Counting corners in partitions

Abstract

A partition of a positive integer $n$ is a non-increasing sequence of positive integers whose sum is $n$. It may be represented by a Ferrers diagram. These diagrams contain corners which are points of degree two. We define corners of types $(a,b)$, $(a+,b)$ and $(a+,b+)$, and also define the size of a corner. Via a generating function, we count corners of each type and corners of size $m$. We also find asymptotics for the number of corners as $n$ tends to infinity.

1 Introduction

A partition of a positive integer $n$ is a non-increasing sequence of positive integers whose sum is $n$, and the number of such partitions is denoted by $P(n)$ (we define $P(0)=1$). A standard result is

$$\begin{aligned} \sum _{j\ge 0}P(j)x^j=\frac{1}{\prod _{j\ge 1}(1-x^j)}. \end{aligned}$$

Many researchers, for example George Andrews, have focused on the number of partitions satisfying certain conditions (for example, see [1–5, 8, 10, 11] and references therein). For instance, if $Q(n)$ is the number of partitions of $n$ with distinct parts and $Q(0):=1$, then we have the generating function

$$\begin{aligned} \sum _{j\ge 0}Q(j)x^j=\prod _{j\ge 0}(1+x^j). \end{aligned}$$

In this paper, we define a new statistic, namely a corner for such partitions. This is related both to descents and the number of occurrences of a part (of fixed size). It is simple to describe corners in terms of the associated Ferrers diagrams.

The Ferrers diagram of an integer partition proves to be a useful tool for visualizing partitions. It is constructed by stacking left-justified rows of cells, where the number of cells in the $i$th row corresponds to the size of the $i$th part.

A corner of a partition $\pi$ is a point of degree two in the corresponding Ferrers diagram. We denote the number of corners of $\pi$ by $cor(\pi )$. For example, if $\pi =4422111$, then $cor(\pi )=6$, see Fig. 1. Moreover, we define $cor_k(\pi )$ to be the number of corners at line $y=k$ in the Ferrers diagram of $\pi$, where the topmost horizontal line of the Ferrers diagram is the line $y=0$. For example, for the partition illustrated below, $cor_0(\pi )=2$ (corners $E$ and $F$), $cor_2(\pi )=1$ (corner $C$) and so on. More generally, we are interested in several types of corners. We say that a corner is of type $(a,b)$ if it is at the bottom right of a specific maximal $a\times b$ rectangle (where $b$ is its length and $a$ its height). For such a maximal rectangle, there are no cells below it and no cells to its right. Thus, corner $B$ is at the bottom right of the 2 by 1 rectangle, with cells marked by X. So that $B$ is a $(2,1)$ corner. Here, we shall only consider corners which are at the bottom-right extremities of such $b$ by $a$ rectangles. So for convenience, we ignore the 3 corners $D$, $E$ and $F$ at levels $x=0$ and $y=0$. Thus, the partition in Fig. 1 has corners $C$ of type $(2,2)$, $B$ of type $(2,1)$ and $A$ of type $(3,1)$.

Fig. 1

The 6 corners of $\pi =4422111$, with $A$, $B$ and $C$ of type $(a,b)$

We observe that a corner of type $(a,b)$ in the Ferrers diagram of $\pi$ corresponds to consecutive parts $c\,,\,c-b$ in the partition $\pi$ such that the multiplicity of $c$ is exactly $a$ and the last occurrence of $c$ is followed by a part of size $c-b$. We also consider corners of type $(a+,b)$; these are corners of type $(j,b)$ for any $j \ge a$. Similarly, we define corners of type $(a+,b+)$. Finally, we define the size of corners of type $(a,b)$ to be $a+b$.

We summarize some of the main results obtained in this paper in Table 1.

Table 1 Summary of main results

2 Corners at level $k$

Let $P_m=P_m(x,t_0,t_1,\ldots )$ be the generating function for the number of partitions of $n$ where the first part is $m$ and $t_k$ marks the number of corners at level $k$ ($y=k$, measured downwards with $y=0$ at the top). Each partition $\pi$ with first part $m$ can be written as $\pi =m\pi '$, where $\pi '$ is either empty or its largest part is less than or equal to $m$. Therefore we obtain

$$\begin{aligned} P_m(x,t_0,t_1,\ldots )=&\,x^m\bigg (\frac{t_0^2}{t_1^2}P_m(x,t_1,t_2,\ldots )\\&+\frac{t_0^2}{t_1}\sum _{j=1}^{m-1}P_j(x,t_1,t_2,\ldots )+t_0^2t_1^2\bigg ). \end{aligned}$$

Define $P(z,x,t_0,t_1,\ldots ):=\sum _{m\ge 1}P_m(x,t_0,t_1,\ldots )z^m$. Multiplying the above equation by $z^m$ and summing over $m\ge 1$, we derive the following result.

Proposition 2.1

The generating function $P(z,x,t_0,t_1,\ldots )$ satisfies the following functional equation:

$$\begin{aligned} P(z,x,t_0,t_1,\ldots )&= \frac{t_0^2}{t_1^2}P(xz,x,t_1,t_2,\ldots )+\frac{t_0^2t_1^2xz}{1-xz}\\&+\frac{t_0^2xz}{t_1(1-xz)}P(xz,x,t_1,t_2,\ldots ). \end{aligned}$$

Now, we are ready to study the total number of corners in a partition. Let $P(z,x,t):=P(z,x,t,t,\ldots )$. Proposition 2.1 with $t_j=t$ for all $j\ge 0$ gives

$$\begin{aligned} P(z,x,t)=P(xz,x,t)+\frac{t^4xz}{1-xz}+\frac{txz}{(1-xz)}P(xz,x,t), \end{aligned}$$

which is equivalent to

$$\begin{aligned} P(z,x,t)=\frac{t^4xz}{1-xz}+\left( 1+\frac{xzt}{1-xz}\right) P(xz,x,t). \end{aligned}$$

By iterating this equation for $z=1,x,x^2,\ldots$, and solving for $P(1,x,t)$, we obtain the following result.

Theorem 2.2

The generating function for the number of partitions of $n$ according to the number of corners counted by $t$ is given by

$$\begin{aligned} P(1,x,t)=t^4\sum _{j\ge 1}\frac{x^j}{1-x^j}\prod _{i=1}^{j-1}\frac{1-(1-t)x^i}{1-x^i}. \end{aligned}$$

We shall need the lemma below for the work that follows:

Lemma 2.3

For all $a,b\ge 1$,

$$\begin{aligned} (a) \quad \sum _{j\ge 0}\frac{x^{jb}}{\prod _{i=1}^j(1-x^i)}=\frac{1}{\prod _{i\ge b}\big (1-x^i\big )} \end{aligned}$$

and

$$\begin{aligned} (b) \quad \sum _{j\ge 0}x^{jb}\prod _{i=j+1}^{j+a}\big (1-x^i\big )=\frac{\prod _{i=1}^{a}\big (1-x^i\big )}{\prod _{i=b}^{b+a}\big (1-x^i\big )}. \end{aligned}$$

Proof

(a) We have

$$\begin{aligned} f(b):&=\sum _{j\ge 0}\frac{x^{jb}}{\prod _{i=1}^j\big (1-x^i\big )}=1+\sum _{j\ge 0}\frac{x^{(j+1)b}}{\prod _{i=1}^{j+1}\big (1-x^i\big )}\\&=1+\sum _{j\ge 0}\frac{x^{(j+1)b}-x^{(j+1)b+j+1}+x^{(j+1)b+j+1}}{\prod _{i=1}^{j+1}\big (1-x^i\big )}\\&=1+\sum _{j\ge 0}\frac{x^{(j+1)b}}{\prod _{i=1}^{j}\big (1-x^i\big )}+\sum _{j\ge 1}\frac{x^{j(b+1)}}{\prod _{i=1}^j\big (1-x^i\big )}=x^bf(b)+f(b+1), \end{aligned}$$

which implies that $f(b)=\frac{1}{1-x^b}f(b+1)$. We use the fact that $f(1)=\frac{1}{\prod _{i\ge 1}(1-x^i)}$ to complete the proof of the first identity.

