q-Binomial Sums toward Euler’s Pentagonal Number Theorem

Abstract

We examine a class of q-binomial sums whose evaluations lead to several finite forms of Euler’s pentagonal number theorem.

1 Introduction and Motivation

Let $\mathbb {Z}$ and $\mathbb {N}$ be the sets of integers and natural numbers with $\mathbb {N}_0=\{0\}\cup \mathbb {N}$. For an indeterminate x, the shifted factorial of order $n\in \mathbb {Z}$ with the base q is defined by $(x;q)_0\equiv 1$ and

$$\begin{aligned}(x;q)_n={\left\{ \begin{array}{ll} (1-x)(1-qx)\cdots (1-q^{n-1}x),&{}n>0;\\ \dfrac{(-x)^nq^{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }}{(1-q/x)(1-q^2/x)\cdots (1-q^{-n}x)},&n<0. \end{array}\right. } \end{aligned}$$

Then the Gaussian binomial coefficient can be expressed as

$$\begin{aligned} \left[ \begin{array}{c}m\\ n\end{array}\right] =\frac{(q;q)_m}{(q;q)_n(q;q)_{m-n}} =\frac{(q^{m-n+1};q)_n}{(q;q)_n} \quad \text {where}\quad m,\,n\in \mathbb {N}_0. \end{aligned}$$

By means of the package “qEKHAD,” Ekhad and Zeilberger [8] discovered in 1996 the following elegant q-binomial identity:

$$\begin{aligned} \sum _{k=0}^{\lfloor {\frac{m}{2}}\rfloor } (-1)^k\left[ \begin{array}{c}m-k\\ k\end{array}\right] q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) } ={\left\{ \begin{array}{ll} (-1)^nq^{\frac{n(3n-1)}{2}},&{}m=3n;\\ (-1)^nq^{\frac{n(3n+1)}{2}},&{}m=3n+1;\\ 0,&{}m=3n-1. \end{array}\right. } \end{aligned}$$

Recently, Liu [10] provided an induction proof and studied a few variants of it. As done by Warnaar [11], this identity can be deduced as a limiting case from some known cubic q-series formula (see also Chen–Chu [2] and Chu [5]). In this paper, we shall examine the following q-binomial sum extended by two extra integer parameters $\lambda$ and $\sigma$:

$$\begin{aligned} \varOmega _m(\lambda ,\sigma ) :=\sum _{k=0}^{\lfloor {\frac{m}{2}}\rfloor } (-1)^k\left[ \begin{array}{c}m-k\\ k\end{array}\right] (q^{1+m-k};q)_{\lambda } q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k\sigma }. \end{aligned}$$

(1)

In the next section, four main theorems will be proved that evaluate explicitly $\varOmega _m(\lambda ,\sigma )$ according to the signs of $\lambda$ and $\sigma$. Under the base change “$q\rightarrow q^{-1}$,” these identities will be shown in Sect. 3 to be finite forms of Euler’s pentagonal number theorem, including those important ones due to by Berkovich–Garvan [1], Cigler [6], Liu [10] and Warnaar [11].

Throughout the paper, the multivariate notations for shifted factorials in base q will be abbreviated to

$$\begin{aligned} \left[ A,B,\cdots ,C;q\right] _{n} ~&=\left( A;q\right) _{n}\left( B;q\right) _{n}\cdots \left( C;q\right) _{n},\\ \left[ \begin{array}{cccc}\alpha ,\beta ,\cdots ,\gamma \\ A,B,\cdots ,C\end{array}{\Big |q}\right] _n&=\frac{\left( \alpha ;q\right) _{n}\left( \beta ;q\right) _{n}\cdots \left( \gamma ;q\right) _{n}}{\left( A;q\right) _{n}\left( B;q\right) _{n}\cdots \left( C;q\right) _{n}}. \end{aligned}$$

Following Gasper and Rahman [9], the unilateral and bilateral basic hypergeometric series (shortly as q-series) are defined, respectively, by

$$\begin{aligned} {_{1+r}\phi _{s}}\left[ \begin{array}{cccc}a_0,~a_1,~\cdots ,~a_r\\ ~b_1,~\cdots ,~b_s\end{array}{\Big |q;z}\right]&=\sum _{k=0}^{\infty } ~\left[ \begin{array}{cccc}a_0,~a_1,~\cdots ,~a_r\\ q,~b_1,~\cdots ,~b_s\end{array}{\Big |q}\right] _k z^k,\\ {_{r}\psi _{s}} \left[ \begin{array}{cccc}a_1,~a_2,~\cdots ,~a_r\\ b_1,~b_2,~\cdots ,~b_s\end{array}{\Big |q;z}\right]&=\sum _{k=-\infty }^{\infty } \left[ \begin{array}{ccccc}a_1,~a_2,~\cdots ,~a_r\\ b_1,~b_2,~\cdots ,~b_s\end{array}{\Big |q}\right] _k z^k. \end{aligned}$$

In order to reduce lengthy expressions, we shall make use of the following notations. As usual, the logical function is defined by $\chi (\text {true})=1$ and $\chi (\text {false})=0$. For a real number x, the greatest integer not exceeding x will be denoted by $\lfloor {x}\rfloor$. When m is a natural number, $i\equiv _mj$ stands for that “i is congruent to j modulo m.”

