The Janjić–Petković Inset Counting Function: Riordan Array Properties and a Thermodynamic Application

Abstract

Let $q_1+\cdots +q_n+m$ objects be arranged in n rows with $q_1,\dots ,q_n$ objects and one last row with m objects. The Janjić–Petković counting function denotes the number of $(n+k)$-insets, defined as subsets containing $n+k$ objects such that at least one object is chosen from each of the first n rows, generalizing the binomial coefficient that is recovered for $q_1$ = … = $q_n$ = 1, as then only the last row matters. Here, we discuss two explicit forms, combinatorial interpretations, recursion relations, an integral representation, generating functions, convolutions, special cases, and inverse pairs of summation formulas. Based on one of the generating functions, we show that the Janjić–Petković counting function, like the binomial coefficients that it generalizes, may be regarded as a Riordan array, leading to additional identities. As an application to a physical system, we calculate the heat capacity of a many-body system for which the configurations are constrained as described by the Janjić–Petković counting function, resulting in a modified Schottky anomaly.

1. Introduction

Suppose that $q_1+\dots +q_n+m$ objects are arranged in n rows with $q_1,\dots ,q_n$ objects and one last row with m objects. The Janjić–Petković counting function $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\{q_1,\dots ,q_n\}}}\right)$ denotes the number of $(n+k)$-insets, defined in this context as subsets containing $n+k$ objects such that at least one object is chosen from each of the first n rows, as in Figure 1. This function was introduced by Janjić in Ref. [1] and developed further in Refs. [2,3]. It reduces to the binomial coefficient $\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)$ for $q_1$ = … = $q_n$ = 1, as then k objects are chosen only from the m objects in the last row. Note that the order of the numbers $q_1,\dots q_n$ does not matter in $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\{q_1,\dots ,q_n\}}}\right)$. By convention, $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\{q_1,\dots ,q_n\}}}\right)$ = 0 for k < 0 or $n+k$ > $m+q_1+\dots +q_n$. From the many properties and applications of $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\{q_1,\dots ,q_n\}}}\right)$ presented in Refs. [1,2,3], we will start from the special case n = 1 (see (3) in [1], (6) in [2]),

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m,1}{k,q}}\right)& =\left({\displaystyle \genfrac{}{}{0pt}{}{m+q}{k+1}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k+1}}\right),\hfill \end{array}$$

(1)

and, for $n\ge 2$, the recursion relation (see (13) in [2])

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\{q_1,\dots ,q_n\}}}\right)& =\sum _{p_n=0}^{q_n-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m+p_n,n-1}{k,\{q_1,\dots ,q_{n-1}\}}}\right).\hfill \end{array}$$

(2)

Combinatorially, (1) means that, for n = 1, one chooses $n+k$ = $k+1$ out of a total of $m+q$ objects, of which the choices only from the last row do not count and are therefore subtracted. The combinatorial meaning of (2) was given in Ref. [2] and may be paraphrased as follows: in row n, let $r_n$ (1 ≤ $r_n$ ≤ $q_n$) be the index of the selected object with the highest index; then, $n+k-1$ objects remain to be selected, namely at least one object each from the first $n-1$ rows, as well as $m+r_n-1$ objects from the last row and the rest of row n, for which there are $\left({\displaystyle \genfrac{}{}{0pt}{}{m+r_n-1,n-1}{k,\{q_1,\dots ,q_{n-1}\}}}\right)$ choices. Summing this over $p_n$ = $r_n-1$ = $0\dots q_n-1$ yields $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\{q_1,\dots ,q_n\}}}\right)$.

Below, we use (1) and (2) to establish several properties and applications related to $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\{q_1,\dots ,q_n\}}}\right)$. In Section 2, two explicit forms and an integral representation are obtained and checked against the recursion relations of Refs. [1,2]. In Section 3, a generating function appearing in [3] is obtained, leading to several convolution identities involving the number $p_{Q_n}\left(m\right)$ of partitions of m into parts from $Q_n$. In [2], the special set $Q_n$ = $\{1,\dots ,n\}$ was related to Euler’s pentagonal number theorem, which we extend in Section 4 to the case where this set contains gaps. Using the generating function from Section 3, in Section 5, further relations for $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\{q_1,\dots ,q_n\}}}\right)$ are obtained by regarding it as a Riordan array. In Section 6, we obtain another type of generating function and use it to obtain inverse relations for a set of linear equations. Finally, in Section 7, the thermodynamic properties of a many-body system with restricted configurations as in $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\{q_1,\dots ,q_n\}}}\right)$ are discussed, leading to a generalized Schottky anomaly in the heat capacity.

For brevity, similarly to [3], we write

$$\begin{array}{cccccccc}\hfill Q_n& =\{q_1,\dots ,q_n\},\hfill & \hfill {\mathcal{Q}}_n& =\sum _{j=1}^n{q}_j,\hfill & \hfill \sum _{p<Q_n}& =\sum _{p_1=0}^{q_1-1}\dots \sum _{p_n=0}^{q_n-1},\hfill & \hfill {\mathcal{P}}_n& =\sum _{j=1}^n{p}_j.\hfill \end{array}$$

(3)

Note also that, if a list Q has more than n elements, then, by convention, $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q}}\right)$ = $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\{q_1,\dots ,q_n\}}}\right)$, where $\{q_1,\dots ,q_n\}$ are the first n elements of Q.

2. Explicit Forms, Row Extension, and Integral Representation

Proposition 1 

(Explicit form). For integers $n,q_1,\dots q_n$ $\ge 1$, $m,k$ $\ge 0$,

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)& =\sum _{p<Q_n}\left({\displaystyle \genfrac{}{}{0pt}{}{m+{\mathcal{P}}_n}{k}}\right).\hfill \end{array}$$

(4)

Proof. 

For n = 1, we use ${\sum }_{r=0}^s\left({\displaystyle \genfrac{}{}{0pt}{}{r}{k}}\right)$ = $\left({\displaystyle \genfrac{}{}{0pt}{}{s+1}{k+1}}\right)$ for k ≥ 0 (proven by induction in s, or by multiplying both sides with $x^{k+1}$ and summing over $k\ge 0$) to rewrite (1) as

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m,1}{k,q_1}}\right)& =\left({\displaystyle \genfrac{}{}{0pt}{}{m+q_1}{k+1}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k+1}}\right)=\sum _{r=m}^{q_1+m-1}\left({\displaystyle \genfrac{}{}{0pt}{}{r}{k}}\right)=\sum _{p_1=0}^{q_1-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m+p_1}{k}}\right),\hfill \end{array}$$

(5)

while, for n ≥ 2, we apply (2) $n-1$ times and insert (5),

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)& =\sum _{p_n=0}^{q_n-1}\dots \sum _{p_2=0}^{q_2-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m+p_n+\dots +p_2,1}{k,q_1}}\right)\hfill \\ & =\sum _{p_n=0}^{q_n-1}\dots \sum _{p_1=0}^{q_1-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m+p_n+\dots +p_1}{k}}\right),\hfill \end{array}$$

(6)

completing the proof of (4). □

Combinatorially, the explicit form (4) can be understood in a similar fashion as (2): in each of the rows j = $1\dots n$, let $r_j$ (1 ≤ $r_j$ ≤ $q_j$) be the index of the selected object with the highest index; then, k objects remain to be selected from the $m+r_1+\dots +r_n-n$ objects in the last row and the rest of the rows $1\dots n$, for which there are $\left({\displaystyle \genfrac{}{}{0pt}{}{m+r_1+\dots +r_n-n}{k}}\right)$ choices. Summing this over all $p_j$ = $r_j-1$ = $0\dots q_j-1$ yields $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)$.