(b) Next we prove the second identity using induction on $a$. For $a=1$, we have

$$\begin{aligned} \sum _{j\ge 0}x^{jb}\prod _{i=j+1}^{j+1}\big (1-x^i\big )=\frac{1-x}{\big (1-x^b\big )\big (1-x^{b+1}\big )}. \end{aligned}$$

Now, assume that the claim holds for all $1,2,\ldots ,a$, and let us prove it for $a+1$. Firstly, we have

$$\begin{aligned} g(a,b):&=\sum _{j\ge 0}x^{jb}\prod _{i=j+1}^{j+a}(1-x^i)\\&=\sum _{j\ge 0}\Big (x^{jb}-x^{jb+j+a+1}+x^{jb+j+a+1}\Big )\prod _{i=j+1}^{j+a}\big (1-x^i\big )\\&=\sum _{j\ge 0}x^{jb}\prod _{i=j+1}^{j+a+1}(1-x^i)+x^{a+1}\sum _{j\ge 0}x^{j(b+1)}\prod _{i=j+1}^{j+a}(1-x^i)\\&=g(a+1,b)+x^{a+1}g(a,b+1). \end{aligned}$$

Thus, by the induction hypothesis, we have

$$\begin{aligned} g(a+1,b)&=\frac{\prod _{i=1}^{a}\big (1-x^i\big )}{\prod _{i=b}^{b+a}\big (1-x^i\big )}-x^{a+1}\frac{\prod _{i=1}^{a}\big (1-x^i\big )}{\prod _{i=b+1}^{b+a+1}\big (1-x^i\big )}\\&=\frac{\prod _{i=1}^{a+1}\big (1-x^i\big )}{\prod _{i=b}^{b+a+1}\big (1-x^i\big )}\left( \frac{1-x^{a+b+1}}{1-x^{a+1}}-\frac{x^{a+1}(1-x^b)}{1-x^{a+1}}\right) \\&=\frac{\prod _{i=1}^{a+1}\big (1-x^i\big )}{\prod _{i=b}^{b+a+1}\big (1-x^i\big )}, \end{aligned}$$

which completes the induction. $\square$

We now find the derivative of the generating function $t^{-4}P(1,x,t)$ with respect to $t$ and thereafter substitute $t=1$:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}(t^{-4}P(1,x,t)){\Big |}_{t=1}=&\,\sum _{j \ge 1} \frac{x^j}{1-x^j}\sum _{i=1}^{j-1}\frac{x^i}{\prod _{m=1}^{j-1}\big (1-x^m\big )}\\ =&\,\frac{x}{1-x}\sum _{j\ge 0}\frac{x^j}{\prod _{m=1}^{j-1}\big (1-x^m\big )}\\&\,-\frac{1}{1-x}\sum _{j \ge 0} \frac{x^{2j}}{\prod _{m=1}^{j-1}\big (1-x^m\big )}-1\quad \hbox {by Lemma } 2.3\\ =&\,\frac{x}{1-x}\frac{1}{\prod _{i \ge 1}\big (1-x^i\big )}-\frac{1}{1-x}\frac{1}{\prod _{i \ge 2}\big (1-x^i\big )}-1\\ =&-1+\frac{2x-1}{(1-x)\prod _{j\ge 1}(1-x^j)}. \end{aligned}$$

Theorem 2.4

Thus, the generating function for the total number of corners is

$$\begin{aligned} -3+\dfrac{3-2x}{(1-x)\displaystyle \prod _{j \ge 1}\left( 1-x^j\right) }. \end{aligned}$$

This implies that the total number of corners in all partitions of $n$ is given by

$$\begin{aligned} 3P(n)+\sum _{j=1}^nP(n-j). \end{aligned}$$

Remark 2.5

For any given partition $\pi$, the total number of corners is equal to the number of distinct part sizes in $\pi$ plus 3. Distinct part sizes in partitions have previously been studied in [7, 10].

By using Theorem 2.2 and the $q$-binomial theorem, we obtain

Theorem 2.6

The generating function for the number of partitions with exactly $m+4$ corners is given by

$$\begin{aligned} \sum _{j\ge 1}\frac{x^j}{\prod _{i=1}^{j}\left( 1-x^i \right) }\sum _{i=0}^{j-1}(-1)^{i-m}x^{\left( {\begin{array}{c}i+1\\ 2\end{array}}\right) }{\Big [}\begin{matrix}j-1\\ i\end{matrix}{\Big ]}_x\left( {\begin{array}{c}i\\ m\end{array}}\right) . \end{aligned}$$

Another application of our general result in Proposition 2.1 is to study the total number of corners at line $y=2k$ for any $k\ge 0$. In order to do that, we define

$$\begin{aligned} Q(z,x,t)=P(z,x,t,1,t,1,\ldots ) \quad \mathrm and \quad Q'(z,x,t)=P(z,x,1,t,1,t,\ldots ). \end{aligned}$$

Proposition 2.1 shows

$$\begin{aligned} Q(z,x,t)=\frac{t^2xz}{1-xz}+\frac{t^2}{1-xz}Q'(xz,x,t) \end{aligned}$$

and

$$\begin{aligned} Q'(z,x,t)=\frac{1}{t^2}Q(xz,x,t)+\frac{t^2xz}{1-xz}+\frac{xz}{t(1-xz)}Q(xz,x,t). \end{aligned}$$

Therefore,

$$\begin{aligned} Q(z,x,t)=\frac{t^2xz(1+t^2x-x^2z)}{(1-xz)(1-x^2z)}+\frac{1+(t-1)x^2z}{(1-xz)(1-x^2z)}Q(x^2z,x,t). \end{aligned}$$

Iterating infinitely many times, we obtain

$$\begin{aligned} Q(z,x,t)=t^2xz\sum _{j\ge 0}\left( x^{2j}(1+t^2x-x^{2j+2}z)\frac{\prod _{i=1}^{j}(1+(t-1)x^{2i}z)}{\prod _{i=1}^{2j+2}(1-x^iz)}\right) . \end{aligned}$$

Thus, we may state the following result.

