# Polynomization of the Bessenrodt–Ono type inequalities for $A$-partition
functions

Krystian Gajdzica  Institute of Mathematics
Faculty of Mathematics and Computer Science
Jagiellonian University in Cracow<EMAIL_ADDRESS>,
Bernhard Heim  Faculty of Mathematical and Natural Sciences
Mathematical Institute
University of Cologne
Weyertal 86–90
50931 Cologne
Germany<EMAIL_ADDRESS>Lehrstuhl A für Mathematik
RWTH Aachen University
52056 Aachen
Germany<EMAIL_ADDRESS>and  Markus Neuhauser  Kutaisi
International University
5/7
Youth Avenue
Kutaisi
4600 Georgia  Lehrstuhl A für Mathematik
RWTH Aachen University
52056 Aachen
Germany<EMAIL_ADDRESS>
###### Abstract.

For an arbitrary set or multiset $A$ of positive integers, we associate the
$A$-partition function $p_{A}(n)$ (that is the number of partitions of $n$
whose parts belong to $A$). We also consider the analogue of the $k$-colored
partition function, namely, $p_{A,-k}(n)$. Further, we define a family of
polynomials $f_{A,n}(x)$ which satisfy the equality $f_{A,n}(k)=p_{A,-k}(n)$
for all $n\in\mathbb{Z}_{\geq 0}$ and $k\in\mathbb{N}$. This paper concerns
the polynomization of the Bessenrodt–Ono type inequality for $f_{A,n}(x)$:

$\displaystyle f_{A,a}(x)f_{A,b}(x)>f_{A,a+b}(x),$

where $a$ and $b$ are arbitrary positive integers; and delivers some efficient
criteria for its solutions. Moreover, we also investigate a few basic
properties related to both functions $f_{A,n}(x)$ and $f_{A,n}^{\prime}(x)$.

###### Key words and phrases:

partition; restricted partition function; unrestricted partition function;
polynomization; Bessenrodt–Ono inequality.

###### 2020 Mathematics Subject Classification:

Primary 05A17, 11P82; Secondary 05A20.

##  1\. Introduction

Let $A$ be a multiset of positive integers and finite mutiplicities $\mu(a)$.
We consider a sequence of polynomials $\\{f_{A,n}(x)\\}_{n}$:

(1.1) $\sum_{n=0}^{\infty}f_{A,n}(x)\,q^{n}:=\prod_{a\in
A}\left(\frac{1}{1-q^{a}}\right)^{x}.$

The $n$th polynomial evaluated at positive integers $k$ provide the
$k$-colored $A$-partition function $p_{A,-k}(n)=f_{A,n}(k)$. Interesting
multisets are given by $A_{I}:=\\{1\\}$, $A_{II}:=\\{1,\ldots,1\\}$ with
$\mu(1)=m$, $A_{III}:=\mathbb{N}$, and $A_{IV}:=\\{1,2,2,3,3,3,4\ldots\\}$
with $\mu(a)=a$. Therefore,
$p_{A_{III},-1}=f_{A_{III},n}(1)=\mathrm{p}\left(n\right)$ the usual partition
function, and more general $p_{A_{III},-k}=f_{A_{III},n}(k)$ the $k$-colored
partition function. Note that

$f_{A_{I},n}(x)=\frac{1}{n!}\sum_{k=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{k}x^{k},$

where the coefficients are the unsigned Stirling numbers of the first kind.
Further, $f_{A_{II},n}(x)=f_{A_{I},n}(m\,x)$ and
$f_{A_{IV},n}(1)=\operatorname{pp}\left(n\right)$, where
$\operatorname{pp}\left(n\right)$ denotes the plane partition function.

In this paper we prove the following Bessenrodt–Ono type [9] of inequalities.
Let $a,b\in\mathbb{N}$ with $a+b>2$. Let $A$ be a multiset with $1\in A$. We
have

(1.2) $f_{A,a}(x)\,f_{A,b}(x)>f_{A,a+b}(x)$

  * (i)

for $x\geq 3$ and $A$ an ordinary set. Note that
$f_{A,1}(3)^{2}-f_{A,2}(3)=0$. We refer to Theorem 1.3.

  * (ii)

for $x\geq 5$ and $A$ a multiset with $\mu(a)\leq a$ for all all $a\in A$.
Note that $f_{A,1}(5)^{2}-f_{A,2}(5)=0$. We refer to Theorem 1.4.

Note that (ii) implies (i) for $x\geq 5$. Moreover, let $A$ be a set, $B$ the
multiset derived from $A$, where the elements of $B$ are the elements of $A$
with multiplicity $m$. Then

(1.3) $f_{B,n}(x)=f_{A,n}(m\,x).$

Let us recall some basic definitions and known results. For a non-negative
integer $n$, the partition function $\mathrm{p}\left(n\right)$ enumerates all
possible sequences

$\lambda=(\lambda_{1},\lambda_{2},\ldots,\lambda_{j})$

of positive integers such that
$\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{j}\geq 1$ and

$n=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{j}.$

Elements $\lambda_{i}$ are called the parts of the partition $\lambda$. For
example, there are $5$ partitions of $4$: $(4)$, $(3,1)$, $(2,2)$, $(2,1,1)$
and $(1,1,1,1)$. Thus, $\mathrm{p}\left(4\right)=5$. It should be pointed out
that $\mathrm{p}\left(n\right)=0$ if $n$ is negative, and
$\mathrm{p}\left(0\right)=1$—as there is only the empty partition in this
case. In the 1740’s Euler discovered the generating function for
$\mathrm{p}\left(n\right)$

$\sum_{n=0}^{\infty}\mathrm{p}\left(n\right)q^{n}=\prod_{i=1}^{\infty}\frac{1}{1-q^{i}}.$

There is a wealth of literature devoted to the partition theory. For a general
introduction to the subject, we encourage the reader to see Andrews’ books [4,
5] as well as [1, 24, 31, 34].

However, it also motivated to investigate other similar problems for
combinatorial sequences. For instance, in 2016 Bessenrodt and Ono [9] proved
the following result.

######  Theorem 1.1 (Bessenrodt, Ono).

Let $a$ and $b$ be natural numbers such that $a+b>9$. Then

$\mathrm{p}\left(a\right)\mathrm{p}\left(b\right)>\mathrm{p}\left(a+b\right).$

Their proof is based on asymptotic estimates due to Rademacher [32] and Lehmer
[27]. It is worth noting that there are also other proofs. Alanazi, Gagola,
and Munagi [2] show the above theorem in a combinatorial manner. On the other
hand, Heim and Neuhauser [21] present the reasoning via induction on $a+b$ to
obtain the aforementioned property.

It should be pointed out that Theorem 1.1 is not only art of art’s sake, but
it allows us to determine the value of

$\max{\mathrm{p}\left(n\right)}=\max{\left(\mathrm{p}\left(\bm{\lambda}\right):\bm{\lambda}\text{
is a partition of }n\right)},$

where
$\mathrm{p}\left(\bm{\lambda}\right)=\prod_{i=1}^{j}\mathrm{p}\left(\lambda_{i}\right)$
for $\bm{\lambda}=(\lambda_{1},\lambda_{2},\ldots,\lambda_{j})$. For more
details, we encourage to see Bessenrodt and Ono’s article [9].

Since 2016, a lot of people have been examining analogous inequalities for
various types of partition functions, among others, Chern, Fu, and Tang [10];
Beckwith and Bessenrodt [8]; Hou and Jagadeesan [25]; Males [29]; Dawsey and
Masri [13]; Heim, Neuhauser, and Tröger [23]; Gajdzica [16].

However, it turns out that we can also pass from the discrete task to the
continuous one by considering the appropriate family of polynomials—that is
Gian–Carlo Rota’s advice to study problems in combinatorics and number theory
in terms of roots of polynomials. In 1955, Newman [30] began studying a family
of polynomials $P_{n}(x)$ with exceptional properties. These polynomials are
defined via the recursive formula $P_{0}(x):=1$ and

$P_{n}(x):=\frac{x}{n}\sum_{j=1}^{n}\sigma(j)P_{n-j}(x)$

for $n\geq 1$, where $\sigma(m):=\sum_{\ell\mid m}\ell$ is the sum of divisors
of $m\in\mathbb{N}$. Each polynomial $P_{n}(x)$ is of degree $n$, and is
integer-valued. Further, these functions were investigated by Heim et al. [19,
20, 22]. In particular, they exhibited a practical formula for the derivatives
$P_{n}^{\prime}(x)$ for all $n\in\mathbb{N}$ [19], namely,

$P_{n}^{\prime}(x)=\sum_{j=1}^{n}\frac{\sigma(j)}{j}P_{n-j}(x).$

Moreover, Heim and Neuhauser [20] showed that the equality

$P_{n}(k)=p_{-k}(n)$

is true for all positive integers $k$ and $n$, where $p_{-k}\left(n\right)$
denotes the number of $k$-colored partitions of $n$—in other words the
generating function corresponding to $p_{-k}(n)$ is given by

$\sum_{n=0}^{\infty}p_{-k}(n)q^{n}=\prod_{i=1}^{\infty}\frac{1}{(1-q^{i})^{k}}.$

Furthermore, Heim, Neuhauser, and Tröger [22] proved the Bessenrodt–Ono type
inequality for polynomials $P_{n}(x)$. More specifically, they obtained the
following result.

######  Theorem 1.2 (Heim, Neuhauser, Tröger).

Let $a$ and $b$ be positive integers such that $a+b>2$, and let $x>2$. Then,
we have that

$P_{a}(x)P_{b}(x)>P_{a+b}(x).$

If $x=2$, then the above inequality is valid for all $a+b>4$.

It should be pointed out that there are more results of that kind for other
families of polynomials. For example, Heim, Neuhauser, and Tröger [23]
considered a family polynomials associated to the so-called plane partition
function $\operatorname{pp}\left(n\right)$. That is the number of partitions
of $n$ such that each part $j$ might appear in $j$ distinct colors. The
generating function for $\operatorname{pp}\left(n\right)$ takes the form

$\sum_{n=0}^{\infty}\operatorname{pp}\left(n\right)q^{n}=\prod_{i=1}^{\infty}\frac{1}{(1-q^{i})^{i}}.$

For more information about plane partitions and the plane partition function,
we refer the reader to [4, Section 12], [35, Section 7.20], and [26]. Further,
Li [28] obtained the Bessenrodt–Ono type inequalities for the overpartition
function $\overline{p}(n)$ and its polynomization. Let us recall that the
overpartition function $\overline{p}(n)$ counts the number of partitions of
$n$ such that the last occurrence of each distinct part may be overlined. In
that case the corresponding generating function is given by

$\sum_{n=0}^{\infty}\overline{p}(n)q^{n}=\prod_{i=1}^{\infty}\frac{1+q^{i}}{1-q^{i}}.$

For more details related to overpartitions and the overpartition function, see
[12].

