1 Introduction
Euler’s pentagonal number theorem plays an important role in the theory of partitions. It gives
$$ \begin{align} (q;q)_\infty=\sum_{j=0}^\infty(-1)^jq^{\,j(3j+1)/2}(1-q^{2j+1}). \end{align} $$
In (1.1) and for the rest of this article, we use the notation
$$ \begin{align*} (x_1, x_2, \ldots, x_k;q)_m &:=\prod_{n=0}^{m-1} (1-x_1q^n)(1-x_2q^n)\cdots(1-x_kq^n),\\ (x_1,x_2, \ldots, x_k;q)_\infty &:=\prod_{n=0}^\infty (1-x_1q^n)(1-x_2q^n)\cdots(1-x_kq^n),\\ { L \brack K}_q&:=\left\{ \begin{aligned} &0&&\textrm{if }K<0\textrm{ or } K>L,\\ &\frac{(q;q)_L}{(q;q)_K(q;q)_{L-K}} &&\text{otherwise}, \end{aligned} \right. \end{align*} $$
and we require
$|q|<1$
for absolute convergence.
Andrews and Merca [Reference Andrews and Merca2] proved the truncated pentagonal number theorem
$$ \begin{align} &\frac{1}{(q)_\infty}\sum_{j=0}^{k-1}(-1)^{\,j}q^{\,j(3j+1)/2}(1-q^{2j+1})\nonumber \\& \quad = 1+(-1)^{k-1}\sum_{n=1}^{\infty}\frac{q^{(k+1)n+k(k-1)/2}}{(q;q)_n}\begin{bmatrix} n-1 \\ k-1 \end{bmatrix}_q. \end{align} $$
Using (1.2), they proved a new partition theorem. Recall that a partition of an integer n is a nonincreasing sequence of positive integers whose sum is n. Let
$p(n)$
denote the number of partitions of n.
Theorem 1.1 [Reference Andrews and Merca2, Theorem 1.1].
For
$n>0, k\geq 1$
, let
$$ \begin{align*} M_k(n):=(-1)^{k-1}\sum_{j=0}^{k-1}(-1)^{\,j}(p(n-j(3j+1)/2)- p(n-j(3j+5)/2-1)). \end{align*} $$
Then
$M_k(n)$
is the number of partitions of n in which k is the least integer that is not a part and there are more parts
$> k$
than there are
$< k$
.
The first result of this paper is the following new partition theorem, which is an analogue of Theorem 1.1.
Theorem 1.2. For
$n, k\geq 0$
, let
$$ \begin{align*} N_k(n):=(-1)^k\bigg\{p(n)+\sum_{j=1}^{k}(-1)^{\,j}(p(n-j(3j-1)/2) +p(n-j(3j+1)/2))\bigg\}. \end{align*} $$
Then
$N_k(n)$
is the number of partitions of n with every integer
$\leq k$
appearing as a part, and in which the first part larger than k appears at least
$k+1$
times.
For example,
$N_2(16)=6$
and the partitions in question are
$4+4+4+2+1+1$
,
$4+3+3+3+2+1,3+3+3+3+2+1+1,3+3+3+2+2+2+1$
,
$3+3+3+2+2+1+1+1,3+3+3+2+1+1+1+1+1.$
To prove Theorem 1.2, we need a new truncated pentagonal number theorem. Note that the pentagonal number theorem is also given by
$$ \begin{align} (q;q)_\infty=1+\sum_{j=1}^\infty(-1)^jq^{\,j(3j-1)/2}(1+q^{\,j}). \end{align} $$
Then we have the truncated version of (1.3).
