Rogers-Ramanujan恒等式是分拆理论和组合学中最著名的恒等式.推导此类恒等式最常用的方法是Bailey引理, Bailey^[6], Slater^[12-13]和Andrews^[2-3]先后给出了详细的阐述.在文献[5]和[8]中,双边Bailey对被引入以及建立了相应的双边Bailey引理.在本文里,我们利用双边Bailey引理和迭代技巧得到一类新的多重Rogers-Ramanujan恒等式.

为了方便,本文总是假设$|q|<1$.我们使用如下标准的符号

$(a)_\infty=(a;q)_\infty=\prod\limits_{k=0}^{\infty}(1-aq^k), $

$(a)_m=(a;q)_m=\frac{(a;q)_\infty}{(aq^m;q)_\infty}, $

其中$a$为任意复数, $m$为任意整数.为了简便起见,我们记

$(a_1, \cdots, a_k;q)_n=(a_1;q)_n\cdots(a_k;q)_n, $

其中$n$为整数或者无穷.

$q$ -二项式系数定义为

$\left[{n\atop k}\right]_q=\frac{(q)_n}{(q)_k(q)_{n-k}}, $

我们假定$\left[{n\atop k}\right]_q=0$,当$k<0$或者$k>n$.我们再引入下述两个基本恒等式.

Jacobi三重积恒等式(参见文献[10, (1.6.1)])

(1.1) $ \begin{eqnarray}\label{1-1} \sum\limits_{n=-\infty}^{\infty}(-1)^{n}q^{n\choose 2}z^{n}=(q, z, q/z;q)_\infty \end{eqnarray} $

是重要的级数-乘积恒等式之一,在文献[9, 15-16]中有具体的应用.

$q$-Pfaff-Saalschutz求和公式(参见文献[10, (1.7.2)])

(1.2) $\begin{eqnarray}\label{1-2} _3\phi_2\left[{a, b, q^{-n}\atop c, abq^{1-n}/c};q, q\right]=\frac{(c/a, c/b;q)_n}{(c, c/ab;q)_n}.\end{eqnarray}$

文献[4]中, Andrews给出Bailey对的定义如下:若序列$(\alpha_n(a, q), \beta_n(a, q))$满足下列关系

(1.3) $\begin{eqnarray}\label{1-3}\beta_n(a, q)=\sum\limits_{j=0}^n\frac{\alpha_j(a, q)}{(q;q)_{n-j}(aq;q)_{n+j}}, \forall n\in{\Bbb N}, \end{eqnarray}$

则称$(\alpha_n(a, q), \beta_n(a, q))$是关于$(a, q)$的Bailey对.

称序列$(\alpha_n(a, q), \beta_n(a, q))$是关于$(a, q)$的双边Bailey对,若满足下列关系

(1.4) $\begin{eqnarray}\label{1-4}\beta_n(a, q)=\sum\limits_{j\leq{n}}\frac{\alpha_j(a, q)}{(q;q)_{n-j}(aq;q)_{n+j}}, \ \ \ \forall n\in{\Bbb Z}.\end{eqnarray}$

引理1.1(双边Bailey引理)  若$(\alpha_n(a, q), \beta_n(a, q))$是关于$(a, q)$的双边Bailey对,则$(\alpha_n'(a, q), \beta_n'(a, q))$也是,其中

$\alpha_n'(a, q)=\frac{(\rho_1, \rho_2)_{n}}{(aq/\rho_1, aq/\rho_2)_{n}}(\frac{aq}{\rho_1\rho_2})^n\alpha_n(a, q), $

$\beta_n'(a, q)=\sum\limits_{j\leq{n}}\frac{(\rho_1, \rho_2)_{j}(aq/\rho_1\rho_2)_{n-j}} {(aq/\rho_1, aq/\rho_2)_{n}(q)_{n-j}}(\frac{aq}{\rho_1\rho_2})^j\beta_j(a, q). $

假设序列$\alpha_n(a, q)$和$\beta_n(a, q)$满足一定的收敛条件使得两边的无穷级数都绝对收敛.