2 Main Theorems and Proofs

Due to the q-shifted factorial of oder $\lambda$ appearing in the binomial sum $\varOmega _m(\lambda ,\sigma )$, we can easily check that $\varOmega _m(\lambda ,\sigma )$ is well-defined for all the $m\in \mathbb {N}_0$ when ${\lambda \ge 0}$. Instead, $\varOmega _m(\lambda ,\sigma )$ is defined for $m\in \mathbb {N}$ but with $m\ge -2\lambda$ when ${\lambda <0}$. In this section, we shall present four main summation theorems in accordance with the signs of $\sigma$ and $\lambda$.

2.1 Generating Function for $\Omega _n(0,0)$

As a warm up, we start with the following classical result:

$$\begin{aligned} \sum _{0\le k\le m/2}(-1)^k\left( {\begin{array}{c}m-k\\ k\end{array}}\right) =(-1)^{\lfloor {\frac{m}{3}}\rfloor }\chi (m\not \equiv _32). \end{aligned}$$

(2)

Its generating function can be manipulated as follows:

$$\begin{aligned} \sum _{m\ge 0}x^m\sum _{0\le k\le m/2}(-1)^k\left( {\begin{array}{c}m-k\\ k\end{array}}\right)&=\sum _{k\ge 0}(-x^2)^k\sum _{m\ge 2k}\left( {\begin{array}{c}m-k\\ k\end{array}}\right) x^{m-2k}\\&=\sum _{k\ge 0}\frac{(-x^2)^k}{(1-x)^{k+1}} =\frac{1}{(1-x)(1+\frac{x^2}{1-x})}\\&=\frac{1}{1-x+x^2}=\frac{1+x}{1+x^3}. \end{aligned}$$

Then the binomial identity displayed in (2) is confirmed by extracting the coefficient of $x^m$ across the above equations. $\square$

Observe that the above generating function can also be reformulated as

$$\begin{aligned} \frac{1}{1-x+x^2}=\sum _{k\ge 0}x^k(1-x)^k. \end{aligned}$$

This suggests a new proof for $\varOmega _m(0,0)$ by examining the following q-analogue

$$\begin{aligned} \phi (x):=\sum _{k=0}^{\infty }x^k(x;q)_k, \end{aligned}$$

which is convergent for $0<|x|<1$. According to Euler’s q-binomial theorem

$$\begin{aligned} (x;q)_k=\sum _{i=0}^k(-1)^i\left[ \begin{array}{c}k\\ i\end{array}\right] q^{\left( {\begin{array}{c}i\\ 2\end{array}}\right) }x^i, \end{aligned}$$

we have that

$$\begin{aligned} {[}x^m]\phi (x)=\sum _{0\le i\le m/2}(-1)^i \left[ \begin{array}{c}m-i\\ i\end{array}\right] q^{\left( {\begin{array}{c}i\\ 2\end{array}}\right) }, \end{aligned}$$

which implies that $\phi (x)$ is the generating function of the sequence $\{\varOmega _m(0,0)\}_{m\ge 0}.$

In order to find an explicit formula for $\varOmega _m(0,0)$, we examine the bilateral series

$$\begin{aligned} \psi (x)&:=\sum _{k=-\infty }^{\infty }x^k(x;q)_k\\&=\phi (x)+\sum _{k=-\infty }^{-1}x^k(x;q)_k\\&=\phi (x)+\sum _{k=1}^{\infty }(-1)^k \frac{q^{\left( {\begin{array}{c}k+1\\ 2\end{array}}\right) }x^{-2k}}{(q/x;q)_k}. \end{aligned}$$

Therefore, $\psi (x)$ coincides with $\phi (x)$ for the coefficients of nonnegative powers of the variable x when they are expanded into Laurent and Maclaurin series.

According to Ramanujan’s $_1\psi _1$-series (cf. [3] and [9, II-29])

$$\begin{aligned} {_1\psi _1}\left[ \begin{array}{c}a\\ c\end{array}{\Big |q;z}\right] =\left[ \begin{array}{c}q,c/a,az,q/az\\ c,q/a,z,c/az\end{array}{\Big |q}\right] _{\infty }, \end{aligned}$$

we have the closed form evaluation

$$\begin{aligned} \psi (x)&={_1\psi _1}\left[ \begin{array}{c}x\\ 0\end{array}{\Big |q;x}\right] =\left[ \begin{array}{c}q,x^2,q/x^2\\ x,~q/x\end{array}{\Big |q}\right] _{\infty }\\&=\left[ q,-x,-q/x;q\right] _{\infty }\left[ qx^2,q/x^2;q^2\right] _{\infty }. \end{aligned}$$

Now expanding $\psi (x)$ by the quintuple product identity (see Chu [4] and Cooper [7])

$$\begin{aligned} \psi (x)=\sum _{n=-\infty }^{\infty }(-1)^nq^{3\left( {\begin{array}{c}n\\ 2\end{array}}\right) +n}(1+q^nx)x^{3n} \end{aligned}$$

we recover the following significant identity.