In this paper we deal with the Bessenrodt–Ono type inequalities for the so-
called $A$-partition function and its polynomization. Let $A$ be an arbitrary
(finite or infinite) set or multiset of positive integers. By an $A$-partition
of a non-negative integer $n$, we mean any partition
$\lambda=(\lambda_{1},\lambda_{2},\ldots,\lambda_{j})$ of $n$ such that
$\lambda_{i}\in A$ for each $1\leq i\leq j$. For the sake of clarity, we also
assume that two $A$-partitions of $n$ are considered the same if there is only
a difference in the order of their parts. The $A$-partition function
$p_{A}(n)$ enumerates all possible $A$-partitions of $n$. Additionally, we set
$p_{A}(n)=0$ if $n$ is negative, and $p_{A}(0)=1$. For example, if
$A=\\{1,2,{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}2},3,{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3}\\}$,
then $p_{A}(4)=8$ with:
$({\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3},1)$,
$(3,1)$,
$({\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}2},{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}2})$,
$({\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}2},2)$,
$(2,2)$,
$({\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}2},1,1)$,
$(2,1,1)$, and $(1,1,1,1)$. The generating function for $p_{A}(n)$ satisfies
the equality

(1.4) $\sum_{n=0}^{\infty}p_{A}(n)q^{n}=\prod_{a\in A}\frac{1}{1-q^{a}}.$

There is an abundance of literature devoted to $A$-partitions, we refer the
reader to, for instance, [3, 6, 7, 11, 14, 15, 17, 33, 36, 37].

It is proven in the next section that one can assign unique polynomials
$f_{A,n}(x)$ to the given $A$-partition function in such a way that
$f_{A,n}(k)=p_{A,-k}(n)$ holds for every natural number $k$ and non-negative
integer $n$, where $p_{A,-k}(n)$ is, similarly to the case of the classical
partition function, the $k$-colored $A$-partition function.

Further, we proceed to the main part of the paper. Using the law of induction
and the analogous methods to those presented in Heim, Neuhauser, and Tröger’s
paper [22], we prove the following result.

######  Theorem 1.3.

For any set $A\subseteq\mathbb{N}$ with $1\in A$, the inequality

$f_{A,a}(x)f_{A,b}(x)>f_{A,a+b}(x)$

holds for all $x\geq 3$ and $a,b\in\mathbb{N}$ except $a=b=1$. For $a=b=1$, we
have that

$f_{A,1}(3)f_{A,1}(3)\geq f_{A,2}(3)$

with equality whenever $2\in A$ and

$f_{A,1}(x)f_{A,1}(x)>f_{A,2}(x)$

for all $x>3$.

It is worth pointing out that the above property significantly extends Theorem
1.2—in a sense that instead of taking $\mathbb{N}$, we can consider any finite
or infinite set. For example, one may apply it to obtain the polynomization of
the Bessenrodt–Ono type inequality for the $k$-regular partition function for
every $k\geq 2$. That is an extension of Beckwith and Bessenrodt’s theorem,
see [8, Theorem 2.1].

Moreover, we also show a similar result for some family of multisets $A$. More
precisely, we figure out the following.

######  Theorem 1.4.

If $A$ is such a multiset of natural numbers that $1$ occurs exactly once in
$A$, and each number $j$ appears at most $j$ times in $A$, then the inequality

$f_{A,a}(x)f_{A,b}(x)>f_{A,a+b}(x)$

holds for all $x\geq 5$ and $a,b\in\mathbb{N}$ except $a=b=1$. For $a=b=1$, we
have that

$f_{A,1}(5)f_{A,1}(5)\geq f_{A,2}(5)$

with equality whenever $2$ appears exactly two times in $A$ and

$f_{A,1}(x)f_{A,1}(x)>f_{A,2}(x)$

for every $x>5$.

On the other hand, the above theorem generalizes Heim, Neuhauser, and Tröger’s
result devoted to the plane partition function [23, Theorem 1.3].

This manuscript is organized as follows. In Section 2, we introduce necessary
notations, define the appropriate family of polynomials $f_{A,n}(x)$
associated to the fixed $A$-partition function and show some of their basic
properties. Section 3 deals with the Bessenrodt–Ono inequalities for both
$k$-colored $A$-partition functions $p_{A,-k}(n)$ and the polynomials
$f_{A,n}(x)$ for such sets $A$ that $1\in A$ ($\mu(a)=1$ for each $a\in A$).
Finally, Section 4 examines the analogues problems to those from Section 3 for
such multisets $A$ that $1\in A$ and $\mu(j)\leq j$ for every
$j\in\mathbb{N}.$

##  2\. Preliminaries

At first, let us fix some notations and conventions. By $\mathbb{R}$,
$\mathbb{N}$, and $\mathbb{Z}_{\geq 0}$ we denote the set of real numbers, the
set of positive integers, and the set of non-negative integers, respectively.
For an arbitrary multiset of positive integers $A$, $\mu(a)$ denotes the
number of appearances of the number $a$ in $A$. Further, for a fixed finite or
infinite unbounded set or multiset of positive integers $A$, the $A$-partition
function $p_{A}(n)$ is the number of $A$-partitions of $n$ (actually, if $A$
is infinite, we have to require that $\mu(a)<\infty$ for every $a\in A$,
otherwise there will be infinitely many numbers $m$ such that
$p_{A}(m)=\infty$). In such a setting, we also introduce a family of
polynomials associated to the $A$-partition function via the power series
expansion (1.1).

It is straightforward to see that $f_{A,0}(x)=1$ and $f_{A,n}(1)=p_{A}(n)$.
Furthermore, if we consider the $k$-colored $A$-partition function
$p_{A,-k}(n)$, then we have that $p_{A,-k}(n)=f_{A,n}(k)$ for all
$k\in\mathbb{N}$ and $n\in\mathbb{Z}_{\geq 0}$. Moreover, there are recurrence
formulae for the polynomials $f_{A,n}(x)$ and $f_{A,n}^{\prime}(x)$, as well.

######  Proposition 2.1.

For a fixed set (or multiset) $A$ of positive integers, the following
identities

(2.1) $f_{A,n}(x)=\frac{x}{n}\sum_{j=1}^{n}\sigma_{A}(j)f_{A,n-j}(x)$

and

(2.2) $f_{A,n}^{\prime}(x)=\sum_{j=1}^{n}\frac{\sigma_{A}(j)}{j}f_{A,n-j}(x),$

are true for all $n\in\mathbb{N}$, where $\sigma_{A}(i)$ is a sum of those
divisors of $i$ that belong to $A$.

###### Proof.

If we denote the left hand side of the equality $(\ref{def:polynomization})$
by $F(q,x)$ and compare the appropriate coefficients standing next to $q^{n}$
of both

$\displaystyle\frac{\partial F(q,x)}{\partial
q}=\frac{\partial\log{(F(q,x))}}{\partial q}F(q,x)$

and

$\displaystyle\frac{\partial F(q,x)}{\partial
x}=\frac{\partial\log{(F(q,x))}}{\partial x}F(q,x),$

we obtain the required property. ∎

Now, it is not difficult to notice that if $1\in A$, then $f_{A,n}(x)$ is a
polynomial of degree $n$. Otherwise, we have that $f_{A,1}(x)=0$. In order to
omit such a confusion, we assume that henceforth $1\in A$ and $1$ occurs
exactly one time in $A$. In addition, that assumption asserts that $\gcd A=1$
which is a necessary and sufficient condition to have the Bessenrodt–Ono type
inequality for $p_{A}(n)$ when $A$ is a finite set or multiset, see [16].
Next, we introduce an additional notation, namely,

(2.3) $\displaystyle\Delta_{A,n}(x):=f_{A,n+1}(x)-f_{A,n}(x).$

The following characterization follows from the above discussion.

######  Theorem 2.2.

Let $n\in\mathbb{N}$ and $x\in\mathbb{R}$ with $x\geq 1$. Then

$f_{A,n}(x)\leq f_{A,n+1}(x)\hskip 14.22636pt\text{and}\hskip 14.22636pt1\leq
f_{A,n}^{\prime}(x)<f_{A,n+1}^{\prime}(x).$

For $x>1$, we additionally have that

$f_{A,n}(x)<f_{A,n+1}(x).$

###### Proof.

At first, we show that $f_{A,n}(x)\leq f_{A,n+1}(x)$. The proof is by
induction on $n$ under the assumption that $x\geq 1$. Let us notice that
$f_{A,1}(x)=x$ and $f_{A,2}(x)=x(x+(1+2s_{2}))/2$, where $s_{2}$ is the number
of $2$s appearing in the multiset $A$. Thus, we get that

$\displaystyle\Delta_{A,1}(x)\geq\frac{x}{2}(x-1)\geq 0$

for every $x\geq 1$. Now, we assume that the statement is valid for all $x\geq
1$ and $1\leq m\leq n-1$, and check its correctness for $m=n$. The identity
(2.2) asserts that

$f_{A,n+1}^{\prime}(x)>\sum_{j=1}^{n}\frac{\sigma_{A}(j)}{j}f_{A,n+1-j}(x)\geq\sum_{j=1}^{n}\frac{\sigma_{A}(j)}{j}f_{A,n-j}(x)>f_{A,n}^{\prime}(x).$

Thus, we have that the inequality $f_{A,n+1}^{\prime}(x)>f_{A,n}^{\prime}(x)$
is satisfied for every $x\geq 1$. One can also notice that

$f_{A,n+1}(1)=p_{A}(n+1)=p_{A}(n)+p_{A\setminus\\{1\\}}(n+1)\geq f_{A,n}(1).$

Hence, the function $\Delta_{A,n}^{\prime}(x)$ is positive for every $x\geq
1$, and the inequality $f_{A,n+1}(x)\geq f_{A,n}(x)$ holds for all such $x$,
as required. The second part of the statement follows directly from the above
discussion. ∎

Now, we are ready to investigate the Bessenrodt–Ono inequality for the
polynomials $f_{A,n}(x)$. We divide our consideration into a few section
depending on the structure of $A$.

##  3\. The Bessenrodt–Ono inequality for an arbitrary set $A$ with $1\in A$

In this section, we assume that $A\subseteq\mathbb{N}$ and $A$ is a
set—meaning that each number $j$ might occur at most once in $A$. Moreover, we
also require that $1\in A$. At the beginning, let us consider a very useful
example when $A$ is a singleton.

######  Lemma 3.1.

For the singleton $A_{I}=\\{1\\}$, the equality

$f_{A_{I},n}(x)=\frac{1}{n!}\prod_{i=0}^{n-1}(x+i)=\binom{x+n-1}{n}$

holds for every $n\in\mathbb{Z}_{\geq 0}$.