在双边Bailey引理中,令$\rho_1\rightarrow\infty$, $\rho_2\rightarrow\infty$,则我们有

${\rm (S1)}\qquad\left\{\begin{array}{lr}\alpha_n'(a, q)=a^{n}q^{n^{2}}\alpha_{n}(a, q), &(1.5)\\ \beta_n'(a, q)=\sum\limits_{k\leq{n}}\frac{a^{k}q^{k^{2}}}{(q;q)_{n-k}}\beta_{k}(a, q). \hskip 3.4cm &(1.6)\end{array}\right.$

在文献[9]中, Jouhet指出,当$a=q^{m}$, $m\in{\Bbb N}$时, (1.4)式中右边的无穷级数是有限的,因此称这样的双边Bailey对$(\alpha_n(q^{m}, q), \beta_n(q^{m}, q))$为移位Bailey对. Jouhet还给出了下述一对移位Bailey对.

引理1.2^{[9,命题2.1]}  设$m\in{\Bbb N}$, $(\alpha_n(q^{m}, q), \beta_n(q^{m}, q))$是移位Bailey对,其中

(1.7) $\begin{equation}\label{1-5}\alpha_n(q^{m}, q)=(-1)^{n}q^{n\choose2}, \end{equation}$

(1.8) $\begin{equation}\label{1-6}\beta_n(q^{m}, q)=(-1)^{n}q^{n\choose2}(q)_m\left[{m+n\atop m+2n}\right]_q.\end{equation}$

在文献[7]中, Bressoud, Ismail和Stanton提出了几个新的Bailey链.在这一节里,我们给出文献[7,定理2.2]和[7,定理2.3]的下列双边推广.

定理2.1  若$({\alpha}_{n}(a, q), {\beta}_{n}(a, q))$是关于$(a, q)$的Bailey对,则$({\alpha}_{n}'(a, q), {\beta}_{n}'(a, q))$是关于$(a^4, q^4)$的Bailey对,其中

(2.1) $\begin{equation}\label{2-1}\beta_n'(a, q)=\sum\limits_{k\leq{n}}\frac{(-Bq;q^2)_{k}(qa^2/B;q^2)_{2n-k}}{(-a^2q^2;q^2)_{2n}(a^4q^2/B^2;q^4)_{n}(q^4;q^4)_{n-k}}a^{2k}B^{-k}q^{k^2}\beta_k(a^2, q^2), \end{equation}$

(2.2) $\begin{equation}\label{2-2}\alpha_n'(a, q)=\frac{(-Bq;q^2)_n}{(-qa^2/B;q^2)_n}a^{2n}B^{-n}q^{{n^2}}\alpha_{n}(a^2, q^2), \end{equation}$

这里假定相关的级数绝对收敛.

证  假设${\alpha}_{n}(a, q)$已成立,将(2.2)式代入等式(1.4)中,然后交换求和顺序,应用$q$-Pfaff-Saalschutz求和公式(1.2)并且令$a\rightarrow -q^{-2n+2r}$, $b\rightarrow -Bq^{1+2r}$, $c\rightarrow a^2q^{4r+2}$, $q\rightarrow q^{2}$, $n\rightarrow n-r$,可得此定理.

在定理2.1中令$B\rightarrow\infty$,则有

${\rm (S2)}\qquad\left\{\begin{array}{ll}\alpha_n'(a, q)=a^{2n}q^{2{n^2}}\alpha_{n}(a^2, q^2), \\[2mm]\beta_n'(a, q)=\sum\limits_{k\leq{n}}\frac{a^{2k}q^{2{k^2}}}{(-a^2q^2;q^2)_{2n}(q^4;q^4)_{n-k}}\beta_{k}(a^2, q^2).\end{array}\right.$

在定理2.1令$B\rightarrow 0$,则有

${\rm (S3)}\qquad\left\{\begin{array}{ll}\alpha_n'(a, q)=\alpha_{n}(a^2, q^2), \\[2mm]\beta_n'(a, q)=\sum\limits_{k\leq{n}}\frac{(-1)^{n-k}q^{2{n^2}+2{k^2}-4nk}}{(-a^2q^2;q^2)_{2n}(q^4;q^4)_{n-k}}\beta_{k}(a^2, q^2).\end{array}\right.$

定理2.2  若$({\alpha}_{n}(a, q), {\beta}_{n}(a, q))$是关于$(a, q)$的双边Bailey对,则$({\alpha}_{n}'(a, q), {\beta}_{n}'(a, q))$是关于$(a^3, q^3)$的双边Bailey对,其中

${\rm (T1)}\qquad\left\{\begin{array}{lr}\label{2-3}\alpha_n'(a, q)=a^{n}q^{n^{2}}\alpha_{n}(a, q), &(2.3)\\ \beta_n'(a, q)=\frac{1}{(a^3q^3;q^3)_{2n}}\sum\limits_{k\leq{n}}\frac{(aq;q)_{3n-k}}{(q^3;q^3)_{n-k}}a^{k}q^{k^{2}}\beta_k(a, q), \hskip 1.5cm &(2.4)\end{array}\right.$

这里假定相关的级数绝对收敛.