Theorem 1.3. For
$k\geq 0$
,
$$ \begin{align} &\frac{1}{(q;q)_\infty}\bigg\{1+\sum_{j=1}^{k}(-1)^{\,j}q^{\,j(3j-1)/2}(1+q^{\,j})\bigg\}\nonumber\\& \quad = 1+(-1)^k\frac{q^{{k(k+1)}/{2}}}{(q;q)_k}\sum_{j=0}^{\infty}\frac{q^{(j+k+1)(k+1)}}{(q^{\,j+k+1};q)_\infty}. \end{align} $$
We remark that letting
$k\rightarrow \infty $
in (1.4) gives (1.3). Andrews and Merca’s work inspired a wide study on truncated identities and their partition-theoretic interpretations (see, for example, [Reference Chern and Xia4, Reference He, Ji and Zang6–Reference Merca10, Reference Wang and Yee12–Reference Yee17]). In particular, Guo and Zeng [Reference Guo and Zeng5] obtained truncated forms of two identities of Gauss (see [Reference Andrews1, page 23]): that is,
$$ \begin{align} \frac{(q;q)_\infty}{(-q;q)_\infty}&= 1+2\sum_{j=1}^\infty(-1)^jq^{\,j^2}, \end{align} $$
$$ \begin{align} \frac{(q^2;q^2)_\infty}{(-q;q^2)_\infty}&=\sum_{j=0}^\infty (-q)^{\,j(j+1)/2}.\ \kern1pt\,\quad \end{align} $$
They found that
$$ \begin{align} &\frac{(-q)_\infty}{(q)_\infty}\bigg(1+2\sum_{j=1}^{k}(-1)^{\,j}q^{\,j^2}\bigg) \nonumber \\ & \quad = 1+(-1)^k\sum_{n=k+1}^{\infty}\frac{q^{(k+1)n}(-1;q)_{n-k}(-q;q)_k}{(q)_n}\begin{bmatrix} n-1 \\ k-1 \end{bmatrix}_q,\quad\qquad\end{align} $$
$$ \begin{align} &\frac{(-q;q^2)_\infty}{(q^2;q^2)_\infty}\sum_{j=1}^{k}(-1)^{\,j}q^{2j^2-j}(1+q^{2j}) \nonumber \\& \quad = 1+(-1)^{k-1}\sum_{n=k}^{\infty}\frac{q^{2(k+1)n-k}(-q;q^2)_{k}(-q;q^2)_{n-k}}{(q^2;q^2)_n}\begin{bmatrix} n-1 \\ k-1 \end{bmatrix}_{q^2}. \end{align} $$
Partition-theoretic interpretations of the two truncated sums in (1.7) and (1.8) are obtained in [Reference Andrews and Merca3]. Rewrite (1.5) and (1.6) as
$$ \begin{align} \frac{(q;q)_\infty}{(-q;q)_\infty}&= \sum_{j=0}^\infty(-1)^jq^{\,j^2}(1-q^{2j+1}),\ \quad \end{align} $$
$$ \begin{align} \frac{(q^2;q^2)_\infty}{(-q;q^2)_\infty}&=1+\sum_{j=1}^\infty (-1)^jq^{2j^2-j}(1+q^{2j}). \end{align} $$
Wang and Yee [Reference Wang and Yee13] obtained the following truncated version of (1.9): for
$k\geq 0$
,
$$ \begin{align*} &\frac{(-q;q)_\infty}{(q;q)_\infty}\sum_{j=0}^{k}(-1)^{\,j}q^{\,j^2}(1-q^{2j+1})\\& \quad = 1+(-1)^k q^{-k-1}\sum_{n=k+1}^{\infty}\frac{q^{(k+2)n}(-1;q)_{n-k-1}(-q;q)_{k+1}}{(q;q)_n}\begin{bmatrix} n-1 \\ m\nonumber \end{bmatrix}_q. \end{align*} $$
We give a truncated version of (1.10).