###### Proof.

To prove the required property, it is enough to systematically utilize the
formula $(\ref{1})$. ∎

It is worth pointing out that the above polynomials are naturally connected
with a well-known combinatorial sequence. The unsigned Stirling number of the
first kind $\genfrac{[}{]}{0.0pt}{}{k}{l}$ is the number of permutations of
$k$ elements with exactly $l$ cycles. Equivalently, it may be defined as the
appropriate coefficient of the rising factorial (Pochhammer function), namely

$\displaystyle
x^{\overline{n}}:=x(x+1)\cdots(x+n-1)=\sum_{i=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{i}x^{i}.$

For more information about Stirling numbers, we refer the reader to [18,
Chapter 6]. Therefore, it is clear that the equality

$\displaystyle
f_{A_{I},n}(x)=\frac{1}{n!}\sum_{i=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{i}x^{i}$

is valid for all $n\in\mathbb{Z}_{\geq 0}$.

Furthermore, let us notice that for any other set (or multiset)
$B\subseteq\mathbb{N}$ with $1\in B$, we have that

$f_{A_{I},n}(x)\leq f_{B,n}(x)$

for all $n\geq 0$ and $x\geq 0$—that is a direct consequence of the equality
(2.1). Moreover, we may easily extend Lemma 3.1 to any finite multiset of
ones.

######  Corollary 3.2.

For an arbitrary positive integer $m$ and the multiset
$A_{II}=\\{1,\ldots,1\\}$ with $\mu(1)=m$, the equality

$f_{A_{II},n}(x)=f_{A_{I},n}(mx)$

is satisfied for every $n\in\mathbb{Z}_{\geq 0}$.

Before we go to the main part of this section, let us take a look on a
concrete family of polynomials $f_{A,n}(x)$. For instance, if we set
$A=\\{1,2,3,4,5\\}$, then one can easily calculate the corresponding
polynomials $f_{A,n}(x)$ for small values of $n$. We present all the cases for
$1\leq n\leq 7$ in Table 1.

$n$ | $f_{A,n}(x)$
---|---
$1$ | $x$
$2$ | $\frac{1}{2}x(x+3)$
$3$ | $\frac{1}{6}x(x+1)(x+8)$
$4$ | $\frac{1}{24}x(x+1)(x+3)(x+14)$
$5$ | $\frac{1}{120}x(x+3)(x+6)\left(x^{2}+21x+8\right)$
$6$ | $\frac{1}{720}x(x+2)(x+5)\left(x^{3}+38x^{2}+289x+72\right)$
$7$ | $\frac{1}{5040}x\left(x^{6}+63x^{5}+1225x^{4}+9345x^{3}+28294x^{2}+25872x+720\right)$
Table 1. The polynomials $f_{A,n}(x)$ for $A=\\{1,2,3,4,5\\}$ and $1\leq n\leq
7$.

In order to gain some intuition related to the behaviour of the polynomials
$f_{A,a}(x)f_{A,b}(x)-f_{A,a+b}(x)$, one can determine their roots. Figure 1
shows all their complex roots of $1\leq a,b\leq 10$.

Figure 1. Complex roots of the polynomials $f_{A,a}(x)f_{A,b}(x)-f_{A,a+b}(x)$
for $A=\\{1,2,3,4,5\\}$ and $1\leq a,b\leq 10$.

Now, our aim is to find a Bessenrodt–Ono type inequality for the $3$-colored
$A$-partition function—that is $p_{A,-3}(n)=f_{A,n}(3)$. That result will play
a central role in the proof of the main theorem of this section. To deal with
the task, we show the reasoning by induction on $a+b$. Actually, we combine
methods recently utilized by Heim, Neuhauser, and Tröger in [21, 22].

######  Theorem 3.3.

For every set $A\subseteq\mathbb{N}$ with $1\in A$, the inequality

$f_{A,a}(3)f_{A,b}(3)>f_{A,a+b}(3)$

holds for all pairs $(a,b)\in\mathbb{N}^{2}$ except for $(1,1)$. In that case,
we have that

$f_{A,1}(3)f_{A,1}(3)\geq f_{A,2}(3)$

with equality whenever $2\in A$.

###### Proof.

At first, let us notice that for each $n\geq 0$, we have
$f_{A,n}(3)f_{A,0}(3)=f_{A,n}(3)$. We prove the statement by induction on
$a+b$. Let us assume that $a+b=2$. Since

$\displaystyle f_{A,2}(3)=\begin{cases}9,&\text{if }2\in A,\\\ 6,&\text{if
}2\not\in A,\end{cases}$

it is straightforward to see that

$\displaystyle f_{A,1}(3)f_{A,1}(3)=9\geq f_{A,2}(3).$

On the other hand, if $a+b=3$, then

$\displaystyle f_{A,3}(3)=\begin{cases}22,&\text{if }2,3\in A,\\\ 19,&\text{if
}2\in A\text{ and }3\not\in A,\\\ 13,&\text{if }2\not\in A\text{ and }3\in
A,\\\ 10,&\text{if }2,3\not\in A,\end{cases}$

and the inequality

$f_{A,1}(3)f_{A,2}(3)>f_{A,3}(3)$

follows. Next, we check the validity of the statement for all pairs $(a,b)$
such that $a,b\leq 14$ and possible choices for the set $A$. The procedure
which allows us to deal with this task is specifically described in Lemma 3.4.
Now, we suppose that the statement is valid for all pairs
$(a,b)\in\mathbb{N}^{2}$ with $3\leq a+b\leq n$, and check whether it is also
true for $a+b=n+1.$ Without loss of generality we may assume that $a\geq b\geq
1$ and $a\geq 15$. Employing the formula (2.1) to the difference
$f_{A,a}(3)f_{A,b}(3)-f_{A,a+b}(3)$, we get

$\displaystyle\frac{3}{a}\sum_{j=1}^{a}\sigma_{A}(j)f_{A,a-j}(3)f_{A,b}(3)-\frac{3}{a+b}\sum_{i=1}^{a+b}\sigma_{A}(i)f_{A,a+b-i}(3)$
$\displaystyle>$
$\displaystyle\frac{3}{a}\sum_{j=1}^{a}\sigma_{A}(j)f_{A,a-j}(3)f_{A,b}(3)-\frac{3}{a+b}\sum_{i=1}^{a}\sigma_{A}(i)f_{A,a-i}(3)f_{A,b}(3)$
$\displaystyle\phantom{\frac{3}{a}\sum_{j=1}^{a}\sigma_{A}(j)f_{A,a-j}(3)f_{A,b}(3)}-\frac{3}{a+b}\sum_{i=1}^{b}\sigma_{A}(a+i)f_{A,b-i}(3)$
$\displaystyle=$
$\displaystyle\frac{3}{a(a+b)}\left(b\sum_{j=1}^{a}\sigma_{A}(j)f_{A,a-j}(3)f_{A,b}(3)-a\sum_{i=1}^{b}\sigma_{A}(a+i)f_{A,b-i}(3)\right)$

where the inequality follows directly from the induction hypothesis which is
applied to $f_{A,a+b-i}(3)$ for each $i$ between $1$ and $a$. Thus, it is
enough to prove that the expression in the bracket is non-negative. To make
the text more transparent, let us set

$\displaystyle\kappa_{a}:=\sum_{j=1}^{a}\sigma_{A}(j)f_{A,a-j}(3).$

We obtain that

$\displaystyle
b\kappa_{a}f_{A,b}(3)-a\sum_{i=1}^{b}\sigma_{A}(a+i)f_{A,b-i}(3)$
$\displaystyle=$ $\displaystyle
3\kappa_{a}\sum_{i=1}^{b}\sigma_{A}(i)f_{A,b-i}(3)-a\sum_{i=1}^{b}\sigma_{A}(a+i)f_{A,b-i}(3)$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{b}\left(3\kappa_{a}\sigma_{A}(i)-a\sigma_{A}(a+i)\right)f_{A,b-i}(3).$

Since $1\leq\sigma_{A}(n)\leq\sigma(n)\leq n(1+\ln{n})$ for every $n\geq 1$
and $a\geq b$, we further have that

$\displaystyle 3\kappa_{a}-2a^{2}(1+\ln{(2a)})$ $\displaystyle\geq
3\sum_{j=0}^{a-1}f_{A,j}(3)-2a^{2}(1+\ln{(2a)})$ ($\star$) $\displaystyle\geq
3\binom{a+2}{3}-2a^{2}(1+\ln{(2a)}),$

where the last inequality is a consequence of Lemma 3.1. Hence, it is enough
to show that the function

$\displaystyle\Psi(x):=(x+1)(x+2)-4x(1+\ln{(2x)})$

is non-negative for all $x\geq 15$; and it is not a difficult problem to
solve. The proof is complete by the law of induction. ∎

For the sake of clarity, let us complete the proof by dealing with the
remaining cases for $1\leq a,b\leq 14$.

######  Lemma 3.4.

For any set $A\subseteq\mathbb{N}$ such that $1\in A$, the inequality

$f_{A,a}(3)f_{A,b}(3)>f_{A,a+b}(3)$

holds for all such pairs $a,b\in\mathbb{N}$ that $a,b\leq 14$ and $a+b>2$.

###### Proof.

In order to prove the lemma, we carry out appropriate computations in Wolfram
Mathematica [38]. At the beginning, we assume that $a,b\leq 10$, $a+b\geq 3$,
and consider all possible subsets $A$ of the set $\\{1,2,\ldots,a+b\\}$ such
that $1\in A$. Observe that, in fact, it is enough to consider the
aforementioned subsets. Otherwise, if $j\in A$ and $j>a+b$, then $j$ can not
appear as a part of the partitions of $a$, $b$ and $a+b$. Hence, any such $j$
does not influence the values of $f_{A,a}(3)$, $f_{A,b}(3)$, and
$f_{A,a+b}(3)$. Considering all possible choices of $A$, and numbers $a$ and
$b$ such that $a,b\leq 10$, we check the correctness of the statement for
every such a triple. Since in all of these cases the set $A$ is finite, it
follows from Bell’s theorem (see, [7, 16]) that the function $f_{A,n}(3)$ is a
quasi-polynomial, namely, the expression of the form

$f_{A,n}(3)=c_{l}(n)n^{l}+c_{l-1}(n)n^{l-1}+\cdots+c_{0}(n),$

where $l=3|A|-1$ and all of these coefficients $c_{i}(n)$ depend on the
residue class of $n\mkern 4.0mu({\operator@font mod}\mkern
6.0mu\operatorname{lcm}(A))$. Thus, we may explicitly determine its values and
directly verify the claim.