证  同定理2.1的证明一样,应用等式(1.4)和$q$-Pfaff-Saalschutz求和公式(1.2),并且令$a\rightarrow \omega q^{-n+r}$, $b\rightarrow \omega^2q^{-n+r}$, $c\rightarrow aq^{2r+1}$, $n\rightarrow n-r$,这里$\omega$是三次单位原根,化简即得.

在这一节里,我们从移位Bailey对(1.7)和(1.8)式出发,利用文献[1]中的迭代技巧去得到一系列新的多重Rogers-Ramanuan恒等式.

定理3.1  对所有的$k\in{{\Bbb N}^{*}}$, $m\in{\Bbb N}$, $1\leq i\leq k$,我们有

(3.1) $\begin{eqnarray}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_k\leq n_{k-1}\leq\cdots \leq n_1}}\frac{q^{{2{n^{2}_1}}+\cdots+{2{n^{2}_{i-1}}}+{{n^{2}_i}}+\cdots+{{n^{2}_k}}+m(2{n_1}+\cdots+2{n_{i-1}}+n_i+\cdots+n_k)}}{(q^2;q^2)_{n_1-n_2}\cdots(q^2;q^2)_{n_{i-1}-n_{i}}(q)_{n_{i}-n_{i+1}}\cdots(q)_{n_{k-1}-n_k}}\nonumber\\&&\times\frac{1}{(-q;q)_{2{n_{i-1}}+m}}(-1)^{n_{k}}q^{n_{k}\choose2}\left[{m+n_{k}\atop m+2n_{k}}\right]_q\nonumber\\&=&\frac{(q^{2k+2i-1}, q^{(k+i-1)(m+1)}, q^{(k+i-1)(1-m)+1};q^{2k+2i-1})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  若Bailey链表示如下

$({\alpha}^{(k)}, {\beta}^{(k)})\rightarrow\cdots\rightarrow({\alpha}^{(i)}, {\beta}^{(i)})\rightarrow({\alpha}^{(i-1)}, {\beta}^{(i-1)})\rightarrow\ldots\rightarrow({\alpha}^{(0)}, {\beta}^{(0)}), $

其中${\alpha}^{(j)}={\alpha}_n^{(j)}, {\beta}^{(j)}={\beta}_n^{(j)}$.迭代(S1) $i-1$ ($i\geq1$)次,并且将其中的$a$替换成$q^{4m}$,则有

(3.2) $\begin{eqnarray}\label{lattice1-1}{\alpha}_n^{(0)}(q^{m}, q)=q^{4(i-1)n^2+4(i-1)mn}{\alpha}_n^{(i-1)}(q^m, q), \end{eqnarray} $

(3.3) $\begin{eqnarray}\label{lattice1-2} {\beta}_n^{(0)}(q^m, q)=\sum\limits_{{n_{i-1}\leq n_{i-2}\leq\cdots\leq n_1\leq n}} \frac{q^{4(n^{2}_1+\cdots+n^{2}_{i-1})+4m(n_1+\cdots+n_{i-1})}}{(q^4;q^4)_{n-n_{1}}\cdots(q^4;q^4)_{n_{i-2}-n_{i-1}}} {\beta}_{n_{i-1}}^{(i-1)}(q^m, q). \end{eqnarray} $

应用(S2),可以得到

(3.4) $\begin{eqnarray}\label{lattice1-3} {\alpha}_n^{(i-1)}(q^{m}, q)=q^{2{n^2}+2mn}{\alpha}_n^{(i)}(q^{2m}, q^2), \end{eqnarray}$

