can be decreased to 1, with equality only for $z=1$ and 
for $z=e^{\pm 2\pi i/3}$ when $3|n$.
\item[(b)] When $2\pi/3\leq|\arg{z}|\leq\pi$, then the lower bound in 
\eqref{7.7} can be increased to 1.
\end{itemize}
\end{remark}

We conclude this section with a result on real zeros.

\begin{proposition}\label{prop:7.6}
For any $n\geq 0$, the polynomial $f_n(z)$ has
\leavevmode
\begin{itemize}
\item[(a)] exactly one positive real zero, $z_1=1$;
\item[(b)] exactly one negative real zero $-1-\frac{\log{n+1}}{n+1}&lt;z_2&lt;-1$ 
when $n$ is odd, where the left inequality holds for $n\geq 17$;
\item[(c)] no negative real zeros when $n$ is even.
\end{itemize}
As a consequence, when $k$ is odd, $\overline{P}_{3k}(z)$ has exactly the one 
real zero given by (b) with $n=3k$, and when $k$ is even, it has no real zeros.
\end{proposition}

\begin{proof}
(a) $z_1=1$ is obviously a zero. But since the sequence of coefficients of
$f_n(z)=z^{n+3}+z^{n+2}-z^2-1$ has exactly one sign change, by Descartes's Rule
of Signs (see, e.g., \cite[pp.~439--443]{He}) there can be only this one
positive zero.

(b) For odd $n$, $f_n(z)$ is positive for sufficiently large negative $z$, 
while $f_n(-1)=-1$. Hence there is a zero $z_2&lt;-1$, and by \eqref{7.7} it must
lie to the right of $-1-\log(n+1)/(n+1)$, for $n\geq 17$. Since
$f_n(-z)=z^{n+3}-z^{n+2}-z^2-1$, there is just one sign change in the
coefficient sequence of $f_n(-z)$, and again by Descartes's Rule of Signs there
can be no further negative real zeros.

(c) When $n$ is even, we consider the derivative
\begin{equation}\label{7.11}
f'_n(z) = z\cdot\left((n+3)z^{n+1}+(n+2)z^n-2\right),
\end{equation}
and we set $g_n(z):=(n+3)z^{n+1}+(n+2)z^n$. It is easy to see that, at least for
negative $z$, this polynomial has a unique maximum $z_0:=-n(n+2)/((n+1)(n+3))$,
and
\[
g_n(z_0) = z_0^n\left(-\frac{n(n+2)}{n+1}+n+2\right)= z_0^n\cdot\frac{n+2}{n+1}.
\]
It is easy to verify that
\[
\frac{n(n+2)}{(n+1)(n+3)}\cdot\frac{n+2}{n+1}&lt;1\quad\hbox{for all}\quad n\geq 0,
\]
and so we have $g_n(z_0)&lt;1$ and thus $g_n(z_0)-2&lt;0$ for all even $n$. Therefore
by \eqref{7.11} we have $f'_n(z)&gt;0$ for all $z&lt;0$. Since $f_n(0)=-1$, this means
that $f_n(z)&lt;0$ for all $z\leq 0$, which implies part (c).

Finally, the last statement of the proposition follows from \eqref{7.5} and the
fact that $\overline{P}_{3k}(z)$ cannot have positive zeros since it has only 
nonnegative coefficients.
\end{proof}

\section{Overpartitions and colored partitions}\label{sec:8}

In this final section we consider the polynomials $P_n(z)$, as defined in 
\eqref{1.3}, in the framework of binary and $b$-ary partitions \cite{CSe, DE9}, 
where $b\geq 2$ is an integer base, and more generally in connection with 
overpartitions and colored partitions.

\subsection{Binary and $b$-ary overpartitions}
R{\o}dseth and Sellers 
\cite{RS} introduced and studied $b$-ary overpartitions, in analogy to ordinary 
overpartitions that had been introduced a little earlier by Corteel and 
Lovejoy \cite{CL}.

A {\it $b$-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 \cite{RS}.

\begin{example}\label{ex:1}
\cite[p.~346]{RS}. The binary overpartitions of $n=4$ are

\noindent
$4,\;\overline{4},\;2+2,\;\overline{2}+2,\;2+1+1,\;2+\overline{1}+1,\;
\overline{2}+1+1,\;\overline{2}+\overline{1}+1,\; 1+1+1+1,\;
\overline{1}+1+1+1.$
Thus $\overline{S}_2(4)=10$.
\end{example}

As seen 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. 

The concept of a $b$-ary overpartition can be restricted in different ways,
one of which is to restrict by an integer $\lambda$ the number of times a 
non-overlined power of $b$ may occur. Then the generating function is
\begin{equation}\label{8.2}
\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),
\end{equation}
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. 

\begin{example}\label{ex:2}
Let $b=\lambda=2$. Then \eqref{8.2} becomes
\begin{equation}\label{8.3}
1+2q+4q^2+5q^3+8q^4+10q^5+13q^6+14q^7+18q^8+21q^9+26q^{10}+\cdots
\end{equation}
Thus, in particular, $\overline{S}_2^2(4)=8$, which is consistent with 
Example~\ref{ex:1}, where all but the last two binary overpartitions are 
counted by $\overline{S}_2^2(4)$.
\end{example}

In the recent paper \cite{DE10} 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.