As mentioned at the beginning, $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)$ is nonzero if $m+{\mathcal{Q}}_n$ ≥ $n+k$. However, it should be noted that, if, as in (1) or (4), $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)$ is expressed in terms of binomial coefficients, then some of them may not contribute because their upper argument is smaller than the lower one. For example, for n = 1, $\left({\displaystyle \genfrac{}{}{0pt}{}{m,1}{k,q}}\right)$ is nonzero if $m+q$ ≥ $1+k$, but the second binomial coefficient in (1) is zero for k > m, which can happen for q > 1, e.g.,

$$\left({\displaystyle \genfrac{}{}{0pt}{}{m,1}{m,2}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{m+2}{m+1}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{m}{m+1}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{m+2}{m+1}}\right)=m+2.$$

(7)

From (4) follow several properties of $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)$ that are known from Refs. [1,2], such as the special cases (see Section 1 in [2])

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\{1,\dots ,1\}}}\right)& =\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right),\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{0,\{q_1,\dots ,q_n\}}}\right)& =\prod _{j=1}^n{q}_j.\hfill \end{array}$$

(9)

The recursion relation for the binomial coefficients,

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m+1}{k+1}}\right)& =\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k+1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right),\hfill \end{array}$$

(10)

when inserted in (4), leads to a similar relation for the generalized coefficients (see (9) in [2]):

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m+1,n}{k+1,Q_n}}\right)& =\sum _{p<Q_n}\left[\left({\displaystyle \genfrac{}{}{0pt}{}{m+1+{\mathcal{P}}_n}{k+1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{m+{\mathcal{P}}_n}{k}}\right)\right]\hfill \\ & =\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k+1,Q_n}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right).\hfill \end{array}$$

(11)

Further recursion relations are obtained from (4) by exchanging the order of summations, such as (see (4) in [1], (10) in [2])

$$\begin{array}{ccc}\hfill \sum _{j=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{0,n}{k-j,Q_n}}\right)& =\sum _{p<Q_n}\sum _{j=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{{\mathcal{P}}_n}{k-j}}\right)=\sum _{p<Q_n}\left({\displaystyle \genfrac{}{}{0pt}{}{m+{\mathcal{P}}_n}{k}}\right)\hfill & \hfill \\ & =\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right),\hfill \end{array}$$

(12)

where the inner sum in the first line is a Chu–Vandermonde sum. For $n\ge 2$, we also have (see (16) in [1], (13) in [2])

$$\begin{array}{cc}\hfill \sum _{j=1}^{q_n}\left({\displaystyle \genfrac{}{}{0pt}{}{q_n}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{m,n-1}{k+1-j,Q_{n-1}}}\right)& =\sum _{p<Q_{n-1}}\sum _{j=1}^{q_n}\left({\displaystyle \genfrac{}{}{0pt}{}{q_n}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{m+P_{n-1}}{k+1-j}}\right)\hfill \\ & =\sum _{p<Q_{n-1}}\left[\left({\displaystyle \genfrac{}{}{0pt}{}{q_n+m+P_{n-1}}{k+1}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{m+P_{n-1}}{k+1}}\right)\right]\hfill \\ & =\sum _{p<Q_{n-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+P_{n-1},1}{k,q_n}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right),\hfill \end{array}$$

(13)

where the inner sum in the first line is a Chu–Vandermonde sum with the first term missing, while (5) and (4) are used in the third line.

On the other hand, a different type of recursion relation is obtained from (4) if we add r objects to row n, i.e., if we increase $q_n$ by r.

Proposition 2 

(Row extension). For integers $n,q_1,\dots q_n,r$ $\ge 1$, $m,k$ $\ge 0$,

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\{q_1,\dots ,q_{n-1},q_n+r\}}}\right)& =\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{m+q_n,n}{k,\{q_1,\dots ,q_{n-1},r\}}}\right)\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}& =\left({\displaystyle \genfrac{}{}{0pt}{}{m+r,n}{k,Q_n}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\{q_1,\dots ,q_{n-1},r\}}}\right).\hfill \end{array}$$

(15)

Proof. 

Using (4) and splitting the sum in two ways,

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\{q_1,\dots ,q_{n-1},q_n+r\}}}\right)& =\sum _{p<Q_{n-1}}\left[\sum _{p_n=0}^{q_n-1}+\sum _{p_n=q_n}^{q_n+r-1}\right]\left({\displaystyle \genfrac{}{}{0pt}{}{m+{\mathcal{P}}_n}{k}}\right)\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}& =\sum _{p<Q_{n-1}}\left[\sum _{p_n=0}^{r-1}+\sum _{p_n=r}^{q_n+r-1}\right]\left({\displaystyle \genfrac{}{}{0pt}{}{m+{\mathcal{P}}_n}{k}}\right),\hfill \end{array}$$

(17)

gives (14) and (15), respectively.  □

The combinatorial meaning of (14) is the following. One can first ignore the r additional objects in the extended row, which gives $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)$ choices, and then select $(n+k)$-insets considering only the r additional objects in row n (but grouping the other $q_n$ objects in this row with the m unrestricted objects in the last row), which gives an additional $\left({\displaystyle \genfrac{}{}{0pt}{}{m+q_n,n}{k,\{q_1,\dots ,q_{n-1},r\}}}\right)$ choices. In (15), the roles of $q_n$ and r are interchanged.

Proposition 3 

(Integral representation). For integers $n,q_1,\dots q_n$ $\ge 1$, $m,k$ $\ge 0$,

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)& =\underset{-\pi }{\overset{\pi }{\int }}{\displaystyle \frac{dt}{2\pi }}{(1+e^{it})}^m{e}^{-i(n+k)t}\prod _{j=1}^n[{(1+e^{it})}^{q_j}-1].\hfill \end{array}$$

(18)

Proof. 

We use an integral representation of the binomial coefficients in (4),

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)& =\underset{-\pi }{\overset{\pi }{\int }}{\displaystyle \frac{\mathrm{d}t}{2\pi }}{(1+{\mathrm{e}}^{\mathrm{i}t})}^m{\mathrm{e}}^{-\mathrm{i}kt},\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)& =\sum _{p<Q_n}\left({\displaystyle \genfrac{}{}{0pt}{}{m+{\mathcal{P}}_n}{k}}\right)=\underset{-\pi }{\overset{\pi }{\int }}{\displaystyle \frac{\mathrm{d}t}{2\pi }}\sum _{p<Q_n}{(1+{\mathrm{e}}^{\mathrm{i}t})}^{m+{\mathcal{P}}_n}{\mathrm{e}}^{-\mathrm{i}kt},\hfill \end{array}$$

(20)

and perform n geometric sums (setting ${\mathrm{e}}^{\mathrm{i}t}$ = x ≠ 0),

$$\begin{array}{cc}\hfill \sum _{p_j=0}^{q_j-1}{(1+x)}^{p_j}& ={\displaystyle \frac{{(1+x)}^{q_j}-1}{(1+x)-1}},\hfill \end{array}$$

(21)

which results in (18) after replacing x = ${\mathrm{e}}^{\mathrm{i}t}$.  □

The integral representation (18) reduces to (19) for $q_1$ = … = $q_n$ = 1, in agreement with (8). Next, we use it to obtain a different type of explicit form of the counting function.