Theorem 2.7

The generating function $Q(1,x,t)$, where $t$ marks the number of corners at even levels, is given by

$$\begin{aligned} Q(1,x,t)=&\,t^2x\sum _{j\ge 0}\frac{x^{2j}\prod _{i=1}^{j}\big (1+(t-1)x^{2i}\big )}{\prod _{i=1}^{2j+1}\big (1-x^i\big )}\\&+t^4x^2\sum _{j\ge 0}\frac{x^{2j}\prod _{i=1}^{j}\big (1+(t-1)x^{2i}\big )}{\prod _{i=1}^{2j+2}\big (1-x^i\big )}. \end{aligned}$$

For the total number of such corners, we calculate

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}Q(1,x,t){\Big |}_{t=1}=&\,2\sum _{j\ge 0}\frac{x^{2j+1}}{\prod _{i=1}^{2j+1}\big (1-x^i\big )} +x\sum _{j\ge 0}\frac{x^{2j}\sum _{i=1}^{j}x^{2i}}{\prod _{i=1}^{2j+1}\big (1-x^i\big )}\\&+4\sum _{j\ge 0}\frac{x^{2j+2}}{\prod _{i=1}^{2j+2}\big (1-x^i\big )} +x^2\sum _{j\ge 0}\frac{x^{2j}\sum _{i=1}^{j}x^{2i}}{\prod _{i=1}^{2j+2}\big (1-x^i\big )}, \end{aligned}$$

which implies

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}Q(1,x,t){\Big |}_{t=1}&=\frac{2-x^2}{1-x^2}\sum _{j\ge 0}\frac{x^{2j+1}}{\prod _{i=1}^{2j+1}\big (1-x^i\big )} +\frac{3-2x^2}{1-x^2}\sum _{j\ge 0}\frac{x^{2j}}{\prod _{i=1}^{2j}\big (1-x^i\big )}-3. \end{aligned}$$

(1)

Note that $\sum _{j\ge 0}\frac{x^jy^j}{\prod _{i=1}^j\big (1-x^i\big )}=\frac{1}{\prod _{j\ge 1}\big (1-yx^j\big )}$. By splitting the respective sums above into those with the largest part odd (and generating function $\prod _{j \ge 1}\frac{1}{1-x^j}-\prod _{j \ge 1} \frac{1}{1+x^j}$), or the largest part even (with generating function $\prod _{\mathrm{j} \ge 1}\frac{1}{1-x^j}+\prod _{\mathrm{j} \ge 1} \frac{1}{1+x^j}$), we obtain

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}Q(1,x,t)\mid _{t=1}&=\frac{5-3x^2}{2\big (1-x^2\big )\prod _{j\ge 1}\big (1-x^j\big )}+\frac{1}{2\prod _{j\ge 1}\big (1+x^j\big )}-3. \end{aligned}$$

(2)

Using the facts that $\frac{1}{\prod _{j\ge 0}\big (1-x^j\big )}=\sum _{j\ge 0}P(j)x^j$ and $\frac{1}{\prod _{j\ge 0}\big (1+x^j\big )}=\sum _{j\ge 0}P'(j)x^j$ (see sequence A081362 in [9]), we find

Theorem 2.8

Let $n\ge 1$, then the total number of corners at even level $2k$, where $k\ge 0$, in all partitions of $n$ is given by

$$\begin{aligned} \frac{1}{2}(5P(n)+P'(n))+\sum _{j\ge 1}P(n-2j). \end{aligned}$$

3 Corners of type $(a,b)$

In this section, we study partitions according to the number of corners of type $(a,b)$. Let

$$\begin{aligned} P_{k;a,b}(x):=P_{k;a,b}(x,q) \end{aligned}$$

be the generating function for the number of partitions of $n$ according to the number of corners of type $(a,b)$ with first part of size $k$, where $x$ marks the size of the partition and $q$ the number of such corners. Then

$$\begin{aligned} P_{k;a,b}(x)=\frac{x^k}{1-x^k}\sum _{j=0}^{k-1}P_{j;a,b}(x)+(q-1)x^{ka}P_{k-b;a,b}(x), \end{aligned}$$

where $P_{0;a,b}(x):=1$. This is equivalent to

$$\begin{aligned} P_{k;a,b}(x)=x^k\sum _{j=0}^kP_{j;a,b}(x)+(q-1)x^{ka}(1-x^k)P_{k-b;a,b}(x)\delta _{k\ge b}\quad \hbox {for }k \ge 0. \end{aligned}$$

Define $P_{a,b}(x,y,q):=\sum _{k\ge 0}P_{k;a,b}(x)y^k$. By multiplying the last recurrence by $y^k$ and summing over $k\ge 0$, we get the following result.

Theorem 3.1

The generating function $P_{a,b}(x,y,q)$ satisfies

$$\begin{aligned} P_{a,b}(x,y,q)=&\frac{1}{1-xy}P_{a,b}(x,xy,q)\\&+(q-1)y^bx^{ab}\big (P_{a,b}\big (x,x^ay,q\big )-x^bP_{a,b}\big (x,x^{a+1}y,q\big )\big ). \end{aligned}$$

Now we will find an explicit formula for the generating function $P_{1,b}(x,y,q)$. To do that we need the following lemma.

Lemma 3.2

Let $\mathcal {A}_0=\{1\}$ and $\mathcal {A}_1=\{a_0\}$. For $n\ge 2$, we define $\mathcal {A}_n$ to be the set

$$\begin{aligned} \mathcal {A}_{n-1}\cdot a_{n-1}\cup \mathcal {A}_{n-2}\cdot b_{n-2}, \end{aligned}$$

where $B\cdot x=\{\pi \cdot x\mid \pi \in B\}$. Assume that $\alpha _i=a_i\alpha _{i+1}+b_i\alpha _{i+2}$ for all $i\ge 0$. Then

$$\begin{aligned} \alpha _0=\left( \sum _{\pi \in \mathcal {A}_{n+1}} \pi \right) \alpha _{n+1} +\left( \sum _{\pi \in \mathcal {A}_n}\pi \right) b_{n}\alpha _{n+2}, \end{aligned}$$

for all $n\ge 0$.

Proof

We proceed by induction on $n$. For $n=0$, the statement reduces to $\alpha _0=a_0\alpha _1+b_0\alpha _2$, which holds. Assume that the claim holds for $n$, and let us prove it for $n+1$. By the induction hypothesis, we have

$$\begin{aligned} \alpha _0&=\left( \sum _{\pi \in \mathcal {A}_{n+1}} \pi \right) \alpha _{n+1} +\left( \sum _{\pi \in \mathcal {A}_n}\pi \right) b_{n}\alpha _{n+2}\\&=\left( \sum _{\pi \in \mathcal {A}_{n+1}} \pi \right) \big (a_{n+1}\alpha _{n+2}+b_{n+1}\alpha _{n+3}\big )+\left( \sum _{\pi \in \mathcal {A}_n}\pi \right) b_{n}\alpha _{n+2}\\&=\left( \sum _{\pi \in \mathcal {A}_{n+1}} \pi a_{n+1}+\sum _{\pi \in \mathcal {A}_n}\pi b_n\right) \alpha _{n+2}+\left( \sum _{\pi \in \mathcal {A}_{n+1}} \pi \right) b_{n+1}\alpha _{n+3}, \end{aligned}$$

which, by the definition of the sets $\mathcal {A}_n$, implies

$$\begin{aligned} \alpha _0&=\left( \sum _{\pi \in \mathcal {A}_{n+2}} \pi \right) \alpha _{n+2}+\left( \sum _{\pi \in \mathcal {A}_{n+1}} \pi \right) b_{n+1}\alpha _{n+3}. \end{aligned}$$

This completes the induction step. $\square$

Clearly, the set $\mathcal {A}_n$ in the statement of the above Lemma 3.2 can be described bijectively as all sequences $\pi _0\pi _1\cdots \pi _s$ such that $\pi _0=0$, $\pi _s\in \{n-1,n-2\}$ and $\pi _j-\pi _{j-1}=1,2$ for all $j=1,2,\ldots ,s$. We denote all such sequences by $\mathcal {B}_n$. For each word $\pi =\pi _0\pi _1\cdots \pi _j\in \mathcal {B}_n$, we define

$$\begin{aligned} J(\pi )&=\{i|\pi _{i+1}-\pi _{i}=2 \text{ with } i=0,1,\ldots ,j-1, \text{ or } \pi _i=\pi _j=n-2\},\\ I(\pi )&=\{i|\pi _{i+1}-\pi _i=1 \text{ with } i=0,1,\ldots ,j-1, \text{ or } \pi _i=\pi _j=n-1\}. \end{aligned}$$