Theorem 1.4. For
$k\geq 0$
,
$$ \begin{align} &\frac{(-q;q^2)_\infty}{(q^2;q^2)_\infty}\bigg\{1+\sum_{j=1}^{k}(-1)^{\,j}q^{2j^2-j}(1+q^{2j})\bigg\} \nonumber \\& \quad = 1+(-1)^k\frac{(-q;q^2)_{k+1}}{(q^2;q^2)_{k}}\sum_{j=0}^{\infty}\frac{q^{(k+1)(2j+2k+1)}(-q^{2j+2k+3};q^2)_{\infty}}{(q^{2j+2k+2};q^2)_\infty}. \end{align} $$
Letting
$k\rightarrow \infty $
in (1.11), we obtain (1.10). Denote the number of partitions of n without repeated odd parts by
$pod(n)$
. Then we give a partition-theoretic interpretation of the truncated sum in (1.11).
Corollary 1.5. For
$n>0, k\geq 1$
, let
$$ \begin{align*} \mathit{NP}_k(n):=(-1)^k\bigg\{pod(n)+\sum_{j=1}^{k}(-1)^{\,j}(pod(n-j(2j-1)) +pod(n-j(2j+1)))\bigg\}. \end{align*} $$
Then
$\mathit {NP}_k(n)$
is the number of partitions of n in which
$2k+1$
appears
$k+1$
or
$k+2$
times and all other odd parts appear at most once, or
$2k+2$
does not appear, and the first part larger than
$2k+2$
is odd and appears exactly
$k+1$
times, and all other odd parts appear at most once.
For example,
$\mathit {NP}_1(13)=9$
and the partitions in question are
$7+3+3,6+3+3+1$
,
${5+3+3+2,4+3+3+3,4+3+3+2+1,3+3+3+2+2,3+3+2+2+ 2\kern1.4pt{+}\kern1.4pt1}, 5+5+3,5+5+2+1.$
We remark that Corollary 1.5 does not hold when
$k=0$
. For example,
$NP_0(4)=3$
, but there are only two partitions in question, namely,
$3+1,2+1+1.$
The proofs of truncated identities (1.4) and (1.11) are inspired by [Reference Andrews and Merca3]. The Rogers–Fine identity (see (2.2)) plays a key role. The paper is organised as follows. In Section 2, we prove Theorems 1.2 and 1.3. Theorem 1.4 and Corollary 1.5 are obtained in Section 3.
2 Ordinary partitions
Lemma 2.1. For
$\mid \alpha \mid <1$
,
$$ \begin{align} \sum_{n=0}^\infty(\alpha;q)_n\alpha^n =\sum_{n=0}^\infty(-1)^n\alpha^{3n}q^{{n(3n-1)}/{2}}(1+\alpha q^{n}). \end{align} $$
Proof. Recall the Rogers–Fine identity [Reference Rogers11, page 15] (see also [Reference Andrews and Merca3, Equation (1.10)])
$$ \begin{align} \sum_{n=0}^\infty\frac{(\alpha;q)_n\tau^n}{(\beta;q)_n}=\sum_{n=0}^\infty\frac{(\alpha;q)_n(\alpha\tau q/\beta;q)_n \beta^n\tau^nq^{n^2-n}(1-\alpha\tau q^{2n})}{(\beta;q)_n(\tau;q)_{n+1}}. \end{align} $$
Setting
$\tau =\alpha $
and letting
$\beta \rightarrow 0$
in (2.2) yields (2.1).
Proof of Theorem 1.3.
