Properties of Trinomials of Height at least 2

V. Flammang and P. Voutier

This paper is concerned with trinomials of the form $z^{n}+az^{m}+b$, where
$0<m<n$ are relatively prime integers and $a$ and $b$ are non-zero complex
numbers. Typically (but not exclusively), $a$ will be an integer with $|a| \geq 2$,
while $b=\pm 1$. Our main results cover the irreducibility, Mahler measure and
house of such trinomials.

§ INTRODUCTION

Trinomials have long been of interest to algebraists and number theorists.

For example, in 1907, O. Perron [23] proved that the trinomial $z^{n}+az \pm 1$
is irreducible over $\bbQ$ for $|a| \geq 3$ and his results have been generalised
since then. Still, our understanding of the irreducibility of such polynomials is
weak. Theorem <ref> below improves our knowledge when $b=\pm 1$
and we have substantial evidence that the truth is captured in Conjecture <ref>

The diophantine properties of trinomials are also of considerable importance.
C.J. Smyth proved that if $P$ is not a reciprocal polynomial, then the Mahler
measure of $P$ is bounded below by $\theta_{0}=1.324717\ldots$, the real zero
of the trinomial $z^{3}-z-1$. Furthermore, it is known that $\theta_{0}$ is also
the smallest Pisot number. The Mahler measure of $P$ is
also the smallest limit point of the Mahler measure of trinomials in $\bbZ[z]$
and one of the smallest known limit points for general polynomials in $\bbZ[z]$
See Subsections <ref> and <ref> for our results on
the Mahler measure of trinomials and, the related quantity, their house.

§.§ Irreducibility of trinomials $z^{n}+az^{m} \pm 1 \in \bbZ[z]$

We have the following result, generalising the result of Perron cited above.

If $a$, $m$ and $n$ are integers satisfying $n \geq 3$, $0<m<n$, $\gcd(m,n)=1$
and $|a| \geq n^{2}/3$, then $x^{n}+ax^{m}\pm 1$ is irreducible over $\bbQ$.

In fact, from calculations we have performed, it appears that the following
much stronger conjecture is true. As in Perron's result, we appear to have
irreducibility once $|a|$ is greater than a small absolute constant regardless
of the degree of the middle term.

If $a$, $m$ and $n$ are integers satisfying $n \geq 3$, $0<m<n$, $\gcd(m,n)=1$
and $|a| \geq 5$, then $x^{n}+ax^{m}\pm 1$ is irreducible over $\bbQ$.

Furthermore, there are only finitely many reducible polynomials
of the form $x^{n}+ax^{m}\pm 1$ with $|a|=3$, $4$.
They are $x^{8} \pm 3x^{3}-1$, $x^{8} \pm 3x^{5}-1$,
$x^{13}+3x^{4}-1$, $x^{13}-3x^{4}+1$,
$x^{13}-3x^{6}-1$, $x^{13}+3x^{6}+1$,
$x^{13}+3x^{7} \pm 1$,
$x^{13}-3x^{9} \pm 1$,
$x^{14} \pm 4x^{5}-1$ and $x^{14} \pm 4x^{9}-1$.

Note that the condition that $\gcd(m,n)=1$ in both Theorem <ref>
and Conjecture <ref> is necessary. Examples from
Bremner [6] like $x^{33}+67x^{11}+1=\left( x^{3}+x+1 \right) \left(
x^{30}-\cdots -1 \right)$ in his Theorem on pages 153–154 and
$x^{6}+ \left( 4\mu^{4}-4\mu \right)x^{2}-1
= \left( x^{3}+2\mu x^{2}+2\mu^{2}x+1 \right)
\left( x^{3}-2\mu x^{2}+2\mu^{2}x-1 \right)$ for integers $\mu \neq 0,1$
in his Postscript on page 154 demonstrate this.

§.§ Mahler Measure of trinomials $z^{n}+az^{m}+b \in \bbC[z]$

The Mahler measure of a polynomial $\d P(z)=a_{0}z^{n}+ \cdots +a_{n} = a_{0} \prod_{j=1}^{n} \left( z- \alpha_{j} \right) \in \bbC[z]$,
with $a_{0} \neq 0$, as defined by D. H. Lehmer [19] in 1933,

\[
\mHgt(P)= \left| a_{0} \right| \prod_{j=1}^{n} \max \left( 1, \left| \alpha_{j} \right| \right).
\]

Let $\alpha$ be a nonzero algebraic number. The Mahler measure of $\alpha$ is the Mahler measure of its minimal polynomial.

From a result of Kronecker, we know that, if $P$ is the minimal polynomial of an
algebraic integer then $M(P)=1$ if and only if $P$ is a cyclotomic polynomial.
D. H. Lehmer asked: does there exist a constant $c_{0} >0$ such that $M(P) > 1+ c_{0}$
for all $P$ not cyclotomic? The smallest known Mahler measure was found by
D. H. Lehmer himself. It is the Mahler measure of the polynomial
$z^{10}+z^{9}+z^{7}-z^{6}-z^{5}-z^{4}-z^{3}+z+1$ for which the Mahler measure
is $1.176280\ldots$.

A polynomial $P$ is reciprocal if $z^{n}P( 1/z) = P(z)$ and an algebraic
number is reciprocal if its minimal polynomial is reciprocal. As stated
above, C.J. Smyth <cit.> solved Lehmer's problem when $P$ is not reciprocal.
In the general case, the
best known result is due to the second author [35], who proved that if $\alpha$ is
an algebraic integer of degree $n \geq 2$ and not a root of unity, then

\[
\mHgt(\alpha) \geq 1+ \frac{1}{4} \left( \frac{\log \log n}{\log n} \right)^{3}.
\]

In 2013, the first author [11] studied the Mahler measure of trinomials of
height $1$ and gave two criteria to identify those trinomials whose Mahler measure
is less than $1.381356\ldots=\mHgt \left( 1+z_{1}+z_{2} \right)$. In this same
work, she was able to prove a conjecture of Smyth on the Mahler measure of such
trinomials for $n$ sufficiently large compared to $m$. Stankov [32]
was interested in trinomials of the type $z^{n}-az-1$ with $a \in (0,2]$ and he
presented the explicit expression by an integral of the limit of their Mahler
measure when $n$ tends to $\infty$. We generalise this result to any trinomial
of the form $z^{n}+az^{k}+b \in \bbC[z]$ in Theorem <ref>. In 2016,
J-L. Verger-Gaugry [34] studied
the family of trinomials $z^{n}+z-1$ for which he gave the asymptotic expansion
of the Mahler measure as a function of $n$ only. Here, we prove the following

Let $P(z)=z^{n}+az^{m}+b \in \bbC[z]$ with $ab \neq 0$.

* If $|a|-|b| \geq 1$, then $\d \lim_{n \rightarrow \infty} M(P) = |a|$.

* If $|b|-|a| \geq 1$, then $M(P) = |b|$.

* If $|a| + |b| \leq 1$, then $M(P)=1$.

* If $||a| -|b|| < 1 < |a|+ |b|$, then

\[
\lim_{n \rightarrow \infty} M(P) = \exp \left( \frac{1}{2 \pi} \int_{0}^{\gamma} \ln \left( |a|^{2}+2|ab| \cos (t) +|b|^{2} \right) dt \right),
\]

where we put $\gamma=\arccos \left( \dfrac{1-|a|^{2}-|b|^{2}}{2|ab|} \right)$.

As a result of the following theorem, we have exact expressions for $M(P)$ in
all cases in Theorem <ref> except (d).

Let $a$ and $b \neq 0$ be fixed complex numbers with $|a|-|b| \geq 1$.
For all integers $m$ and $n$ with $0<m<n$ and $\gcd(m,n)=1$,

we have

\begin{equation}
\label{eq:mahler-S}
\log M \left( z^{n}+az^{m}+b \right)
= \log |a| - \sum_{k \geq 1} \frac{1}{km} (-1)^{kn} \binom{kn-1}{km-1} \Real \left( b^{-km} (b/a)^{kn} \right).
\end{equation}

This behaviour is quite different from what happens when $a, b=\pm 1$.
In 2007, Duke [10] showed, for $0<m<n$ with $\gcd(m,n)=1$, that

\[
\log M \left( z^{n}+z^{m}+1 \right) = \log M(x+y+1)+\frac{c(n,m)}{n^{2}}
+ O \left( \frac{m}{n^{3}} \right),
\]

where $c(n,m)=-\pi \sqrt{3}/6$ if $3$ divides $m+n$ and $c(n,m)=\pi \sqrt{3}/18$
otherwise. The first author [11] proved that similar results hold for the other
trinomials with $\pm 1$ as their coefficients and used them as mentioned above.