(3.5) $ \begin{eqnarray}\label{lattice1-4} {\beta}_{n_{i-1}}^{(i-1)}(q^m, q)=\sum\limits_{{n_{i}\leq n_{i-1}}} \frac{q^{2{n^{2}_{i}}+2mn_{i}}}{(q^4;q^4)_{n_{i-1}-n_{i}}(-q^{2+2m};q^2)_{2n_{i-1}}} {\beta}_{n_{i}}^{(i)}(q^{2m}, q^2). \end{eqnarray}$

然后再应用(S1) $k-i$次,并且令$q \rightarrow q^2$,有

(3.6) $\begin{eqnarray}\label{lattice1-5} {\alpha}_n^{(i)}(q^{2m}, q^2)=q^{2(k-i)n^2+2(k-i)mn}{\alpha}_n^{(k)}(q^{2m}, q^2), \end{eqnarray} $

(3.7) $\begin{eqnarray}\label{lattice1-6} {\beta}_{n_i}^{(i)}(q^{2m}, q^2)=\sum\limits_{{n_{k}\leq n_{k-1}\leq\cdots\leq n_{i}}} \frac{q^{2(n^{2}_{i+1}+\cdots+n^{2}_{k})+2m(n_{i+1}+\cdots+n_{k})}}{(q^2;q^2)_{n_{i}-n_{i+1}}\cdots(q^2;q^2)_{n_{k-1}-n_{k}}} {\beta}_{n_{k}}^{(k)}(q^{2m}, q^2). \end{eqnarray} $

联立等式(3.2)-(3.7),有

(3.8) $\begin{eqnarray}\label{lattice1-7} {\alpha}_n^{(0)}(q^{m}, q)=q^{(2k+2i-2)n^2+(2k+2i-2)mn}{\alpha}_n^{(k)}(q^{2m}, q^2), \end{eqnarray}$

(3.9) $ \begin{eqnarray} &&{\beta}_n^{(0)}(q^m, q)\nonumber\\&=&\sum\limits_{{n_{k}\leq n_{k-1}\leq\cdots\leq n_1\leq n}} \frac{q^{4(n^{2}_1+\cdots+n^{2}_{i-1})+2(n^{2}_{i}+\cdots+n^{2}_{k})+4m(n_1+\cdots+n_{i-1})+2m(n_{i}+\cdots+n_{k})}} {(q^4;q^4)_{n-n_{1}}\cdots(q^4;q^4)_{n_{i-1}-n_{i}}(q^2;q^2)_{n_{i}-n_{i+1}}\cdots(q^2;q^2)_{n_{k-1}-n_{k}}} \nonumber\\\label{lattice1-8} &&\times\frac{1}{(-q^{2+2m};q^2)_{2n_{i-1}}}{\beta}_{n_{k}}^{(k)}(q^{2m}, q^2). \end{eqnarray} $

将移位Bailey对(1.7)和(1.8)式代入(3.8)-(3.9)式中,然后将得到的移位Bailey对代入等式(1.4),并将(1.4)式中的$a$替换成$q^{4m}$, $q$替换成$q^4$,最后令$n\rightarrow\infty$, $q \rightarrow q^{1/2}$,定理得证.

推论3.1^{[9,定理2.3]}  对所有的$k\in{\Bbb N^{*}}$和$m\in{\Bbb N}$,我们有

(3.10) $\begin{eqnarray}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_k\leq n_{k-1}\leq\cdots \leq n_1}}\frac{q^{n^{2}_{1}+\cdots+{n^{2}_k}+m(n_1+\cdots+n_k)}}{(q)_{n_1-n_2}\cdots(q)_{n_{k-1}-n_k}}(-1)^{n_{k}}q^{n_{k}\choose2}\left[{m+n_{k}\atop m+2n_{k}}\right]_q\nonumber\\\label{3-2}&=&\frac{(q^{2k+1}, q^{k(m+1)}, q^{k(1-m)+1};q^{2k+1})_\infty}{(q)_{\infty}}.\end{eqnarray}$

证  在定理3.1中令$i=1, n_0 \rightarrow\infty$.

注3.1  在推论3.1中令$m=0$,则对所有的$k\in{\Bbb N^{*}}$,我们有

$\begin{eqnarray*}\sum\limits_{{n_{k-1}\leq\cdots \leq n_1}}\frac{q^{n^{2}_{1}+\cdots+{n^{2}_k}}}{(q)_{n_1-n_2}\cdots(q)_{n_{k-1}}}=\frac{(q^{2k+1}, q^{k}, q^{k+1};q^{2k+1})_\infty}{(q)_{\infty}}.\end{eqnarray*}$

该恒等式是文献[2,定理1] $i=k$时的另一种表达形式.