Proposition 4 

(Alternative explicit form). For integers $n,q_1,\dots q_n$ $\ge 1$, $m,k$ $\ge 0$,

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)& =\sum _{p<Q_n}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k-{\mathcal{P}}_n}}\right)\prod _{j=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{q_j}{p_j+1}}\right).\hfill \end{array}$$

(22)

Proof. 

We start from (18), abbreviate ${\mathrm{e}}^{\mathrm{i}t}$ = x for the moment, and expand the binomials in the product,

$$\begin{array}{cc}\hfill \prod _{j=1}^n[{(1+x)}^{q_j}-1]& =\prod _{j=1}^n\sum _{p_j=1}^{q_j}\left({\displaystyle \genfrac{}{}{0pt}{}{q_j}{p_j}}\right)x^{p_j}=x^n\sum _{p<Q_n}{x}^{{\mathcal{P}}_n}\prod _{j=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{q_j}{p_j+1}}\right),\hfill \end{array}$$

(23)

after which we replace x = ${\mathrm{e}}^{\mathrm{i}t}$, integrate over t using (19), and arrive at (22).  □

The combinatorial interpretation of (22) is straightforward: set $r_j$ = $p_j$ + 1 ≥ 1; for j = $1\dots n$, select $r_j$ elements from the $q_j$ elements in row j; then, select the remaining $n+k-{\sum }_{j=1}^n{r}_j$ (= $k-{\mathcal{P}}_n$ ≥ 0) elements from the m elements in the last row.

3. Generating Function for a Fixed Number of Selected Objects and Convolutions with Partition Counts

The following generating function for a fixed number of selected objects appears as Theorem 1 in [3] with a proof by induction. Here, we provide a short alternative proof using the explicit representation (4).

Proposition 5 

(Janjić: Generating functions for fixed number of selected objects $n+k$). For $\left|x\right|$ < 1 and integers $n,q_1,\dots q_n$ $\ge 1$, k ≥ ${\mathcal{Q}}_n-n$,

$$\begin{array}{cc}\hfill F_{k,n,Q}\left(x\right)& =\sum _{\ell =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +s,n}{k,Q_n}}\right)x^{\ell }={\displaystyle \frac{1}{{(1-x)}^{n+k+1}}}\prod _{j=1}^n\left(1-x^{q_j}\right),\hfill \end{array}$$

(24)

with the abbreviation s = $k-{\mathcal{Q}}_n+n$ ≥ 0.

Proof. 

We use (4), omitting the initial terms in the sum for which the binomial coefficient vanishes,

$$\begin{array}{cc}\hfill F_{k,n,Q_n}\left(x\right)& =\sum _{\ell =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +k-{\mathcal{Q}}_n+n,n}{k,Q_n}}\right)x^{\ell }\hfill \\ & =\sum _{p<Q_n}\sum _{\ell ={\mathcal{Q}}_n-n-{\mathcal{P}}_n}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +k-{\mathcal{Q}}_n+n+{\mathcal{P}}_n}{k}}\right)x^{\ell }\hfill \\ & =\sum _{p<Q_n}{x}^{{\mathcal{Q}}_n-n-{\mathcal{P}}_n}\sum _{m=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{m+k}{k}}\right)x^m,\hfill \end{array}$$

(25)

where ${\sum }_{m=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{m+k}{k}}\right)x^k$ = ${(1-x)}^{-k-1}$, and

$$\begin{array}{cc}\hfill \sum _{p<Q_n}{x}^{{\mathcal{Q}}_n-n-{\mathcal{P}}_n}& =\prod _{j=1}^n\sum _{p_j=0}^{q_j-1}{x}^{q_j-1-p_j}=\prod _{j=1}^n{\displaystyle \frac{1-x^{q_j}}{1-x}},\hfill \end{array}$$

(26)

arriving at (24). □

As discussed in [3], the generating function (24) implies the connection of $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)$ to $p_{Q_n}^{\phantom{|}}\left(m\right)$, the number of partitions of the integer m into parts taken from $Q_n$ [4]; by definition, $p_{Q_n}^{\phantom{|}}\left(m\right)$ = 0 for m < 0. These partitions are generated by the generating function $P_{Q_n}\left(x\right)$,

$$\begin{array}{cccc}\hfill p_{Q_n}^{\phantom{|}}\left(m\right)& =\sum _{\begin{array}{c}{s}_1,\dots ,s_n\ge 0\\ m={\sum }_{j=1}^n{s}_j{q}_j\end{array}}1,\hfill & \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}{P}_{Q_n}\left(x\right)& =\sum _{m=0}^{\infty }{p}_{Q_n}^{\phantom{|}}\left(m\right)x^m=\prod _{j=1}^n{\displaystyle \frac{1}{1-x^{q_j}}},\hfill \end{array}$$

(27)

in the denominator of which the product ${\prod }_{j=1}^n(1-x^{q_j})$ appears, which characterizes the rows $Q_n$ and appears in the numerator of $F_{k,n,Q_n}\left(x\right)$. This relation suggests that we should eliminate this product from the two generating functions, providing convolutions of $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)$ and $p_{Q_n}^{\phantom{|}}\left(m\right)$.

Proposition 6 

(Convolutions). For integers $n,q_1,\dots q_n$ $\ge 1$, k ≥ ${\mathcal{Q}}_n-n$, and integer r ≥ 0,

$$\begin{array}{cc}\hfill {(-1)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{r}{m}}\right)& =\sum _{\ell =0}^m\sum _{i=0}^{m-\ell }\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +k-{\mathcal{Q}}_n+n,n}{k,Q_n}}\right)p_{Q_n}^{\phantom{|}}(m-\ell -i){(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n+k+r+1}{i}}\right),\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill {\delta }_{m0}=\left({\displaystyle \genfrac{}{}{0pt}{}{0}{m}}\right)& =\sum _{\ell =0}^m\sum _{i=0}^{m-\ell }\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +k-{\mathcal{Q}}_n+n,n}{k,Q_n}}\right)p_{Q_n}^{\phantom{|}}(m-\ell -i){(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n+k+1}{i}}\right),\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m+r}{m}}\right)& =\sum _{\ell =0}^m\sum _{i=0}^{m-\ell }\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +k-{\mathcal{Q}}_n+n,n}{k,Q_n}}\right)p_{Q_n}^{\phantom{|}}(m-\ell -i){(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n+k-r}{i}}\right),\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{m+n+k}{m}}\right)& =\sum _{\ell =0}^m\phantom{\sum _{i=0}^{m-\ell }}\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +k-{\mathcal{Q}}_n+n,n}{k,Q_n}}\right)p_{Q_n}^{\phantom{|}}(m-\ell ).\hfill \end{array}$$

(31)

Here, (29) is equivalent to [3] (Theorem 2).

Proof. 