§.§ House of trinomials $z^{n}+az^{m} \pm 1 \in \bbZ[z]$

Let $\alpha$ be a nonzero algebraic integer of degree $n$, with conjugates
$\a_{1}=\a,\ldots,\a_{n}$ and minimal polynomial, $P$. The house of $\a$
(and of $P$) is defined by:

\[
\maison{\a}=\d \max_{1 \leq i \leq n} \left| \a_{i} \right|.
\]

We have the inequality: $\maison{\a} \geq \mHgt(\a)^{1/n}$. In 1965, A. Schinzel
and H. Zassenhaus [29] conjectured that there exists a constant $c>0$ such
that if $\a$ is not a root of unity then $\maison{\a} \geq 1 + c/n$. Thanks to
the polynomial $P(x)= x^{n}-2$, we see that $c \leq \log(2)$. In 1985, a result
of C.J. Smyth [30] led D. Boyd [3] to conjecture that $c$ should be
equal to $3/2 \log \t_{0}$ where $\t_{0}=1.324717\ldots$, and that this value
is attained too, via the polynomial $x^{3n}+x^{2n}-1$. The second author [35]
proved that, if $\a$ is an algebraic integer of degree $n \geq 3$, not a root
of unity, then

\[
\maison{\alpha} \geq 1+\frac{1}{2n} ( \log \log n / \log n)^{3}.
\]

In 1991, E.M. Matveev [20] proved that, if $\a$ is an algebraic integer of
degree $n \geq 2$, not a root of unity, then $\maison{\a} \geq \exp \left( \log \left( n +0.5 \right) /n^{2} \right)$.
Until Dimitov's very recent result (see below), the best-known asymptotic result
was given by A. Dubickas [9]:

\[
\maison{\alpha} > 1+\frac{1}{n} \left( 64/\pi^{2}-\epsilon \right) \left( \log\log n/\log n \right)^{3}
\quad \text{for } n>n_{0}(\epsilon).
\]

In 2007, G. Rhin and Q. Wu [26] verified the conjecture of Schinzel and
Zassenhaus with the constant of Boyd up to degree $28$. They also established
that, if $\a$ is an algebraic integer of degree $n \geq 4$, not a root of unity,
then $\maison{\a} \geq \exp \left( 3 \log (n/3)/n^{2} \right)$ for $n \leq 12$,
and $\maison{\a} \geq \exp \left( 3 \log (n/2)/n^{2} \right)$ for $n \geq 13$.
It appears that the
result of [26] improves Matveev for $n \geq 6$. Very recently (Dec. 2019),
Dimitrov [8] has proven the Schinzel-Zassenhaus conjecture with $c=\log(2)/4$.
His Theorem 1 states that when $P(z) \in \bbZ[z]$ is a non-cyclotomic monic polynomial
of degree $n>1$ that is irreducible over $\bbQ$ and $\alpha$ is a zero of $P(z)$,

\[
\maison{\alpha} \geq 2^{1/(4n)}=1+\frac{\log(2)}{4n} + O \left( 1/n^{2} \right).
\]

For trinomials, let $n \geq 2$ and $\theta_{n}$ be the unique real zero in $(0,1)$
of the trinomial $z^{n}+z-1$. By his method of asymptotic expansions of the zeros
mentioned above, J-L. Verger-Gaugry [34] obtained a direct proof of the conjecture
of Schinzel-Zassenhaus for $\theta_{n}^{-1}$, proving that

\[
\maison{\theta_{n}^{-1}} > 1 + \frac{ (\log n) \left( 1 - \frac{ \log \log n}{\log n} \right)}{n}.
\]

Here we focus on trinomials of the form $z^{n}+az^{m} \pm 1$,
where $0<m<n$ are relatively prime integers and $a$ is a positive real number.
replacing $z$ by $-z$, we find that all such polynomials are of the form

$R_{n,m,a}(z)=z^{n}-az^{m}+1$ with $a>0$, $m$ odd and $n$ even;

$S_{n,m,a}(z)=z^{n}+az^{m}-1$ with $a>0$ and $n$ odd; or

$T_{n,m,a}(z)=z^{n}-az^{m}-1$ with $a>0$.

For any positive real number $a \geq 2$ and relatively prime positive integers
$0<m<n$ with $m$ odd and $n$ even, $r_{n,m,a}(z)$ has a real root $r_{n,m,a}^{(1)} \geq 1$

\[
r_{n,m,a}^{(1)} \geq 1 + \frac{\log(a-1)}{n-m}.
\]

Therefore, if we restrict $a$ to be an integer with $a \geq 2$, we have

\[
\maison{R_{n,m,a}} \geq 1 + \frac{\log(a-1)}{n-m}.
\]

For any positive real number $a \geq 2$ and relatively prime positive integers
$0<m<n$ with $m$ even and $n$ odd, $S_{n,m,a}(z)$ has a real root $s_{n,m,a}^{(3)} \leq -1$

\[
\left| s_{n,m,a}^{(3)} \right| \geq 1 + \frac{\log(a-1)}{n-m}.
\]

If we restrict $a$ to be an integer with $a \geq 2$, we have

\[
\maison{S_{n,m,a}} \geq 1 + \frac{\log(a-1)}{n-m}.
\]

Unfortunately, when $m$ and $n$ are both odd, the single real zero of $S_{n,m,a}$
is between $0$ and $1$, while $\maison{S_{n,m,a}}>1$, so we are unable to obtain
a non-trivial lower bound for $\maison{S_{n,m,a}}$ in this case.

For any positive real number $a \geq 2$ and relatively prime positive integers
$0<m<n$, $T_{n,m,a}(z)$ has a real root $t_{n,m,a}^{(1)}>1$ and

\[
t_{n,m,a}^{(1)} > 1 + \frac{\log(a)}{n-m}.
\]

Therefore, if $a$ is restricted to be an integer with $a \geq 2$, we have

\[
\maison{T_{n,m,a}} > 1 + \frac{\log(a)}{n-m}.
\]

An algebraic integer $\alpha$ of degree $n$ is extremal if $\maison{\alpha}$
is the minimum of the houses of the algebraic integers of degree $n$. Denote by
$m(n)$ this minimum. From Smyth's example, $P_{n}(x)=x^{3n}+x^{2n}-1$, given
above (again, see [3, 30]), it is known that $m(n) \leq \theta_{0}^{3/(2n)}
= \maison{P_{n}}$. Here we have

* For all positive integers $a$, $m$ and $n$ with $a \geq 2$, $\gcd(m,n)=1$, $m$
odd and $n$ even, if $R_{n,m,a}$ is irreducible over $\bbQ$, then its zeros
are not extremal.

* For all positive integers $a$, $m$ and $n$ with $a \geq 2$, $\gcd(m,n)=1$, $m$
even and $n$ odd, if $S_{n,m,a}$ is irreducible over $\bbQ$, then its zeros are
not extremal.

* For all positive integers $a$, $m$ and $n$ with $a \geq 2$ and $\gcd(m,n)=1$,
if $T_{n,m,a}$ is irreducible over $\bbQ$, then its zeros are not extremal.

§ PROOF OF THEOREM <REF>

To prove Theorem <ref>, we use the following result of Schinzel.

For positive integers $0<m<n$, put $m_{1}=m/\gcd(m,n)$ and $n_{1}=n/\gcd(m,n)$.

Let $a,b,c \in \bbZ \backslash \{0\}$, $\gcd(a,b,c)=1$. If $ax^{n}+bx^{m}+c$ is
reducible, then at least one of the following four conditions is satisfied:

(a) $|b| \leq |a|^{m_{1}} |c|^{n_{1}-m_{1}}+1$;

(b) $|b| \leq \dfrac{2m_{1}(n_{1}-m_{1})}{\log \left( 2m_{1} (n_{1}-m_{1}) \right)} |a|^{m/n} |c|^{(n-m)/n}$,
$\min \{ |a|, |c| \}=1$ and $\sqrt[p]{\max \{ |a|, |c| \}} \in \bbZ$ for some
prime $p|n_{1}$;

(c) for some $q|\gcd(m,n)$, $q$ a prime or $q=$, $\sqrt[q]{|a|} \in \bbZ$,
$\sqrt[q]{|c|} \in \bbZ$ and if $q=2$, then $(-1)^{n_{1}}ac>0$, while if $q=4$,
then $ac>0$ and $n_{1} \equiv 0 \bmod 2$;

(d) $4|\gcd(m,n)$, $ac>0$, $n_{1} \equiv 1 \bmod 2$ and either
$\sqrt[4]{|a|} \in \bbZ$, $\sqrt[4]{4|c|} \in \bbZ$ or
$\sqrt[4]{4|a|} \in \bbZ$, $\sqrt[4]{|c|} \in \bbZ$.

This is Theorem 9 on pages 12–13 of [28].

To prove Theorem <ref>, we suppose that $x^{n}+ax^{m}\pm 1$ is

We apply Lemma <ref> with $a=1$, $c=\pm 1$ and $b$ equal to our $a$ here.

Condition (a) implies that $|a| \leq 2$.

Since $m_{1} \left( n_{1}-m_{1} \right) \leq n_{1}^{2}/4$ for all $0 \leq m_{1} \leq n_{1}$,
condition (b) implies that $|a|<n^{2}/2$. A quick calculation for small relatively
prime values of $m$ and $n$ shows that we must have $|a|<0.321n^{2}$ for $n \geq 3$.

Since $\gcd(m,n)=1$, conditions (c) and (d) cannot hold.

§ PROOF OF THEOREM <REF>

The proof requires two preliminary results.