推论3.2^{[9,推论2.5]}  对所有的$m\in{\Bbb N}$,有

(3.11) $\begin{eqnarray}\label{3-5}\sum\limits_{j=0}^{\lfloor m/2 \rfloor}(-1)^{j}q^{{j\choose 2}}\left[{m-j\atop j}\right]_{q}=\left\{ \begin{array}{ll} 0, & \hbox{if $m\equiv2$ (mod 3), } \\ (-1)^{\lfloor m/3 \rfloor}q^{m(m-1)/6}, & \hbox{if $m \not\equiv2$ (mod 3).} \end{array} \right.\end{eqnarray}$

证  在推论3.1中令$k=1$.

推论3.3^{[9,推论2.7]}  对所有的$m\in{\Bbb N}$,有

(3.12) $\begin{eqnarray}\label{3-6}\sum\limits_{{j\geq 0}}(-1)^{j}q^{5{j\choose2}-(2m-3)j}\left[{m-j\atop j}\right]_{q}\sum\limits_{{k\geq 0}}\frac{q^{{k^2}+(m-2j)k}}{(q;q)_k}=\frac{(q^5, q^{2+2m}, q^{3-2m};q^5)_\infty}{(q;q)_\infty}.\end{eqnarray}$

证  在推论3.1中令$k=2$.

注3.2  等式(3.12)是第一Rogers-Ramanujan恒等式

$\sum\limits_{n=0}^{\infty}\frac{q^{n^2}}{(q;q)_n}=\frac{(q^5, q^2, q^3;q^5)_{\infty}}{(q;q)_{\infty}}$

的$m$-等价形式.

推论3.4  对所有的$m\in{\Bbb N}$,有

(3.13) $\begin{equation}\label{3-8}\sum\limits_{{-\lfloor m/2 \rfloor\leq n_3\leq n_{2}\leq n_1}}\frac{q^{n^{2}_{1}+{n^{2}_{2}}+{n^{2}_3}+m(n_1+n_2+n_3)}}{(q)_{n_1-n_2}(q)_{n_2-n_3}}(-1)^{n_{3}}q^{n_{3}\choose2}\left[{m+n_{3}\atop m+2n_{3}}\right]_q=\frac{(q^{7}, q^{3m+3}, q^{4-3m};q^{7})_\infty}{(q)_{\infty}}.\end{equation}$

证  在推论3.1中令$k=3$.

在(3.13)式中,当$m=0$时,令$n_1\rightarrow m$, $n_2\rightarrow n$,则有(参见文献[2, p.4083, (1.8)])

$ \begin{eqnarray*} \sum\limits_{{m , n\geq 0}}\frac{q^{{m^2+2mn+2{n^2}}}}{(q;q)_{m}(q;q)_{n}} =\frac{(q^7, q^{3}, q^{4};q^7)_\infty}{(q;q)_\infty}. \end{eqnarray*}$

推论3.5  对所有的$m\in{\Bbb N}$,有

(3.14) $\begin{eqnarray}\label{3-11}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_4\leq n_3\leq n_{2}\leq n_1}}\frac{q^{n^{2}_{1}+{n^{2}_{2}}+{n^{2}_3}+{n^{2}_4}+m(n_1+n_2+n_3+n_4)}}{(q)_{n_1-n_2}(q)_{n_2-n_3}(q)_{n_3-n_4}}(-1)^{n_{4}}q^{n_{4}\choose2}\left[{m+n_{4}\atop m+2n_{4}}\right]_q\nonumber\\&=&\frac{(q^{9}, q^{4m+4}, q^{5-4m};q^{9})_\infty}{(q)_{\infty}}.\end{eqnarray}$

证  在推论3.1中令$k=4$.