Afterwards, we slightly modify our reasoning for all pairs $a$ and $b$ such
that $a\geq b\geq 1$ and $11\leq a\leq 14$. Mainly because the number of
possible choices of the set $A$ rapidly grows with the value of $a+b$, which,
on the other hand, significantly influences the time of the calculations.
Thus, the idea is to assume something more about the set $A$. For instance,
let us suppose that $1,2\in A$, then $f_{A,n}(3)\geq f_{\\{1,2\\},n}(3)$ for
each non-negative integer $n$. However, we can directly derive the formula for
$f_{\\{1,2\\},n}(3)$ (due to Bell’s theorem [7]) which takes the form

$\displaystyle
f_{\\{1,2\\},n}(3)=\begin{cases}\frac{n^{5}}{960}+\frac{3n^{4}}{128}+\frac{19n^{3}}{96}+\frac{25n^{2}}{32}+\frac{173n}{120}+1,&\text{if
}2\mid n,\\\
\frac{n^{5}}{960}+\frac{3n^{4}}{128}+\frac{19n^{3}}{96}+\frac{49n^{2}}{64}+\frac{1249n}{960}+\frac{91}{128},&\text{if
}2\nmid n.\end{cases}$

Now, taking minima of the corresponding coefficients of the above polynomials,
it is straightforward to see that

$\displaystyle
f_{\\{1,2\\},n}(3)\geq\frac{n^{5}}{960}+\frac{3n^{4}}{128}+\frac{19n^{3}}{96}+\frac{49n^{2}}{64}+\frac{1249n}{960}+\frac{91}{128}$

holds for every $n\geq 0$. Therefore, we might replace the lower bound in
($\star$ ‣ 3) in the proof of Theorem 3.3 by the right hand side of the above
inequality, and verify that

$\displaystyle
3\sum_{n=1}^{a-1}\left(\frac{n^{5}}{960}+\frac{3n^{4}}{128}+\frac{19n^{3}}{96}+\frac{49n^{2}}{64}+\frac{1249n}{960}+\frac{91}{128}\right)\geq
2a^{2}(1+\ln{(2a)})$

holds for every $a>4$. Actually, we might present a similar reasoning also for
$f_{\\{1,r\\},n}(3)$, where $r\in\\{3,4,5,6,7\\}$ and obtain that the required
inequality for $f_{\\{1,r\\},n}(3)$ is satisfied for all $a\geq 11$ and $b\geq
1$. Hence, in order to deal with the remaining cases that are those
corresponding to $11\leq a\leq 14$ and $b\leq a$, it is enough to repeat the
above method for all possible subsets $A$ of $\\{1,8,9,\ldots,a+b\\}$. ∎

Before we pass to the central part of this section, let us point out some
consequences of Theorem 3.3.

######  Corollary 3.5.

For any finite (or infinite) multiset $B$ such that $1\in B$ and $\mu(j)=3$
for each number $j\in B$, we have that

$\displaystyle p_{B}(a)p_{B}(b)>p_{B}(a+b)$

is true for all positive integers $a$ and $b$ (except possibly $a=b=1$
whenever $2\in B$).

######  Remark 3.6.

It should be noted that Corollary 3.5 significantly improves Gajdzica’s result
[16, Theorem 5.4] related to the Bessenrodt–Ono inequality for all multisets
$B$ of the form $B=\\{1,1,1,a_{2},a_{2},a_{2},\ldots,a_{k},a_{k},a_{k}\\}$
with $1<a_{2}<\cdots<a_{k}$. More precisely, his theorem asserts that

$\displaystyle p_{B}(a)p_{B}(b)>p_{B}(a+b)$

holds for all
$a,b\geq\frac{2}{3k}\prod_{i=1}^{3k}(1+3ik\cdot\operatorname{lcm}{(a_{2},a_{3},\ldots,a_{k})})+2$.
On the other hand, Corollary 3.5 maintains that the above inequality is valid
for all positive integers $a$ and $b$ (except possibly $a=b=1$ whenever $2\in
B$).

Now, it is time to present the proof of Theorem 1.3.

###  3.1. Proof of Theorem 1.3

###### Proof.

We show the claim by induction on $a+b$. For $a+b=2$, we have that

$\displaystyle f_{A,1}(x)f_{A,1}(x)-f_{A,2}(x)\geq x^{2}-x/2(x+3)=x(x-3)/2,$

which is a consequence of the equality

$\displaystyle f_{A,2}(x)=\begin{cases}x(x+3)/2,&\text{if }2\in A,\\\
x(x+1)/2,&\text{if }2\not\in A.\end{cases}$

For $a+b=3$, on the other hand, one can determine that

$\displaystyle f_{A,3}(x)=\begin{cases}x(x+1)(x+8)/6,&\text{if }2,3\in A,\\\
x\left(x^{2}+9x+2\right)/6,&\text{if }2\in A\text{ and }3\not\in A,\\\
x\left(x^{2}+3x+8\right)/6,&\text{if }2\not\in A\text{ and }3\in A,\\\
x(x+1)(x+2)/6,&\text{if }2,3\not\in A.\end{cases}$

Hence, we get that

$\displaystyle
f_{A,2}(x)f_{A,1}(x)-f_{A,3}(x)=\begin{cases}x(x-2)(x+2)/3,&\text{if }3\in
A,\\\ x(x-1)(x+1)/3,&\text{if }3\not\in A,\end{cases}$

and the claim follows for $a+b=3$. Thus, let us assume that the statement is
valid for all $3\leq a+b\leq n$, and examine whether it is also correct for
$a+b=n+1$ with $a$ and $b$ such that $1\leq b\leq a$ and $a\geq 2$. Theorem
3.3 points out that the inequality $f_{A,a}(3)f_{A,b}(3)>f_{A,a+b}(3)$ is true
for all positive integers $a$ and $b$ such that $a+b>2$. Now, it suffices to
show that the derivative

$\displaystyle\frac{d}{dx}\left(f_{A,a}(x)f_{A,b}(x)-f_{A,a+b}(x)\right)$

is positive for all $x\geq 3$, because this implies that
$f_{A,a}(x)f_{A,b}(x)-f_{A,a+b}(x)>0$. Since
$f_{A,n}(x)f_{A,0}(x)=f_{A,n}(x)$, we have

$\displaystyle
f_{A,a}^{\prime}(x)f_{A,b}(x)+f_{A,a}(x)f_{A,b}^{\prime}(x)-f_{A,a+b}^{\prime}(x)$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{a}\frac{\sigma_{A}(j)}{j}f_{A,a-j}(x)f_{A,b}(x)+\sum_{j=1}^{b}\frac{\sigma_{A}(j)}{j}f_{A,a}(x)f_{A,b-j}(x)-\sum_{j=1}^{a+b}\frac{\sigma_{A}(j)}{j}f_{A,a+b-j}(x)$
$\displaystyle>$
$\displaystyle\sum_{j=1}^{a}\frac{\sigma_{A}(j)}{j}f_{A,a-j}(x)f_{A,b}(x)+\sum_{j=1}^{b}\frac{\sigma_{A}(j)}{j}f_{A,a}(x)f_{A,b-j}(x)$
$\displaystyle-$
$\displaystyle\sum_{j=1}^{a}\frac{\sigma_{A}(j)}{j}f_{A,a-j}(x)f_{A,b}(x)-\sum_{j=1}^{b}\frac{\sigma_{A}(a+j)}{a+j}f_{A,b-j}(x)$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{b}\left(\frac{\sigma_{A}(j)}{j}f_{A,a}(x)-\frac{\sigma_{A}(a+j)}{a+j}\right)f_{A,b-j}(x).$

Therefore, we can just show that

$\displaystyle\frac{\sigma_{A}(j)}{j}f_{A,a}(x)-\frac{\sigma_{A}(a+j)}{a+j}\geq
0$

for each $j$ between $1$ and $b$. Since $a\geq b$ and
$1\leq\sigma_{A}(n)\leq\sigma(n)\leq n(1+\ln{n})$ for each $n\geq 1$, it is
easy to see that the inequality

$\displaystyle\frac{\sigma_{A}(j)}{j}f_{A,a}(x)-\frac{\sigma_{A}(a+j)}{a+j}\geq\frac{1}{a}\binom{x+a-1}{a}-(1+\ln{(2a)})$

is valid for every such a $j$. However, the assumption that $x\geq 3$ asserts
that

$\displaystyle\frac{1}{a}\binom{x+a-1}{a}-(1+\ln{(2a)})\geq(a+1)(a+2)/(2a)-1-\ln{(2a)}.$

Thus, it suffices to prove that the function

$\displaystyle\psi(x):=(x+1)(x+2)-2x(1+\ln{(2x)})$

is positive for all real numbers $x\geq 3$, and this is not difficult to
verify. ∎

Since Theorem 1.3 is a very general result, it delivers us a lot of meaningful
consequences.

######  Corollary 3.7.

For the set $A_{III}=\mathbb{N}$, the inequality

$f_{A_{III},a}(x)f_{A_{III},b}(x)>f_{A_{III},a+b}(x)$

is valid for all $x\geq 3$ and $a,b\in\mathbb{N}$ except $a=b=1$. For $a=b=1$,
we have that

$f_{A_{III},1}(3)f_{A_{III},1}(3)=f_{A_{III},2}(3)\hskip
5.69046pt\text{and}\hskip
5.69046ptf_{A_{III},1}(x)f_{A_{III},1}(x)>f_{A_{III},2}(x)$

for all $x>3$.

One can compare the obtained result with the theorem due to Heim, Neuhauser,
and Tröger [22, Theorem 1.4]. More specifically, they proved that the
inequality

$f_{A_{III},a}(x)f_{A_{III},b}(x)>f_{A_{III},a+b}(x)$

is satisfied for all $x>2$, $a,b\in\mathbb{N}$ and $a+b>2$. For $x=2$, it is
true whenever $a+b>4$. Clearly, Theorem 1.3 has some additional requirements
on the values of $a,b$ and $x$, which is a consequence of its general
character.

######  Remark 3.8.

Let $k\geq 2$ be a natural number and let
$A=\mathbb{N}\setminus\\{jk:j\in\mathbb{N}\\}$. Theorem 1.3 asserts the
polynomization of the Bessenrodt–Ono inequality for the $k$-regular partition
function $p_{k}(n)$ for any such a $k$. Recall that Beckwith and Bessenrodt
[8, Theorem 2.1] proved the Bessenrodt–Ono inequality for $p_{k}(n)$ for
$2\leq k\leq 6$. Hence, to resolve the issue completely, it suffices to find
the analogue of their result for any $k\geq 7$ and investigate the $2$-colored
$k$-regular partition function in that regard.

At the end of this section, let us show the general version of Corollary 3.5.

######  Corollary 3.9.

Let $m>3$ be a fixed natural number, and let $B$ be an arbitrary multiset of
positive integers such that $1\in B$ and $\mu(j)=m$ for every $j\in B$. Then
the inequality

$p_{B}(a)p_{B}(b)>p_{B}(a+b)$

is true for all positive integers $a$ and $b$.

##  4\. The Bessenrodt–Ono type inequality for the A-plane partitions

In this section we have two general assumptions on the multiset $A$. The first
one is that $1$ occurs exactly once in $A$. The second one, on the other hand,
states that each number $j$ might appear at most $j$ times in $A$—in other
words $\mu(j)\leq j$. In particular, observe that if every number $j$ occurs
exactly $j$ times in $A$, then the corresponding $A$-partition function is
just the plane partition function $p_{A}(n)=\operatorname{pp}\left(n\right)$,
and that is the reason for the title of this section. Nevertheless, there is a
myriad possibilities to choose the multiset $A$. We can, for instance,
consider any set $A$ which satisfies the general assumptions from Section 3.
However, we do not need to restrict ourselves only to sets.