From (24) and (27), we have

$$\begin{array}{cc}\hfill {(1-x)}^r& =F_{k,n,Q_n}\left(x\right)P_{Q_n}\left(x\right){(1-x)}^{n+k+r+1},\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{{(1-x)}^{r+1}}}& =F_{k,n,Q_n}\left(x\right)P_{Q_n}\left(x\right){(1-x)}^{n+k-r},\hfill \end{array}$$

(33)

and, taking the coefficients of $x^m$ in each of these, respectively, gives (28), which becomes (29) for r = 0, and (30), which becomes (31) for r = $n+k$. □

4. Linearly Increasing Number of Objects per Row with Gaps and Connection to Pentagonal and Related Numbers

In this section, special sets of $Q_n$ will be considered, namely the first n = $3N+1$ entries (for integer N ≥ 1) of one of the following lists:

$$\begin{array}{cc}\hfill L_2& =\{1,1,2,3,3,4,5,5,\dots ,2j,2j+1,2j+1,\dots \},\hfill \end{array}$$

(34)

$$\begin{array}{cc}\hfill \mathbb{N}=L_3& =\{1,2,3,\dots ,j,\dots \},\hfill \end{array}$$

(35)

$$\begin{array}{cc}\hfill L_4& =\{1,3,4,5,7,8,9,\dots ,4j-1,4j,4j+1,\dots \},\hfill \end{array}$$

(36)

$$\begin{array}{cc}\hfill L_5& =\{1,4,5,6,9,10,11,\dots ,5j-1,5j,5j+1,\dots \},\hfill \end{array}$$

(37)

$$\begin{array}{ccc}\hfill L_r& =\{1,rj-1,rj,rj+1,\dots \},\hfill & \hfill \hspace{1em}\hspace{1em}\hspace{1em}(r\ge 4)\end{array}$$

(38)

i.e., for integer r ≥ 4, there are repeating gaps in $L_r$ of $r-3$ missing entries on the list of consecutive integers (namely $r(j-1)+2$, …, $rj-2$), while, for r = 2, each of the odd entries is once repeated. In other words, $L_r$ is a lacunary list of triples $rj-1,rj,rj+1$ centered around multiples of r. For this set $Q_n$ = $Q_{3N+1}$, the first and last entries are thus $q_1$ = 1 and $q_n$ = $rN+1$, and the sum of all entries is

$$\begin{array}{cc}\hfill {\mathcal{Q}}_n& =1+\sum _{j=1}^N[(rj-1)+rj+(rj+1)]=1+{\displaystyle \frac{3rN(N+1)}{2}}=1+{\displaystyle \frac{r(n-1)(n+2)}{6}}.\hfill \end{array}$$

(39)

The reason for choosing $Q_n$ from these lists is that their entries occur precisely in the power series exponents in a lacunary product identity (see (42) below) that generalizes Euler’s pentagonal number theorem, the connection of which to $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,\mathbb{N}}}\right)$ was already discussed in [3]. To derive this identity, we follow [5] and start from Jacobi’s triple product identity [4], which reads, for $\left|u\right|$ < 1 and v ≠ 0,

$$\begin{array}{cc}\hfill \prod _{j=0}^{\infty }(1-u^{2j+2})(1+u^{2j+1}v)(1+u^{2j+1}{v}^{-1})& =\sum _{m=-\infty }^{\infty }{u}^{m^2}{v}^m.\hfill \end{array}$$

(40)

It reduces to the pentagonal number theorem for u = $x^{3/2}$, v = $-x^{1/2}$, $\left|x\right|$ < 1,

$$\begin{array}{cc}\hfill \prod _{j=1}^{\infty }(1-x^j)& =\sum _{m=-\infty }^{\infty }{(-1)}^m{x}^{m(3m-1)/2},\hfill \end{array}$$

(41)

which is well known for the cancellation of powers of x in the series and the fact it involves only coefficients $-1,0,1$. It may be regarded as the special case (r = 3) of the identity that follows from letting instead u = $x^{r/2}$ and v = $-x^{1-r/2}$ in (40), with integer r ≥ 2,

$$\begin{array}{cc}\hfill & (1-x)\prod _{j=1}^{\infty }(1-x^{rj-1})(1-x^{rj})(1-x^{rj+1})\hfill \\ & =\sum _{m=-\infty }^{\infty }{(-1)}^m{x}^{rm(m-1)/2+m}=1+\sum _{m=1}^{\infty }{(-1)}^m\left(x^{[r{m}^2-(r-2)m]/2}+x^{[r{m}^2+(r-2)m]/2}\right).\hfill \end{array}$$

(42)

For r ≥ 4, the factors with exponents $r(j-1)+2$, …, $rj-2$ are absent in (42), resulting in a lacunary product, while, for r = 2, factors with odd powers occur twice,

$$\begin{array}{cc}\hfill \prod _{j=1}^{\infty }{(1-x^{2j-1})}^2(1-x^{2j})& =1+2\sum _{m=1}^{\infty }{(-1)}^m{x}^{m^2},\hfill \end{array}$$

(43)

and even more powers of x are cancelled in the series than in (41).

The cancellation of summands in the pentagonal number theorem requires an infinite product. If, instead, the product in (41) is finite, i.e., terminating with $(1-x^n)$, then the expansion into a polynomial only agrees with the power series in (41) up to order $x^n$, while, for exponents in the range $n+1$ to $(n+1)n/2$ other than pentagonal numbers $1,5,12,22,\dots$ and coefficients other than $-1,0,1$, we have, e.g.,

$$\begin{array}{cc}\hfill \prod _{j=1}^4(1-x^j)& =1-x-x^2+2x^5-x^8-x^9+x^{10},\hfill \end{array}$$

(44)

and similar cases occur for truncated lacunary products (42).

However, the generating function (24) for our $Q_n$ (i.e., by the first n = $rN+1$ elements of $L_r$) gives us

$$\begin{array}{cc}\hfill {(1-x)}^{n+k+1}{F}_{k,n,Q_n}\left(x\right)& ={(1-x)}^{n+k+1}\sum _{\ell =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +s,n}{k,L_r}}\right)x^{\ell }\hfill \\ & =(1-x){\displaystyle \prod _{j=1}^N}(1-x^{rj-1})(1-x^{rj})(1-x^{rj+1})\hfill \end{array}$$

(45)

provided that k ≥ ${\mathcal{Q}}_n-n$ = $3\left[r\right(N+1)-2]N/2$ = $\left[r\right(n+2)-6](n-1)/6$. Before taking coefficients of $x^i$ for 0 ≤ i ≤ n, we replace the product on the right-hand side with (42), as the extra factors result in a difference only for higher exponents.

Proposition 7 

(Linearly increasing number of objects per row with gaps). For the set $Q_{rN+1}$ = $\{1,\dots ,rN+1\}$ ⊂ $L_r$ with integers r ≥ 2, N ≥ 1, n = $3N+1$, k ≥ ${\mathcal{Q}}_n$ − n (see (39)), 0 ≤ i ≤ n,

$$\begin{array}{c}{\displaystyle \sum _{\ell =0}^i}\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +k-{\mathcal{Q}}_n+n,n}{k,Q_n}}\right){(-1)}^{i+\ell }\left({\displaystyle \genfrac{}{}{0pt}{}{n+k+1}{\ell -i}}\right)\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}=\left\{\begin{array}{cc}1\hfill & ifi=0,\hfill \\ 2{(-1)}^m\hfill & ifr=2,i=m^2\ge 1,\hfill \\ {(-1)}^m\hfill & ifr\ge 3,i=[r{m}^2\pm (r-2)m]/2\ge 1.\hfill \end{array}\right.\hfill \end{array}$$

(46)

Proof. 