Therefore,

$$\begin{aligned} \sum _{\pi \in \mathcal {A}_{n}} \pi =\sum _{\pi \in \mathcal {B}_{n}} \prod _{i\in J(\pi )}b_{\pi _i}\prod _{i\in I(\pi )}a_{\pi _i}. \end{aligned}$$

By Theorem 3.1 with $a=1$, the generating function $P_{a,b}(x,y,q)$ satisfies

$$\begin{aligned} P_{1,b}(x,y,q)=&\left( \frac{1}{1-xy}+(q-1)y^bx^{b}\right) P_{1,b}(x,xy,q)\\&-(q-1)y^bx^{2b}P_{1,b}\big (x,x^2y,q\big ). \end{aligned}$$

Define $\alpha _i=P_{1,b}\big (x,x^iy,q\big )$, $a_i=\frac{1}{1-x^{i+1}y}+(q-1)y^bx^{b(i+1)}$ and $b_i=(1-q)y^bx^{b(i+2)}$; so

$$\begin{aligned} \alpha _i=a_i\alpha _{i+1}+b_i\alpha _{i+2}, \end{aligned}$$

where $\lim _{j\rightarrow \infty }\alpha _j=1$. Hence, by Lemma 3.2 (note that $b_i\rightarrow 0$ when $i\rightarrow \infty$), we have

$$\begin{aligned} \alpha _0=\sum _{\pi \in \mathcal {B}_\infty } \prod _{i\in J(\pi )}b_{\pi _i}\prod _{i\in I(\pi )}a_{\pi _i}, \end{aligned}$$

where $\mathcal {B}_\infty =\lim _{n\rightarrow \infty }\mathcal {B}_n$ which is the set of all sequences $\pi _0\pi _1\pi _2\cdots$ such that $\pi _0=0$ and $\pi _{i+1}-\pi _i$ is 1 or 2. So we can state the following result.

Theorem 3.3

The generating function $P_{1,b}(x,y,q)$ is given by

$$\begin{aligned}&\frac{P_{1,b}(x,y,q)}{\prod _{i\ge 1}\left( \frac{1}{1-x^{i}y}+(q-1)y^bx^{bi}\right) }\\&\quad =\sum _{\pi \in \mathcal {B}_\infty } \frac{(1-q)^{|J(\pi )|}y^{b|J(\pi )|}x^{2b|J(\pi )|+b\sum _{i\in J(\pi )}\pi _i}}{\prod _{i\in J(\pi ),m=1,2}\left( \frac{1}{1-x^{\pi _i+m}y}+(q-1)y^bx^{b(\pi _i+m)}\right) }. \end{aligned}$$

Note that $J(\pi )=\emptyset$ if and only if $\pi =0123\cdots$. Thus the above theorem with $q=1$ becomes

$$\begin{aligned} P_{1,b}(x,y,1)=\prod _{i\ge 1}\frac{1}{1-x^{i}y}, \end{aligned}$$

as is well known.

Also, note that $|J(\pi )|=1$ if and only if there exists $i\ge 0$ such that

$$\begin{aligned} \pi =012\cdots i(i+2)(i+3)(i+4)\cdots , \end{aligned}$$

and $|J(\pi )|=2$ if and only if there exists $i'>i+1\ge 1$ such that

$$\begin{aligned} \pi =012\cdots i(i+2)(i+3)\cdots i'(i'+2)(i'+3)(i'+4)\cdots . \end{aligned}$$

(Similarly, we can characterize all the infinite sequences with $J(\pi )=m$). Then the above theorem shows that

$$\begin{aligned}&\frac{P_{1,b}(x,y,q)}{\prod _{i\ge 1}\left( \frac{1}{1-x^{i}y}+(q-1)y^bx^{bi}\right) }\\&\quad =1+(1-q)y^bx^b\sum _{s\ge 1}\frac{x^{bs}}{\left( \frac{1}{1-x^{s}y}+(q-1)y^bx^{bs}\right) \left( \frac{1}{1-x^{s+1}y}+(q-1)y^bx^{bs+b}\right) }\\&\qquad +\sum _{\pi \in \mathcal {B}_\infty ,\,|J(\pi )|\ge 2} \frac{(1-q)^{|J(\pi )|}y^{b|J(\pi )|}x^{2b|J(\pi )|+b\sum _{i\in J(\pi )}\pi _i}}{\prod _{i\in J(\pi ),m=1,2}\left( \frac{1}{1-x^{\pi _i+m}y}+(q-1)y^bx^{b(\pi _i+m)}\right) }. \end{aligned}$$

Another way to solve the recurrence $\alpha _i=a_i\alpha _{i+1}+b_i\alpha _{i+2}$ in Lemma 3.2 is by using continued fractions. Note that $\frac{\alpha _i}{\alpha _{i+1}}=a_i+\frac{b_i}{\frac{\alpha _{i+1}}{\alpha _{i+2}}}$, for all $i\ge 0$. Therefore,

$$\begin{aligned} \frac{\alpha _i}{\alpha _{i+1}}=a_i+\dfrac{b_i}{a_{i+1}+\dfrac{b_{i+1}}{a_{i+2}+\ddots }}. \end{aligned}$$

So, multiplying over all $i\ge 0$, assuming that $\lim _{j\rightarrow \infty }\alpha _j=1$, we obtain

$$\begin{aligned} \alpha _0=\prod _{i\ge 0}\left( a_i+\dfrac{b_i}{a_{i+1}+\dfrac{b_{i+1}}{a_{i+2}+\ddots }}\right) . \end{aligned}$$

Thus, if we define $\alpha _i=P_{1,b}(x,x^iy,q)$, $a_i=\frac{1}{1-x^{i+1}y}+(q-1)y^bx^{b(i+1)}$ and $b_i=(1-q)y^bx^{(2+i)b}$, we have the following result.

Theorem 3.4

The generating function $P_{1,b}(x,y,q)$ is given by

$$\begin{aligned} \prod _{i\ge 0}\left( \!\frac{1}{1-x^{i+1}y}+(q-1)y^bx^{b(i+1)} \!+\!\frac{(1\!-\!q)y^bx^{(2+i)b}}{\frac{1}{1\!-\!x^{i+2}y}+(q-1)y^bx^{b(i+2)} \!+\!\frac{(1-q)y^bx^{(3+i)b}}{\frac{1}{1-x^{i+3}y}+(q-1)y^bx^{b(i+3)}+\ddots }}\!\right) \!. \end{aligned}$$

Note that by comparing Theorems 3.3 and 3.4, we get the following corollary.