Theorem 1

(Ekhad and Zeilberger [8]: see also [2, 5, 10, 11])

$$\begin{aligned} \begin{aligned}\varOmega _m(0,0)&=(-1)^{\lfloor {\frac{m}{3}}\rfloor } \chi (m\not \equiv _32)q^{\frac{m^2-m}{6}}\\&={\left\{ \begin{array}{ll} (-1)^nq^{\frac{n(3n-1)}{2}},&{}m=3n;\\ (-1)^nq^{\frac{n(3n+1)}{2}},&{}m=3n+1;\\ 0,&{}m=3n-1. \end{array}\right. } \end{aligned} \end{aligned}$$

$\square$

We remark that if extracting the coefficient of $x^{m}$ for $m\in \mathbb {Z}\backslash \mathbb {N}_0$, then we would have the same result:

$$\begin{aligned} {[}x^{m}]\psi (x) =(-1)^{\lfloor {\frac{m}{3}}\rfloor } \chi (m\not \equiv _32) q^{\frac{m^2-m}{6}}. \end{aligned}$$

2.2 $\Omega _n(0,\sigma )$ with $\sigma <0$

We examine further the generation function

$$\begin{aligned} \phi _{\sigma }(x):=\sum _{m\ge 0}x^m\varOmega _m(0,\sigma ) \end{aligned}$$

which results in

$$\begin{aligned} \phi _{\sigma }(x)=\sum _{k=0}^{\infty }x^k(q^{\sigma }x;q)_k. \end{aligned}$$

In fact, we have easily that

$$\begin{aligned} \phi _{\sigma }(x)&=\sum _{m\ge 0}x^m \sum _{0\le k\le m/2}(-1)^k\left[ \begin{array}{c}m-k\\ k\end{array}\right] q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k\sigma }\\&=\sum _{k\ge 0}(-1)^kq^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k\sigma } \sum _{m\ge 2k}x^m\left[ \begin{array}{c}m-k\\ k\end{array}\right] \\&=\sum _{k\ge 0}\frac{(-x^2)^k}{(x;q)_{k+1}}q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k\sigma }. \end{aligned}$$

By introducing an extra variable y, we can rewrite $\phi _{\sigma }(x)$ as below and then reformulate it by Heine’s transformation (cf. Rahman [9, III-2]):

$$\begin{aligned} \phi _{\sigma }(x)&\Rightarrow ~\frac{1}{1-x} {_2\phi _1}\left[ \begin{array}{c}q,y\\ qx\end{array}{\Big |\frac{q^{\sigma }x^2}{y}}\right] \qquad \boxed {y\rightarrow \infty }\\&\Rightarrow ~\frac{(q^{1+\sigma }x^2/y;q)_{\infty }}{(q^{\sigma }x^2/y;q)_{\infty }} {_2\phi _1}\left[ \begin{array}{c}q,q^{\sigma }x\\ q^{1+\sigma }x^2/y\end{array}{\Big |q;x}\right] . \end{aligned}$$

Then the desired generating function follows by letting $y\rightarrow \infty$.

Consider the bilateral series

$$\begin{aligned} \psi _{\sigma }(x)&:=\sum _{k=-\infty }^{\infty }x^k(q^{\sigma }x;q)_k\\&=\phi _{\sigma }(x)+\sum _{k=-\infty }^{-1}x^k(q^{\sigma }x;q)_k\\&=\phi _{\sigma }(x)+\sum _{k=1}^{\infty }(-1)^k \frac{q^{\left( {\begin{array}{c}k+1\\ 2\end{array}}\right) -k\sigma }x^{-2k}}{(q/x;q)_k}. \end{aligned}$$

Therefore, $\psi _\sigma (x)$ coincides with $\phi _\sigma (x)$ for the coefficients of nonnegative powers of the variable x when they are expanded into Laurent and Maclaurin series.