在(3.14)式中,当$m=0$时,令$n_1\rightarrow m$, $n_2\rightarrow n$, $n_3\rightarrow k$,则有

$\begin{eqnarray*}\sum\limits_{{m, n, k\geq 0}}\frac{q^{m^2+2{n^2}+3{k^2}+2mn+2mk+4nk}}{(q;q)_{m}(q;q)_{n}(q;q)_{k}}=\frac{(q^9, q^{4}, q^{5};q^9)_\infty}{(q;q)_\infty}.\end{eqnarray*}$

推论3.6  对所有的$k\in{\Bbb N^{*}}$和$m\in{\Bbb N}$,我们有

(3.15) $\begin{eqnarray}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_k\leq n_{k-1}\leq\cdots \leq n_1}}\frac{q^{2{n^{2}_{1}}+n^{2}_{2}+\cdots+{n^{2}_k}+m(2{n_1}+n_2+\cdots+n_k)}(-1)^{n_{k}}q^{n_{k}\choose2}}{(q^2;q^2)_{n_1-n_2}(q)_{n_2-n_3}\cdots(q)_{n_{k-1}-n_k}(-q;q)_{2n_{1}+m}}\left[{m+n_{k}\atop m+2n_{k}}\right]_q\nonumber\\&=&\frac{(q^{2k+3}, q^{(k+1)(m+1)}, q^{(k+1)(1-m)+1};q^{2k+3})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  在定理3.1中令$i=2$.

推论3.7  对所有的$m\in{\Bbb N}$,有

(3.16) $\begin{eqnarray}\label{3-7}&&\sum\limits_{{j\geq 0}}\frac{(-1)^{j}q^{7{j\choose2}-(3m-4)j}}{(-q;q)_{m-2j}}\left[{m-j\atop j}\right]_{q}\sum\limits_{{k\geq 0}}\frac{q^{2{k^2}+2(m-2j)k}}{(q;q)_k(-q^{1+m-2j};q)_{2k}}\nonumber\\&=&\frac{(q^7, q^{3+3m}, q^{4-3m};q^7)_\infty}{(q^2;q^2)_\infty}.\end{eqnarray}$

证  在推论3.6中令$k=2$.

在(3.16)式中,当$m=0$时,令$n_1\rightarrow k$,则有

$\begin{eqnarray*} \sum\limits_{{k\geq 0}}\frac{q^{2{k^2}}}{(q;q)_k(-q;q)_{2k}} =\frac{(q^7, q^{3}, q^{4};q^7)_\infty}{(q^2;q^2)_\infty}. \end{eqnarray*}$

推论3.8  对所有的$m\in{\Bbb N}$,我们有

(3.17) $\begin{eqnarray}\label{3-9}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_3\leq n_{2}\leq n_1}}\frac{q^{2{n^{2}_{1}}+{n^{2}_{2}}+{n^{2}_3}+m(2{n_1}+n_2+n_3)}(-1)^{n_{3}}q^{n_{3}\choose2}}{(q^2;q^2)_{n_1-n_2}(q)_{n_2-n_3}(-q)_{2{n_1}+m}}\left[{m+n_{3}\atop m+2n_{3}}\right]_q\nonumber\\&=&\frac{(q^{9}, q^{4m+4}, q^{5-4m};q^{9})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  在推论3.6中令$k=3$.

在(3.17)式中,当$m=0$时,令$n_1\rightarrow m$, $n_2\rightarrow n$,则有

$\begin{eqnarray*} \sum\limits_{{m , n\geq 0}}\frac{q^{{2{m^2}+4mn+3{n^2}}}}{(q^2;q^2)_{m}(q;q)_{n}(-q)_{2m+2n}} =\frac{(q^9, q^{4}, q^{5};q^9)_\infty}{(q^2;q^2)_\infty}. \end{eqnarray*}$

推论3.9  对所有的$m\in{\Bbb N}$,有

(3.18) $\begin{eqnarray}\label{3-12}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_4\leq n_3\leq n_{2}\leq n_1}}\frac{q^{2{n^{2}_{1}}+{n^{2}_{2}}+{n^{2}_3}+{n^{2}_4}+m(2{n_1}+n_2+n_3+n_4)}}{(q^2;q^2)_{n_1-n_2}(q)_{n_2-n_3}(q)_{n_3-n_4}(-q)_{2{n_1+m}}}(-1)^{n_{4}}q^{n_{4}\choose2}\left[{m+n_{4}\atop m+2n_{4}}\right]_q\nonumber\\&=&\frac{(q^{11}, q^{5m+5}, q^{6-5m};q^{11})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  在推论3.6中令$k=4$.