Equation (1.3) implies that
$$ \begin{align*} &(q;q)_\infty-\bigg\{1+\sum_{j=1}^{k}(-1)^{\,j}q^{\,j(3j-1)/2}(1+q^{\,j})\bigg\} = \sum_{j=k+1}^{\infty}(-1)^{\,j}q^{\,j(3j-1)/2}(1+q^{\,j}) \nonumber\\ & \quad = (-1)^{k+1}q^{{(k+1)(3k+2)}/{2}} \sum_{n=0}^{\infty}(-1)^{n}q^{n(3n-1)/2+(3k+3)n}(1+q^{n+k+1})\nonumber. \end{align*} $$
Replacing
$\alpha $
by
$q^{k+1}$
in (2.1) gives
$$ \begin{align*} \sum_{j=0}^\infty(q^{k+1};q)_jq^{(k+1)j} =\sum_{n=0}^\infty (-1)^nq^{{n(3n-1)}/{2}+3n(k+1)}(1+q^{n+k+1}). \end{align*} $$
It follows that
$$ \begin{align*} (q;q)_\infty -\bigg\{1+\sum_{j=1}^{k}(-1)^{\,j}q^{\,j(3j-1)/2}(1+q^{\,j})\bigg\} & = (-1)^{k+1}q^{{(k+1)(3k+2)}{2}} \sum_{j=0}^{\infty}(q^{k+1};q)_jq^{(k+1)j} \nonumber\\&= (-1)^{k+1}q^{{(k+1)k}/{2}} \sum_{j=0}^{\infty}(q^{k+1};q)_jq^{(k+1)(j+k+1)}. \nonumber \end{align*} $$
Thus,
$$ \begin{align*} 1-\frac{1}{(q;q)_\infty}\bigg\{1+\sum_{j=1}^{k}(-1)^{\,j}q^{\,j(3j-1)/2}(1+q^{\,j})\bigg\} = (-1)^{k+1}\frac{q^{{k(k+1)}/{2}}}{(q;q)_k}\sum_{j=0}^{\infty}\frac{q^{(j+k+1)(k+1)}}{(q^{\,j+k+1};q)_\infty}. \end{align*} $$
This proves Theorem 1.3.
Proof of Theorem 1.2.
Theorem 1.3 gives
$$ \begin{align*} \sum_{n=1}^\infty N_k(n)q^n=\frac{q^{{k(k+1)}/{2}}}{(q;q)_k}\sum_{j=0}^{\infty}\frac{q^{(j+k+1)(k+1)}}{(q^{\,j+k+1};q)_\infty}. \end{align*} $$
The term
$$ \begin{align*}\frac{q^{{k(k+1)}/{2}}}{(q;q)_k}\end{align*} $$
is the generating function for partitions with parts
$\leq k$
and each integer
$\leq k$
appearing as a part. In addition,
$q^{(j+k+1)(k+1)}$
produces the
$k+1$
appearances of the first part,
$j+k+1$
, larger than k, and
${1}/{(q^{\,j+k+1};q)_\infty }$
generates partitions with parts
$\geq j+k+1$
. Thus,
$N_k(n)$
enumerates the partitions of n in question.
3 Partitions without repeated odd parts
Lemma 3.1. For
$\mid \alpha \mid <1$
,
$$ \begin{align} \sum_{n=0}^\infty\frac{(\alpha;q^2)_n\alpha^n }{(\alpha q;q^2)_n}=\sum_{n=0}^\infty\alpha^{2n}q^{2n^2-n}(1+\alpha q^{2n}). \end{align} $$
Proof of Theorem 1.4.