Applying Ramanujan’s identity for the $_1\psi _1$-series, we get

$$\begin{aligned} \psi _\sigma (x)&={_1\psi _1}\left[ \begin{array}{c}q^{\sigma }x\\ 0\end{array}{\Big |q;x}\right] =\left[ \begin{array}{c}q,q^{\sigma }x^2,q^{1-\sigma }/x^2\\ x,~q^{1-\sigma }/x\end{array}{\Big |q}\right] _{\infty }\\&={\left\{ \begin{array}{ll} \psi (x)\dfrac{(q^{1-\sigma }/x^2;q)_{\sigma }}{(x^2;q)_{\sigma }(q^{1-\sigma }/x;q)_{\sigma }},&{}\sigma >0;\\ \psi (x)\dfrac{(q/x;q)_{-\sigma }(q^{\sigma }x^2;q)_{-\sigma }}{(q/x^2;q)_{-\sigma }},&{}\sigma <0. \end{array}\right. }. \end{aligned}$$

This can be simplified into the following expression

$$\begin{aligned} \psi _\sigma (x)={_1\psi _1}\left[ \begin{array}{c}q^{\sigma }x\\ 0\end{array}{\Big |q;x}\right] ={\left\{ \begin{array}{ll} \psi (x)x^{-\sigma }(x;q)^{-1}_{\sigma },&{}\sigma >0;\\ \psi (x)x^{-\sigma }(q^{\sigma }x;q)_{-\sigma },&{}\sigma <0. \end{array}\right. } \end{aligned}$$

When $\sigma <0$, according to the q-binomial expansion

$$\begin{aligned} x^{-\sigma }(q^{\sigma }x;q)_{-\sigma } =\sum _{k=0}^{-\sigma }(-1)^k\left[ \begin{array}{c}-\sigma \\ k\end{array}\right] q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k\sigma }x^{k-\sigma } \end{aligned}$$

we find the explicit expression.

Theorem 2

($\sigma <0$)

$$\begin{aligned} \varOmega _m(0,\sigma ) =\sum _{k=0}^{-\sigma }(-1)^k\left[ \begin{array}{c}-\sigma \\ k\end{array}\right] q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k\sigma }\varOmega _{\sigma +m-k}(0,0). \end{aligned}$$

The first example reads as follows:

$$\begin{aligned} \varOmega _m(0,-1)={\left\{ \begin{array}{ll} (-1)^nq^{\frac{3n^2-5n}{2}},&{}m=3n;\\ (-1)^nq^{\frac{3n^2-n}{2}},&{}m=3n+1;\\ (-1)^nq^{\frac{3n^2-7n}{2}}(q-q^{1+n}),&{}m=3n-1. \end{array}\right. } \end{aligned}$$

2.3 $\Omega _n(0,\sigma )$ with $\sigma >0$

However, the above method does not work for the case $\sigma >0$ because we would encounter an expression of infinite terms. In this case, we have to make attempts by a different approach. Consider the difference

$$\begin{aligned} \varOmega _{m}(0,\sigma )-\varOmega _{m-1}(0,\sigma )&=\sum _{k=0}^{\lfloor {\frac{m}{2}}\rfloor }(-1)^k \Bigg \{\left[ \begin{array}{c}m-k\\ k\end{array}\right] -\left[ \begin{array}{c}m-k-1\\ k\end{array}\right] \Bigg \} q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k\sigma }\\&=\sum _{k=0}^{\lfloor {\frac{m}{2}}\rfloor }(-1)^k (1-q^k)\left[ \begin{array}{c}m-k-1\\ k\end{array}\right] q^{m-2k+\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k\sigma }. \end{aligned}$$

Replacing the summation index k by $k+1$ and then simplifying the resultant expression, we find the recurrence relation

$$\begin{aligned} \varOmega _{m}(0,\sigma )-\varOmega _{m-1}(0,\sigma ) +q^{m+\sigma -2}\varOmega _{m-2}(0,\sigma -1)=0. \end{aligned}$$

(3)

For $\sigma =1$, we have immediately

$$\begin{aligned} \varOmega _{m}(0,1)=1-\sum _{k=2}^mq^{k-1}\varOmega _{k-2}(0,0) \end{aligned}$$

which can be stated explicitly as (see Liu [10, Theorem 1.1])

$$\begin{aligned} \sum _{k=0}^{\lfloor {\frac{n}{2}}\rfloor }\;(-1)^k\left[ \begin{array}{c}n-k\\ k\end{array}\right] q^{\left( {\begin{array}{c}k+1\\ 2\end{array}}\right) } =\sum _{k=-\lfloor {\frac{n+1}{3}}\rfloor }^{\lfloor {\frac{n}{3}}\rfloor }(-1)^kq^{\frac{3k^2+k}{2}}. \end{aligned}$$

Next for $\sigma =2$, we can proceed recursively with

$$\begin{aligned} \varOmega _{m}(0,2)&=1-\sum _{k=2}^mq^{k}\varOmega _{k-2}(0,1),\\&=1-\sum _{k=2}^mq^{k}\bigg \{1- \sum _{i=2}^{k-2}q^{i-1}\varOmega _{i-2}(0,0)\bigg \}\\&=1-\sum _{k=2}^mq^{k}+\sum _{k=2}^mq^{k} \sum _{i=2}^{k-2}q^{i-1}\varOmega _{i-2}(0,0)\\&=1-\sum _{k=2}^mq^{k}+\sum _{i=2}^{m-2} \varOmega _{i-2}(0,0)\sum _{k=2+i}^{m}q^{k+i-1} \end{aligned}$$

which yields the expression

$$\begin{aligned} \varOmega _{m}(0,2) =1-\frac{q^2-q^{m+1}}{1-q} +\sum _{i=2}^{m-2} \frac{q^{1+2i}-q^{m+i}}{1-q} \varOmega _{i-2}(0,0). \end{aligned}$$

In general, by making use of the induction principle on $\sigma$, we can show the following analytic formula.