在(3.18)式中,当$m=0$时,令$n_1\rightarrow m$, $n_2\rightarrow n$, $n_3\rightarrow k$,则有

$ \begin{eqnarray*} \sum\limits_{{m, n, k\geq 0}}\frac{q^{2{m^2}+3{n^2}+4{k^2}+4mn+4mk+6nk}}{(q^2;q^2)_{m}(q;q)_{n}(q;q)_{k}(-q)_{2{m}+2{n}}} =\frac{(q^{11}, q^{5}, q^{6};q^{11})_\infty}{(q^2;q^2)_\infty}. \end{eqnarray*}$

推论3.10  对所有的$k\in{\Bbb N^{*}}$和$m\in{\Bbb N}$,我们有

(3.19) $\begin{eqnarray}\label{3-4}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_k\leq n_{k-1}\leq\cdots \leq n_1}}\frac{q^{{2{n^{2}_1}}+{2{n^{2}_2}}+{{n^{2}_3}}+\cdots+{{n^{2}_k}}+m(2{n_1}+2{n_2}+n_3+\cdots+n_k)}}{(q^2;q^2)_{n_1-n_2}\cdots(q^2;q^2)_{n_2-n_3}(q)_{n_3-n_4}\cdots(q)_{n_{k-1}-n_k}(-q;q)_{2{n_2}+m}}\nonumber\\&&\times(-1)^{n_{k}}q^{n_{k}\choose2}\left[{m+n_{k}\atop m+2n_{k}}\right]_q\nonumber\\&=&\frac{(q^{2k+5}, q^{(k+2)(m+1)}, q^{(k+2)(1-m)+1};q^{2k+5})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  在定理3.1中令$i=3$.

推论3.11  对所有的$m\in{\Bbb N}$,有

(3.20) $\begin{eqnarray}\label{3-10}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_3\leq n_{2}\leq n_1}}\frac{q^{2{n^{2}_1}+2{n^{2}_2}+{n^{2}_3}+m(2{n_1}+2{n_2}+n_3)}(-1)^{n_{3}}q^{n_{3}\choose2}}{(q^2;q^2)_{n_1-n_2}(q^2;q^2)_{n_2-n_3}(-q)_{2{n_2}+m}}\left[{m+n_{3}\atop m+2n_{3}}\right]_q\nonumber\\&=&\frac{(q^{11}, q^{5m+5}, q^{6-5m};q^{11})_\infty}{(q^2;q^2)_{\infty}}\end{eqnarray}$

证  在推论3.10中令$k=3$.

在(3.20)式中,当$m=0$时,令$n_1\rightarrow m$, $n_2\rightarrow n$,则有

$ \begin{eqnarray*} \sum\limits_{{m , n\geq 0}}\frac{q^{{2{m^2}+4mn+4{n^2}}}}{(q^2;q^2)_{m}(q^2;q^2)_{n}(-q)_{2n}} =\frac{(q^{11}, q^{5}, q^{6};q^{11})_\infty}{(q^2;q^2)_\infty} .\end{eqnarray*}$

推论3.12  对所有的$m\in{\Bbb N}$,有

(3.21) $\begin{eqnarray}\label{3-13}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_4\leq n_3\leq n_{2}\leq n_1}}\frac{q^{2{n^{2}_{1}}+2{n^{2}_{2}}+{n^{2}_3}+{n^{2}_4}+m(2{n_1}+2{n_2}+n_3+n_4)}}{(q^2;q^2)_{n_1-n_2}(q^2;q^2)_{n_2-n_3}(q)_{n_3-n_4}(-q)_{2{n_2+m}}}(-1)^{n_{4}}q^{n_{4}\choose2}\left[{m+n_{4}\atop m+2n_{4}}\right]_q\nonumber\\&=&\frac{(q^{13}, q^{6m+6}, q^{7-6m};q^{13})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  在推论3.10中令$k=4$.

