1 Introduction
A partition
$\lambda $
of a positive integer n is a finite weakly decreasing sequence of positive integers
$\lambda =(\lambda _1,\lambda _2,\ldots ,\lambda _r)$
such that
$\sum _{i=1}^r \lambda _i=n$
. The terms
$\lambda _{i}$
are called the parts of
$\lambda $
and the number of parts of
$\lambda $
is called the length of
$\lambda $
, denoted
$\ell (\lambda )$
. The weight of
$\lambda $
is the sum of its parts, denoted by
$|\lambda |$
. We use
$m(\lambda )$
to denote the largest part of
$\lambda $
. The Ferrers graph of
$\lambda $
is an array of left-justified cells with
$\lambda _{i}$
cells in the ith row. The Durfee square of a partition is the largest possible square contained within the Ferrers graph and anchored in the upper left-hand corner of the Ferrers graph. The conjugate of
$\lambda $
, denoted
$\lambda ^T$
, is the partition whose Ferrers graph is obtained from that of
$\lambda $
by interchanging rows and columns.
A partition
$\lambda $
is said to be self-conjugate if
$\lambda =\lambda ^T$
. It is well known that the number of self-conjugate partitions of n equals the number of partitions of n into distinct odd parts. This result is due to Sylvester [Reference Sylvester5]. Andrews and Ballantine [Reference Andrews and Ballantine2] recently proved that the number of parts in all self-conjugate partitions of n is almost always equal to the number of partitions of n in which no odd part is repeated and there is exactly one even part (possibly repeated).
There is an interesting variation of self-conjugate partitions. Assuming that
$\lambda $
is a partition of n, then
$\lambda $
is said to be nearly self-conjugate if the Ferrers graphs of
$\lambda $
and its conjugate have exactly
$n-1$
cells in common. For example, there are two nearly self-conjugate partitions of
$5$
, which are depicted in Figure 1.
The two nearly self-conjugate partitions of 5.
Let
$\mathrm {nsc}(n)$
count the number of nearly self-conjugate partitions of n. The sequence
$\{\mathrm {nsc}(n)\}_{n\geq 0}$
seems to be first considered by Campbell, who conjectured the generating function of
$\mathrm {nsc}(n)$
(see [Reference Campbell and Sloane4]).
Conjecture 1.1. We have
$$ \begin{align} \sum_{n\geq 0}\mathrm{nsc}(n)q^n=\frac{2q^2}{1-q^2}(-q^3;q^2)_\infty. \end{align} $$
Throughout the paper, we adopt the following q-series notation:
$$ \begin{align*} (a;q)_{0}&=1,\\ (a;q)_{n}&=\prod_{k=1}^{n}(1-aq^{k-1}) \quad \text{for } n\geq 1,\\ (a;q)_\infty&=\prod_{k=1}^\infty (1-aq^{k-1}). \end{align*} $$
To interpret the right-hand side of (1.1), Campbell and Chern (‘Nearly self-conjugate integer partitions’, submitted for publication) introduced symplectic partitions, in which the smallest part equals
$2$
and all others are distinct odd parts. Clearly, the generating function of twice the number of symplectic partitions equals the expression on the right-hand side of (1.1). Campbell and Chern, established a correspondence showing that the number of nearly self-conjugate partitions of n is equal to twice the number of symplectic partitions of n, which confirms Conjecture 1.1.
In this note, we aim to prove Conjecture 1.1 by analysing the structure of nearly self-conjugate partitions and using elementary techniques (see Section 2). In Section 3, we give an equivalent statement of Conjecture 1.1 and provide a combinatorial proof.