Theorem 3

($\sigma >0$)

$$\begin{aligned} \begin{aligned}\varOmega _{m}(0,\sigma )&=\sum _{i=0}^{\min \{\sigma -1,\lfloor {\frac{m}{2}}\rfloor \}} (-1)^i\left[ \begin{array}{c}m-i\\ i\end{array}\right] q^{\left( {\begin{array}{c}i\\ 2\end{array}}\right) +i\sigma }\\&\quad +(-1)^{\sigma }\sum _{j=0}^{m-2\sigma } q^{\frac{\sigma }{2}(3\sigma +2j-1)}\left[ \begin{array}{c}m-j-\sigma -1\\ \sigma -1\end{array}\right] \varOmega _{j}(0,0). \end{aligned} \end{aligned}$$

2.4 $\Omega _n(\lambda ,\sigma )$ with $\lambda >0$

For $\lambda \in \mathbb {N}$, recall again the q-binomial expansion

$$\begin{aligned} (q^{1+m-k};q)_{\lambda }=\sum _{i=0}^{\lambda } (-1)^i\left[ \begin{array}{c}\lambda \\ i\end{array}\right] q^{\left( {\begin{array}{c}i+1\\ 2\end{array}}\right) +mi-ki}. \end{aligned}$$

We can express $\varOmega _m(\lambda ,\sigma )$ as

$$\begin{aligned} \varOmega _m(\lambda ,\sigma )&=\sum _{k=0}^{\lfloor {\frac{m}{2}}\rfloor } (-1)^k\left[ \begin{array}{c}m-k\\ k\end{array}\right] (q^{1+m-k};q)_{\lambda } q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k\sigma }\\&=\sum _{k=0}^{\lfloor {\frac{m}{2}}\rfloor } (-1)^k\left[ \begin{array}{c}m-k\\ k\end{array}\right] q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k\sigma } \sum _{i=0}^{\lambda }(-1)^i\left[ \begin{array}{c}\lambda \\ i\end{array}\right] q^{\left( {\begin{array}{c}i+1\\ 2\end{array}}\right) +mi-ki}\\&=\sum _{i=0}^{\lambda }(-1)^i\left[ \begin{array}{c}\lambda \\ i\end{array}\right] q^{\left( {\begin{array}{c}i+1\\ 2\end{array}}\right) +mi} \sum _{k=0}^{\lfloor {\frac{m}{2}}\rfloor } (-1)^k\left[ \begin{array}{c}m-k\\ k\end{array}\right] q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k(\sigma -i)}. \end{aligned}$$

This is highlighted in the following theorem.

Theorem 4

($\lambda >0$)

$$\begin{aligned} \varOmega _m(\lambda ,\sigma ) =\sum _{i=0}^{\lambda }(-1)^i \left[ \begin{array}{c}\lambda \\ i\end{array}\right] q^{\left( {\begin{array}{c}i+1\\ 2\end{array}}\right) +mi} \varOmega _m(0,\sigma -i). \end{aligned}$$

The two simplest examples are given by

$$\begin{aligned} \varOmega _m(1,0)~&={\left\{ \begin{array}{ll} (-1)^nq^{\frac{n(3n-1)}{2}}(1-q^{n+1}),&{}m=3n;\\ (-1)^nq^{\frac{n(3n+1)}{2}}(1-q^{2n+2}),&{}m=3n+1;\\ (-1)^nq^{\frac{n(3n+1)}{2}}(q-q^{1-n}),&{}m=3n-1; \end{array}\right. }\\ \varOmega _m(1,-1)&={\left\{ \begin{array}{ll} (-1)^nq^{\frac{3n^2-5n}{2}}\Big \{1-q^{2n+1}(1+q-q^{n+1})\Big \},&{}m=3n;\\ (-1)^nq^{\frac{3n^2-n}{2}}\Big \{(1+q)(1-q^{n+1})\Big \},&{}m=3n+1;\\ (-1)^nq^{\frac{3n^2-7n}{2}}\Big \{q-q^{1+n}(1+q-q^{2n+1})\Big \},&{}m=3n-1. \end{array}\right. } \end{aligned}$$

We record also a non-trivial example

$$\begin{aligned} \varOmega _m(1,1)&=\varOmega _m(0,1)-q^{1+m}\varOmega _m(0,0)\\&=\sum _{k=-\lfloor {\frac{m+1}{3}}\rfloor }^{\lfloor {\frac{m}{3}}\rfloor }(-1)^kq^{\frac{3k^2+k}{2}} +(-1)^{\lfloor {\frac{m}{3}}\rfloor }q^{\frac{m^2-m}{6}}\chi (m\not \equiv _32) \end{aligned}$$

whose limiting case $m\rightarrow \infty$ reduces again to Euler’s pentagonal identity.