Corollary 3.5

We have

$$\begin{aligned} P_{1,b}(x,y,q)=&\prod _{i\ge 1}\left( \frac{1}{1-x^{i}y}+(q-1)y^bx^{bi}\right) \\&\times \sum _{\pi \in \mathcal {B}_\infty } \frac{(1-q)^{|J(\pi )|}y^{b|J(\pi )|}x^{2b|J(\pi )|+b\sum _{i\in J(\pi )}\pi _i}}{\prod _{i\in J(\pi ),m=1,2}\left( \frac{1}{1-x^{\pi _i+m}y}+(q-1)y^bx^{b(\pi _i+m)}\right) }\\ =&\prod _{i\ge 0}\left( \frac{1}{1-x^{i+1}y}+(q-1)y^bx^{b(i+1)}\right. \\&\qquad \left. +\frac{(1-q)y^bx^{(2+i)b}}{\frac{1}{1-x^{i+2}y}+(q-1)y^bx^{b(i+2)} +\frac{(1-q)y^bx^{(3+i)b}}{\frac{1}{1-x^{i+3}y}+(q-1)y^bx^{b(i+3)}+\ddots }}\right) . \end{aligned}$$

3.1 Total number of corners of type $(a,b)$

Using Theorem 3.1, we find the generating function for the total number of corners of type $(a,b)$ to be

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{a,b}(x,y,q){\Big |}_{q=1}=&\frac{1}{1-xy}\frac{\mathrm{d}}{\mathrm{d}q}P_{a,b}(x,xy,q){\Big |}_{q=1}\\&+y^bx^{ab}\big (P_{a,b}\big (x,x^ay,1\big )-x^bP_{a,b}\big (x,x^{a+1}y,1\big )\big ). \end{aligned}$$

Using $P_{a,b}(x,y,1)=\frac{1}{\prod _{j\ge 1}(1-x^jy)}$, this is equivalent to

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{a,b}(x,y,q){\Big |}_{q=1}=\frac{1}{1-xy}\frac{\mathrm{d}}{\mathrm{d}q}P_{a,b}(x,xy,q){\Big |}_{q=1}+y^bx^{ab}\frac{1-x^b+x^{a+b+1}y}{\prod _{j\ge a+1}\big (1-x^jy\big )}. \end{aligned}$$

By iterating this an infinite number of times, and noting that $\frac{\mathrm{d}}{\mathrm{d}q}P_{a,b}(x,x^s,q){\Big |}_{q=1} \rightarrow 0$ as $s \rightarrow \infty$, we have

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{a,b}(x,1,q){\Big |}_{q=1}&=\sum _{j\ge a}\frac{x^{jb}}{\prod _{i=1}^{j-a}\big (1-x^i\big )}\frac{1-x^b+x^{j+b+1}}{\prod _{i\ge j+1}\big (1-x^i\big )} \nonumber \\&=\frac{x^{ab}}{\prod _{i\ge 1}\big (1-x^i\big )}\sum _{j\ge 0}x^{jb}\big (1-x^b+x^{j+a+b+1}\big )\prod _{i=j+1}^{j+a}\big (1-x^i\big ). \end{aligned}$$

(3)

Thus, by Lemma 2.3 and Eq. (3),

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{a,b}(x,y,q){\Big |}_{q=1}\!=\! \frac{x^{ab}\big (1\!-\!x^b\big )}{\prod _{i\ge 1}\big (1\!-\!x^i\big )}\left( \frac{\prod _{i=1}^{a} \big (1\!-\!x^i\big )}{\prod _{i=b}^{b+a}\big (1\!-\!x^i\big )} \!+\!x^{a+b+1}\frac{\prod _{i=1}^{a}\big (1\!-\!x^i\big )}{\prod _{i=b}^{b+a+1}\big (1-x^i\big )}\!\right) \!. \end{aligned}$$

This is captured in the following theorem.

Theorem 3.6

The generating function for the total number of corners of type $(a,b)$ in all partitions is given by

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{a,b}(x,1,q){\Big |}_{q=1}=\frac{x^{ab}\prod _{i=1}^{a}\big (1-x^i\big )}{\prod _{i\ge 1}\big (1-x^i\big )\prod _{i=b+1}^{b+a+1}\big (1-x^i\big )}. \end{aligned}$$

Example 3.7

Theorem 3.6 for $b=1$ leads to

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{a,1}(x,1,q){\Big |}_{q=1}&=\frac{x^a(1-x)}{\big (1-x^{a+1}\big )\big (1-x^{a+2}\big )\prod _{i\ge 1}\big (1-x^i\big )}\\&=\frac{1}{x}\left( \frac{1}{1-x^{a+1}}-\frac{1}{1-x^{a+2}}\right) \frac{1}{\prod _{i\ge 1}\big (1-x^i\big )}, \end{aligned}$$

which implies that the total number of corners of type $(a,1)$ in all partitions of $n$ is given by

$$\begin{aligned} \sum _{j\ge 0}(P(n+1-j(a+1))-P(n+1-j(a+2))). \end{aligned}$$

Let $\mathcal {P}_n$ denote the set of all partitions of size $n$. There is a simple correspondence between the total number of corners of type $(a,b)$ in $\mathcal {P}_n$ and the number of parts $\ge b$ of multiplicity $a+1$ in $\mathcal {P}_{n+b}$. We illustrate the bijection in Fig. 2.

Fig. 2

Bijection between number of $(a,b)$ corners in $\mathcal {P}_{n}$ and parts of multiplicity $a+1$ in $\mathcal {P}_{n+b}$

Note, however, that the distributions of corners of type $(a,b)$ and of parts of multiplicity $a+1$ are quite different. For example, for the case where $a=1$, the generating function for partitions with no parts of multiplicity $2$ is $\prod _{i=1}^\infty (\frac{1}{1-x^i}-x^{2i})$, whereas by setting $q=0$ in Theorem 3.4, the generating function for partitions with no corner of type $(1,b)$ is

$$\begin{aligned}&P_{1,b}(x,y,0)\\&\quad =\prod _{i\ge 0}\left( \frac{1}{1-x^{i+1}y}-y^bx^{b(i+1)}+\frac{y^bx^{(2+i)b}}{\frac{1}{1-x^{i+2}y}-y^bx^{b(i+2)} +\frac{y^bx^{(3+i)b}}{\frac{1}{1-x^{i+3}y}-y^bx^{b(i+3)}+\ddots }}\right) . \end{aligned}$$

4 Corners of type $(a+,b)$

We define a corner of type $(a+,b)$ to be a corner of type $(j,b)$ for any $j \ge a$. Let

$$\begin{aligned} P_{k;a+,b}(x):=P_{k;a+,b}(x,q) \end{aligned}$$

be the generating function for the number of partitions according to the number of corners of type $(a+,b)$ with first part $k$. Then for $k\ge b$,

$$\begin{aligned} P_{k;a+,b}(x)&= \big (x^k+x^{2k}+\cdots +x^{k(a-1)}\big )\sum _{j=0}^{k-1}P_{j;a+,b}(x)\\&+\frac{x^{ka}}{1-x^k}\left( (q-1)P_{k-b;a+,b}(x)+\sum _{j=0}^{k-1}P_{j;a+,b}(x)\right) \end{aligned}$$

with $P_{k;a+,b}(x)=\frac{x^k}{(1-x)(1-x^2)\cdots (1-x^k)}$ for $k=0,1,\ldots ,b-1$. Thus,

$$\begin{aligned} P_{k;a+,b}(x)=x^k\sum _{j=0}^{k}P_{j;a+,b}(x)+x^{ka}(q-1)P_{k-b;a+,b}(x)\delta _{k\ge b}. \end{aligned}$$

Define $P_{a+,b}(x,y,q):=\sum _{k\ge 0}P_{k;a+,b}(x,q)y^k$. By multiplying the last recurrence by $y^k$ and summing over $k\ge 0$, we get

$$\begin{aligned} P_{a+,b}(x,y,q)=\frac{1}{1-xy}P_{a+,b}(x,xy,q)+(q-1)x^{ab}y^bP_{a+,b}(x,x^ay,q). \end{aligned}$$

(4)

We solve the recursion in the case where $a=1$.