If $Q$ is a polynomial with complex coefficients, then

\[
\lim_{n \rightarrow \infty} \mHgt \left( z^{n} + Q(z) \right)
= \exp \left( \frac{1}{2 \pi} \int_{0}^{2 \pi} \log \max \left( 1, \left| Q \left( e^{it} \right) \right| \right) dt \right).
\]

D. Boyd <cit.> proved that $\d \lim_{n \rightarrow \infty} \mHgt \left( F \left(z, z^{n} \right) \right)
= \mHgt(F(z, w))$ if $F$ is a polynomial. Thus we have $\d \lim_{n \rightarrow \infty} \mHgt \left( z^{n} +Q(z) \right)
= \mHgt (w+Q(z))$. Now we apply Jensen's formula [15] with respect to the
variable $w$

to obtain the lemma (also see <cit.>).

\[
\d \lim_{n \rightarrow \infty} \mHgt \left( z^{n} + az^{m}+b \right)
= \exp\left( \frac{1}{2 \pi} \int_{0}^{2 \pi} \log \max \left( 1, \left| ae^{it}+b \right| \right) dt \right).
\]

Lemma <ref> gives $\d \lim_{n \rightarrow \infty} \mHgt \left( z^{n} +az^{m} +b \right)
= \exp\left( \frac{1}{2 \pi} \int_{0}^{2 \pi} \log \max \left( 1, \left| ae^{itm}+b \right| \right) dt \right)$.

\begin{align*}
\int_{0}^{2 \pi} \log \max \left( 1, \left| ae^{itm}+b \right| \right) dt
& = \frac{1}{m} \sum_{\ell=0}^{m-1} \int_{2 \pi \ell}^{2 \pi (\ell+1)} \log \max \left(1, \left| ae^{it}+b \right| \right) dt \\
&= \int_{0}^{2 \pi} \log \max \left( 1, \left| ae^{it}+b \right| \right) dt,
\end{align*}

proving the corollary.

We are now able to prove Theorem <ref>. Let $P(z)=z^{n}+az^{m}+b$ be a
trinomial with $a,b \in \bbC$. We will need this preliminary calculation: if
$a =|a|e^{iu}$ and $b=|b|e^{iv}$ then
$\left| ae^{it} +b \right|^{2} = \left| |a|e^{i(t+u-v)} + |b| \right|^{2}
=|a|^{2} + 2|a||b| \cos(t') + |b|^{2}$ where $t'=t+u-v$.

(a) $|a|-|b| \geq 1$:

We have $\left| ae^{it}+b \right|^{2} \geq (|a|-|b|)^{2} \geq 1$. Therefore,
$\max \left( 1, \left| ae^{it}+b \right| \right)=\left| ae^{it}+b \right|$, so
Corollary <ref> gives

\[
\d \lim_{n \rightarrow \infty} \mHgt \left( z^{n} +az^{m} +b \right) = \mHgt (az+b)
= |a| \max (1, |b|/|a|)= \max(|a|,|b|).
\]

(b) $|b|-|a| \geq 1$:

We use Rouché's Theorem (again, see Corollary on page 153 [1]). We want
to prove that all the zeros of $z^{n}P(1/z)=bz^{n}+az^{n-m}+1$
have absolute value at most $1$.

Suppose first that $|b|-|a|>1$. Putting $f(z)=bz^{n}$ and $g(z)=az^{n-m}+1$, we see
that $|g(z)| \leq |a|+1 < |b|=|f(z)|$ on the unit circle, since $|b|-|a|>1$. Hence
$P(z)=f(z)+g(z)$ and $f(z)$ have the same number of zeros inside the unit circle.
Since $f(z)$ has $n$ such zeros, so does $P(z)$. Hence all the zeros of $P$ have
absolute value at least $1$ and $\mHgt(P)=|b|$.

Now suppose that $|b|-|a|=1$ and let $\varepsilon>0$. We now consider $g(z)$ and
$f(z)$ on the circle of radius $1+\varepsilon$ centred at $0$. Now
$|g(z)| \leq |a|(1+\varepsilon)^{n-m}+1=|a|+|a|\varepsilon_{1}+1$ and
for some $0<\varepsilon_{1}<\varepsilon_{2}$.
So $|f(z)|-|g(z)| \geq |a|\varepsilon_{2}+\varepsilon_{2}-|a|\varepsilon_{1}>0$.
Hence within any circle of radius $1+\varepsilon$ centred at $0$, $P(z)=f(z)+g(z)$
and $f(z)$ have the same number of zeros.
Since $f(z)$ has $n$ such zeros, so does $P(z)$. Taking the limit
as $\varepsilon \rightarrow 0$, we see that all the zeros of $P(z)$ have absolute
value at most $1$. Hence $\mHgt(P)=|b|$.

(c) $|a| + |b| \leq 1$:

We use Rouché's Theorem here too and proceed in the same way as above.
We put $f(z)=z^{n}+b$ and $g(z)=az^{m}$, and consider the cases $|a|+|b|<1$ and
$|a|+|b|=1$ separately.

(d) $|a-b|<1<|a+b|$:

We have $\left| ae^{it} +b \right| >1$ iff $t \in (0, \gamma)$ where
$\gamma = \d \arccos \left( \frac{1-|a|^{2}-|b|^{2}}{2|ab|} \right)$. From
Corollary <ref>, we deduce

\[
\lim_{n \rightarrow \infty} \mHgt \left( z^{n} +az^{m} +b \right)
=\exp \left( \frac{1}{2 \pi} \int_{0}^{\gamma} \log \left( |a|^{2}+2|ab| \cos(t)+|b|^{2} \right) dt \right).
\]

§ PROOF OF THEOREM <REF>

We first need a lemma to show that we can interchange an integral and a sum that
arise in our proof.

Let $a$ and $b \neq 0$ be complex numbers with $|a|-|b| \geq 1$ and let
$m$ and $n$ be integers satisfying $0<m<n$. Then

\[
\int_{0}^{2\pi} \sum_{k \geq 1} \frac{1}{k} \left( \frac{e^{int}}{-ae^{imt}-b} \right)^{k} dt
=\sum_{k \geq 1} \int_{0}^{2\pi} \frac{1}{k} \left( \frac{e^{int}}{-ae^{imt}-b} \right)^{k} dt.
\]

According to Fubini's Theorem, this relationship holds if

\[
\int_{0}^{2\pi} \sum_{k \geq 1} \left| \frac{1}{k} \left( \frac{e^{int}}{-ae^{imt}-b} \right)^{k} \right| dt
\, \text{ or } \,
\sum_{k \geq 1} \int_{0}^{2\pi} \left| \frac{1}{k} \left( \frac{e^{int}}{-ae^{imt}-b} \right)^{k} \right| dt
\]

We will use the first condition here:

\[
\int_{0}^{2\pi} \sum_{k \geq 1} \frac{1}{k} \left| \left( \frac{1}{ae^{imt}+b} \right)^{k} \right| dt.
\]

For $|a|-|b|>1$, then $\left| ae^{imt}+b \right|>1$, so
$\sum_{k \geq 1} \frac{1}{k} \left| \left( \frac{1}{ae^{imt}+b} \right)^{k} \right|$
converges and can be bounded from above independently of $t$. Since the integral
is over a bounded set, it is finite and the lemma holds.

For $|a|-|b|=1$, we use the fact that $\sum_{k \geq 1} z^{k}/k=-\log(1-z)$ if
$|z| \leq 1$ and $z \neq 1$. So we need to show that

\[
\int_{0}^{2\pi} \log \left( 1- \left| \frac{1}{ae^{imt}+b} \right| \right) dt < \infty.
\]

The only place where $\left| ae^{imt}+b \right|=1$ is where
$ae^{imt}=\left( |b|+1 \right) (-b)/|b|$. Note that there are $m$ such values of
$t$ satisfying $0 \leq t<2\pi$. Elsewhere $\left| ae^{imt}+b \right|>1$,
so away from these $m$ values of $t$, the integral is bounded. We only need to
show that in all neighbourhoods of each of these $m$ values of $t$, the integral
is also bounded. It suffices to consider only one such value of $t$, $t_{0}$,
and only neighbourhoods on one side of such a $t$. That is to show that

\[
\int_{t_{0}}^{t_{0}+\varepsilon} \log \left( 1- \left| \frac{1}{ae^{it}+b} \right| \right) dt < \infty,
\]

for $\varepsilon>0$.

Suppose that $a$ and $b$ are both real numbers.
Here $\left| ae^{it}+b \right|=\sqrt{a^{2}+b^{2}+2ab\cos(t)}$.

If $a$ and $b$ have different signs, then we need to consider what happens near $t_{0}=0$.
We want a lower bound for
$1- \dfrac{1}{\sqrt{a^{2}+b^{2}+2ab\cos(t)}}$ of the form $ct$ where $c>0$
for $t$ in some interval whose left endpoint is $t_{0}=0$.

There are
two cases to consider: (i) $a>0$, $b<0$ and $a=-b+1$; or (ii) $a<0$, $b>0$ and $a=-b-1$.
In the first case we consider the value of the function
$1- \dfrac{1}{\sqrt{a^{2}+b^{2}+2ab\cos(t)}}
=1- \dfrac{1}{\sqrt{2b^{2}-2b+1+2(1-b)b\cos(t)}}$ at $t=0$ (where it has the
value $0$) and $t=\pi/2$ (where it has the value $1- \dfrac{1}{\sqrt{2b^{2}-2b+1}}$).
This function is also convex for $t \in [0,\pi/2]$ (this follows from the fact
that $2b^{2}-2b+1+2(1-b)b\cos(t)$ is non-negative and increasing in this
interval, approaching $b^{2}+(b-1)^{2}$ as $t \rightarrow \pi/2$, and that
$1-1/\sqrt{x}$ is a convex function). So we have

\[
1- \frac{1}{\sqrt{a^{2}+b^{2}+2ab\cos(t)}}
\geq \frac{2}{\pi} \left( 1- \frac{1}{\sqrt{2b^{2}-2b+1}} \right) t,
\]

for $0 \leq t \leq \pi/2$. That is, we can take $c$ above to be
$\dfrac{2}{\pi} \left( 1- \dfrac{1}{\sqrt{2b^{2}-2b+1}} \right)$.