2.5 Linearization Method for $\lambda <0$

In order to avoid visual confusions, make replacement ${\lambda \rightarrow -\rho }$. Recall the q-Dougall sum (cf. [9, II-21])

$$\begin{aligned} {_6\phi _5}\left[ \begin{array}{rc}q^{-1}/x,\pm \sqrt{q/x},&{}T^{-1},~y,~q^{-\rho }\\ \pm 1/\sqrt{qx},&{}T/x,1/xy,q^{\rho }/x\end{array}{\Big |q;\frac{q^{\rho }T}{xy}}\right] =\frac{(T/xy;q)_{\rho }(1/x;q)_{\rho }}{(T/x;q)_{\rho }(1/xy;q)_{\rho }}. \end{aligned}$$

Letting $T\rightarrow 0$, $x\rightarrow q^m$ and $y\rightarrow q^{-k}$, we can express equivalently the resultant equality as

$$\begin{aligned} 1\equiv \sum _{i=0}^{\rho } q^{(i-k)(i-\rho )}\left[ \begin{array}{c}\rho \\ i\end{array}\right] \frac{1-q^{1+m-2i}}{\langle {q^{1+m-i};q}\rangle _{1+\rho }} \langle {q^k;q}\rangle _i(q^{1+m-k-\rho };q)_{\rho -i}. \end{aligned}$$

Then for $\rho \in \mathbb {N}$, we can manipulate, by putting the above relation inside the $\varOmega _m$-sum, the double series

$$\begin{aligned} \varOmega _m(-\rho ,\sigma )&=\sum _{k=0}^{\lfloor {\frac{m}{2}}\rfloor } (-1)^k\left[ \begin{array}{c}m-k\\ k\end{array}\right] (q^{1+m-k};q)_{-\rho } q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k\sigma }\\&=\sum _{k=0}^{\lfloor {\frac{m}{2}}\rfloor } \frac{(-1)^k\langle {q^{m-k};q}\rangle _k}{(q;q)_k\langle {q^{m-k};q}\rangle _{\rho }} q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k\sigma } \sum _{i=0}^{\rho } q^{(i-k)(i-\rho )}\left[ \begin{array}{c}\rho \\ i\end{array}\right] \\&\quad \times \frac{1-q^{1+m-2i}}{\langle {q^{1+m-i};q}\rangle _{1+\rho }} \langle {q^k;q}\rangle _i(q^{1+m-k-\rho };q)_{\rho -i}\\&=\sum _{i=0}^{\rho }\left[ \begin{array}{c}\rho \\ i\end{array}\right] \frac{1-q^{1+m-2i}}{\langle {q^{1+m-i};q}\rangle _{1+\rho }}\\&\qquad \times \sum _{k=i}^{\lfloor {\frac{m}{2}}\rfloor }(-1)^k \frac{\langle {q^{m-k-i};q}\rangle _{k-i}}{(q;q)_{k-i}} q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k\sigma +(i-k)(i-\rho )}\\&=\sum _{i=0}^{\rho }\left[ \begin{array}{c}\rho \\ i\end{array}\right] \frac{1-q^{1+m-2i}}{\langle {q^{1+m-i};q}\rangle _{1+\rho }}\\&\qquad \times \sum _{j=0}^{\lfloor {\frac{m-2i}{2}}\rfloor }(-1)^{i+j} \left[ \begin{array}{c}m-2i-j\\ j\end{array}\right] q^{\left( {\begin{array}{c}i+j\\ 2\end{array}}\right) +\sigma (i+j)-j(i-\rho )}\\&=\sum _{i=0}^{\rho }(-1)^i\left[ \begin{array}{c}\rho \\ i\end{array}\right] \frac{1-q^{1+m-2i}}{\langle {q^{1+m-i};q}\rangle _{1+\rho }} q^{\left( {\begin{array}{c}i\\ 2\end{array}}\right) +i\sigma }\\&\quad \times \sum _{j=0}^{\lfloor {\frac{m-2i}{2}}\rfloor }(-1)^{j} \left[ \begin{array}{c}m-2i-j\\ j\end{array}\right] q^{\left( {\begin{array}{c}j\\ 2\end{array}}\right) +j(\rho +\sigma )}. \end{aligned}$$

Hence, we have established the explicit formula.