Equate the coefficients of $x^i$ on the left-hand side of (45) and right-hand side of (42). □

5. The Janjić–Petković Counting Function as a Riordan Array

As discussed above, the counting function $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)$ is a generalization of the binomial coefficients $\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)$. As shown in this section, the former may be regarded as a Riordan array, like the latter, which is also suggested the by the Pascal triangle analog (11) discussed in [2], as well as by the generating functions occurring in the previous two sections.

Let us first recall the essential properties of Riordan arrays [6,7,8], using slightly unconventional lettering (i.e., N, K, x instead of the usual n, k, t) for better compatibility with the notation used here for $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)$ and its generating functions. An infinite lower triangular matrix $\mathit{d}$ with entries $d_{N,K}$ = ${\left(\mathit{d}\right)}_{N\ge 0,K\ge 0}$ and generating function

$$\begin{array}{cccccc}\hfill \sum _{N=0}^{\infty }{d}_{N,K}{x}^N& =d\left(x\right)D{\left(x\right)}^K,\hfill & \hfill d\left(0\right)& =d_{0,0}\ne 0,\hfill & \hfill D\left(0\right)& =d_{N(<K),K}=0,\hfill \end{array}$$

(47)

is called a Riordan array and denoted as $\mathit{d}$ $\widehat{=}$ $\left(d\left(x\right),D\left(x\right)\right)$, referring either to the matrix or the pair of formal power series. They form the Riordan group, so called because the matrix multiplication $\mathit{c}$ = $\mathit{a}·\mathit{b}$ of two Riordan arrays $\mathit{a}$ and $\mathit{b}$ gives again a Riordan array,

$$\begin{array}{cc}\hfill c_{N,K}& =\sum _{j=0}^{\infty }{a}_{N,j}{b}_{j,K},\hfill \end{array}$$

(48)

$$\begin{array}{cc}\hfill \left(c\right(x),C(x\left)\right)& \triangleq \left(a\left(x\right),A\left(x\right)\right)\circ \left(b\left(x\right),B\left(x\right)\right)=\left(a\left(x\right)b\left(A\right(x\left)\right),B\left(A\right(x\left)\right)\right),\hfill \end{array}$$

(49)

with the identity element given by unit array $\mathbb{1}$ $\widehat{=}$ $\left(1,x\right)$ with ones on the diagonal as the only nonzero entries. A useful case corresponds to the matrix times vector, i.e., for a Riordan array $\mathit{d}$,

$$\begin{array}{cccccc}\hfill w_N& =\sum _{K=0}^N{d}_{N,K}{v}_K,\hfill & \hfill v\left(x\right)& =\sum _{K=0}^{\infty }{v}_K{x}^K,\hfill & \hfill \sum _{N=0}^{\infty }{w}_N{x}^N& =d\left(x\right)v\left(D\right(x\left)\right).\hfill \end{array}$$

(50)

Furthermore, the matrix inverse ${\mathit{d}}^{-1}$ $\widehat{=}$ ${\left(d\left(x\right),D\left(x\right)\right)}^{-1}$ is also a Riordan array,

$$\begin{array}{cc}\hfill \sum _{j=0}^{\infty }{\left(\mathit{d}\right)}_{N,j}{\left({\mathit{d}}^{-1}\right)}_{j,K}& ={\left(\mathbb{1}\right)}_{N,K},\hfill \end{array}$$

(51)

$$\begin{array}{cc}\hfill {\left(d\left(x\right),D\left(x\right)\right)}^{-1}& =\left({\displaystyle \frac{1}{d\left(\overline{D}\right(x\left)\right)}},\overline{D}\left(x\right)\right),\hfill \end{array}$$

(52)

$$\begin{array}{cc}\hfill \left(d\left(x\right),D\left(x\right)\right)\circ {\left(d\left(x\right),D\left(x\right)\right)}^{-1}& ={\left(d\left(x\right),D\left(x\right)\right)}^{-1}\circ \left(d\left(x\right),D\left(x\right)\right)=(1,x).\hfill \end{array}$$

(53)

where $\overline{D}\left(x\right)$ is the compositional inverse of $D\left(x\right)$, defined by x = $\overline{D}\left(D\right(x\left)\right)$ = $D\left(\overline{D}\right(x\left)\right)$. The binomial coefficients are a well-known example of a Riordan array:

$$\begin{array}{cccccc}\hfill {\left(\mathit{b}\right)}_{N,K}& =\left({\displaystyle \genfrac{}{}{0pt}{}{N}{K}}\right),\hfill & \hfill \sum _{N=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{N}{K}}\right)x^N& ={\displaystyle \frac{1}{{(1-x)}^{K+1}}},\hfill & \hfill \mathit{b}& \widehat{=}\left({\displaystyle \frac{1}{1-x}},{\displaystyle \frac{x}{1-x}}\right).\hfill \end{array}$$

(54)

Let us now return to the counting function $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)$, which is found to fit within the Riordan array framework.

Proposition 8 

(Janjić–Petković counting function as Riordan array). For integers $N,K$ ≥ 0, $n,q_1,\dots q_n$ $\ge 1$, we have the Riordan arrays

$$\begin{array}{cc}\hfill \mathit{f}& \widehat{=}(f\left(x\right),F\left(x\right))=\left({\displaystyle \frac{{\prod }_{j=1}^n\left(1-x^{q_j}\right)}{{(1-x)}^{n+1}}},{\displaystyle \frac{x}{1-x}}\right),\hfill \end{array}$$

(55)

$$\begin{array}{cc}\hfill {\mathit{f}}^{-1}& \widehat{=}\left({\displaystyle \frac{1}{f\left(\overline{F}\right(x\left)\right)}},\overline{F}\left(x\right)\right)=\left({\displaystyle \frac{1}{{(1+x)}^{n+1}{\prod }_{j=1}^n\left(1-x^{q_j}/{(1+x)}^{q_j}\right)}},{\displaystyle \frac{x}{1+x}}\right),\hfill \end{array}$$

(56)

$$\begin{array}{cc}\hfill {\left(\mathit{f}\right)}_{N,K}& =\left({\displaystyle \genfrac{}{}{0pt}{}{N-{\mathcal{Q}}_n+n,n}{K,Q_n}}\right),\hfill \end{array}$$

(57)

$$\begin{array}{cc}\hfill {\left({\mathit{f}}^{-1}\right)}_{N,K}& =\sum _{i=0}^{N-K}{p}_{Q_n}^{\phantom{|}}(N-K-i)\left({\displaystyle \genfrac{}{}{0pt}{}{N+n}{i}}\right){(-1)}^i,\hfill \end{array}$$

(58)

for fixed n and $Q_n$, with ${\left({\mathit{f}}^{-1}\right)}_{N,K}$ = ${\left({\mathit{f}}^{-1}\right)}_{N,K}$ = 0 if K > N. Furthermore,

$$\begin{array}{cc}\hfill {\delta }_{N,M}& =\sum _{K=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{N-{\mathcal{Q}}_n+n,n}{K,Q_n}}\right)\sum _{i=0}^{K-M}{p}_{Q_n}^{\phantom{|}}(K-M-i)\left({\displaystyle \genfrac{}{}{0pt}{}{K+n}{i}}\right){(-1)}^i\hfill \end{array}$$

(59)

$$\begin{array}{c}\hspace{1em}\hspace{1em}\hspace{1em}=\sum _{K=0}^N\left({\displaystyle \genfrac{}{}{0pt}{}{K-{\mathcal{Q}}_n+n,n}{M,Q_n}}\right)\sum _{i=0}^{N-K}{p}_{Q_n}^{\phantom{|}}(N-K-i)\left({\displaystyle \genfrac{}{}{0pt}{}{N+n}{i}}\right){(-1)}^i.\hfill \end{array}$$

(60)

Proof. 