2 Analytic proof
Let
$\mathscr {N}$
denote the set of nearly self-conjugate partitions and
$\mathscr {D}$
denote the set of partitions into distinct parts. Let
$\mathscr {G}$
denote the set of gap-free partitions (partitions in which every integer less than the largest part must appear at least once) and
$\mathscr {G}_k$
denote the subset of
$\mathscr {G}$
consisting of the partitions with the largest part being k. Given a partition
$\lambda \in \mathscr {G}$
, we define
$$ \begin{align*} \mathrm{rep}(\lambda)=|\{i:i\ \text{appears more than once in}\ \lambda\ \text{and}\ i\ \text{is not the largest part in}\ \lambda\}|. \end{align*} $$
That is,
$\mathrm {rep}(\lambda )$
counts the number of repeated parts less than the largest part of
$\lambda $
. For example, if
$\lambda =(4,4,3,2,2,1,1)$
, then
$\mathrm {rep}(\lambda )=2$
. For a partition
$\lambda \in \mathscr {D}$
, we use
$c(\lambda )$
to denote the number of separate sequences of consecutive integers of
$\lambda $
. For example, if
$\mu =(8,7,5,4,3,1)$
, then
$c(\,\mu )=3$
, counting the sequences
$$ \begin{align*}(8,7),(5,4,3),(1).\end{align*} $$
Remark 2.1.
$c(\lambda )$
serves as an important statistic in Sylvester’s refinement of Euler’s partition identity [Reference Andrews and Eriksson3, page 88], namely, the number of odd partitions of n with exactly k different parts equals the number of distinct partitions of n into k separate sequences of consecutive integers.
We first give an auxiliary result.
Lemma 2.2. We have
$$ \begin{align*} \sum_{\lambda \in \mathscr{N}}q^{|\lambda|}=2\sum_{\lambda \in \mathscr{G}}(\mathrm{rep}(\lambda)+1)q^{2|\lambda|+1-m(\lambda)}. \end{align*} $$
Proof. Recall the Frobenius symbol [Reference Andrews and Eriksson3] of
$\lambda $
, which is a two-rowed array
$$ \begin{align*}\left(\begin{matrix}a_1&a_2&\cdots&a_k\\b_1&b_2&\cdots&b_k\end{matrix}\right) \end{align*} $$
with
$a_1>a_2>\cdots >a_k\geq 0$
and
$b_1>b_2>\cdots >b_k\geq 0$
, where
$a_i$
(respectively,
$b_i$
) counts the number of cells to the right of (respectively, below) the ith diagonal entry of
$\lambda $
in its Ferrers graph and k is the size of the Durfee square of
$\lambda $
. Then,
$$ \begin{align*} |\lambda|=\sum_{i=1}^{k}a_i+\sum_{i=1}^{k}b_i+k. \end{align*} $$
Now we add one to each term of the Frobenius symbol of
$\lambda $
to get a modified Frobenius symbol,
$$ \begin{align*} \left( \begin{matrix} \alpha_1&\alpha_2&\cdots&\alpha_k\\ \beta_1&\beta_2&\cdots&\beta_k \end{matrix} \right) \end{align*} $$
with
$\alpha _1>\alpha _2>\cdots >\alpha _k\geq 1$
and
$\beta _1>\beta _2>\cdots >\beta _k\geq 1$
.
If
$\lambda \in \mathscr {N}$
, there exists exactly one j such that
$|\alpha _j-\beta _j|=1$
and
$\alpha _i=\beta _i$
for all
$i\neq j$
. Assuming that
$\alpha _j-\beta _j=1$
,
$$ \begin{align*} |\lambda|=2\sum_{i=1}^{k}\beta_i+1-k. \end{align*} $$
Conversely, for a partition
$\beta =(\,\beta _1,\beta _2,\ldots ,\beta _k)\in \mathscr {D}$
, there exists exactly
$c(\,\beta )$
separate sequences of consecutive integers. For each i with
$1\leq i\leq c(\,\beta )$
, let
$\beta _{j_i}$
be the first integer in the ith sequence of consecutive integers in
$\beta $
. We have
$$ \begin{align*}\beta_1=\beta_{j_1}>\beta_{j_2}>\cdots>\beta_{j_{c(\,\beta)}}.\end{align*} $$
We now construct a partition
$\alpha ^i=(\alpha ^i_1,\alpha ^i_2,\ldots ,\alpha ^i_{k})$
for each i with
$1\leq i\leq c(\,\beta )$
by
$$ \begin{align*} \alpha^i_r=\begin{cases} \beta_r+1 \quad& \mbox{if}\ r=j_i, \\[5pt] \beta_r \quad& \mbox{otherwise.} \end{cases} \end{align*} $$
Then, we have a modified Frobenius symbol with the first row being
$\alpha ^i$
and the second row being
$\beta $
, which corresponds to a nearly self-conjugate partition. Thus, in total, there are
$c(\,\beta )$
nearly self-conjugate partitions with such modified Frobenius symbols.