Theorem 4.1

The generating function $P_{1+,b}(x,y,q)$ is given by

$$\begin{aligned} P_{1+,b}(x,y,q)=\prod _{i\ge 1}\left( \frac{1}{1-x^iy}+(q-1)(x^iy)^b\right) . \end{aligned}$$

Moreover, the generating function for the number of partitions without corners of types $(1+,b)$ is given by

$$\begin{aligned} P_{1+,b}(x,1,0)=\prod _{i\ge 1}\left( \frac{1}{1-x^i}-x^{ib}\right) . \end{aligned}$$

Proof

By (4) with $a=1$, we have

$$\begin{aligned} P_{1+,b}(x,y,q)=\left( \frac{1}{1-xy}+(q-1)(xy)^b\right) P_{1+,b}(x,xy,q), \end{aligned}$$

which, by repeated iteration and using $\lim _{i\rightarrow \infty }P_{1+,b}(x,x^iy,q)=1$, implies

$$\begin{aligned} P_{1+,b}(x,y,q)=\prod _{i\ge 1}\left( \frac{1}{1-x^iy}+(q-1)(x^iy)^b\right) , \end{aligned}$$

as required. $\square$

4.1 Total number of corners of type $(a+,b)$

Differentiating Eq. (4) with respect to $q$, then substituting $q=1$ and using $P_{a+,b}(x,y,1)=\frac{1}{\prod _{j\ge 1}(1-x^jy)}$, we obtain

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{a+,b}(x,y,q){\Big |}_{q=1}=\frac{1}{1-xy}\frac{\mathrm{d}}{\mathrm{d}q}P_{a+,b}(x,xy,q){\Big |}_{q=1}+x^{ab}y^bP_{a+,b}(x,x^ay,1). \end{aligned}$$

This implies

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{a+,b}(x,y,q){\Big |}_{q=1}=\frac{x^{ab}y^b}{\prod _{j\ge a+1}(1-x^jy)}+\frac{1}{1-xy}\frac{\mathrm{d}}{\mathrm{d}q}P_{a+,b}(x,xy,q){\Big |}_{q=1}. \end{aligned}$$

Iterating infinitely many times, we obtain

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{a+,b}(x,y,q){\Big |}_{q=1}=\sum _{j\ge 0}\frac{x^{(a+j)b}y^b}{\prod _{i=1}^j\big (1-x^iy\big )\prod _{i\ge a+1+j}\big (1-x^iy\big )}. \end{aligned}$$

By Lemma 2.3, this leads to the following result.

Theorem 4.2

The generating function for the total number of corners of type $(a+,b)$ in all partitions of $n$ is given by

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{a+,b}(x,1,q){\Big |}_{q=1}=\frac{x^{ab}\prod _{i=1}^a\big (1-x^i\big )}{\prod _{i\ge 1}\big (1-x^i\big )\prod _{i=b}^{a+b}\big (1-x^i\big )}. \end{aligned}$$

Example 4.3

Theorem 4.2 for $a=1$ shows that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{1+,b}(x,1,q){\Big |}_{q=1}=\frac{1}{\prod _{i\ge 1}\big (1-x^i\big )}\left( \frac{1}{1-x^b}-\frac{1}{1-x^{b+1}}\right) . \end{aligned}$$

Thus, the total number of corners of type $(1+,b)$ in all the partitions of $n$ is given by

$$\begin{aligned} \sum _{j\ge 1}(P(n-bj)-P(n-(b+1)j)). \end{aligned}$$

We note that a simple bijection between a Ferrers diagram and its transpose implies that the above results also count corners of type $(a,b+)$.

We also note that the number of corners of type $(a+,b)$ in the partitions of $n$ equals the number $\ge b$ of parts of size $a+1$ in the partitions of $n+b$. We illustrate the bijection in Fig. 3.

Fig. 3

Bijection between number of $(a+,b)$ corners in $\mathcal {P}_n$ and parts in $\mathcal {P}_{n+b}$

First, we add $b$ cells to the first part of size $j-b$ and then conjugate the $a+1$ parts of size $j$; thereafter we rearrange all parts (in the right-hand side drawing in Fig. 3) in descending order.

Example 4.4

For $a=2$, the corners of type $(2+,1)$ in the partition $2211$ maps to the two different partitions: $322$ and $331$. Also, the corner of type $(2+,1)$ in the partition $111111$ (resp. $21111$, $3111$, $411$) maps to the partition $31111$ (resp. $3211$, $331$, $43$).

5 Corners of type $(a+,b+)$

Let $P_{k;a+,b+}(x):=P_{k;a+,b+}(x,q)$ be the generating function for the number of partitions of $n$ according to the number of corners of type $(a+,b+)$ with first part $k$. Then for $k\ge b$,

$$\begin{aligned} P_{k;a+,b+}(x)&= \Big (x^k+x^{2k}+\cdots +x^{k(a-1)}\Big )\sum _{j=0}^{k-1}P_{j;a+,b+}(x)\\&+\frac{x^{ka}}{1-x^k}\left( q\sum _{j=0}^{k-b}P_{j;a+,b+}(x)+\sum _{j=k-b+1}^{k-1}P_{j;a+,b+}(x)\right) \end{aligned}$$

with $P_{k;a+,b+}(x)=\frac{x^k}{(1-x)(1-x^2)\cdots (1-x^k)}$ for $k=0,1,\ldots ,b-1$. Thus,

$$\begin{aligned} P_{k;a+,b+}(x)=x^k\sum _{j=0}^{k}P_{j;a+,b+}(x)+(q-1)x^{ka}\sum _{j=0}^{k-b}P_{j;a+,b+}(x). \end{aligned}$$

Define $P_{a+,b+}(x,y,q):=\sum _{k\ge 0}P_{k;a+,b+}(x,q)y^k$. Multiplying the last recurrence by $y^k$ and summing over $k\ge 0$, we get

$$\begin{aligned} P_{a+,b+}(x,y,q)=\frac{1}{1-xy}P_{a+,b+}(x,xy,q)+\frac{(q-1)x^{ab}y^b}{1-x^ay}P_{a+,b+}(x,x^ay,q). \end{aligned}$$

(5)

For $a=1$, we obtain the following result.

Theorem 5.1

The generating function $P_{1+,b+}(x,y,q)$ is

$$\begin{aligned} P_{1+,b+}(x,y,q)=\prod _{i\ge 1}\frac{1+(q-1)\big (x^iy\big )^b}{1-x^iy}. \end{aligned}$$

Note that $P_{a+,b+}(x,y,1)=\frac{1}{\prod _{j\ge 1}(1-x^jy)}$. By differentiating Eq. (5) with respect to $q$ and then substituting $q=1$, we obtain

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{a+,b+}(x,y,q){\Big |}_{q=1}=&\frac{1}{1-xy}\frac{\mathrm{d}}{\mathrm{d}q}P_{a+,b+}(x,xy,q){\Big |}_{q=1}\\&+\frac{x^{ab}y^b}{1-x^ay}P_{a+,b+}(x,x^ay,1), \end{aligned}$$

which implies

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{a+,b+}(x,y,q){\Big |}_{q=1}=\frac{1}{1-xy}\frac{\mathrm{d}}{\mathrm{d}q}P_{a+,b+}(x,xy,q){\Big |}_{q=1}+\frac{x^{ab}y^b}{\prod _{j\ge a}\big (1-x^jy\big )}. \end{aligned}$$

Iterating infinitely many times, we obtain

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{a+,b+}(x,y,q){\Big |}_{q=1}=\sum _{j\ge 0}\frac{x^{(a+j)b}y^b}{\prod _{i\ge a+j}\big (1-x^iy\big )\prod _{i=1}^{j}\big (1-x^iy\big )}. \end{aligned}$$

By Lemma 2.3b, this implies the following result.