Theorem 5

($\rho >0$)

$$\begin{aligned} \varOmega _m(-\rho ,\sigma ) =\sum _{i=0}^{\rho }(-1)^i\left[ \begin{array}{c}\rho \\ i\end{array}\right] \frac{1-q^{1+m-2i}}{\langle {q^{1+m-i};q}\rangle _{1+\rho }} q^{\left( {\begin{array}{c}i\\ 2\end{array}}\right) +i\sigma } ~\varOmega _{m-2i}(0,\rho +\sigma ). \end{aligned}$$

A couple of simple formulae are given below, where the first one is due to Liu [10, Lemma 2.3]:

$$\begin{aligned} \varOmega _m(-1,0)~&={\left\{ \begin{array}{ll} (-1)^nq^{\frac{n(3n-1)}{2}}\dfrac{1+q^n}{1-q^m},&{}m=3n;\\ (-1)^nq^{\frac{n(3n+1)}{2}}\dfrac{1}{1-q^m},&{}m=3n+1;\\ (-1)^nq^{\frac{n(3n-1)}{2}}\dfrac{1}{1-q^m},&{}m=3n-1; \end{array}\right. }\\ \varOmega _m(-1,-1)&={\left\{ \begin{array}{ll} (-1)^nq^{\frac{3n^2-5n}{2}}\dfrac{1+q^{2n}}{1-q^{3n}},&{}m=3n;\\ (-1)^nq^{\frac{3n^2+n}{2}}\dfrac{1}{1-q^{3n+1}},&{}m=3n+1;\\ (-1)^nq^{\frac{3n^2-n}{2}}\dfrac{q^{1-3n}}{1-q^{3n-1}},&{}m=3n-1. \end{array}\right. } \end{aligned}$$

3 Finite Forms of Euler’s Pentagonal Theorem

By making use of reciprocal relations

$$\begin{aligned} \left[ \begin{array}{c}m-k\\ k\end{array}\right] \Big |_{q\rightarrow q^{-1}}&=\left[ \begin{array}{c}m-k\\ k\end{array}\right] q^{k(2k-m)},\\ (q^{1+m-k};q)_{\lambda }\Big |_{q\rightarrow q^{-1}}&=(-1)^{\lambda }(q^{1+m-k};q)_{\lambda } q^{-\left( {\begin{array}{c}\lambda +1\\ 2\end{array}}\right) -\lambda (m-k)}; \end{aligned}$$

and then writing $m=3n+\varepsilon$ with $n\in \mathbb {N}_0$ and $\varepsilon =0,1,2$, we find that under the base change $q\rightarrow q^{-1}$, the q-binomial sum $\varOmega _m(\lambda ,\sigma )$ becomes

$$\begin{aligned}&\varOmega _{3n+\varepsilon }(\lambda ,\sigma )\Big |_{q\rightarrow q^{-1}} =(-1)^{\lambda }q^{-\left( {\begin{array}{c}\lambda +1\\ 2\end{array}}\right) -\lambda (3n+\varepsilon )}\\&\quad \times \sum _{k=0}^{\lfloor {\frac{3n+\varepsilon }{2}}\rfloor } (-1)^k\left[ \begin{array}{c}3n-k+\varepsilon \\ k\end{array}\right] (q^{1+3n-k+\varepsilon };q)_{\lambda } q^{\frac{3k^2+k}{2}+k(\lambda -\sigma -\varepsilon -3n)}. \end{aligned}$$

Replacing further k by $n+k$, we get the following transformation

$$\begin{aligned} \varOmega _{3n+\varepsilon }(\lambda ,\sigma )\Big |_{q\rightarrow q^{-1}} =(-1)^{n+\lambda } q^{n-n\sigma -\frac{\lambda +n}{2}(1+\lambda +2\varepsilon +3n)} \mathbf {W}_n(\lambda ,\lambda -\sigma -\varepsilon ,\varepsilon ), \end{aligned}$$

(4)

where $\mathbf {W}_n$ is the bilateral sum defined by

$$\begin{aligned} \mathbf {W}_n(\lambda ,\sigma ,\varepsilon ) =\sum _{k=-n}^{\lfloor {\frac{n+\varepsilon }{2}}\rfloor } (-1)^k\left[ \begin{array}{c}2n-k+\varepsilon \\ n+k\end{array}\right] (q^{1+\varepsilon +2n-k};q)_{\lambda } q^{\frac{3k^2+k}{2}+k\sigma }. \end{aligned}$$

(5)

By applying Jacobi’s triple product identity, we can evaluate the limit

$$\begin{aligned} \lim _{n\rightarrow \infty }\mathbf {W}_n(\lambda ,\sigma ,\varepsilon ) =\sum _{k=-\infty }^{\infty }\frac{(-1)^k}{(q;q)_{\infty }} q^{3\left( {\begin{array}{c}k\\ 2\end{array}}\right) +k(2+\sigma )} =\frac{\left[ q^3,q^{1-\sigma },q^{2+\sigma };q^3\right] _{\infty }}{(q;q)_{\infty }}. \end{aligned}$$

Therefore, for all the $\varepsilon =0,1,2$ and $\lambda ,\sigma \in \mathbb {Z}$ subject to $\sigma \not \equiv _31$, the q-binomial sum $\mathbf {W}_n(\lambda ,\sigma ,\varepsilon )$ results always in a finite analogue of Euler’s pentagonal number theorem that has consequently infinite number of finite forms.