In the special case of restricted $b$-ary overpartitions with $\lambda=2$, these
polynomials 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
\cite[Def.~2.4]{DE10}, we define
\begin{equation}\label{8.4}
\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).
\end{equation}
Comparing this with \eqref{8.2}, we immediately get, for any base $b\geq 2$,
\begin{equation}\label{8.5}
\overline{S}_b^2(n)=\Omega_{b,T}^{(1,2)}(n;1,1,1),\qquad n=0, 1, 2,\ldots,
\end{equation}
where the triple $T$ is arbitrary. The following result establishes a 
connection to the polynomials $P_n(x)$ defined in \eqref{1.3}, and gives a 
combinatorial interpretation.

\begin{proposition}\label{prop:8.1}
For any $n\geq 0$ we have
\begin{equation}\label{8.6}
P_n(x) = \Omega_{2,T}^{(1,2)}(n;x,1,1),
\end{equation}
where $T=(2,s,t)$, with $s$ and $t$ arbitrary. Furthermore,
\leavevmode
\begin{itemize}
\item[(a)] $P_n(1)$ is the number of restricted binary overpartitions of $n$,
with $\lambda=2$.
\item[(b)] The coefficient of $x^j$ in $P_n(x)$ is the number of overlined
parts that sum to~$j$.
\end{itemize}
\end{proposition}

\begin{proof}
With $b=r=2$ and $y=z=1$, the right-hand side of \eqref{8.4} becomes
\[
\prod_{j=0}^\infty\left(1+x^{2^j}q^{2^j}\right)
\prod_{j=0}^\infty\left(1+q^{2^j}+q^{2^{j+1}}\right).
\]
By the uniqueness of the binary expansion, the first product gives
\[
\prod_{j=0}^\infty\left(1+(xq)^{2^j}\right) = \sum_{k=0}^\infty(xq)^k
=\frac{1}{1-xq}.
\]
The identity \eqref{8.6} now follows from Proposition~\ref{prop:2.6} by
equating coefficients of $q^n$.

(a) This follows from combining \eqref{8.6} with \eqref{8.5}.

(b) When we remove the overlined parts that sum to $j$ from a binary 
overpartition of $n$, we are left with a restricted binary partition of $n-j$
that can have at most $\lambda=2$ equal parts. But this is a hyperbinary 
expansion of $n-j$, and we know (see, e.g., \cite[p.~470]{Re}) that this 
number is $s(n-j+1)$. This fact, together with the definition \eqref{1.3},
proves part (b).
\end{proof}

\subsection{Generalizations and colored partitions}
By considering \eqref{8.6} in conjunction with \eqref{8.4}, it
becomes clear that the polynomials $P_n(x)$ can be generalized in several 
different directions. We could take different bases $b$, or different values
of $\lambda$, or both. For instance, when $b=\lambda\geq 2$ and $r=2$, we get 
(in a simplified notation) 
\begin{equation}\label{8.7}
\sum_{n=0}^{\infty}P^{(b)}_n(x)q^n
=\prod_{j=0}^\infty\left(1+x^{2^j}q^{b^j}\right)
\left(1+q^{b^j}+q^{2\cdot b^j}+\cdots+q^{b\cdot b^j}\right).
\end{equation}
Then the polynomials $P^{(b)}_n(x)$ are analogues of $P_n(x)$, with
the $b$-generalized Stern numbers (as defined in \cite{De}; see also \cite{DE9})
in place of the Stern numbers $s(n)$.

Another level of generalization would be to consider polynomial analogues of
restricted colored $b$-ary partitions, as investigated in \cite{DE10}. As a
specific example we mention 
\begin{equation}\label{8.8}
\sum_{n=0}^{\infty}{\widetilde P}_n(x)q^n
=\prod_{j=0}^\infty\left(1+x^{2^j}q^{2^j}\right)\left(1+q^{2^j}\right)
\left(1+q^{2^j}\right)\left(1+q^{2^j}+q^{2^{j+1}}\right).
\end{equation}
In analogy to Proposition~\ref{prop:8.1}(b), ${\widetilde P}_n(1)$ is the number
of ways $n$ can be written as a sum of powers of 2, where each part can be in
one of four colors, with the added conditions that each part in the first 
three colors can occur only once, while a part in the fourth color can occur
at most twice.

The generating function \eqref{8.8} also shows that the coefficients of 
${\widetilde P}_n(x)$ can be obtained by starting with the Stern sequence $s(n)$
and forming the sequence of partial sums twice:
\begin{center}
\begin{tabular}{rrrrrrrrrrr}
1 &amp; 1 &amp; 2 &amp; 1 &amp; 3 &amp; 2 &amp; 3 &amp; 1 &amp; 4 &amp; 3 &amp; $\ldots$ \\
1 &amp; 2 &amp; 4 &amp; 5 &amp; 8 &amp;10 &amp;13 &amp;14 &amp;18 &amp;21 &amp; $\ldots$ \\ 
1 &amp; 3 &amp; 7 &amp;12 &amp;20 &amp;30 &amp;43 &amp;57 &amp;75 &amp;96 &amp; $\ldots$
\end{tabular}
\end{center}
so that
\[
{\widetilde P}_n(x)=x^n+3x^{n-1}+7x^{n-2}+12x^{n-3}+20x^{n-4}+30x^{n-5}+
43x^{n-6}+\cdots,
\]
where we leave the details to the reader.

Each of the sequences of polynomials $P^{(b)}_n(x)$, ${\widetilde P}_n(x)$,
or others of this type, would lead to questions and results that are similar in
nature to the contents of this paper.


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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B83; Secondary 30C15, 12D05.

\noindent \emph{Keywords: }
Stern sequence, polynomial, zero distribution, polynomial factor, 
irreducibility, $b$-ary overpartition.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000045}, \seqnum{A001045}, \seqnum{A002487}, 
\seqnum{A011655}, \seqnum{A048573}, and \seqnum{A174868}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  July 14 2021;
revised version received  November 24 2021.
Published in {\it Journal of Integer Sequences},  November 25 2021.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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\end{document}


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