Setting N = $\ell +k$ and K = k in (24), we obtain a generating function of the form (47),

$$\begin{array}{cc}\hfill x^K{F}_{K,n,Q_n}\left(x\right)& =\sum _{\ell =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +K-{\mathcal{Q}}_n+n,n}{k,Q_n}}\right)x^{\ell +K}=\sum _{N=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{N-{\mathcal{Q}}_n+n,n}{K,Q_n}}\right)x^N\hfill \\ & ={\displaystyle \frac{x^K}{{(1-x)}^{n+K+1}}}\prod _{j=1}^n\left(1-x^{q_j}\right)=f\left(x\right)F{\left(x\right)}^K,\hfill \end{array}$$

(61)

where we use the fact that $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)$ = 0 for $m+{\mathcal{Q}}_n$ <$n+k$ to start the second summation at N = 0. This establishes (55) and (57). Then, (56) follows by applying (53), using the fact that the compositional inverse of $F\left(x\right)$ = $x/(1-x)$ is $\overline{F}\left(x\right)$ = $x/(1+x)$. Then, from (56), together with the generating functions (27) and (54), we have

$$\begin{array}{cc}\hfill \sum _{N=0}^{\infty }{\left({\mathit{f}}^{-1}\right)}_{N,K}{x}^N& ={\displaystyle \frac{\overline{F}{\left(x\right)}^K}{f\left(\overline{F}\right(x\left)\right)}}={\displaystyle \frac{x^K}{{(1+x)}^{n+K+1}{\prod }_{j=1}^n\left(1-x^{q_j}/{(1+x)}^{q_j}\right)}}\hfill \\ & ={\displaystyle \frac{x^K{P}_{Q_n}(x/(1+x))}{{(1+x)}^{n+K+1}}}=\sum _{m=0}^{\infty }{p}_{Q_n}^{\phantom{|}}\left(m\right){\displaystyle \frac{x^{K+m}}{{(1+x)}^{n+K+m+1}}}\hfill \\ & =\sum _{m,i=0}^{\infty }{p}_{Q_n}^{\phantom{|}}\left(m\right)x^{K+m+i}\left({\displaystyle \genfrac{}{}{0pt}{}{i+n+K+m}{i}}\right){(-1)}^i\hfill \\ & =\sum _{N=K}^{\infty }{x}^N\sum _{i=0}^{N-K}{p}_{Q_n}^{\phantom{|}}(N-K-i)\left({\displaystyle \genfrac{}{}{0pt}{}{N+n}{i}}\right){(-1)}^i,\hfill \end{array}$$

(62)

and taking the coefficients yields (58). Then, (59) and (60) follow from (53),

$$\begin{array}{cc}\hfill {\delta }_{N,M}& =\sum _{K=0}^{\infty }{\left(\mathit{f}\right)}_{N,K}{\left({\mathit{f}}^{-1}\right)}_{K,M}=\sum _{K=0}^{\infty }{\left({\mathit{f}}^{-1}\right)}_{N,K}{\left(\mathit{f}\right)}_{K,M},\hfill \end{array}$$

(63)

upon inserting (57) and (58). □

The Riordan array property also entails the following inverse pair of summation formulas involving the counting function $\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)$.

Proposition 9 

(Inverse relations I). For fixed integers $n,q_1,\dots q_n$ $\ge 1$, the following sets of linear equations for two sequences $\left\{a_N\right\}$ and $\left\{b_N\right\}$ (for all N ≥ 0) are equivalent:

$$\begin{array}{cc}\hfill b_N& =\sum _{K=0}^N\left({\displaystyle \genfrac{}{}{0pt}{}{N-{\mathcal{Q}}_n+n,n}{K,Q_n}}\right)a_K,\hfill \end{array}$$

(64)

$$\begin{array}{cc}\hfill \iff a_N& =\sum _{K=0}^N\left[\sum _{i=0}^{N-K}{p}_{Q_n}^{\phantom{|}}(N-K-i)\left({\displaystyle \genfrac{}{}{0pt}{}{N+n}{i}}\right){(-1)}^i\right]b_K.\hfill \end{array}$$

(65)

Proof. 

The claim corresponds to $b_N$ = ${\sum }_{K=0}^N{\left(\mathit{f}\right)}_{N,K}{a}_K$⇔$a_N$ = ${\sum }_{K=0}^N{\left({\mathit{f}}^{-1}\right)}_{N,K}{b}_K$, which corresponds to (63). □

6. Generating Function for Fixed Number of Rows, Inverse Pair of Summation Formulas, and Restricted Power Set

The representation (4) also provides a generating function for a fixed number of rows.

Proposition 10 

(Generating functions for fixed number of rows m). For integers $n,q_1,\dots ,$ $q_n$ $\ge 1$, m $\ge 0$,

$$\begin{array}{cc}\hfill G_{m,n,Q_n}\left(x\right)& =\sum _{k=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)x^{n+k}={(1+x)}^m\prod _{j=1}^n[{(1+x)}^{q_j}-1],\hfill \end{array}$$

(66)

where the sum in fact terminates, i.e., 0 ≤ k ≤ $m+{\mathcal{Q}}_n-n$.

Proof. 

For x = 0, Equation (66) is trivially true (because n ≥ 1). Hence, we assume that x ≠ 0 and find from (4), using the summation (21),

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)x^k& =\sum _{k=0}^{\infty }\sum _{p<Q_n}\left({\displaystyle \genfrac{}{}{0pt}{}{m+{\mathcal{P}}_n}{k}}\right)x^k=\sum _{p<Q_n}{(1+x)}^{m+{\mathcal{P}}_n}\hfill \\ & ={(1+x)}^m\prod _{j=1}^n\left[\sum _{p_j=0}^{q_j-1}{(1+x)}^{p_j}\right]={(1+x)}^m\prod _{j=1}^n\left[{\displaystyle \frac{{(1+x)}^{q_j}-1}{(1+x)-1}}\right],\hfill \end{array}$$

(67)

establishing (66) for all x. □

For the example in Figure 1, we find from (66) that

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{3,4}{k,\{5,3,4,1\}}}\right)x^k=x^{-3}{G}_{3,4,\{5,3,4,1\}}\hfill \\ & ={(x+1)}^3{\displaystyle \frac{{(x+1)}^3-1}{x}}{\displaystyle \frac{{(x+1)}^4-1}{x}}{\displaystyle \frac{{(x+1)}^5-1}{x}}\hfill \\ & ={(x+1)}^3(x+2)(x^2+2x+2)(x^2+3x+3)(x^4+5x^3+10x^2+10x+5)\hfill \\ & =60+450x+1580x^2+3445x^3+5202x^4+5731x^5+4720x^6\hfill \\ & +2925x^7+1352x^8+454x^9+105x^{10}+15x^{11}+x^{12},\hfill \end{array}$$

(68)

the coefficients of which are listed in the figure caption.