$\int_{0}^{\varepsilon} \log(ct) dt =\varepsilon \log (c\varepsilon) - \varepsilon$,

\[
\lim_{\varepsilon \rightarrow 0} \int_{0}^{\varepsilon} \log(ct) dt
=\lim_{\varepsilon \rightarrow 0} \varepsilon \log (c\varepsilon) - \varepsilon
=\lim_{\varepsilon \rightarrow 0} \frac{\log (c\varepsilon)}{1/\varepsilon}
=\lim_{\varepsilon \rightarrow 0} \frac{1/\left( c\varepsilon \right)}{-1/\varepsilon^{2}}
=-\lim_{\varepsilon \rightarrow 0} \varepsilon/c=0,
\]

by L'Hôpital's rule. Thus

\[
\int_{t_{0}}^{t_{0}+\varepsilon} \log \left( 1- \left| \frac{1}{ae^{it}+b} \right| \right) dt < \infty,
\]

and so, as argued above,

\[
\int_{0}^{2\pi} \log \left( 1- \left| \frac{1}{ae^{imt}+b} \right| \right) dt < \infty.
\]

The other cases, namely when $a<0$, $b>0$ and $a=-b-1$; and when
$a$ and $b$ have the same sign are treated in the very same way, proving the lemma
when $a, b \in \bbR$.

Furthermore, a more complicated version of this same argument can be used to
prove the lemma for any $a, b \in \bbC$ satisfying the conditions of the lemma.

The idea of the proof of Theorem <ref> is the same as the one used by
the first author in the study of trinomials of height $1$ (see [11]).

Using Jensen's formula, we put

\[
\lambda_{n,m,a}=\log M \left( z^{n}+az^{m}+b \right)
= \dfrac{1}{2\pi} \int_{0}^{2\pi} \log \left| e^{int}+ae^{imt}+b \right| dt.
\]

Note that $\lambda_{n,m,a}=\log M \left( -z^{n}-az^{m}-b \right)$ holds, so we
work with $-z^{n}-az^{m}-b$ here. This change simplifies slightly what follows.

Using Jensen's formula again and since $|a|>|b|$ (from our hypothesis that
$|a|-|b| \geq 1$), we have

\[
\dfrac{1}{2\pi}\int_{0}^{2\pi} \log \left| -ae^{imt}-b \right| dt
=\log M \left( -az^{m}-b \right)
=\log |a|.
\]

\begin{align*}
\lambda_{n,m,a}-\log |a|
&= \frac{1}{2\pi}
\int_{0}^{2\pi} \log \left| -e^{int}-ae^{imt}-b \right| - \log \left| -ae^{imt}-b \right| dt \\
&= \frac{1}{2\pi}
\int_{0}^{2\pi} \log \left| 1-\frac{e^{int}}{-ae^{imt}-b} \right| dt.
\end{align*}

Writing any $0 \neq x \in \bbC$ as $x=re^{i\theta}$ where $r>0$ and $|\theta| \leq \pi$,
we have $\log(x)=i\theta+\log(r)=i\theta+\log|r|$, so $\log |x| = \Real( \log(x))$
– we have written this using the principal value of the logarithm function, but
it holds for any branch of the logarithm function.

\[
\lambda_{n,m,a}-\log |a|
= \frac{1}{2\pi}
\Real \int_{0}^{2\pi} \log \left( 1-\frac{e^{int}}{-ae^{imt}-b} \right) dt.
\]

Applying Lemma <ref> and the series expansion

\[
\log \left( 1-\frac{e^{int}}{-ae^{imt}-b} \right) \\
=- \sum_{k \geq 1} \frac{1}{k} \left( \frac{e^{int}}{-ae^{imt}-b} \right)^{k},
\]

we obtain

\begin{equation}
\label{eq:lambda-diff-S}
\lambda_{n,m,a}-\log |a|
= -\frac{1}{2\pi} \sum_{k \geq 1} \frac{1}{k}
\Real \underbrace{\int_{0}^{2\pi} e^{inkt} \left( -ae^{imt}-b \right)^{-k} dt}_{I_{k}}.
\end{equation}

From $z=e^{it}$, we have $dz=ie^{it}dt$, so

\[
I_{k} = -i\int_{|z|=1} z^{kn-1} \left( -az^{m}-b \right)^{-k} dz.
\]

We evaluate this integral using the Residue theorem:

\[
I_{k} = -i (2\pi i) \sum_{r: r^{m}=-b/a} \res \left( z^{kn-1} \left( -az^{m}-b \right)^{-k}; r \right),
\]

where the sum is over all $m$-th roots of $-b/a$ and $\res \left( f(z);z_{0} \right)$
is the residue of a function $f(z)$ at $z=z_{0}$.

Now observe that $z^{kn-1} \left( -az^{m}-b \right)^{-k}$ has one more pole when
we consider this function over the Riemann sphere. It has a pole at $\infty$ too.
Furthermore, the sum of the residues of a function over the Riemann sphere
equals $0$ <cit.>, so

\begin{equation}
\label{eq:I-as-residue-S}
I_{k} = -2\pi \res \left( z^{kn-1} \left( -az^{m}-b \right)^{-k}; \infty \right),
\end{equation}

It is also known <cit.> that

\[
\res \left( f(z); \infty \right)
= -\res \left( \frac{1}{z^{2}}f \left( \frac{1}{z} \right); 0 \right),
\]

so we compute this quantity now. With
$f(z)=z^{kn-1} \left( -az^{m}-b \right)^{-k}$, we have

\[
\frac{1}{z^{2}}f \left( \frac{1}{z} \right)
= \frac{z^{-1-kn+km}}{\left( -a-bz^{m} \right)^{k}}.
\]

The negative binomial series $\left( -a-bz^{m} \right)^{-k}$ equals

\[
\sum_{i=0}^{\infty} \binom{-k}{i} (-b)^{i}z^{im}(-a)^{-k-i}
=(-1)^{k} \sum_{i=0}^{\infty} \binom{-k}{i} b^{i}z^{im}a^{-k-i}.
\]

To get the the residue, we need the coefficient for $z^{kn-km}$ term in this sum.
I.e., $i=k(n-m)/m$. If $m \nmid kn$, then there is no such coefficient and the residue
is $0$. Otherwise, the coefficient of the $z^{kn-km}$ term is

\begin{align*}
\binom{-k}{k(n-m)/m} b^{k(n-m)/m} a^{-k-k(n-m)/m}
&=(-1)^{k(n-m)/m} \binom{k+k(n-m)/m-1}{k(n-m)/m} b^{k(n-m)/m} a^{-kn/m} \\
&=(-1)^{k(n-m)/m} \binom{kn/m-1}{kn/m-k} b^{k(n-m)/m} a^{-kn/m},
\end{align*}

since $\d \binom{-r}{s}=(-1)^{s} \binom{r+s-1}{s}$ for positive integers $r$ and
$s$. In fact, recall that $\gcd(m,n)=1$, so $m |kn$ if and only if $m|k$.
Therefore, from this expression and $\binom{kn/m-1}{kn/m-k}=\binom{kn/m-1}{k-1}$,
we obtain

\[
\res \left( z^{kn-1} \left( -az^{m}-b \right)^{-k}; \infty \right)
= \left\{
\begin{array}{ll}
-\displaystyle(-1)^{kn/m} \binom{kn/m-1}{k-1} b^{k(n-m)/m}a^{-kn/m} & \text{if $m|k$}, \\
0 & \text{otherwise.}
\end{array}
\right.
%the (-1)^{k(-m)/m} term gets killed off by the (-1)^k on the RHS of the
%negative binomail series expansion.
\]

Applying this expression to (<ref>) and then (<ref>),
we obtain

\[
\lambda_{n,m,a}-\log |a|
= -\sum_{k \geq 1} \frac{1}{km} (-1)^{kn} \binom{kn-1}{km-1} \Real \left( b^{-km} (b/a)^{kn} \right),
\]

which is equation (<ref>).

§ PROOFS OF THEOREMS <REF>–<REF> AND COROLLARY <REF>

We start with a lemma about the location of the real zeros of the trinomials we
are considering.

Let $a \geq 2$ be a real number, $m$ and $n$ are positive relatively prime integers
with $0<m<n$.

* Suppose also that $m$ odd and $n$ even.
$R_{n,m,a}(z)$ has two real zeros, $r_{n,m,a}^{(1)}$ and
$r_{n,m,a}^{(2)}$ satisfying
$r_{n,m,a}^{(1)}>1$ if $a>2$,
$r_{n,m,a}^{(1)}=1$ if $a=2$,
and $0<r_{n,m,a}^{(2)}<1$.

* Suppose also that $n$ is odd.
If $m$ is odd, then $S_{n,m,a}(z)$ has one real zero.
If $m$ is even. then $S_{n,m,a}(z)$ has three real zeros.
There is always one real zero, $0<s_{n,m.a}^{(1)}<1$.
If $m$ is even, then there is another real zero, $-1<s_{n,m,a}^{(2)}<0$ and a
third real zero, $s_{n,m,a}^{(3)}$.
If $a>2$, then $s_{n,m,a}^{(3)}<-1$.
If $a=2$, then $s_{n,m,a}^{(3)}=-1$.