If
$\beta _j-\alpha _j=1$
, we can exchange
$\alpha $
and
$\beta $
in the Frobenius symbol and obtain the same conclusion. Consequently, we can conclude that
$$ \begin{align*} \sum_{\lambda \in \mathscr{N}}q^{|\lambda|}=2\sum_{\beta \in \mathscr{D}}c(\,\beta)q^{2|\beta|+1-\ell(\,\beta)}. \end{align*} $$
For a partition
$\lambda \in \mathscr {D}$
, its conjugate
$\lambda ^T$
must be a gap-free partition and it follows that
$\ell (\lambda )=m(\lambda ^T)$
and
$c(\lambda )=\mathrm {rep}(\lambda ^T)+1$
. Thus,
$$ \begin{align*} \sum_{\lambda \in \mathscr{D}}c(\lambda)q^{2|\lambda|+1-\ell(\lambda)} =\sum_{\lambda \in \mathscr{G}}(\mathrm{rep}(\lambda)+1)q^{2|\lambda|+1-m(\lambda)}. \end{align*} $$
This completes the proof.
We are now in a position to prove (1.1). By standard combinatorial arguments,
$$ \begin{align*} \sum_{\lambda \in \mathscr{G}_k}z^{\mathrm{rep}(\lambda)+1}q^{|\lambda|} &=\sum_{j\geq0}q^{\,jk}zq^{{k(k+1)}/{2}}\prod_{i=1}^{k-1}(1+zq^{i}+zq^{2i}+\cdots)\\ &=zq^{{k(k+1)}/{2}}\frac{\prod_{i=1}^{k-1}(1-q^{i}+zq^{i})}{(q;q)_{k}}. \end{align*} $$
Differentiating the above equation with respect to z, putting
$z=1$
and replacing q by
$q^2$
,
$$ \begin{align*} \sum_{\lambda \in \mathscr{G}_k}(\mathrm{rep}(\lambda)+1)q^{2|\lambda|} &=\frac{1}{(q^{2};q^{2})_{k}}\bigg(q^{k(k+1)}+q^{k(k+1)}\sum_{i=1}^{k-1}q^{2i}\bigg) \notag\\ &=\frac{q^{k(k+1)}}{(q^{2};q^{2})_{k}}\sum_{i=0}^{k-1}q^{2i} =\frac{q^{k(k+1)}(1-q^{2k})}{(q^{2};q^{2})_{k}(1-q^{2})}. \end{align*} $$
By Lemma 2.2,
$$ \begin{align} \sum_{\lambda \in \mathscr{N}}q^{|\lambda|}&=2\sum_{\lambda \in \mathscr{G}}(\mathrm{rep}(\lambda)+1)q^{2|\lambda|+1-m(\lambda)}\notag\\ &=2\sum_{k\geq 1}q^{1-k}\sum_{\lambda \in \mathscr{G}_k}(\mathrm{rep}(\lambda)+1)q^{2|\lambda|} \notag\\&=2\sum_{k\geq 1}\frac{q^{1-k}q^{k(k+1)}(1-q^{2k})}{(q^{2};q^{2})_{k}(1-q^{2})} \notag \\ &=2\sum_{k\geq 1}\frac{q^{k^{2}+1}(1-q^{2k})}{(q^{2};q^{2})_{k}(1-q^{2})} \notag \\ &=\frac{2q^{2}}{1-q^{2}}\sum_{k\geq 1}\frac{q^{k^{2}-1}}{(q^{2};q^{2})_{k-1}}. \end{align} $$
Recall Euler’s identity [Reference Andrews1, page 19],
$$ \begin{align*} (-zq;q)_\infty=\sum_{k\geq 0} \frac{z^kq^{k(k+1)/2}}{(q;q)_k}. \end{align*} $$
Replacing q by
$q^2$
and z by q,
$$ \begin{align*} (-q^3;q^2)_\infty =\sum_{k \geq 0} \frac{q^{k(k+2)}}{(q^2;q^2)_k}. \end{align*} $$
From (2.1) and the above identity,
$$ \begin{align*} \sum_{\lambda \in \mathscr{N}}q^{|\lambda|}= \frac{2q^{2}}{1-q^{2}}\sum_{k\geq 0}\frac{q^{k^2+2k}}{(q^{2};q^{2})_{k}} =\frac{2q^{2}}{1-q^{2}}(-q^{3};q^{2})_{\infty}. \end{align*} $$
This completes the proof.