Let us assume that $A=\\{1,2,2,3,5,5,5\\}$. The corresponding polynomials
$f_{A,n}(x)$ might be easily calculated by employing the formula $(\ref{1})$.
Table 2 presents these polynomials for $1\leq n\leq 7$.

$n$ | $f_{A,n}(x)$
---|---
$1$ | $x$
$2$ | $\frac{1}{2}x(x+5)$
$3$ | $\frac{1}{6}x\left(x^{2}+15x+8\right)$
$4$ | $\frac{1}{24}x\left(x^{3}+30x^{2}+107x+30\right)$
$5$ | $\frac{1}{120}x\left(x^{4}+50x^{3}+455x^{2}+550x+384\right)$
$6$ | $\frac{1}{720}x\left(x^{5}+75x^{4}+1285x^{3}+4725x^{2}+5194x+960\right)$
$7$ | $\frac{1}{5040}x\left(x^{6}+105x^{5}+2905x^{4}+22575x^{3}+49294x^{2}+55440x+720\right)$
Table 2. The polynomials $f_{A,n}(x)$ for $A=\\{1,2,2,3,5,5,5\\}$ and $1\leq
n\leq 7$.

Once again, one can ask about the localization of the complex roots of the
polynomials $f_{A,a}(x)f_{A,b}(x)-f_{A,a+b}(x)$. Figure 2 shows all their
complex roots for $1\leq a,b\leq 10$.

Figure 2. Complex roots of the polynomials $f_{A,a}(x)f_{A,b}(x)-f_{A,a+b}(x)$
for $A=\\{1,2,2,3,5,5,5\\}$ and $1\leq a,b\leq 10$.

Now, we wish to go to the main part of the section and find out the analogue
of Theorem 3.3. However, in that case we consider the $5$-colored
$A$-partition function which agrees with the result due to Heim, Neuhauser,
and Tröger [23] for the classical plane partition function.

######  Theorem 4.1.

If $A$ is such a multiset of natural numbers that $1$ occurs exactly once in
$A$, and each number $j$ appears at most $j$ times in $A$. Then, the
inequality

$f_{A,a}(5)f_{A,b}(5)>f_{A,a+b}(5)$

holds for all $a,b\in\mathbb{N}$ with $a+b>1$. For $a+b=1$, we have that

$f_{A,1}(5)f_{A,1}(5)\geq f_{A,2}(5)$

with the equality whenever $2$ occurs exactly two times in $A$.

###### Proof.

As before, the proof is by the induction on $a+b$. If $a+b=2$, then it is not
difficult to check that

$\displaystyle f_{A,2}(5)=\begin{cases}25,&\text{if }2\text{ appears exactly
two times in }A,\\\ 20,&\text{if }2\text{ appears exactly one time in }A,\\\
15,&\text{if }2\not\in A,\end{cases}$

and, thus,

$\displaystyle f_{A,1}(5)f_{A,1}(5)=25\geq f_{A,2}(5),$

where the equality occurs if and only if $2$ appears exactly two times in $A$.
Let us also verify the statement for $a+b=3$. In that case, it is not
difficult to check that

$\displaystyle f_{A,3}(5)\leq\begin{cases}100,&\text{if }2\text{ appears
exactly two times in }A,\\\ 75,&\text{if }2\text{ appears exactly one time in
}A,\\\ 50,&\text{if }2\not\in A.\end{cases}$

Therefore, we immediately obtain that

$f_{A,2}(5)f_{A,1}(5)>f_{A,3}(5),$

as required. Now, we use the same methods as in Lemma 3.4 to verify the claim
for all $a,b\leq 8$ and possible choices for $A$. In fact, it turns out that
the statement is valid for all the cases. Hence, let us assume that the claim
is true for all pairs $a$ and $b$ such that $3\leq a+b\leq n$, and check
whether it is also true for $a+b=n+1.$ Without loss of generality we may
additionally require that $a\geq b\geq 1$ and $a\geq 9$. Once again, applying
the formula (2.1) to the difference $f_{A,a}(5)f_{A,b}(5)-f_{A,a+b}(5)$, we
get

$\displaystyle\frac{5}{a}\sum_{j=1}^{a}\sigma_{A}(j)f_{A,a-j}(5)f_{A,b}(5)-\frac{5}{a+b}\sum_{i=1}^{a+b}\sigma_{A}(i)f_{A,a+b-i}(5)$
$\displaystyle>$
$\displaystyle\frac{5}{a}\sum_{j=1}^{a}\sigma_{A}(j)f_{A,a-j}(5)f_{A,b}(5)-\frac{5}{a+b}\sum_{i=1}^{a}\sigma_{A}(i)f_{A,a-i}(5)f_{A,b}(5)$
$\displaystyle\phantom{\frac{5}{a}\sum_{j=1}^{a}\sigma_{A}(j)f_{A,a-j}(5)f_{A,b}(5)}-\frac{5}{a+b}\sum_{i=1}^{b}\sigma_{A}(a+i)f_{A,b-i}(5)$
$\displaystyle=$
$\displaystyle\frac{5}{a(a+b)}\left(b\sum_{j=1}^{a}\sigma_{A}(j)f_{A,a-j}(5)f_{A,b}(5)-a\sum_{i=1}^{b}\sigma_{A}(a+i)f_{A,b-i}(5)\right),$

where the second line follows from the induction hypothesis applied to
$f_{A,a+b-i}(5)$ for each $i$ between $1$ and $a$. Thus, it suffices to show
that the expression in the bracket is non-negative. Let us put

$\displaystyle\kappa_{a}:=\sum_{j=1}^{a}\sigma_{A}(j)f_{A,a-j}(5).$

We see that

$\displaystyle
b\kappa_{a}f_{A,b}(5)-a\sum_{i=1}^{b}\sigma_{A}(a+i)f_{A,b-i}(5)$
$\displaystyle=$ $\displaystyle
5\kappa_{a}\sum_{i=1}^{b}\sigma_{A}(i)f_{A,b-i}(5)-a\sum_{i=1}^{b}\sigma_{A}(a+i)f_{A,b-i}(5)$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{b}\left(5\kappa_{a}\sigma_{A}(i)-a\sigma_{A}(a+i)\right)f_{A,b-i}(5).$

Since $1\leq\sigma_{A}(n)\leq\sigma_{2}(n)\leq n(2n-1)$ for each $n\geq 1$ and
$a\geq b$, we further simplify the above to

$\displaystyle 5\kappa_{a}-2a^{2}(4a-1)\geq
5\sum_{j=0}^{a-1}f_{A,j}(5)-2a^{2}(4a-1)\geq 5\binom{a+4}{5}-2a^{2}(4a-1),$

where the last inequality is the consequence of Lemma 3.1. Therefore, it is
enough to prove that the function

$\displaystyle\Psi(x):=(x+1)(x+2)(x+3)(x+4)-48x(4x-1)$

is non-negative for all $x\geq 9$ which is not a difficult task. This ends the
proof by the law of induction. ∎

There are a few direct consequences of Theorem 4.1. Since they are very
similar to Corollary 3.5 and Remark 3.6, we omit them and proceed to the proof
of Theorem 1.4.

###  4.1. Proof of Theorem 1.4

###### Proof.

We show the claim by induction on $a+b$. For $a+b=2$, we have that

$\displaystyle f_{A,1}(x)f_{A,1}(x)-f_{A,2}(x)\geq x^{2}-x/2(x+5)=x(x-5)/2,$

which is a consequence of the equality

$\displaystyle f_{A,2}(x)=\begin{cases}x(x+5)/2,&\text{if }2\text{ appears
exactly two times in }A,\\\ x(x+3)/2,&\text{if }2\text{ appears exactly one
time in }A,\\\ x(x+1)/2,&\text{if }2\not\in A.\end{cases}$

For $a+b=3$, on the other hand, one can verify that

$\displaystyle
f_{A,2}(x)f_{A,1}(x)-f_{A,3}(x)=\begin{cases}x\left(x^{2}-10\right)/3,&\text{if
}3\text{ occurs three times in }A,\\\ x\left(x^{2}-7\right)/3,&\text{if
}3\text{ occurs two times in }A,\\\ x(x-2)(x+2)/3,&\text{if }3\text{ occurs
one time in }A,\\\ x(x-1)(x+1)/3,&\text{if }3\not\in A,\end{cases}$

and we see that the claim is true for $a+b=3$. Thus, let us suppose that the
statement is valid for all $3\leq a+b\leq n$, and investigate whether it is
also correct for $a+b=n+1$ with $a$ and $b$ such that $1\leq b\leq a$ and
$a\geq 2$. Theorem 4.1 asserts that the inequality
$f_{A,a}(5)f_{A,b}(5)>f_{A,a+b}(5)$ is true for all positive integers $a$ and
$b$ such that $a+b>2$. Now, it suffices to show that the derivative

$\displaystyle\frac{d}{dx}\left(f_{A,a}(x)f_{A,b}(x)-f_{A,a+b}(x)\right)$

is positive for all $x\geq 5$, because it maintains that
$f_{A,a}(x)f_{A,b}(x)-f_{A,a+b}(x)>0$. Since
$f_{A,n}(x)f_{A,0}(x)=f_{A,n}(x)$, we get

$\displaystyle
f_{A,a}^{\prime}(x)f_{A,b}(x)+f_{A,a}(x)f_{A,b}^{\prime}(x)-f_{A,a+b}^{\prime}(x)$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{a}\frac{\sigma_{A}(j)}{j}f_{A,a-j}(x)f_{A,b}(x)+\sum_{j=1}^{b}\frac{\sigma_{A}(j)}{j}f_{A,a}(x)f_{A,b-j}(x)-\sum_{j=1}^{a+b}\frac{\sigma_{A}(j)}{j}f_{A,a+b-j}(x)$
$\displaystyle>$
$\displaystyle\sum_{j=1}^{a}\frac{\sigma_{A}(j)}{j}f_{A,a-j}(x)f_{A,b}(x)+\sum_{j=1}^{b}\frac{\sigma_{A}(j)}{j}f_{A,a}(x)f_{A,b-j}(x)$
$\displaystyle-$
$\displaystyle\sum_{j=1}^{a}\frac{\sigma_{A}(j)}{j}f_{A,a-j}(x)f_{A,b}(x)-\sum_{j=1}^{b}\frac{\sigma_{A}(a+j)}{a+j}f_{A,b-j}(x)$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{b}\left(\frac{\sigma_{A}(j)}{j}f_{A,a}(x)-\frac{\sigma_{A}(a+j)}{a+j}\right)f_{A,b-j}(x).$

Therefore, it suffices to prove that

$\displaystyle\frac{\sigma_{A}(j)}{j}f_{A,a}(x)-\frac{\sigma_{A}(a+j)}{a+j}\geq
0$

for each $j$ between $1$ and $b$. Since $a\geq b$ and
$1\leq\sigma_{A}(n)\leq\sigma_{2}(n)\leq n(2n-1)$ for every $n\geq 1$, it is
clear that

$\displaystyle\frac{\sigma_{A}(j)}{j}f_{A,a}(x)-\frac{\sigma_{A}(a+j)}{a+j}\geq\frac{1}{a}\binom{x+a-1}{a}-4a+1$

is valid for each such a $j$. However, the assumption that $x\geq 5$ points
out that

$\displaystyle\frac{1}{a}\binom{x+a-1}{a}-4a+1\geq(a+1)(a+2)(a+3)(a+4)/(24a)-4a+1.$

Now, we just examine when the function

$\displaystyle\psi(x):=(x+1)(x+2)(x+3)(x+4)-96x^{2}+24x$

is positive. One can easily check that it is positive for every positive real
number $x$. This completes the proof by the law of induction. ∎

It is worth pointing out that Theorem 1.4 agrees with the result of Heim,
Neuhauser, and Tröger [23, Theorem 1.3], for the polynomization of the plane
partition function $\operatorname{pp}\left(n\right)$. Actually, Theorem 1.4 is
a general property which might be successfully applied for various multisets
$A$. In particular, it extends Theorem 1.3 for all $x\geq 5$.