* If $n$ is odd, then $T_{n,m,a}(z)$ has three real zeros if $m$ is odd and one
real zero if $m$ is even. There is always one zero, $t_{n,m.a}^{(1)}>1$. If $m$
is odd, then there is another real zero, $-1 \leq t_{n,m,a}^{(2)}<0$ and a third
real zero, $t_{n,m,a}^{(3)} \leq -1$.

If $n$ is even, then $T_{n,m,a}(z)$ has two real zeros. One, $t_{n,m,a}^{(1)}>1$
and the other, $-1 < t_{n,m,a}^{(2)}<0$.

We prove only part (c), as the proofs of parts (a) and (b) are identical.

Using Descartes' Sign Rule, we find there is one positive real zero, since
there is one sign change among the coefficients of $T_{n,m,a}(z)$. Since
$T_{n,m,a}(1)=-a$ and $\lim_{z \rightarrow +\infty} T_{n,m,a}(z)=+\infty$, this
zero is strictly greater than $1$.

Applying Descartes' Sign Rule to $T_{n,m,a}(-z)=(-1)^{n}z^{n}+(-1)^{m-1}az^{m}-1$,
there are two negative zeros if $m$ and $n$ are both odd; no negative zeros if
$n$ is odd and $m$ is even; and one negative zero if $n$ is even (in which case,
$m$ is odd).

We have $T_{n,m,a}(-1)=(-1)^{n}-(-1)^{m}a-1$ equals $(-1)^{m-1}a=a$ if $n$ is even
(recalling that $m$ is odd in this case)
and $(-1)^{m-1}a-2$ if $n$ is odd. For $n$ even and since $T_{n,m,a}(0)=-1$, it
follows that the unique negative real zero is strictly between $-1$ and $0$.

Lastly, for $n$ odd and $m$ odd, we have $T_{n,m,a}(0)=-1$ and $T_{n,m,a}(-1)=a-2$.
So if $a>2$ and since $\lim_{z \rightarrow -\infty} T_{n,m,a}(z)=-\infty$, there
must be one in $(-1,0)$ and another in $(-\infty, -1)$. For $a=2$, there is a
zero at $z=-1$ and the second negative zero turns out to be less than $-1$ if
$m>n/2$ or in $(-1,0)$ if $m<n/2$.

§.§ Proof of Theorem <ref>

When $n$ is even and $m$ is odd, from Lemma <ref>(a),
$r_{n,m,a}^{(1)}$ is the unique real zero of the trinomial $R_{n,m,a}$ which
satisfies $r_{n,m,a}^{(1)} \geq 1$ and put $r_{n,m,a}^{(1)}=1+t$ with $t \geq 0$.
From the expression for $R_{n,m,a}$, we  we see that $(1+t)^{n}-a(1+t)^{m}+1=0$, so
$(1+t)^{n}-a(1+t)^{m}+1+(1+t)^{m} \geq 1$, i.e., $(1+t)^{n}-(a-1)(1+t)^{m} \geq 0$.

\[
(1 + t)^{n} \geq (a-1)(1 + t)^{m}.
\]

Let $t_{0}$ be the largest real number such that

\[
\left(1 + t_{0} \right)^{n} = (a-1) \left( 1 + t_{0} \right)^{m},
\]

then $t \geq t_{0}$ and $t_{0}=\exp \left( \log(a-1)/(n-m) \right)-1$. Using the
Taylor expansion of $\exp(x)$, $t_{0} \geq \log(a-1)/(n-m)$ and the theorem follows.

§.§ Proof of Theorem <ref>

When $n$ is odd and $m$ is even, recall from Lemma <ref>(b)
that $s_{n,m,a}^{(3)}$ is the unique real zero of the trinomial $S_{n,m,a}$ which
satisfies $s_{n,m,a}^{(3)} \leq -1$ and put $s_{n,m,a}^{(3)}=-(1+t)$ with $t \geq 0$.
From the expression for $S_{n,m,a}$, we see that $(1+t)^{n}-a(1+t)^{m}+1=0$, so
as above

\[
(1 + t)^{n} \geq (a-1)(1 + t)^{m}
\]

and the proof follows as above too.

§.§ Proof of Theorem <ref>

From Lemma <ref>(c), $t_{n,m,a}^{(1)}$ is the unique real
zero of $T_{n,m,a}$ which satisfies $t_{n,m,a}^{(1)}>1$ and put $t_{n,m,a}^{(1)}=1+t$
with $t>0$. From the expression for $T_{n,m,a}$, we know that

\[
(1 + t)^{n}>a(1 + t)^{m}.
\]

\[
\left(1 + t_{0} \right)^{n} = a \left( 1 + t_{0} \right)^{m},
\]

then $t>t_{0}$ and $t_{0}=\exp \left( \log(a)/(n-m) \right)-1$. Using the Taylor
expansion of $\exp(x)$, $t_{0}>\log(a)/(n-m)$ and our result follows.

§.§ Proof of Corollary <ref>

We prove only part (c) as the proof for the other two parts is identical.

It is clear that $m(n) \leq 2^{1/n}$.
We have $T_{n,m,a} \left( 2^{1/n} \right)= 1- a 2^{m/n} < 0$, since $a \geq 2$.
Since $T_{n,m,a}(z)>0$ for all $z>\theta_{m,n,a}$, we deduce $t_{n,m,a}^{(1)} > 2^{1/n}$,
which implies the non-extremality of $t_{n,m,a}^{(1)}$.

[1]
L. Ahlfors.
Complex Analysis, 3rd ed., McGraw-Hill (1979).

[3]
D. Boyd.
The maximal modulus of an algebraic integer,
Math. Comp. 45 (1985), 243–245.

[4]
D. W. Boyd.
Speculations concerning the range of Mahler's measure,
Canad. Math. Bull. 24 no. 4 (1981), 453–469.

[5]
D. W. Boyd, M. J. Mossinghoff,
Small Limit Points of Mahler's Measure,
Exp. Math. 14 no. 4 (2005), 403–414.

Speculations concerning the range of Mahler's measure,
Canad. Math. Bull. 24 no. 4 (1981), 453–469.

[6]
A. Bremner.
On trinomials of type $x^{n}+Ax^{m} +1$,
Math. Scand. 49 no. 2 (1981), 145–155.

[7]
H. Cartan.
Elementary Theory of Analytic Functions of One or Several Complex Variables,
Dover (1995).

[8]
V. Dimitrov.
A proof of the Schinzel-Zassenhaus conjecture on polynomials,
<https://arxiv.org/abs/1912.12545v1>, Dec. 2019.

[9]
A. Dubickas.
On a conjecture of A. Schinzel and H. Zassenhaus,
Acta Arith. 63 no. 1 (1993), 15–20.

[10]
W. Duke.
A combinatorial problem related to Mahler measure,
Bull. Lond. Math. Soc. 39(5) (2007), 741–748.

[11]
V. Flammang.
The Mahler measure of trinomials of height $1$,
J. Aust. Math. Soc. 96 no. 2 (2014), 231–243.

[15]
J.L.V.W. Jensen.
Sur un nouvel et important théorème de la théorie des fonctions,
Acta Math. 22 (1899), 359–364.

[19]
D. H. Lehmer.
Factorization of certain cyclotomic functions,
Ann. of Math. (2) 34 no. 3 (1933), 461–479.

[20]
E. M. Matveev.
On the cardinality of algebraic integers,
Math. Notes 49 (1991), 437–438.

[23]
O. Perron.
Neue Kriterien für die Irreduzibilität algebraischer Gleischungen,
J. reine angew. Math. 132 (1907), 288–307.

[26]
G. Rhin, Q. Wu.
On the smallest value of the maximal modulus of an algebraic integer,
Math. Comp. 76 (258) (2007), 1025–1038.

[28]
A. Schinzel.
On reducible trinomials,
Dissertationes Mathematicae 329 (1993).

[29]
A. Schinzel, H. Zassenhaus.
A refinement of two theorems of Kronecker,
Michigan Math. J. 12 (1965), 81–85.

[30]
C. J. Smyth.
On the product of the conjugates outside the unit circle of an algebraic integer,
Bull. London Math. Soc. 3 (1971), 169–175.

[32]
D. Stankov.
Boyd's conjecture,
<https://arxiv.org/abs/arXiv:1401.1688v2>, March 2014.

[34]
J.-L. Verger-Gaugry.
On the conjecture of Lehmer, limit Mahler measure of trinomials and asymptotic expansions,
Unif. Distrib. Theory 11 no. 1 (2016), 79–139.

[35]
P. Voutier.
An effective lower bound for the height of algebraic numbers,
Acta Arith. 74 no. 1 (1996), 81–95.

V. Flammang:

UMR CNRS 7502. IECL, Université de Lorraine, site de Metz,

Département de Mathématiques, UFR MIM,

Ile du Saulcy, CS 50128. 57045 METZ cedex 01. FRANCE

E-mail address<EMAIL_ADDRESS>
P. Voutier:

London, UK

E-mail address: Paul.Voutier@gmail.com


# Some identities involving degenerate $r$-Stirling numbers

Taekyun Kim  Department of Mathematics, Kwangwoon University, Seoul 139-701,
Republic of Korea<EMAIL_ADDRESS>and  DAE SAN KIM  Department of Mathematics,
Sogang University, Seoul 121-742, Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

Recently, Kim-Kim studied the unsigned degenerate $r$-Stirling number of the
first kind and the degenerate $r$-Stirling number of the second kind,
respectively of which are the degenerate versions of the unsigned $r$-Stirling
numbers of the first kind and those of the $r$-Stirling numbers of the second
kind. The aim of this paper is to derive some identities involving such
special numbers from the inverse relations for them.