Theorem 5.2

The generating function for the total number of corners of type $(a+,b+)$ in all partitions is given by

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{a+,b+}(x,1,q){\Big |}_{q=1}=\frac{x^{ab}\prod _{i=1}^{a-1}\big (1-x^i\big )}{\prod _{i\ge 1}(1-x^i)\prod _{i=b}^{a+b-1}\big (1-x^i\big )}. \end{aligned}$$

Example 5.3

Theorem 5.2 for $a=1$ shows that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}q}P_{1+,b+}(x,1,q){\Big |}_{q=1}=\frac{x^b}{(1-x^b)\prod _{i\ge 1}\big (1-x^i\big )}. \end{aligned}$$

Thus, the total number of corners of type $(1+,b+)$ in all partitions of $n$ is given by

$$\begin{aligned} \sum _{j\ge 1}P(n-bj). \end{aligned}$$

6 Corners of size $m$

We shall define the size of a corner of type $(a,b)$ to be $m$ when $m=a+b$. In this section, we find the generating function that counts the total number of corners of size $m$ in partitions of $n$. For a corner of size $m$, where $m$ is fixed, there are $p$ parts of size $k$ followed by a part of size $k+p-m$, for $1 \le p \le m-1$. We denote the generating function by $P_{k,m}(x):=P_{k,m}(x,q)$, where $q$ marks such corners, as illustrated in Fig. 4.

Fig. 4

Corner of size $m$

Clearly, the generating function satisfies the following equation:

$$\begin{aligned} P_{k,m}=\frac{x^k}{1-x^k}\sum _{j=0}^{k-1}P_{j,m}+(q-1)\sum _{p=1}^{m-1}x^{kp}P_{k+p-m,m}. \end{aligned}$$

Multiplying by $(1-x^k)$ and simplifying,

$$\begin{aligned} P_{k,m}=x^k\sum _{j=0}^{k}P_{j,m}+(q-1)(1-x^k)\sum _{p=1}^{m-1}x^{kp}P_{k+p-m,m}. \end{aligned}$$

(6)

We now define $P_{m}(x,y,q):=\sum _{k \ge 0}P_{k,m}(x,y,q)y^k$. Thus after multiplying (6) by $y^k$, summing over all $k$ and interchanging the order of summation, we have

$$\begin{aligned} P_{m}(x,y,q) =&\sum _{j=0}^\infty P_{j,m}\frac{(xy)^j}{1-xy}+(q-1) \sum _{p=1}^{m-1}\big (x^py\big )^{m-p}\sum _{r \ge 0}\big (x^py\big )^{r-p+m}P_{r,m} \nonumber \\&-(q-1)\sum _{p=1}^{m-1}\sum _{r\ge 0} \big (x^{1+p}y\big )^{r-p+m} \, P_{r,m}\qquad (\mathrm where r=k+p-m) \nonumber \\ =&\frac{P_m(x,xy,q)}{1-xy}+(q-1)\sum _{p=1}^{m-1}P_m\big (x,x^py,q\big )\big (x^py\big )^{m-p}\nonumber \\&-(q-1)\sum _{p=1}^{m-1}P_m\big (x,x^{p+1}y,q\big )\big (x^{p+1}y\big )^{m-p}. \end{aligned}$$

(7)

Following the previous procedure for finding the total number of corners, we compute

$$\begin{aligned} \frac{{\mathrm{d}}P_m(x,y,q)}{{\mathrm{d}}q}{\Big |}_{q=1}=\,&\frac{1}{1-xy}\frac{{\mathrm{d}}P_m(x,xy,q)}{{\mathrm{d}}q}{\Big |}_{q=1}+\sum _{p=1}^{m-1}P_m\big (x,x^py,1\big )\big (x^py\big )^{m-p}\nonumber \\&-\sum _{p=1}^{m-1}P_m\big (x,x^{p+1}y,1\big )\big (x^{p+1}y\big )^{m-p}. \end{aligned}$$

(8)

The generating function for all partitions is $P_m(x,y,1)=\frac{1}{\prod _{j\ge 1}(1-x^jy)}$. Therefore

$$\begin{aligned} \frac{{\mathrm{d}}P_m(x,y,q)}{{\mathrm{d}}q}{\Big |}_{q=1} =\,&\frac{1}{1-xy}\frac{{\mathrm{d}}P_m(x,xy,q)}{{\mathrm{d}}q}{\Big |}_{q=1}\nonumber \\&+\sum _{p=1}^{m-1}\frac{1-x^{m-p}\big (1-x^{1+p}yv)}{\prod _{j \ge 1}\big (1-x^{j+p}y\big )}\,\big (x^py\big )^{m-p}. \end{aligned}$$

(9)

After infinitely many iterations and noticing that $\frac{{\mathrm{d}} P_m(x,x^s,q)}{{\mathrm{d}}q} \rightarrow 0$ as $s \rightarrow \infty$, we obtain our result

Theorem 6.1

The generating function for the total number of corners of size $m$ is

$$\begin{aligned} \frac{{\mathrm{d}}P_m(x,1,q)}{{\mathrm{d}}q}{\Big |}_{q=1}&=\sum _{i=0}^\infty \frac{1}{\prod _{l=1}^i \big (1-x^l\big )}\sum _{p=1}^{m-1} \frac{\big (x^{p+i}\big )^{m-p}\big (1-x^{m-p}+x^{m+1+i}\big )}{\prod _{j \ge 1} \big (1-x^{p+i+j}\big )}\end{aligned}$$

(10)

$$\begin{aligned}&=\frac{1}{\prod _{l=1}^\infty \big (1-x^l\big )}\sum _{p=1}^{m-1}\frac{x^{p(m-p)}\prod _{i=1}^p \big (1-x^i\big )}{\prod _{i=m-p+1}^{m+1} \big (1-x^i\big )}. \end{aligned}$$

(11)

Here, Eq. (11) follows from Eq. (10) by using Lemma 2.3.

7 Asymptotics

To find asymptotic estimates for the various types of corners studied above, we need to study generating functions of the form $P(x)F(x)$, where $P(x)$ is the generating function for the number of partitions. In the paper [6], the authors show how such asymptotic expansions can be obtained in a quasi-automatic way from expansions of $F(x)$ around $x = 1$. For the convenience of the reader, we state the relevant results from [6] that we need:

Theorem 7.1

Suppose that the function $F(z)$ satisfies

$$\begin{aligned} |F(z)| =\mathcal {O}( e^{C/(1-|z|)^{\eta }}) \text { as } |z| \rightarrow 1 \text { for some } C > 0 \text { and } \eta < 1, \end{aligned}$$

(12)

and $F(e^{-t}) = a t^b + \mathcal {O}(f(|t|))$ as $t \rightarrow 0$, $\mathfrak {R}t > 0$, for real numbers $a,b$. Then one has

$$\begin{aligned} \frac{1}{p(n)} [x^n] P(x)F(x) =&\, a \left( \frac{2\pi }{\sqrt{24n-1}} \right) ^b \times \frac{I_{|b+3/2|}\left( \sqrt{\frac{2\pi ^2}{3} \left( n - \frac{1}{24} \right) } \right) }{I_{3/2}\left( \sqrt{\frac{2\pi ^2}{3} \left( n - \frac{1}{24} \right) } \right) } \\&+ \mathcal {O}\left( \exp \left( -n^{1/2-\epsilon }\right) + f\left( \frac{\pi }{\sqrt{6n}} + \mathcal {O}\left( n^{-1/2-\epsilon }\right) \right) \right) \end{aligned}$$

as $n \rightarrow \infty$ for any $0 < \epsilon < \frac{1 - \eta }{2}$, where $I_{\nu }$ denotes a modified Bessel function of the first kind.