By transforming the formulae from $\varOmega _m(\lambda ,\sigma )$ into $\mathbf {W}_n(\lambda ,\sigma ,\varepsilon )$, we recover the following known identities:

• Berkovich and Garvan [1]:

  $$\begin{aligned} \mathbf {W}_n(0,0,0)=\sum _{k=-n}^{\lfloor {\frac{n}{2}}\rfloor } (-1)^k\left[ \begin{array}{c}2n-k\\ n+k\end{array}\right] q^{\frac{3k^2+k}{2}}=1. \end{aligned}$$

• Warnaar [11]:

  $$\begin{aligned} \mathbf {W}_n(0,-1,1)=\sum _{k=-n}^{\lfloor {\frac{n+1}{2}}\rfloor } (-1)^k\left[ \begin{array}{c}1+2n-k\\ n+k\end{array}\right] q^{\frac{3k^2-k}{2}}=1. \end{aligned}$$

• Liu [10, Theorem 1.2]:

  $$\begin{aligned}&\mathbf {W}_n(-1,-1,0)=\sum _{k=-n}^{\lfloor {\frac{n}{2}}\rfloor } (-1)^k\left[ \begin{array}{c}2n-k\\ n+k\end{array}\right] \frac{q^{\frac{3k^2-k}{2}}}{1-q^{2n-k}} =\frac{1+q^n}{1-q^{3n}},\\&\mathbf {W}_n(-1,0,-1)=\sum _{k=-n}^{\lfloor {\frac{n-1}{2}}\rfloor } (-1)^k\left[ \begin{array}{c}2n-k-1\\ n+k\end{array}\right] \frac{q^{\frac{3k^2+k}{2}}}{1-q^{2n-k-1}} =\frac{1}{1-q^{3n-1}}. \end{aligned}$$

• By combining (4) with Theorem 2, we get, for $\sigma >0$, the explicit formula

  $$\begin{aligned}\mathbf {W}_n(0,\sigma ,0)&=\sum _{k=-n}^{\lfloor {\frac{n}{2}}\rfloor } (-1)^k\left[ \begin{array}{c}2n-k\\ n+k\end{array}\right] q^{\frac{3k^2+k}{2}+k\sigma }\\&=\sum _{i=0}^{\sigma }(-1)^i\left[ \begin{array}{c}\sigma \\ i\end{array}\right] q^{\left( {\begin{array}{c}k+1\\ 2\end{array}}\right) -\frac{2}{3}\left( {\begin{array}{c}k+\sigma +1\\ 2\end{array}}\right) +nk}\varOmega _{k+i+1}(0,0). \end{aligned}$$

  This expression should be attributed to Cigler [6], even though the factor $q^{-\frac{2}{3}\left( {\begin{array}{c}k+\sigma +1\\ 2\end{array}}\right) }$ is missing in his Theorem 1.

Six further variants are highlighted as examples below.

$$\begin{aligned} \mathbf {W}_n(0,0,1)&=\sum _{k}(-1)^k\left[ \begin{array}{c}1+2m-k\\ m+k\end{array}\right] q^{\frac{k(3k+1)}{2}}=1,\\ \mathbf {W}_n(0,0,2)&=\sum _{k}(-1)^k\left[ \begin{array}{c}2+2m-k\\ m+k\end{array}\right] q^{\frac{k(3k+1)}{2}}=1-q^{2+2m},\\ \mathbf {W}_n(0,0,3)&=\sum _{k}(-1)^k\left[ \begin{array}{c}3+2m-k\\ m+k\end{array}\right] q^{\frac{k(3k+1)}{2}}=1-q^{2m+2}(1+q+q^2)+q^{3m+4};\\ \mathbf {W}_n(0,-1,2)&=\sum _{k}(-1)^k\left[ \begin{array}{c}2+2m-k\\ m+k\end{array}\right] q^{\frac{k(3k-1)}{2}}=1-q^{m+1},\\ \mathbf {W}_n(0,-1,3)&=\sum _{k}(-1)^k\left[ \begin{array}{c}3+2m-k\\ m+k\end{array}\right] q^{\frac{k(3k-1)}{2}}=1-q^{m+1}-q^{m+2},\\ \mathbf {W}_n(0,-1,4)&=\sum _{k}(-1)^k\left[ \begin{array}{c}4+2m-k\\ m+k\end{array}\right] q^{\frac{k(3k-1)}{2}}=1-q^{m+1}(1+q+q^2)+q^{3m+5}. \end{aligned}$$

They resemble somehow Schur’s well-known formula

$$\begin{aligned} 1=\sum _{k}(-1)^k\left[ \begin{array}{c}2n\\ n+\lfloor {\frac{3k}{2}}\rfloor \end{array}\right] q^{\frac{3k^2-k}{2}}. \end{aligned}$$