###### Key words and phrases:

degenerate $r$-Stirling numbers of the second kind; unsigned degenerate
$r$-Stirling numbers of the first kind; inverse relation

###### 2010 Mathematics Subject Classification:

11B73; 11B83

##  1\. Introduction

Carlitz initiated a study of degenerate versions of some special polynomials
and numbers, namely the degenerate Bernoulli and Euler polynomials and numbers
(see [3]). In recent years, we have witnessed that some mathematicians have
explored various degenerate versions of many special polynomials and numbers
by using various tools like combinatorial methods, generating functions,
differential equations, umbral calculus techniques, $p$-adic analysis, special
functions, operator theory, probability theory, and analytic number theory.
These degenerate versions include the degenerate Stirling numbers of the first
and second kinds, which appear very frequently when we study degenerate
versions of some special numbers and polynomials (see [5-11]).

The unsigned $r$-Stirling number of the first kind ${n\brack k}_{r}$ is the
number of permutations of the set $[n]=\\{1,2,3,\dots,n\\}$ with exactly $k$
disjoint cycles in such a way that the numbers $1,2,\dots,r$ are in distinct
cycles, while the $r$-Stirling number of the second kind ${n\brace k}_{r}$
counts the number of partitions of the set $[n]$ into $k$ non-empty disjoint
subsets in such a way that the numbers $1,2,\dots,r$ are in distinct subsets.
We remark that Border [2] studied the combinatorial and algebraic properties
of the $r$-Stirling numbers.

The unsigned degenerate $r$-Stirling numbers of the first kind ${n\brack
k}_{r,\lambda}$ are degenerate versions of the unsigned $r$-Stirling numbers
of the first kind ${n\brack k}_{r}$ and the degenerate $r$-Stirling numbers of
the second kind ${n\brace k}_{r,\lambda}$ are degenerate versions of the
$r$-Stirling numbers of the second kind ${n\brace k}_{r}$. They can be viewed
also as natural extensions of the unsigned degenerate Stirling numbers of the
first kind ${n\brack k}_{\lambda}$ and the degenerate Stirling numbers of the
second kind ${n\brace k}_{\lambda}$ which were introduced earlier (see
[6,10,11]). The aim of this paper is to derive from the inverse relations for
the degenerate $r$-Stirling numbers some identities involving such numbers and
some special numbers which are given by the evaluations at $r$ of the fully
degenerate Bernoulli polynomials, the degenerate two variable Fubini
polynomials, the degenerate Euler polynomials and the degenerate poly-
Bernoulli polynomials.

The outline of this paper is as follows. In Section 1, we recall the
degenerate exponential functions, the degenerate logarithms, the degenerate
Bernoulli polynomials, the fully degenerate Bernoulli polynomials, the
degenerate harmonic numbers, the unsigned degenerate Stirling numbers of the
first kind, the degenerate Stirling numbers of the second kind, the unsigned
degenerate $r$-Stirling numbers of the first kind, and the degenerate
$r$-Stirling numbers of the second kind. In Section 2, we derive the
orthogonality relations between the unsigned degenerate $r$-Stirling numbers
of the first kind and the degenerate $r$-Stirling numbers of the second kind,
and thereby get their inverse relations. By invoking these inverse relations
we show some identities involving such degenerate $r$-Stirling numbers and
some special numbers which are given by the evaluations at $r$ of
aforementioned special polynomials.

For any $\lambda\in\mathbb{R}$, the degenerate falling factorial sequence is
defined by

(1)
$(x)_{0,\lambda}=1,\quad(x)_{n,\lambda}=x(x-\lambda)(x-2\lambda)\cdots(x-(n-1)\lambda),\quad(n\geq
1).$

Note that $\displaystyle\lim_{\lambda\rightarrow
0}(x)_{n,\lambda}=x^{n},\displaystyle$ (see [7-9]). The degenerate exponential
functions are defined by

(2)
$e_{\lambda}^{x}(t)=\sum_{n=0}^{\infty}(x)_{n,\lambda}\frac{t^{n}}{n!},\quad(\mathrm{see}\
[6,10,11]).$

For $x=1$, we let $e_{\lambda}(t)=e_{\lambda}^{1}(t)$.

Let $\log_{\lambda}t$ be the degenerate logarithm which is the compositional
inverse of $e_{\lambda}(t)$. Then we have

(3)
$\log_{\lambda}(1+t)=\sum_{n=1}^{\infty}\frac{\lambda^{n-1}(1)_{n,1/\lambda}}{n!}t^{n},\quad(\mathrm{see}\
[6]).$

Note that
$e_{\lambda}\big{(}\log_{\lambda}(1+t)\big{)}=\log_{\lambda}\big{(}e_{\lambda}(1+t)\big{)}=1+t$.

Carlitz considered the degenerate Bernoulli polynomials given by

(4)
$\frac{t}{e_{\lambda}(t)-1}e_{\lambda}^{x}(t)=\sum_{n=0}^{\infty}B_{n,\lambda}(x)\frac{t^{n}}{n!},\quad(\mathrm{see}\
[3]).$

Note that $\displaystyle\lim_{\lambda\rightarrow
0}B_{n,\lambda}(x)=B_{n}(x)\displaystyle$, where $B_{n}(x)$ are the ordinary
Bernoulli polynomials given by

$\frac{t}{e^{t}-1}e^{xt}=\sum_{n=0}^{\infty}B_{n}(x)\frac{t^{n}}{n!},\quad(\mathrm{see}\
[1-13]).$

Kim-Kim considered the fully degenerate Bernoulli polynomials
$\beta_{n,\lambda}(x)$ which are given by

(5) $\frac{\log(1+\lambda
t)}{\lambda(e_{\lambda}(t)-1)}e_{\lambda}^{x}(t)=\sum_{n=0}^{\infty}\beta_{n,\lambda}(x)\frac{t^{n}}{n!},\quad(\mathrm{see}\
[9]).$

Note that $\displaystyle\lim_{\lambda\rightarrow
0}\beta_{n,\lambda}(x)=B_{n}(x),\ (n\geq 0)\displaystyle$.

The degenerate harmonic numbers are defined by Kim-Kim as

(6) $H_{0,\lambda}=1,\quad
H_{n,\lambda}=\sum_{k=1}^{n}\frac{1}{\lambda}\binom{\lambda}{k}(-1)^{k-1},\quad(n\geq
1),\quad(\mathrm{see}\ [8]).$

Note that $\displaystyle\lim_{\lambda\rightarrow
0}H_{n,\lambda}=H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}\displaystyle$.

From (6), we have

(7)
$-\frac{1}{1-t}\log_{\lambda}(1-t)=\sum_{n=1}^{\infty}H_{n,\lambda}t^{n},\quad(\mathrm{see}\
[8]).$

The degenerate Stirling numbers of the first kind are defined by

(8) $(x)_{n}=\sum_{k=0}^{n}S_{1,\lambda}(n,k)(x)_{k,\lambda},\quad(n\geq
0),\quad(\mathrm{see}\ [6]).$

Note that $\displaystyle\lim_{\lambda\rightarrow
0}S_{1,\lambda}(n,k)=S_{1}(n,k)\displaystyle$ are the Stirling numbers of the
first kind. The unsigned degenerate Stirling numbers of the first kind are
defined by Kim-Kim as

(9) $\langle x\rangle_{n}=\sum_{k=0}^{n}{n\brack k}_{\lambda}\langle
x\rangle_{k,\lambda},\quad(n\geq 0),\quad(\mathrm{see}\ [6,10]),$

where

$\displaystyle\langle x\rangle_{0}=1,\quad\langle
x\rangle_{n}=x(x+1)\cdots(x+n-1),$ $\displaystyle\langle
x\rangle_{0,\lambda}=1,\quad\langle
x\rangle_{n,\lambda}=x(x+\lambda)\cdots(x+(n-1)\lambda),\quad(n\geq 1).$

Note that $\displaystyle{n\brack k}_{\lambda}=(-1)^{n-k}S_{1,\lambda}(n,k),\
(n,k)\geq 0\displaystyle$.