Nevertheless, there is still a wide class of multisubsets for which we can
utilize neither Theorem 1.3 nor Theorem 1.4, and that gap to our current
knowledge will be investigated in the future.

## Acknowledgements

The first author would like to thank Piotr Miska, Maciej Ulas, and Błażej
Żmija for their valuable comments and helpful suggestions. His research was
funded by both a grant of the National Science Centre (NCN), Poland, no.
UMO-2019/34/E/ST1/00094 and a grant from the Faculty of Mathematics and
Computer Science under the Strategic Program Excellence Initiative at the
Jagiellonian University in Kraków.

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# The spin-flip-induced quadrupole resonance in odd-$A$ exotic atoms

Fredrik P. Gustafsson  Physics Department, CERN, 1211 Geneva 23, Switzerland
Daniel Pęcak  Faculty of Physics, Warsaw University of Technology, Ulica
Koszykowa 75, PL-00662 Warsaw, Poland Institute of Physics, Polish Academy of
Sciences, Aleja Lotnikow 32/46, PL-02668 Warsaw, Poland Tomasz Sowiński
Institute of Physics, Polish Academy of Sciences, Aleja Lotnikow 32/46,
PL-02668 Warsaw, Poland

###### Abstract

We examine the possible existence of quadrupole resonances in exotic atoms
containing odd-$A$ nuclei. We find that the spin-flip of the de-exciting
exotic particles can induce a resonance, altering the orbital angular momentum
of the nucleus by one quanta. This process results in an excited nucleus with
a suppression in X-ray photon emissions during the exotic atom cascade. We
provide specific cases of antiprotonic atoms of stable elements where this
resonance effect is expected to occur. The study of this phenomena may provide
insight into the strong interaction of deeply bound antiprotonic states in the
proximity of the unpaired nucleon, and serve as a tool for probing short-lived
excited nuclear states.

##  I Introduction

Exotic atoms are formed by the substitution of an electron with an exotic
negatively charged particle such as an antiproton, muon, pion or kaon. These
exotic particles, owing to their greater mass, are deeply bound within the
electron cloud, positioning them in close proximity to the atomic nucleus.
This renders them highly susceptible to short-range interactions, such as the
weak and strong nuclear forces, making them invaluable tools for probing
fundamental interactions Cohen (2004); Devons and Duerdoth (1969). While
exotic atoms are typically short-lived, either due to the intrinsic lifetime
of the exotic particle or resulting from the annihilation with nucleons in the
nucleus, as seen with antiprotons, spectroscopic analysis of these atoms can
offer a unique insights into nuclear properties like mass, charge radii, and
shape Acker (1966); De Wit  _et al._ (1966); Green  _et al._ (1988); Wycech
_et al._ (1993); Hori  _et al._ (2011); Yang  _et al._ (2023). A notable
feature of exotic atoms is their significantly higher transition energies
compared to conventional electronic systems. These energies often lie in the
range of keV to MeV, as opposed to the eV scale typical of electrons Gotta
_et al._ (2008). Interestingly, these energies are comparable with energy
scales of nuclear phenomena. In some cases, the energy of a nuclear electric
quadrupole (E2) transition and the X-ray transition of an exotic atom can be
sufficiently close to create a resonance effect Leon (1976); Green  _et al._
(1988); Wycech  _et al._ (1990); Kłos  _et al._ (2004). This resonance
suppresses the emission of an X-ray photon from the cascade, resulting from
the direct coupling with the quadrupole excitation of the nucleus. This
quadrupole resonance effect have mainly been observed in exotic atoms
containing spin-0 nuclei with bound pions, kaons, and antiprotons Devons and
Duerdoth (1969); Leon (1976).

Studies of this resonance effect in antiprotonic atoms have provided insight
of the influence of the shift and width caused by the strong interaction
influence of deeply bound states, which are not accessible through
conventional spectroscopy measurements of direct x-rays from the antiproton
cascade Kłos  _et al._ (2004). So far, cases of this resonance effect have
only been studied for collective quadrupole excitations of the nucleus, such
as rotations and vibrations, leading to a change of the nuclear spin by two
quanta, as observed in nuclei with spin-0 ground states. In this article, we
extend phenomenologically the study of the nuclear resonance effect to odd-$A$
nuclei, with the focus on antiprotonic atoms. We demonstrate that, in this
case, the spin-flip of the antiproton results in the excitation of the nuclear
spin by one quanta typically associated with single-particle excitation of the
unpaired nucleons. Furthermore, we identify candidates of stable nuclei where
we anticipate this spin-flip-induced nuclear resonance effect to occur.
Finally we discuss potential scenarios where this phenomena could be used as a
probe for nuclear structure and strong interaction effects.

##  II The system

The most convenient way of considering corrections coming from the quadrupole
interactions is to write the Hamiltonian describing the system of deformed
nucleus and orbiting antiproton as

${\cal H}={\cal H}_{R}+{\cal H}_{r}+{\cal H}_{Q},$ (1)

where ${\cal H}_{R}$ is the Hamiltonian acting solely in the nucleus subspace
while ${\cal H}_{r}$ is the Dirac Hamiltonian for the antiproton in the
spherically symmetric electric potential, $V(r)=-Z/r$. Its bound eigenstates
$|njm\kappa\rangle$ are conveniently labeled with the principal quantum number
$n$, the total angular momentum $j$, its projection $m$, and the sign of the
relativistic quantum number $\kappa=\pm 1$. Although in principle the combined
integer $\ell=j+\kappa/2$ is not a good quantum number, for interpretation
purposes it can be viewed as an orbital angular momentum number. In the non-
relativistic limit (Pauli approximation), corresponding spinor wavefunctions
have a form Bethe and Salpeter (1957); Drake (2006)

$\displaystyle\langle\boldsymbol{r}|njm+\rangle=$ $\displaystyle
R^{n}_{j+\text{\textonehalf}}(r)\left(\begin{array}[]{c}\sqrt{\frac{j-m+1}{2j+2}}\mathrm{Y}_{j+\text{\textonehalf}}^{m-\text{\textonehalf}}(\boldsymbol{\vartheta})\\\
\sqrt{\frac{j+m+1}{2j+2}}\mathrm{Y}_{j+\text{\textonehalf}}^{m+\text{\textonehalf}}(\boldsymbol{\vartheta})\end{array}\right),$
(2c) $\displaystyle\langle\boldsymbol{r}|njm-\rangle=$ $\displaystyle
R^{n}_{j-\text{\textonehalf}}(r)\left(\begin{array}[]{c}\sqrt{\frac{j+m}{2j}}\mathrm{Y}_{j-\text{\textonehalf}}^{m-\text{\textonehalf}}(\boldsymbol{\vartheta})\\\
\sqrt{\frac{j-m}{2j}}\mathrm{Y}_{j-\text{\textonehalf}}^{m+\text{\textonehalf}}(\boldsymbol{\vartheta})\end{array}\right),$
(2f)

where $R^{n}_{\ell}(r)$ encodes radial probability amplitude while
$\mathrm{Y}_{\ell}^{m}(\boldsymbol{\vartheta})$ is a spherical harmonic
function. In the following, we consider the most typical experimental scenario
when the antiproton substitutes one of the inner most electrons, enters to the
Rydberg-like orbit with corresponding principal number $n\approx 40$, and then
undergoes spontaneous transitions to lower $n$ states emitting X-ray photons
Cohen (2004); Gotta  _et al._ (2008). Since Rydberg orbits are spatially well-
localised we assume that $\ell$ and $m$ are as large as possible, i.e., they
are close to the principal quantum number $n$.

Quantum description of the nucleus is not so straightforward, especially when
odd-$A$ nuclei are considered. Therefore we work within the simplest possible
framework capturing the essence of the quadrupole resonance phenomena in
odd-$A$ systems. This gives us a path to explain in details all aspects of the
phenomena, to specify conditions for its observability, and to estimate its
basic measurable features. Of course, one can use more sophisticated nuclear
models, especially for cases when quadrupole moments of the nucleus are
essentially dependent on its quantum state, to obtain more accurate
predictions de Toledo  _et al._ (1977); de Toledo  _et al._ (1979); Jia  _et
al._ (2007); Al-Khudair (2009); Deleanu  _et al._ (2013). Nevertheless, the
general scheme outlined below will not change.

Our framework is based on the fundamental assumption that the density of
nuclear matter, represented by $\rho(\boldsymbol{R})$, exhibits minimal
dependence on its quantum state. Then deformations of the nucleus are
characterized by the nuclear quadrupole tensor:

${\cal
Q}_{2\mu}=\sqrt{\frac{16\pi}{5}}\int\mathrm{d}^{3}\boldsymbol{R}\,\rho(\boldsymbol{R})\,R^{2}\,\mathrm{Y}_{2}^{\mu}({\boldsymbol{\Theta}})$
(3)

while its rotational quantum states are described within the Bohr-Mottelson
collective rotational model framework Bohr (1976); Mottelson (1976); Naqvi
_et al._ (1995); Rowe  _et al._ (2016). Assuming additionally that the nucleus
is axially symmetric one finds that the quadrupole tensor has only one
nonvanishing element ${\cal Q}_{20}$ equal to the intrinsic quadrupole moment
${\cal Q}_{0}/e$. Moreover, all the eigenstates of ${\cal H}_{R}$ are
expressed explicitly in terms of Wigner D-matrices

$\langle\boldsymbol{\Theta}|JKM\rangle=\sqrt{\frac{2J+1}{16\pi^{2}}}\Big{[}\xi_{K}D^{J}_{MK}(\boldsymbol{\Theta})\\\
+(-1)^{J+K}\xi_{\bar{K}}D^{J}_{M-K}(\boldsymbol{\Theta})\Big{]}.$ (4)

Here we assumed that $J$ is half-integer since we focus on odd-$A$ elements,
similarly as its projections $K,M$.