The product appearing in (66) is reminiscent of the denominator of $P_{Q_n}\left(x\right)$ in (27). This is expressed in the following pair of summation formulas.

Proposition 11 

(Inverse relations II). For fixed integers $n,q_1,\dots ,q_n$ $\ge 1$, the following sets of linear equations for two sequences $\left\{a_k\right\}$ and $\left\{b_k\right\}$ (for all k ≥ 0) are equivalent:

$$\begin{array}{cc}\hfill b_k& ={(-1)}^k\sum _{m\ge 0}\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)a_m\hfill \\ \hfill \iff a_k& ={(-1)}^k\sum _{m\ge 0}{b}_m\left[\sum _{s=0}^k{(-1)}^s{p}_{Q_n}\left(s\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n+m}{k-s}}\right)\right].\hfill \end{array}$$

(69)

Proof. 

In terms of the generating functions $A\left(x\right)$ = ${\sum }_{k\ge 0}$ $a_k{x}^k$ and $B\left(x\right)$ = ${\sum }_{k\ge 0}$ $b_k{x}^k$, we have from (66)

$$\begin{array}{cc}\hfill B\left(x\right)& =x^{-n}A(1-x)\prod _{j=1}^n[1-{(1-x)}^{q_j}].\hfill \end{array}$$

(70)

Replacing x with $1-x$, this is equivalent to

$$\begin{array}{cc}\hfill A\left(x\right)& ={(1-x)}^{n}B(1-x)P_{Q_n}\left(x\right)\hfill \\ & ={\displaystyle \sum _{m=0}^{\infty }}{b}_m{\displaystyle \sum _{r=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{n+m}{r}}\right){(-x)}^r{\displaystyle \sum _{s=0}^{\infty }}p{Q}_n(s)x^s\\ & ={\displaystyle \sum _{k=0}^{\infty }}{x}^k{\displaystyle \sum _{m=0}^{\infty }}{b}_m{\displaystyle \sum _{s=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{n+m}{k-s}}\right){(-1)}^{k-s}p{Q}_n(s),\end{array}$$

(71)

and the claim follows by taking the coefficients of $x^k$. □

Note that, for $Q_n$ = $\{1,\dots ,1\}$, the equivalence (69) reduces to the standard binomial inversion formula [9]

$$\begin{array}{cc}\hfill b_k& ={(-1)}^k\sum _{m\ge 0}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)a_m\iff a_k={(-1)}^k\sum _{m\ge 0}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)b_m.\hfill \end{array}$$

(72)

As a simple application, the total number of $(n+k)$-insets, setting x = 1 in (66), is

$$\begin{array}{cc}\hfill \sum _{k=0}^{m+{\mathcal{Q}}_n-n}\left({\displaystyle \genfrac{}{}{0pt}{}{m,n}{k,Q_n}}\right)& =(2^{q_1}-1)\cdots (2^{q_n}-1)2^m.\hfill \end{array}$$

(73)

Combinatorially, this means that, in each row, each object can either be selected or not, but zero objects must not be selected from any of the first n rows. Therefore, this restricted power set has fewer than the $2^{m+q_1+\dots +q_n}$ elements of the unrestricted power set.

7. Thermodynamic Properties of a Many-Body System with Janjić–Petković Interaction: A Generalized Schottky Anomaly

Statistical mechanics describes the macroscopic properties of many-body systems in thermodynamic equilibrium [10]. One of the simplest such systems is given by L noninteracting quantum particles, each of which may be in a state 0 or 1 with respective energies ${ϵ}_0$ and ${ϵ}_1$, e.g., a collection of spins-${\displaystyle \frac{1}{2}}$ with Zeeman coupling to a magnetic field. The Hamiltonian for such a system reads

$$\begin{array}{cc}\hfill \widehat{H}& =\sum _{i=1}^L[{ϵ}_1{\widehat{n}}_i+{ϵ}_0(1-{\widehat{n}}_i)],\hfill \end{array}$$

(74)

where each number operator ${\widehat{n}}_i$ has eigenvalues 0 and 1. The canonical partition function at temperature T for this system, with the abbreviation $\beta$ = $1/\left(k_{\mathrm{B}}T\right)$ and the Boltzmann constant $k_{\mathrm{B}}$, is

$$\begin{array}{cc}\hfill Z_L\left(T\right)& =\mathrm{Tr}{\mathrm{e}}^{-\beta \widehat{H}}\hfill \\ & =\sum _{n_1=0,1}\dots \sum _{n_L=0,1}{\mathrm{e}}^{-\beta {\displaystyle \sum _{i=1}^L}[{ϵ}_1{n}_i+{ϵ}_0(1-n_i)]}={\left(\sum _{n=0,1}{\mathrm{e}}^{-\beta [{ϵ}_{1}n+{ϵ}_0(1-n)]}\right)}^L\hfill \\ & ={({\mathrm{e}}^{-\beta {ϵ}_1}+{\mathrm{e}}^{-\beta {ϵ}_0})}^L\hfill \\ & ={\displaystyle \sum _{\ell =0}^L}\left({\displaystyle \genfrac{}{}{0pt}{}{L}{\ell }}\right){\mathrm{e}}^{-\beta [{ϵ}_1\ell +{ϵ}_0(L-\ell )]}.\hfill \end{array}$$

(75)

where the trace is over all microstates. Here, the two types of summations correspond to either choosing each of the L particles to be in state 0 or 1 or alternatively choosing ℓ particles to be in state 0 and the remaining $L-\ell$ particles to be in state 0, for which there are $\left({\displaystyle \genfrac{}{}{0pt}{}{L}{\ell }}\right)$ possibilities. Although this is one of the most elementary thermodynamic model systems, it nevertheless has observable consequences, e.g., for the heat capacity $C_L\left(T\right)$, given by the derivative with respect to T of the average internal energy $E_L\left(T\right)$,

$$\begin{array}{cc}\hfill E_L\left(T\right)& ={\displaystyle \frac{1}{Z_L\left(T\right)}}\mathrm{Tr}\widehat{H}{\mathrm{e}}^{-\beta \widehat{H}}=-{\displaystyle \frac{\partial }{\partial \beta }}ln{Z}_L\left(T\right)\hfill \\ & =L{\displaystyle \frac{{ϵ}_1{\mathrm{e}}^{-\beta {ϵ}_1}+{ϵ}_0{\mathrm{e}}^{-\beta {ϵ}_0}}{{\mathrm{e}}^{-\beta {ϵ}_1}+{\mathrm{e}}^{-\beta {ϵ}_0}}}\hfill \\ & =L({ϵ}_0+e_1\left(T\right)),\hfill \end{array}$$

(76)

$$\begin{array}{cc}\hfill C_L\left(T\right)& ={\displaystyle \frac{\partial E}{\partial T}}=L{c}_1\left(T\right),\hfill \end{array}$$

(77)

which are both extensive in L. Here, $e_1\left(T\right)$ and $c_1\left(T\right)$ are the energy and heat capacity contributions of a single particle, respectively,

$$\begin{array}{cc}\hfill e_1\left(T\right)& ={\displaystyle \frac{\Delta }{1+{\mathrm{e}}^{\beta \Delta }}},\hfill \end{array}$$

(78)

$$\begin{array}{cc}\hfill c_1\left(T\right)& ={\displaystyle \frac{\partial e_1}{\partial T}}=k_{\mathrm{B}}{\displaystyle \frac{{\Delta }^2/{\left(k_{\mathrm{B}}T\right)}^2}{4{cosh}^2\frac{\Delta }{2k_{\mathrm{B}}T}}},\hfill \end{array}$$

(79)

with the energy gap $\Delta$ = ${ϵ}_1-{ϵ}_0$. The specific heat $c_1\left(T\right)$ = $C_L\left(T\right)/L$ is shown in Figure 2. The maximum of $c_1\left(T\right)$ as a function of T is known as the Schottky anomaly [10], a feature that is qualitatively different from the monotonous increase typical for continuum systems. It is due to the limited number of available states and is observed, e.g., in paramagnetic materials or materials with paramagnetic impurities.