As the inversion formula of (8), the degenerate Stirling numbers of the second
kind are defined by Kim-Kim as

(10) $(x)_{n,\lambda}=\sum_{k=0}^{n}{n\brace k}_{\lambda}(x)_{k},\quad(n\geq
0),\quad(\mathrm{see}\ [6]).$

Note that $\displaystyle\lim_{\lambda\rightarrow 0}{n\brace
k}_{\lambda}={n\brace k}\displaystyle$ are the Stirling numbers of the second
kind which are defined by

(11) $x^{n}=\sum_{k=0}^{n}{n\brace k}(x)_{k},\quad(n\geq
0),\quad(\mathrm{see}\ [12]).$

From (9) and (10), we note that

(12)
$\frac{1}{k!}\big{(}-\log_{\lambda}(1-t)\big{)}^{k}=\sum_{n=k}^{\infty}{n\brack
k}_{\lambda}\frac{t^{n}}{n!},\quad(k\geq 0),$

and

(13)
$\frac{1}{k!}\big{(}e_{\lambda}(t)-1\big{)}^{k}=\sum_{n=k}^{\infty}{n\brace
k}_{\lambda}\frac{t^{n}}{n!},\quad(\mathrm{see}\ [6]).$

Let $r$ be the nonnegative integer. Then the unsigned degenerate $r$-Stirling
numbers of the first kind are defined by Kim-Kim as

(14) $\langle x+r\rangle_{n}=\sum_{k=0}^{n}{n+r\brack k+r}_{r,\lambda}\langle
x\rangle_{k,\lambda},\quad(n\geq 0),\quad(\mathrm{see}\ [6,10,11]).$

From (14), we have

(15)
$\frac{1}{(1-t)^{r}}\frac{1}{k!}\big{(}-\log_{\lambda}(1-t)\big{)}^{k}=\sum_{n=k}^{\infty}{n+r\brack
k+r}_{r,\lambda}\frac{t^{n}}{n!}.$

Note that $\displaystyle\lim_{\lambda\rightarrow 0}{n+r\brack
k+r}_{r,\lambda}={n+r\brack k+r}_{r}\displaystyle$ are the unsigned
$r$-Stirling numbers of the first kind which are defined by

$\frac{1}{(1-t)^{r}}\frac{1}{k!}\big{(}-\log(1-t)\big{)}^{k}=\sum_{n=k}^{\infty}{n+r\brack
k+r}_{r}\frac{t^{n}}{n!},\quad(\mathrm{see}\ [10]).$

In view of (10), the degenerate $r$-Stirling numbers of the second kind are
defined by

(16) $(x+r)_{n,\lambda}=\sum_{k=0}^{n}{n+r\brace
k+r}_{r,\lambda}(x)_{k},\quad(\mathrm{see}\ [6,10,11]).$

By (16), we easily get

(17)
$e_{\lambda}^{r}(t)\frac{1}{k!}\big{(}e_{\lambda}(t)-1\big{)}^{k}=\sum_{n=k}^{\infty}{n+r\brace
k+r}_{r,\lambda}\frac{t^{n}}{n!},\quad(\mathrm{see}\ [6,10,11]).$

##  2\. Some identities involving degenerate $r$-Stirling numbers

In this section, we show orthogonality and inverse relations for the unsigned
degenerate $r$-Stirling numbers of the first kind and the degenerate
$r$-Stirling numbers of the second kind. Then, by using such inverse
relations, we derive some identites involving the degenerate $r$-Stirling
numbers and some special numbers which are given by the evaluations at $r$ of
several special polynomials.

From (14), we note that

(18) $\displaystyle\langle x+r\rangle_{n}$
$\displaystyle=\sum_{k=0}^{n}{n+r\brack k+r}_{r,\lambda}\langle
x\rangle_{k,\lambda}=\sum_{k=0}^{n}{n+r\brack
k+r}_{r,\lambda}(-1)^{k}(-x)_{k,\lambda}$
$\displaystyle=\sum_{k=0}^{n}{n+r\brack
k+r}_{r,\lambda}(-1)^{k}(-x-r+r)_{k,\lambda}$
$\displaystyle=\sum_{k=0}^{n}{n+r\brack
k+r}_{r,\lambda}(-1)^{k}\sum_{j=0}^{k}{k+r\brace j+r}_{r,\lambda}(-x-r)_{j}$
$\displaystyle=\sum_{k=0}^{n}\sum_{j=0}^{k}{n+r\brack
k+r}_{r,\lambda}{k+r\brace j+r}_{r,\lambda}(-1)^{k-j}\langle x+r\rangle_{j}$
$\displaystyle=\sum_{j=0}^{n}\bigg{(}\sum_{k=j}^{n}{n+r\brack
k+r}_{r,\lambda}{k+r\brace j+r}_{r,\lambda}(-1)^{k-j}\bigg{)}\langle
x+r\rangle_{j}.$

By comparing the coefficients on both sides of (18), we obtain the first
orthogonality relation of the following theorem. The proof of the second one
is similar.

######  Theorem 1.

For $n\geq 0$, we have the following orthogonality relations:

$\displaystyle\sum_{k=j}^{n}(-1)^{n-k}{n+r\brack k+r}_{r,\lambda}{k+r\brace
j+r}_{r,\lambda}=\delta_{n,j},$
$\displaystyle\sum_{k=j}^{n}(-1)^{k-j}{n+r\brace k+r}_{r,\lambda}{k+r\brack
j+r}_{r,\lambda}=\delta_{n,j},$

where $\delta_{n,j}$ is the Kronecker’s delta.

######  Theorem 2.

For $n\geq 0$, we have the following inverse relations:

$a_{n,\lambda}=\sum_{k=0}^{n}(-1)^{n-k}{n+r\brack
k+r}_{r,\lambda}b_{k,\lambda}\ \Longleftrightarrow\
b_{k,\lambda}=\sum_{k=0}^{n}{n+r\brace k+r}_{r,\lambda}a_{k,\lambda}.$

###### Proof.

$(\Longrightarrow)$ Assume that

$a_{n,\lambda}=\sum_{k=0}^{n}(-1)^{n-k}{n+r\brack
k+r}_{r,\lambda}b_{k,\lambda}.$

Then we have

$\displaystyle\sum_{k=0}^{n}{n+r\brace k+r}_{r,\lambda}a_{k,\lambda}$
$\displaystyle=\sum_{k=0}^{n}{n+r\brace
k+r}_{r,\lambda}(-1)^{k}\sum_{j=0}^{k}(-1)^{j}{k+r\brack
j+r}_{r,\lambda}b_{j,\lambda}$
$\displaystyle=\sum_{j=0}^{n}b_{j,\lambda}\sum_{k=j}^{n}(-1)^{k-j}{n+r\brace
k+r}_{r,\lambda}{k+r\brack j+r}_{r,\lambda}$ $\displaystyle=b_{n,\lambda}.$

$(\Longleftarrow)$ Let $\displaystyle b_{n,\lambda}=\sum_{k=0}^{n}{n+r\brace
k+r}_{r,\lambda}a_{k,\lambda}\displaystyle$. Then we note that

$\displaystyle\sum_{k=0}^{n}(-1)^{n-k}{n+r\brack
k+r}_{r,\lambda}b_{k,\lambda}$
$\displaystyle=\sum_{k=0}^{n}(-1)^{n-k}{n+r\brack
k+r}_{r,\lambda}\sum_{j=0}^{k}{k+r\brace j+r}_{r,\lambda}a_{j,\lambda}$
$\displaystyle=\sum_{j=0}^{n}a_{j,\lambda}\sum_{k=j}^{n}(-1)^{n-k}{n+r\brack
k+r}_{r,\lambda}{k+r\brace j+r}_{r,\lambda}$ $\displaystyle=a_{n,\lambda}.$

∎

In [9], it is shown that

(19) $\beta_{n,\lambda}(r)=\sum_{k=0}^{n}(-1)^{k}{n+r\brace
k+r}_{r,\lambda}\frac{k!}{k+1}.$

Therefore, by Theorem 2 and (19), we obtain the following corollary.

######  Corollary 3.

For $n\geq 0$, we have

$\frac{n!}{n+1}=\sum_{k=0}^{n}(-1)^{k}{n+r\brack
k+r}_{r,\lambda}\beta_{k,\lambda}(r).$

In [5], the degenerate two variable Fubini polynomials are given by

(20)
$\frac{1}{1-x(e_{\lambda}(t)-1)}e_{\lambda}^{y}(t)=\sum_{n=0}^{\infty}F_{n,\lambda}(x|y)\frac{t^{n}}{n!}.$

We observe that

(21) $\displaystyle\frac{1}{1-x(e_{\lambda}(t)-1)}e_{\lambda}^{r}(t)$
$\displaystyle=\sum_{k=0}^{\infty}x^{k}k!\frac{1}{k!}\big{(}e_{\lambda}(t)-1\big{)}^{k}e_{\lambda}^{r}(t)$
$\displaystyle=\sum_{k=0}^{\infty}x^{k}k!\sum_{n=k}^{\infty}{n+r\brace
k+r}_{r,\lambda}\frac{t^{n}}{n!}$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{k=0}^{n}x^{k}k!{n+r\brace
k+r}_{r,\lambda}\frac{t^{n}}{n!}.$

From (20) and (21), we note that

(22) $F_{n,\lambda}(x|r)=\sum_{k=0}^{n}x^{k}k!{n+r\brace k+r}_{r,\lambda}.$

Therefore, by Theorem 2 and (22), we obtain the following theorem.

######  Theorem 4.

For $n\geq 0$, we have

$x^{n}n!=\sum_{k=0}^{n}(-1)^{n-k}{n+r\brack
k+r}_{r,\lambda}F_{k,\lambda}(x|r).$

In [3], Carlitz introduced the degenerate Euler polynomials
$\mathcal{E}_{n,\lambda}(x)$ which are given by

(23)
$\frac{2}{e_{\lambda}(t)+1}e_{\lambda}^{x}(t)=\sum_{n=0}^{\infty}\mathcal{E}_{n,\lambda}(x)\frac{t^{n}}{n!}.$

By (20), we get

(24) $\displaystyle\sum_{n=0}^{\infty}F_{n,\lambda}\big{(}-\frac{1}{2}\
\big{|}\ r\big{)}\frac{t^{n}}{n!}$
$\displaystyle=\frac{1}{1+\frac{1}{2}(e_{\lambda}(t)-1)}e_{\lambda}^{r}(t)$
$\displaystyle=\frac{2}{e_{\lambda}(t)+1}e_{\lambda}^{r}(t)=\sum_{n=0}^{\infty}\mathcal{E}_{n,\lambda}(r)\frac{t^{n}}{n!}.$

Thus, we have

$\mathcal{E}_{n,\lambda}(r)=F_{n,\lambda}\big{(}-\frac{1}{2}\ \big{|}\
r\big{)}=\sum_{k=0}^{n}\big{(}-\frac{1}{2}\big{)}^{k}k!{n+r\brace
k+r}_{r,\lambda}.$

######  Theorem 5.

For $n\geq 0$, we have

$\mathcal{E}_{n,\lambda}(r)=\sum_{k=0}^{n}\big{(}-\frac{1}{2}\big{)}^{k}k!{n+r\brace
k+r}_{r,\lambda}.$

From Theorem 2 and Theorem 5, we note that

(25) $\big{(}-\frac{1}{2}\big{)}^{n}n!=\sum_{k=0}^{n}(-1)^{n-k}{n+r\brack
k+r}_{r,\lambda}\mathcal{E}_{k,\lambda}(r).$

Therefore, by (25), we obtain the following theorem.