By (1.10),
$$ \begin{align*} &\frac{(q^2;q^2)_\infty}{(-q;q^2)_\infty}-\bigg\{1+\sum_{j=1}^{k}(-1)^{\,j}q^{2j^2-j}(1+q^{2j})\bigg\} =\sum_{j=k+1}^{\infty}(-1)^{\,j}q^{2j^2-j}(1+q^{2j}) \\ &\quad=(-1)^{k+1}q^{(2k+1)(k+1)}\sum_{n=0}^{\infty}(-1)^{n}q^{2n^2+(4k+3)n}(1+q^{2n+2k+2}). \end{align*} $$
Replacing q by
$-q$
and setting
$\alpha =q^{2k+2}$
in (3.1) gives
$$ \begin{align*} \sum_{j=0}^\infty\frac{(q^{2k+2};q^2)_jq^{(2k+2)j} }{( -q^{2k+3};q^2)_j}=\sum_{n=0}^{\infty}(-1)^{n}q^{2n^2+(4k+3)n}(1+q^{2n+2k+2}). \end{align*} $$
Then
$$ \begin{align*} \frac{(q^2;q^2)_\infty}{(-q;q^2)_\infty}&-\bigg\{1+\sum_{j=1}^{k}(-1)^{\,j}q^{2j^2-j}(1+q^{2j})\bigg\} =(-1)^{k+1}\sum_{j=0}^{\infty}\frac{(q^{2k+2};q^2)_jq^{(2j+2k+1)(k+1)}}{(-q^{2k+3};q^2)_j}. \end{align*} $$
It follows that
$$ \begin{align*} 1-&\frac{(-q;q^2)_\infty}{(q^2;q^2)_\infty}\bigg\{1+\sum_{j=1}^{k}(-1)^{\,j}q^{2j^2-j}(1+q^{2j})\bigg\}\nonumber\\ &= (-1)^{k+1}\frac{(-q;q^2)_{k+1}}{(q^2;q^2)_{k}}\sum_{j=0}^{\infty}\frac{q^{(k+1)(2j+2k+1)}(-q^{2j+2k+3};q^2)_{\infty}}{(q^{2j+2k+2};q^2)_\infty}.\nonumber \end{align*} $$
This proves Theorem 1.4.
Proof of Corollary 1.5.
By Theorem 1.4,
$$ \begin{align*} \sum_{n=1}^\infty \mathit{NP}_k(n)q^n&=\frac{(-q;q^2)_{k+1}}{(q^2;q^2)_{k}}\sum_{j=0}^{\infty}\frac{q^{(k+1)(2j+2k+1)}(-q^{2j+2k+3};q^2)_{\infty}}{(q^{2j+2k+2};q^2)_\infty} \\&=\frac{(-q;q^2)_{k+1}}{(q^2;q^2)_{k}}\sum_{j=1}^{\infty}\frac{q^{(k+1)(2j+2k+1)}(-q^{2j+2k+3};q^2)_{\infty}}{(q^{2j+2k+2};q^2)_\infty} + \frac{q^{(2k+1)(k+1)}(-q;q^2)_\infty}{(q^2;q^2)_\infty}. \end{align*} $$
(i) The term
$$ \begin{align*}\frac{q^{(2k+1)(k+1)}(-q;q^2)_\infty}{(q^2;q^2)_\infty}\end{align*} $$
is the generating function for partitions in which
$2k+1$
appears
$k+1$
or
$k+2$
times and all other odd parts appear at most once.
(ii) In
$$ \begin{align*}\frac{(-q;q^2)_{k+1}}{(q^2;q^2)_{k}}\sum_{j=1}^{\infty}\frac{q^{(k+1)(2j+2k+1)}(-q^{2j+2k+3};q^2)_{\infty}}{(q^{2j+2k+2};q^2)_\infty},\end{align*} $$
the expression preceding the summation is the generating function for partitions with parts
$\leq 2k+1$
and odd parts distinct, the expression inside the sum is the generating function for partitions in which the first part,
$2j+2k+1$
, appears exactly
$k+1$
times, and all other odd parts appear at most once. Thus,
$$ \begin{align*}\frac{(-q;q^2)_{k+1}}{(q^2;q^2)_{k}}\sum_{j=1}^{\infty}\frac{q^{(k+1)(2j+2k+1)}(-q^{2j+2k+3};q^2)_{\infty}}{(q^{2j+2k+2};q^2)_\infty}\end{align*} $$
is the generating function of partitions in which
$2k+2$
does not appear, and the first part larger than
$2k+2$
is odd and appears exactly
$k+1$
times, and all other odd parts appear at most once.
Moreover, the sets of partitions of n described in (i) and (ii) are disjoint when
$k\geq 1$
. Thus,
$\mathit {NP}_k(n)$
enumerates the partitions of n in question.