It is also shown in [6] that the quotient of modified Bessel functions simplifies, with $h = |b+3/2|-1/2$ and $m = \sqrt{\frac{2\pi ^2}{3} \left( n - \frac{1}{24} \right) }$, to

$$\begin{aligned} \frac{I_{|b+3/2|}\left( \sqrt{\frac{2\pi ^2}{3} \left( n - \frac{1}{24} \right) } \right) }{I_{3/2}\left( \sqrt{\frac{2\pi ^2}{3} \left( n - \frac{1}{24} \right) } \right) } = \frac{m}{m-1} \times \sum _{j=0}^h \frac{(h+j)!}{j!(h-j)!} \left( - \frac{1}{2m} \right) ^j + \mathcal {O}(e^{-2m}).\nonumber \\ \end{aligned}$$

(13)

In our applications, we apply (13) together with Theorem 7.1. Thereafter, we use computer algebra to obtain asymptotic formulae in terms of powers of $n$.

Firstly, in the case of the total number of corners of type $(a,b)$, we have $\frac{\mathrm{d}}{\mathrm{d}q}P_{a,b}(x,y,q){\Big |}_{q=1}=P(x)F_{a,b}(x)$, where

$$\begin{aligned} F_{a,b}(x)=\frac{x^{ab}\prod _{i=1}^{a}(1-x^i)}{\prod _{i=b+1}^{b+a+1}(1-x^i)}. \end{aligned}$$

To apply Theorem 7.1, we need to consider the Taylor expansion of $F_{a,b}(e^{-t})$:

$$\begin{aligned} F_{a,b}(e^{-t})&=\frac{e^{-t a b}\prod _{i=1}^{a}(1-e^{-it})}{\prod _{i=1}^{a+1}(1-e^{(i+b)t})} \nonumber \\&=\frac{1-abt}{t}\frac{\prod _{i=1}^{a}(i-i^2t/2)}{\prod _{i=1}^{a+1}((i+b)-(i+b)^2 t/2)}+\mathcal {O}(t). \end{aligned}$$

(14)

Now

$$\begin{aligned} \prod _{i=1}^{a}(i-i^2t/2)=a!(1-ta(a+1)/4)+\mathcal {O}(t^2) \end{aligned}$$

(15)

and

$$\begin{aligned} \prod _{i=1}^{a+1}((i\!+\!b)\!-\!(i+b)^2 t/2)\!=\!\frac{(a+b+1)!}{b!}(1-t(a+1)(2b+a+2)/4)+\mathcal {O}(t^2).\nonumber \\ \end{aligned}$$

(16)

Substituting these into (14) gives

$$\begin{aligned} F_{a,b}(e^{-t})=\frac{a!b!}{(a+b+1)!}\left( \frac{1}{t}-\frac{ab-a-b-1}{2} \right) +\mathcal {O}(t). \end{aligned}$$

(17)

Using this in Theorem 7.1 leads to

Theorem 7.2

The average number of corners of type $(a,b)$ in a random partition of $n$ is

$$\begin{aligned} \frac{ a! b!\sqrt{6n}}{\pi (a+b+1)!}-\frac{a! b! \left( \pi ^2 a b-\pi ^2 a-\pi ^2 b-\pi ^2-6\right) }{2 \pi ^2 (a+b+1)!} + \mathcal {O}(n^{-1/2}). \end{aligned}$$

In the case of the total number of corners of type $(a+,b)$, we have $\frac{\mathrm{d}}{\mathrm{d}q}P_{a,b}(x,y,q){\Big |}_{q=1}=P(x)F_{a,b}(x)$, where

$$\begin{aligned} F_{a,b}(x)=\frac{x^{ab}\prod _{i=1}^{a}(1-x^i)}{\prod _{i=b}^{b+a}(1-x^i)}. \end{aligned}$$

From this we find

$$\begin{aligned} F_{a,b}(e^{-t})=\frac{a!(b-1)!}{(a+b)!}\left( \frac{1}{t}-\frac{b(a-2)}{2} \right) +\mathcal {O}(t). \end{aligned}$$

(18)

Using this in Theorem 7.1 leads to

Theorem 7.3

The average number of corners of type $(a+,b)$ in a random partition of $n$ is

$$\begin{aligned} \frac{ a! (b-1)!\sqrt{6n}}{\pi (a+b)!}-\frac{a! (b-1)! \left( \pi ^2 a b-2 \pi ^2 b-6\right) }{2 \pi ^2 (a+b)!} + \mathcal {O}(n^{-1/2}). \end{aligned}$$

Next for the total number of corners of type $(a+,b+)$, we have

$$\begin{aligned} F_{a,b}(x)=\frac{x^{ab}\prod _{i=1}^{a-1}(1-x^i)}{\prod _{i=b}^{b+a-1}(1-x^i)}. \end{aligned}$$

This yields

$$\begin{aligned} F_{a,b}(e^{-t})=\frac{(a-1)!(b-1)!}{(a+b-1)!}\left( \frac{1}{t}-\frac{ab}{2} \right) +\mathcal {O}(t). \end{aligned}$$

(19)

Using this in Theorem 7.1 leads to

Theorem 7.4

The average number of corners of type $(a+,b+)$ in a random partition of $n$ is

$$\begin{aligned} \frac{ (a-1)! (b-1)! \sqrt{6n}}{\pi (a+b-1)!}-\frac{(a-1)! (b-1)! \left( \pi ^2 a b-6\right) }{2 \pi ^2 (a+b-1)!} + \mathcal {O}(n^{-1/2}). \end{aligned}$$

For the total number of corners treated in Theorem 2.4, we use $F(x)=\frac{2x-1}{1-x}+4$ in Theorem 7.1 to obtain

Theorem 7.5

The average number of corners in a random partition of $n$ is

$$\begin{aligned} \frac{\sqrt{6n}}{\pi } + \frac{6+5\pi ^2}{2\pi ^2} + \mathcal {O}(n^{-1/2}). \end{aligned}$$

For the total number of corners at even level, the contribution of $\frac{1}{2\prod _{j\ge 1}(1+x^j)}$ to (2) is asymptotically negligible; therefore, we may take $F(x)=\frac{5-3x^2}{2(1-x^2)}$ in Theorem 7.1 to obtain

Theorem 7.6

The average number of corners at even level in a random partition of $n$ is

$$\begin{aligned} \frac{ \sqrt{6n}}{2\pi }+\frac{3}{2 \pi ^2}+2 + \mathcal {O}(n^{-1/2}). \end{aligned}$$

Finally, for the total number of corners of size $m$ we take

$$\begin{aligned} F(x)&=\sum _{a=1}^{m-1}\frac{x^{a(m-a)}\prod _{i=1}^a (1-x^i)}{\prod _{i=m-a+1}^{m+1} (1-x^i)}, \end{aligned}$$

in Theorem 7.1, with the aid of the expansions (17) applied to cases $(a,m-a)$ with $1\le a\le m-1$, to obtain

Theorem 7.7

The average number of corners of size $m$ in a random partition of $n$ is

$$\begin{aligned} \frac{1}{m+1}\sum _{a=1}^{m-1} {\left( {\begin{array}{c}m\\ a\end{array}}\right) }^{-1}\left( \frac{\sqrt{6 n} }{\pi }+\frac{\pi ^2 \left( a^2-a m+m+1\right) +6}{2 \pi ^2}\right) + \mathcal {O}(n^{-1/2}). \end{aligned}$$