3 Equivalent statement
Multiplying both sides of (1.1) by
$1-q^{2}$
and comparing the coefficients of
$q^n$
,
$$ \begin{align*} \mathrm{nsc}(n)-\mathrm{nsc}(n-2)=2\mathrm{do}_{\geq3}(n-2), \end{align*} $$
where
$\mathrm {do}_{\geq 3}(n)$
denotes the number of partitions of n into distinct odd parts with the smallest part at least
$3$
.
Based on the classical bijection [Reference Sylvester5] between the partitions of n into distinct odd parts and the self-conjugate partitions of n, Campbell and Chern (‘Nearly self-conjugate integer partitions’, submitted for publication) established the following result.
Lemma 3.1 (Campbell and Chern)
The number of nearly self-conjugate partitions of n equals twice the number of partitions of n with exactly one even part such that the differences between parts are at least
$2$
.
Let
$\mathscr {O}(n)$
denote the set of partitions of n with exactly one even part and the differences between parts being at least
$2$
. Thus, Conjecture 1.1 could be restated as follows.
Proposition 3.2. For
$n\geq 2$
,
$$ \begin{align*} |\mathscr{O}(n)|-|\mathscr{O}(n-2)|=\mathrm{do}_{\geq3}(n-2). \end{align*} $$
Proof. We first introduce an injection
$\varphi : \mathscr {O}(n-2)\rightarrow \mathscr {O}(n)$
for every
$n\geq 2$
. Assuming that
$\lambda =(\lambda _1,\lambda _{2},\ldots )\in \mathscr {O}(n-2)$
, we define
$\varphi (\lambda )=\mu \in \mathscr {O}(n)$
as follows.
Case 1: If
$\lambda $
has only one part, then this unique part must be even. Hence, we can assume that
$\lambda =(2m)$
and define
$\mu =(2m+2)$
.
Case 2: If
$\ell (\lambda )\geq 2$
and the smallest part is odd, we assume that
$$ \begin{align*}\lambda=(\lambda_1,\ldots,\lambda_{i-1},2m,\lambda_{i+1},\lambda_{i+2},\ldots).\end{align*} $$
We define
$\mu =(\lambda _1,\ldots ,\lambda _{i-1},2m+1,\lambda _{i+1}+1,\lambda _{i+2},\ldots )$
. It is easy to see that
$\ell (\,\mu )=\ell (\lambda )\geq 2$
and the largest part of
$\mu $
is odd.
Case 3: If
$\ell (\lambda )\geq 2$
and the smallest part is even, we suppose that
$$ \begin{align*}\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_{i-1},2m)\end{align*} $$
and define
$\mu =(\lambda _1+1,\lambda _2,\ldots ,\lambda _{i-1},2m+1)$
. It is clear that
$\ell (\,\mu )=\ell (\lambda )\geq 2$
and the largest part of
$\mu $
is even and the smallest part of
$\mu $
is greater than or equal to 3.
We observe that each partition in
$\mathscr {O}(n)\!\setminus \!\varphi (\mathscr {O}(n-2))$
is a partition
$\mu =(\kern1pt\mu _1,\mu _2,\ldots ,\mu _r)$
with
$r\geq 2$
,
$\mu _1$
being even and
$\mu _r=1$
. Obviously,
$\mu _1-\mu _2\geq 3$
. Subtracting
$1$
from the largest part and removing the smallest part, we get a partition into distinct odd parts with each part at least
$3$
, which is counted by
$\mathrm {do}_{\geq 3}(n-2)$
. This completes the proof.
Acknowledgement
The authors would like to thank the referee for helpful comments and suggestions.