######  Theorem 6.

For $n\geq 0$, we have

$\frac{n!}{2^{n}}=\sum_{k=0}^{n}(-1)^{k}{n+r\brack
k+r}_{r,\lambda}\mathcal{E}_{k,\lambda}(r).$

Recently, Kim-Kim introduced the degenerate polylogarithm function of index
$k$ which is given by

(26)
$\mathrm{Li}_{k,\lambda}(x)=\sum_{n=1}^{\infty}\frac{(-\lambda)^{n-1}(1)_{n,1/\lambda}}{(n-1)!n^{k}}x^{n},\quad(k\in\mathbb{Z}),\quad(\mathrm{see}\
[6]).$

In [9], the degenerate poly-Bernoulli polynomials of index $k$ are defined by

(27)
$\frac{\mathrm{Li}_{k,\lambda}(1-e_{\lambda}(-t))}{1-e_{\lambda}(-t)}e_{\lambda}^{-x}(-t)=\sum_{n=0}^{\infty}\beta_{n,\lambda}^{(k)}(x)\frac{t^{n}}{n!}.$

For $p\in\mathbb{Z}$, we have

(28)
$\beta_{n,\lambda}^{(p)}(-r)=(-1)^{n}\sum_{k=0}^{n}\frac{\lambda^{k}(1)_{k+1,1/\lambda}}{(k+1)^{p}}{n+r\brace
k+r}_{r,\lambda}.$

From Theorem 2 and (2), we have

(29) $\displaystyle\frac{\lambda^{n}(1)_{n+1,1/\lambda}}{(n+1)^{p}}$
$\displaystyle=\sum_{k=0}^{n}(-1)^{n-k}{n+r\brack
k+r}_{r,\lambda}(-1)^{k}\beta_{k,\lambda}^{(p)}(-r)$
$\displaystyle=\sum_{k=0}^{n}(-1)^{n}{n+r\brack
k+r}_{r,\lambda}\beta_{k,\lambda}^{(p)}(-r).$

Therefore, by (29), we obtain the following theorem.

######  Theorem 7.

For $n\geq 0$, we have

$\frac{(-\lambda)^{n}(1)_{n+1,1/\lambda}}{(n+1)^{p}}=\sum_{k=0}^{n}{n+r\brack
k+r}_{r,\lambda}\beta_{k,\lambda}^{(p)}(-r).$

##  3\. Further remark

The degenerate Bernoulli polynomials of the second kind $b_{n,\lambda}(x)$ are
defined by

(30)
$\frac{t}{\log_{\lambda}(1+t)}(1+t)^{x}=\sum_{n=0}^{\infty}b_{n,\lambda}(x)\frac{t^{n}}{n!}.$

In view of (7), we consider the degenerate hyperharmonic numbers which are
defined by

(31) $H_{n,\lambda}^{(1)}=H_{n,\lambda},\quad
H_{n,\lambda}^{(r)}=\sum_{k=1}^{n}H_{k,\lambda}^{(r-1)},\,\,(n\geq 1,r\geq
2),\quad H_{0,\lambda}^{(r)}=0,\,\,(r\geq 2).$

Assume that $r\geq 2$ and that
$-\frac{1}{(1-t)^{r-1}}\log_{\lambda}(1-t)=\sum_{n=1}^{\infty}H_{n,\lambda}^{(r-1)}t^{n}$.
From (31), we note that

(32) $\displaystyle\sum_{n=1}^{\infty}H_{n,\lambda}^{(r)}t^{n}$
$\displaystyle=\sum_{n=1}^{\infty}\sum_{k=1}^{n}H_{k,\lambda}^{(r-1)}t^{n}=\frac{1}{1-t}\sum_{k=1}^{\infty}H_{k,\lambda}^{(r-1)}t^{k}$
$\displaystyle=\frac{1}{1-t}\bigg{(}-\frac{1}{(1-t)^{r-1}}\log_{\lambda}(1-t)\bigg{)}=-\frac{1}{(1-t)^{r}}\log_{\lambda}(1-t).$

Therefore, by the induction step in (32), we obtain the following theorem.

######  Proposition 8.

Let $r$ be a positive integer. Then we have

$-\frac{1}{(1-t)^{r}}\log_{\lambda}(1-t)=\sum_{n=1}^{\infty}H_{n,\lambda}^{(r)}t^{n}.$

Now, by using (30) and Proposition 8, we observe that

(33) $\displaystyle t$
$\displaystyle=\frac{-t}{\log_{\lambda}(1-t)}(1-t)^{r}\frac{-\log_{\lambda}(1-t)}{(1-t)^{r}}$
$\displaystyle=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}b_{k,\lambda}(r)t^{k}\sum_{l=0}^{\infty}H_{l,\lambda}^{(r)}t^{l}$
$\displaystyle=\sum_{n=0}^{\infty}\bigg{(}\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}b_{k,\lambda}(r)H_{n-k,\lambda}^{(r)}\bigg{)}t^{n}.$

Comparing the coefficients on both sides of (33), we have

(34)
$\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}b_{k,\lambda}(r)H_{n-k,\lambda}^{(r)}=\delta_{n,1}.$

##  4\. Conclusion

The unsigned degenerate $r$-Stirling numbers of the first kind ${n\brack
k}_{r,\lambda}$ are degenerate versions of the unsigned $r$-Stirling numbers
of the first kind ${n\brack k}_{r}$ and the degenerate $r$-Stirling numbers of
the second kind ${n\brace k}_{r,\lambda}$ are degenerate versions of the
$r$-Stirling numbers of the second kind ${n\brace k}_{r}$. We derived from the
inverse relations for the degenerate $r$-Stirling numbers some identities
involving such numbers and some special numbers which are given by the
evaluations at $r$ of the fully degenerate Bernoulli polynomials, the
degenerate two variable Fubini polynomials, the degenerate Euler polynomials
and the degenerate poly-Bernoulli polynomials.

It is one of our future projects to continue to explore degenerate versions of
some special numbers and polynomials and their applications not only in
mathematics but also in other disciplines like statistics, physics,
engineering and social sciences.

## References

  * [1] Araci, S. _A new class of Bernoulli polynomials attached to polyexponential functions and related identities,_ Adv. Stud. Contemp. Math. (Kyungshang) 31 (2021), no. 2, 195-204.
  * [2] Broder, A. Z. _The $r$-Stirling numbers,_ Discrete Math. 49 (1984), no. 3, 241-259.
  * [3] Carlitz, L. _Degenerate Stirling, Bernoulli and Eulerian numbers,_ Utilitas Math. 15 (1979), 51-88.
  * [4] Comtet, L. _Advanced combinatorics. The art of finite and infinite expansions._ Revised and enlarged edition. D. Reidel Publishing Co., Dordrecht, 1974.
  * [5] Kim, D. S.; Jang, G.-W.; Kwon, H.-I.; Kim, T. _Two variable higher-order degenerate Fubini polynomials,_ Proc. Jangjeon Math. Soc. 21 (2018), no. 1, 5-22.
  * [6] Kim, D. S.; Kim, T. _A note on a new type of degenerate Bernoulli numbers,_ Russ. J. Math. Phys. 27 (2020), no. 2, 227-235.
  * [7] Kim, T.; Kim, D. S. _Some identities on truncated polynomials associated with degenerate Bell polynomials,_ Russ. J. Math. Phys. 28 (2021), no. 3, 342-355.
  * [8] Kim, T.; Kim, D. S. _On some degenerate differential and degenerate difference operator,_ Russ. J. Math. Phys. 29 (2022), no. 1, 37–47.
  * [9] Kim, T.; Kim, D. S. _Some formulas for fully degenerate Bernoulli numbers and polynomials,_ arxiv(2022).
  * [10] Kim, T.; Kim, D. S.; Lee, H.; Park, J.-W. _A note on degenerate $r$-Stirling numbers,_ J. Inequal. Appl. 2020, Paper No. 225, 12 pp.
  * [11] Kim, T.; Yao, Y.; Kim, D. S.; Jang, G.-W. _Degenerate $r$-Stirling numbers and $r$-Bell polynomials,_ Russ. J. Math. Phys. 25 (2018), no. 1, 44-58.
  * [12] Roman, S. _The umbral calculus,_ Pure and Applied Mathematics, 111. Academic Press, Inc. [Harcourt Brace Jovanovich, Publishers], New York, 1984.
  * [13] Simsek, Y. _Identities and relations related to combinatorial numbers and polynomials,_ Proc. Jangjeon Math. Soc. 20 (2017), no. 1, 127-135.