##  III Quadrupole interactions

The remaining part of the Hamiltonian, ${\cal H}_{Q}$, dominated by the
quadrupole interactions, essentially accounts for all contributions not
captured by the point-like approximation of the nucleus. Within our framework,
this Hamiltonian can be decomposed into terms that act independently on the
nucleus and antiproton sectors, as follows Leon (1976)

${\cal H}_{Q}=-\sqrt{\frac{\pi}{5}}Z{\cal
Q}_{0}\mathrm{Y}_{2}^{0}({\boldsymbol{\Theta}})\,r^{-3}\,\mathrm{Y}_{2}^{0}(\boldsymbol{\vartheta}),$
(5)

where we measure all energies and lengths in natural units of the Coulomb
problem $\alpha^{2}m_{\bar{p}}c^{2}\approx 50\,\mathrm{keV}$ and
$\hbar/(\alpha m_{\bar{p}}c)\approx 28.8\,\mathrm{fm}$, while $\alpha\approx
1/137$ is the fine-structure constant.

It is clear that, as long as purely spherically symmetric interaction is
considered (vanishing ${\cal H}_{Q}$), product states of orbiting antiproton
$|njm\kappa\rangle$, and nucleus $|JKM\rangle$ diagonalize the Hamiltonian
${\cal H}_{0}={\cal H}_{R}+{\cal H}_{r}$. Since quadrupole interactions
preserve total angular momentum, it is convenient to work in the basis of
combined states of well-defined total angular momentum ${\cal F}$
($|j-J|\leq{\cal F}\leq j+J$) and its projection ${\cal M}$ as

$|{\cal F}{\cal M};nj\kappa JK\rangle=(-1)^{j+J+{\cal M}}\sqrt{2{\cal F}+1}\\\
\times\\!\sum_{m=-j}^{j}\left(\begin{array}[]{ccc}j&J&{\cal F}\\\ m&{\cal
M}\\!-\\!m&-{\cal M}\end{array}\right)|njm\kappa\rangle|JK({\cal
M}\\!-\\!m)\rangle.$ (6)

In this basis, all the matrix elements of the quadrupole Hamiltonian can be
calculated using the decomposition rule De Wit  _et al._ (1966); Green  _et
al._ (1988)

$\langle{\cal F}{\cal M};nj\kappa JK|{\cal H}_{Q}|{\cal F}^{\prime}{\cal
M}^{\prime};n^{\prime}j^{\prime}\kappa^{\prime}J^{\prime}K^{\prime}\rangle=\delta_{\cal
M\cal M^{\prime}}\delta_{\cal F\cal F^{\prime}}\\\
\times(-1)^{1+j^{\prime}+J+\cal F}\sqrt{\frac{\pi}{5}}Z{\cal
Q}_{0}\left\\{\begin{array}[]{ccc}j&j^{\prime}&2\\\ J^{\prime}&J&{\cal
F}\end{array}\right\\}\langle nj|r^{-3}|n^{\prime}j^{\prime}\rangle\\\
\times\langle
JKM|\mathrm{Y}_{2}^{0}(\boldsymbol{\Theta})|J^{\prime}K^{\prime}M^{\prime}\rangle\langle
jm\kappa|{\mathrm{Y}}_{2}^{0}(\boldsymbol{\vartheta})|j^{\prime}m^{\prime}\kappa^{\prime}\rangle.$
(7)

Now it is clear that to obtain the transition amplitudes one can calculate
contributing parts independently in nuclear and electronic sectors. However,
due to the conservation of total angular momentum $\cal F$ and its projection
$\cal M$, final contributions must combine to meet this constrain.

###  III.1 Transitions in nuclear sector

Utilizing explicit expressions for the eigenfunctions of the collective
rotational model (4) one finds straightforwardly matrix elements of the
quadrupole moment operator in the nuclear sector

$\langle
JKM|\mathrm{Y}_{2}^{0}(\boldsymbol{\Theta})|J^{\prime}K^{\prime}M^{\prime}\rangle=\\\
\sqrt{\frac{5}{4\pi}}\sqrt{(2J+1)(2J^{\prime}+1)}\,\delta_{MM^{\prime}}\delta_{KK^{\prime}}\\\
\times\left(\begin{array}[]{ccc}J&2&J^{\prime}\\\
M&0&-M\end{array}\right)\left(\begin{array}[]{ccc}J&2&J^{\prime}\\\
K&0&-K\end{array}\right).$ (8)

Quantitatively this matrix element is nuclear-model dependent. However,
qualitatively it captures principled properties of the quadrupole coupling
between nuclear states. Therefore, if needed, its value can be tuned to
experimental values by a simple dimensionless multiplicative factor without
changing the physical picture. Here, we continue with the simple assumption of
a constant quadrupole moment, the influence of changing quadrupole moment
between states is discussed later in the discussion section.

Having this expression in hand, one deduces that quadrupole interactions can
trigger only well-specified transition of the nucleus. For example, if the
even-$A$ element is considered and the nucleus remains in its ground state
$|0^{+})$, then the only non-vanishing transition amplitude (equal to
$1/\sqrt{4\pi}$) couples to the first excited state of the same band
$|2^{+})$. In this particular case the angular momentum of the nucleus
increases by two quanta ($\Delta J=2$) and, due to the conservation of ${\cal
F}$, it must be compensated by a decrease of the angular momentum of an
orbiting particle by the same amount. This is indeed possible since quadrupole
electric field inherently couples electronic states of orbiting particle with
$\Delta\ell=2$ (see the next subsection for details).

Situation is slightly different when odd-$A$ elements are considered. Then the
nucleus, being a fermionic particle, has non-vanishing half-integer spin with
single-particle excitations which typically result in the increase of spin by
one quanta. Fortunately, it is quite easy to verify that the matrix element
(8) is also non-zero in these cases. As an example, Ruthenium-101 nucleus with
ground state $M=K=J=5/2^{+}$ and excited state $J^{\prime}=7/2^{+}$ is equal
to $\sqrt{125/(288\pi)}$, i.e., is of the same order as the matrix element for
even-$A$ nuclei.

###  III.2 Transitions of orbiting particle

Figure 1: Schematic chart of possible decays from electronic states of an
orbiting antiproton triggered by quadrupole interactions, having orbital
angular momentum $\ell$. Note that in addition to the typical transitions with
$\Delta j=0$ and $\Delta j=-2$ caused by amplitudes (9a) and (9k), transitions
that change the total particle spin by $\Delta j=-1$ and $\Delta j=-3$ are
also possible due to the simultaneous change of the orbital angular momentum
with $\Delta\ell=-2$ and the flip of particle spin. They are controlled by
amplitudes (9u).

Analysis of possible transitions in orbiting particle sector is similar. By
taking the central field solutions of Dirac equation for orbiting antiproton
(2), one finds angular matrix elements of the electronic part of the
quadrupole interactions

$\displaystyle\langle jm+$
$\displaystyle|\mathrm{Y}_{20}(\boldsymbol{\vartheta})|j^{\prime}m^{\prime}+\rangle=$
(9a)
$\displaystyle(-1)^{m-\text{\textonehalf}}\sqrt{\frac{5}{4\pi}}\delta_{mm^{\prime}}\left(\begin{array}[]{ccc}j\\!+\\!\text{\textonehalf}&2&j^{\prime}\\!+\\!\text{\textonehalf}\\\
0&0&0\end{array}\right)$ (9d)
$\displaystyle\times\left[\sqrt{(j\\!-\\!m\\!+\\!1)(j^{\prime}\\!-\\!m\\!+\\!1)}\left(\begin{array}[]{ccc}j\\!+\\!\text{\textonehalf}&2&j^{\prime}\\!+\\!\text{\textonehalf}\\\
-m\\!+\\!\text{\textonehalf}&0&m-\\!\text{\textonehalf}\end{array}\right)\right.$
(9g)
$\displaystyle-\left.\sqrt{(j\\!+\\!m\\!+\\!1)(j^{\prime}\\!+\\!m\\!+\\!1)}\left(\begin{array}[]{ccc}j\\!+\\!\text{\textonehalf}&2&j^{\prime}\\!+\\!\text{\textonehalf}\\\
-m\\!-\\!\text{\textonehalf}&0&m\\!+\\!\text{\textonehalf}\end{array}\right)\right],$
(9j) $\displaystyle\langle jm-$
$\displaystyle|\mathrm{Y}_{20}(\boldsymbol{\vartheta})|j^{\prime}m^{\prime}-\rangle=$
(9k)
$\displaystyle(-1)^{m-\text{\textonehalf}}\sqrt{\frac{5}{4\pi}}\delta_{mm^{\prime}}\left(\begin{array}[]{ccc}j\\!-\\!\text{\textonehalf}&2&j^{\prime}\\!-\\!\text{\textonehalf}\\\
0&0&0\end{array}\right)$ (9n)
$\displaystyle\times\left[\sqrt{(j\\!+\\!m)(j^{\prime}\\!+\\!m)}\left(\begin{array}[]{ccc}j\\!-\\!\text{\textonehalf}&2&j^{\prime}\\!-\\!\text{\textonehalf}\\\
-m\\!+\\!\text{\textonehalf}&0&m-\\!\text{\textonehalf}\end{array}\right)\right.$
(9q)
$\displaystyle-\left.\sqrt{(j\\!-\\!m)(j^{\prime}\\!-\\!m)}\left(\begin{array}[]{ccc}j\\!-\\!\text{\textonehalf}&2&j^{\prime}\\!-\\!\text{\textonehalf}\\\
-m\\!-\\!\text{\textonehalf}&0&m\\!+\\!\text{\textonehalf}\end{array}\right)\right],$
(9t) $\displaystyle\langle jm+$
$\displaystyle|\mathrm{Y}_{20}(\boldsymbol{\vartheta})|j^{\prime}m^{\prime}-\rangle=$
(9u)
$\displaystyle(-1)^{m+\text{\textonehalf}}\sqrt{\frac{5}{4\pi}}\delta_{mm^{\prime}}\left(\begin{array}[]{ccc}j\\!+\\!\text{\textonehalf}&2&j^{\prime}\\!-\\!\text{\textonehalf}\\\
0&0&0\end{array}\right)$ (9x)
$\displaystyle\times\left[\sqrt{(j\\!-\\!m\\!+\\!1)(j^{\prime}\\!+\\!m)}\left(\begin{array}[]{ccc}j\\!+\\!\text{\textonehalf}&2&j^{\prime}\\!-\\!\text{\textonehalf}\\\
-m\\!+\\!\text{\textonehalf}&0&m-\\!\text{\textonehalf}\end{array}\right)\right.$
(9aa)
$\displaystyle+\left.\sqrt{(j\\!+\\!m\\!+\\!1)(j^{\prime}\\!-\\!m)}\left(\begin{array}[]{ccc}j\\!+\\!\text{\textonehalf}&2&j^{\prime}\\!-\\!\text{\textonehalf}\\\
-m\\!-\\!\text{\textonehalf}&0&m\\!+\\!\text{\textonehalf}\end{array}\right)\right].$
(9ad)

Essentially, there are three different types of non-vanishing matrix elements
(schematically presented in Fig. 1). The first two involve transitions between
states having the same sign of relativistic quantum number $\kappa$. In these
cases, the antiproton experiences a small shift of the energy (matrix element
with $\Delta j=0$) or undergoes transition with $\Delta j=-2$ keeping the spin
projection unchanged. During this transition, a whole change of the angular
momentum comes from the orbital change $\Delta\ell=-2$. As mentioned in
previous subsection, due to the conservation of ${\cal F}$, this transition
must be accompanied by a corresponding change of angular momentum of the
nucleus, $\Delta J=+2$. For exactly this reason, this transition is typically
considered for E2 resonance of even-$A$ nuclei between $|0^{+})$ and $|2^{+})$
internal states Leon (1976); Wycech  _et al._ (1990, 1993); Kłos  _et al._
(2004). But of course it may be also considered in odd-$A$ systems.