Let us now consider a slightly different setting that will require use of the Janjić–Petković counting function. Suppose that we cannot freely choose ℓ out of L particles to have energy ${ϵ}_1$ as before. Rather, the L particles are arranged into n groups (e.g., magnetic domains), each with size $q_1,\dots ,q_n$, and at least one particle must be in state 1 in each of these domains, e.g., because the particles are strongly (magnetically) interacting. We also suppose that there is a nonmagnetic domain of m = L−${\mathcal{Q}}_n$ ≥ 0 particles that are not constrained in this way. The canonical partition function for a system with this Janjić–Petković interaction is then

$$\begin{array}{cc}\hfill Z_{L,Q_n}\left(T\right)& =\sum _{\ell =n}^L\left({\displaystyle \genfrac{}{}{0pt}{}{L-{\mathcal{Q}}_n,n}{\ell -n,Q_n}}\right){\mathrm{e}}^{-\beta [{ϵ}_1\ell +{ϵ}_0(L-\ell )]}\hfill \\ & ={\mathrm{e}}^{-\beta {ϵ}_{0}L}\sum _{k=0}^{L-n}\left({\displaystyle \genfrac{}{}{0pt}{}{L-{\mathcal{Q}}_n,n}{k,Q_n}}\right){\left({\mathrm{e}}^{-\beta \Delta }\right)}^{n+k},\hfill \end{array}$$

(80)

which corresponds to a generating function for a fixed number of rows (66),

$$\begin{array}{cc}\hfill Z_{L,Q_n}\left(T\right)& ={\mathrm{e}}^{-\beta {ϵ}_{0}L}{G}_{L-{\mathcal{Q}}_n,n,Q_n}\left({\mathrm{e}}^{-\beta \Delta }\right)\hfill \\ & ={\mathrm{e}}^{-\beta {ϵ}_{0}L}{(1+{\mathrm{e}}^{-\beta \Delta })}^{L-{\mathcal{Q}}_n}\prod _{j=1}^n[{(1+{\mathrm{e}}^{-\beta \Delta })}^{q_j}-1]\hfill \\ & =Z_L\left(T\right)\prod _{j=1}^n[1-{(1+{\mathrm{e}}^{-\beta \Delta })}^{-q_j}].\hfill \end{array}$$

(81)

The average internal energy $E_{L,Q_n}\left(T\right)$ in this case can be written as

$$\begin{array}{cc}\hfill E_{L,Q_n}\left(T\right)& =-{\displaystyle \frac{\partial }{\partial \beta }}ln{Z}_{L,Q_n}\left(T\right)=E_L\left(T\right)+{\displaystyle \frac{\Delta }{1+{\mathrm{e}}^{\beta \Delta }}}\sum _{j=1}^n{\displaystyle \frac{q_j}{{(1+{\mathrm{e}}^{-\beta \Delta })}^{q_j}-1}}\hfill \\ & =L{ϵ}_0+(L-{\mathcal{Q}}_n)e_1\left(T\right)+\Delta \sum _{j=1}^n{q}_j{\mathcal{E}}_{q_j}\left({\displaystyle \frac{e_1\left(T\right)}{\Delta }}\right),\hfill \end{array}$$

(82)

$$\begin{array}{cc}\hfill {\mathcal{E}}_q\left(x\right)& ={\displaystyle \frac{x}{1-{(1-x)}^q}},\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \end{array}$$

(83)

where the unconstrained magnetic domain of size m = $L-{\mathcal{Q}}_n$ contributes energy $m{e}_1\left(T\right)$, while the constrained magnetic domains for each $q_j$ contribute energy $\Delta q_j{\mathcal{E}}_{q_j}$, written in terms of an auxiliary function ${\mathcal{E}}_q\left(T\right)$; note that ${\mathcal{E}}_1\left(x\right)$ = 1. We then obtain, for the heat capacity $C_{L,Q_n}\left(T\right)$ of the constrained system,

$$\begin{array}{cc}\hfill C_{L,Q_n}\left(T\right)& ={\displaystyle \frac{\partial E_{L,Q_n}\left(T\right)}{\partial T}}=(L-{\mathcal{Q}}_n)c_1\left(T\right)+\sum _{j=1}^n{q}_j{c}_1^{\left(q_j\right)}\left(T\right),\hfill \end{array}$$

(84)

$$\begin{array}{cc}\hfill c_1^{\left(q_j\right)}\left(T\right)& =c_1\left(T\right){\mathcal{E}}_q^{\prime }\left({\displaystyle \frac{1}{1+{\mathrm{e}}^{\frac{\Delta }{k_{\mathrm{B}}T}}}}\right),{\mathcal{E}}_q^{\prime }\left(x\right)={\displaystyle \frac{1}{1-{(1-x)}^q}}-{\displaystyle \frac{qx{(1-x)}^{q-1}}{{\left(1-{(1-x)}^q\right)}^2}},\hfill \end{array}$$

(85)

where m = $L-{\mathcal{Q}}_n$ unconstrained contributions $c_1\left(T\right)$ are supplemented by contributions $c_1^{\left(q_j\right)}\left(T\right)$ for each $q_j$. The functions $c_1^{\left(q\right)}\left(T\right)$ are shown in Figure 2 for several q ≥ 2. They increase with q, with each $c_1^{\left(q\right)}\left(T\right)$ featuring a Schottky-like maximum as a function of temperature T, expressing the restricted availability of states. Note that $c_1^{\left(1\right)}\left(T\right)$ = 0, i.e., magnetic domains with $q_j$ = 1 must be in state 1 and thus cannot contribute to the heat capacity. On the other hand, we have ${lim}_{q\to \infty }$ ${\mathcal{E}}_q^{\prime }\left(x\right)$ = 1, as its argument lies in the interval 0 < x < 1 for T > 0; hence, $c_1^{\left(q\right)}\left(T\right)$ approaches the unrestricted case $c_1\left(T\right)$, because the constraint becomes irrelevant in this limit.

Finally, let us illustrate the physical meaning of the generalized Schottky anomaly due to the terms $c_1^{\left(q_j\right)}\left(T\right)$ in (84). Suppose, for this purpose, that there is no nonmagetic domain, m = L−${\mathcal{Q}}_n$ = 0, and all $q_j$ = q are equal, so that L = $nq$ with n ≥ 1 and q ≥ 2. In this case, $c_1^{\left(q\right)}\left(T\right)$ = $C_{L,Q_n}\left(T\right)/L$ equals the specific heat of the system, i.e., we may regard $c_1^{(L/n)}\left(T\right)$ as replacing $c_1\left(T\right)$ if the system consists of n interacting domains of equal size $L/n$.