The third possibility exists between states of opposite $\kappa$. Then,
although the antiproton undergoes transition with change of the orbital
angular momentum $\Delta\ell=\\-2$, the total angular momentum change is
$\Delta j=-3$ or $\Delta j=-1$ due to simultaneous flipping of antiproton
spin. The latter is also possible without changing orbital angular momentum,
$\Delta\ell=0$, and leads to mixing of different spin states. Correspondingly,
in the presence of these transitions, the nucleus must undergo excitation with
spin changed by three or one quanta. In the generic scenario considered here
for odd-$A$ elements where $\Delta J=1$ only transitions with $\Delta j=-1$
are allowed. Thus, this transition become typical for odd-$A$ elements. It is
worth noticing here, that this particular transitions are possible since there
is a spin-orbit coupling between electronic states of orbiting particle. Thus,
its amplitude is evidently smaller when compared to the spin-preserving
transitions. However, since in odd-$A$ anti-protonic atoms the spin-preserving
channels are locked, this transition is the leading term forced by quadrupole
interaction.

From the above discussion it follows that independently of the parity of the
nucleus only the transitions with $\Delta\ell=2$ are allowed. Therefore, if we
consider scenario of an exotic particle orbiting in a high Rydberg orbit,
i.e., $\ell\sim n-1$ and $m\sim\ell$, the most probable transitions is
$|n,\ell=n-1\rangle\rightarrow|n-2,\ell^{\prime}=n-3\rangle$ with
$m=m^{\prime}=n-3$. This observation is directly related to the fact that the
prefactor originating in the radial matrix element $\langle
nj|r^{-3}|n^{\prime}j^{\prime}\rangle$ quickly decays with increasing
difference $\Delta n=n-n^{\prime}$. Here, one should also note that this
factor calculated for the most typical scenario,$\langle
nj|r^{-3}|n-2,j-1\rangle$, quickly decays with increasing $n$ (much faster
than dipolar factor $\langle nj|r^{-1}|n-1,j-1\rangle$). Therefore, overall
quadrupole transition amplitudes with large initial $n$ are strongly
suppressed thus we only consider $n\leq 8$ orbitals.

##  IV The resonance

Figure 2: General scheme for matching the energetic condition for quadrupole resonance using the example of an antiprotonic Ruthenium-101. The $7/2^{+}$ excited state $939\,\mathrm{keV}$ above the $5/2$ ground state matches the energy difference between electronic states of orbiting antiproton $|n=7,\ell=6\rangle$ and $|n=5,\ell=4\rangle$. Due to required total angular momentum conservation, the resonance is fully controlled by spin-flipping transitions (9u). In this case the states $|7,11/2,m,+;5/2,K,M\rangle$ and $|5,9/2,m,-;7/2,K,M\rangle$ are resonantly coupled by quadrupole interactions for any allowed quantum numbers $m$, $K$, and $M$ which are conserved.  Table 1:  Examples of stable odd-$A$ antiprotonic atoms where the spin-flip-induced quadrupole resonance. The nuclear transition energies were acquired from International Atomic Energy Agency (2023) and the antiproton transitions were calculated using. In parenthesis, we indicate tentatively assigned nuclear spins of the excited states. Follow text for more details. isotope |  xZx |  xNx | nuclear spin | $\boldsymbol{\Delta E_{N}}$ | $\boldsymbol{\Delta E_{\bar{P}}}$ | $\boldsymbol{n}$
---|---|---|---|---|---|---
|  |  |  xxgroundxx |  xxexcitedxx | (keV) | (keV) |  xxinitialxx |  xxfinalxx
|  |  | NUCLEUS | ANTIPROTON
101Ru | 44 | 57 | 5/2 | (7/2) | $938.65$ | $939.40$ | 7 | 5
111Cd | 48 | 63 | 1/2 | 3/2 | $1115.57$ | $1119.20$ | 7 | 5
123Sb | 51 | 72 | 7/2 | (9/2) | $1260.80$ | $1264.81$ | 7 | 5
165Ho | 67 | 98 | 7/2 | (9/2) | $2178.00$ | $2189.96$ | 7 | 5
169Tm | 69 | 100 | 1/2 | (3/2) | $2312.2$ | $2323.38$ | 7 | 5
183W | 74 | 109 | 1/2 | (3/2) | $2667.8$ | $2667.47$ | 7 | 5
203Tl | 81 | 122 | 1/2 | (3/2) | $1988.88$ | $1987.73$ | 8 | 6

The matrix elements of the quadrupole Hamiltonian are rather small when
compared with typical energy spacings between coupled electronic states, which
is of the order of $\Delta E_{\bar{P}}=E_{nj}-E_{n-2,j-2}$, where $j=\ell\pm
1/2=n-1\pm 1/2$. Thus, the correction is mainly reflected in a relatively
small mixing of states in the same (${\cal F},{\cal M}$)-subspace. However,
when the energy difference between relevant electronic states of orbiting
anti-proton is close to the excitation energy of the nucleus, $\Delta E_{N}$,
one can suspect resonant behaviour triggered by this coupling. For even-$A$
elements, this behaviour is quite well studied and explored experimentally
Leon (1976); Green  _et al._ (1988); Wycech  _et al._ (1990); Kłos  _et al._
(2004). In contrary, for odd-$A$ elements, due to non-trivial interplay
between orbital and spin degrees of motion, observation of the resonance has
not yet been explored. An exemplary case of mentioned ruthenium 101Ru is
displayed in Fig. 2. It turns out that the energy difference between
$|n=7,\ell=6\rangle$ and $|n=5,\ell=4\rangle$ states is very close to the
separation energy between nucleus ground-state $|5/2^{+})$ and the $|7/2^{+})$
excited state at $939\,\mathrm{keV}$. Moreover, due to the required total
angular momentum conservation, this coupling is controlled solely by spin-
flipping transitions (9u) (the two others are forbidden by selection rules)
and the initial states $|7,11/2,m,+;5/2,K,M\rangle$ and and final states
$|5,9/2,m,-;7/2,K,M\rangle$ are resonantly coupled by quadrupole interactions
for any quantum numbers $m$, $K$, and $M$ that obey angular momentum
projection conservation. The coupling matrix element $\langle initial|{\cal
H}_{Q}|final\rangle/Q_{0}$ calculated for $m=4,K=M=5/2$ to 190 $eV/b$ thus is
of the same order as the energy difference $\Delta E_{N}-\Delta E_{\bar{P}}$,
we therefore expect the resonance effect to be of significant relevance here.

Let us mention here, that in fact, the scenario discussed above for
antiprotonic ruthenium is not unique and in principle may be observed for
different stable and unstable odd-$A$ isotopes. In Table. 1 we list candidates
of odd-$A$ stable elements for which the energy matching for quadrupole
transition is comparable with the corresponding quadrupole coupling matrix
element. More precisely, we assume that the effects of quadrupole resonance
can be experimentally relevant if the coupling matrix element is not less than
$5\%$ of the energy difference between coupled states. Obviously, due to the
quick decaying of the quadrupole matrix element $\langle
nj|r^{-3}|n-2,j^{\prime}\rangle$ with increasing $n$, the most interesting are
these elements for which the resonant $n$ is small, since then the energy
matching may be less accurate. On the other hand, when $n$ is sufficiently
small, the corresponding antiprotonic orbit is in close proximity to the
surface of the nucleus. In this case, the resonant condition is strongly
affected by the influence of the strong interaction which is neglected in
these studies.

Thus, precise measurements of the spin-flip-induced resonance effect could
provide valuable insight into the strong interaction shifts and width of
antiproton orbits in the close proximity to the nucleus containing the
unpaired nucleon. The emergence of this resonance effect measured along
isotopic chains may give further insight into the influence of the unpaired
nucleon on the nuclear periphery. Furthermore, if the spin and moments of the
nuclear ground-state is well understood then the observation of this resonance
effect may prove itself as a complementary tool for decoupling the properties
of the short-lived excited nuclear state. For example, the spin of the upper
level nuclear states of the proposed transitions have only been tentatively
assigned, see Table. 1. Hence, the observation of this resonance effect could
provide information of the nuclear spin and nuclear moments of these short-
lived excited states. However, in order to decouple the strong interaction
effects more detailed calculations are required which is beyond the scope of
this work.

For completeness, let us also mention that the same spin-flip-induced
quadrupole resonance could be in principle observed for other exotic atoms
containing heavy fermionic particles. The most natural candidates are muonic
atoms. However, in practice, in these cases, the energy scales of electronic
transitions are significantly smaller, thus fulfilling the resonance condition
is much more demanding. This means probably that only antiprotonic (or
heavier) atoms may serve as appropriate candidates for observing spin-flipping
resonance.

##  V Final remarks

This work extends the study of the nuclear resonance effect of exotic atoms by
focusing on odd-$A$ nuclei with half-integer ground-state spin. We show that
$\Delta J=1$ nuclear excitations can be directly triggered by $\Delta\ell=2$
transitions of the decaying antiproton undergoing spin-flip. We present cases
of odd-A antiprotonic atoms where the spin-flip-induced quadrupole resonance
effect is expected to occur. The observation of this effect could reveal the
strong interaction influence on deeply bound antiproton orbitals on nuclei
with unpaired nucleons and potentially serve as a tool for benchmarking
nuclear structure of short-lived excited nuclear states.

The rapid development of antiproton facilities and the upcoming availability
of antiprotons that can be transported would allow greater access of
antiprotons to the nuclear physics community which will allow for new probes
for nuclear physics, such as the one proposed in this work.

###### Acknowledgements.

The authors are very grateful to Michael Doser, Piotr Magierski, Leszek
Próchniak, and Katarzyna Wrzosek-Lipska for their insightful comments and
inspiring questions. The authors are also indebted for many valuable
discussions with AEgIS Collaboration members. The work by T. S. and D. P. was
supported as part of a project funded by the Polish Ministry of Education and
Science on the basis of agreement no. 2022/WK/06.

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