在(3.21)式中,当$m=0$时,令$n_1\rightarrow m$, $n_2\rightarrow n$, $n_3\rightarrow k$,则有

$\begin{eqnarray*} \sum\limits_{{m, n, k\geq 0}}\frac{q^{2{m^2}+4{n^2}+5{k^2}+4mn+4mk+8nk}}{(q^2;q^2)_{m}(q^2;q^2)_{n}(q;q)_{k}(-q)_{2{n}+2{k}}} =\frac{(q^{13}, q^{6}, q^{7};q^{13})_\infty}{(q^2;q^2)_\infty} .\end{eqnarray*}$

推论3.13  对所有的$m\in{\Bbb N}$,有

(3.22) $\begin{eqnarray}\label{3-14}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_4\leq n_3\leq n_{2}\leq n_1}}\frac{q^{2{n^{2}_{1}}+2{n^{2}_{2}}+2{n^{2}_3}+{n^{2}_4}+m(2{n_1}+2{n_2}+2{n_3}+n_4)}}{(q^2;q^2)_{n_1-n_2}(q^2;q^2)_{n_2-n_3}(q^2;q^2)_{n_3-n_4}(-q)_{2{n_3+m}}}\nonumber\\&&\times (-1)^{n_{4}}q^{n_{4}\choose2}\left[{m+n_{4}\atop m+2n_{4}}\right]_q\nonumber\\&=&\frac{(q^{15}, q^{7m+7}, q^{8-7m};q^{15})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  在定理3.1中令$k=4, i=4$.

在(3.22)式中,当$m=0$时,令$n_1\rightarrow m$, $n_2\rightarrow n$, $n_3\rightarrow k$,则有

$\begin{eqnarray*} \sum\limits_{{m, n, k\geq 0}}\frac{q^{2{m^2}+4{n^2}+6{k^2}+4mn+4mk+8nk}}{(q^2;q^2)_{m}(q^2;q^2)_{n}(q^2;q^2)_{k}(-q)_{2{k}}} =\frac{(q^{15}, q^{7}, q^{8};q^{15})_\infty}{(q^2;q^2)_\infty}. \end{eqnarray*}$

定理3.2  对所有$k\in{\Bbb N^{*}}$, $m\in{\Bbb N}$和$2\leq i\leq k$,我们有

(3.23) $\begin{eqnarray}\label{3-15}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_k\leq n_{k-1}\leq\cdots \leq n_1}}\frac{q^{2{n^{2}_1}+\cdots+2{n^{2}_{i-2}}+3{n^{2}_{i-1}}+{{n^{2}_i}}+\cdots+{{n^{2}_k}}+m(2{n_1}+\cdots+2{n_{i-1}}+n_{i+1}+\cdots+n_k)-2{n_{i-1}n_{i}}}}{(q^2;q^2)_{n_1-n_2}\cdots(q^2;q^2)_{n_{i-1}-n_{i}}(q)_{n_{i}-n_{i+1}}\cdots(q)_{n_{k-1}-n_k}}\nonumber\\&&\times\frac{(-1)^{n_{i-1}-n_i}}{(-q;q)_{2{n_{i-1}}+m}}(-1)^{n_{k}}q^{n_{k}\choose2}\left[{m+n_{k}\atop m+2n_{k}}\right]_q\nonumber\\&=&\frac{(q^{2k+2i-3}, q^{(k+i-2)(m+1)}, q^{(k+i-2)(1-m)+1};q^{2k+2i-3})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  若Bailey链表示如下

$({\alpha}^{(k)}, {\beta}^{(k)})\rightarrow\cdots\rightarrow({\alpha}^{(i)}, {\beta}^{(i)})\rightarrow({\alpha}^{(i-1)}, {\beta}^{(i-1)})\rightarrow\ldots\rightarrow({\alpha}^{(0)}, {\beta}^{(0)}), $

其中${\alpha}^{(j)}={\alpha}_n^{(j)}, {\beta}^{(j)}={\beta}_n^{(j)}$.迭代(S1) $i-1$($i\geq1$)次,并且将其中的$a$替换成$q^{4m}$,则有

(3.24) $\begin{eqnarray}\label{lattice2-1} {\alpha}_n^{(0)}(q^{m}, q)=q^{4(i-1)n^2+4(i-1)mn}{\alpha}_n^{(i-1)}(q^m, q), \end{eqnarray} $

(3.25) $\begin{eqnarray}\label{lattice2-2} {\beta}_n^{(0)}(q^m, q)=\sum\limits_{{n_{i-1}\leq n_{i-2}\leq\cdots\leq n_1\leq n}} \frac{q^{4(n^{2}_1+\cdots+n^{2}_{i-1})+4m(n_1+\cdots+n_{i-1})}}{(q^4;q^4)_{n-n_{1}}\cdots(q^4;q^4)_{n_{i-2}-n_{i-1}}} {\beta}_{n_{i-1}}^{(i-1)}(q^m, q). \end{eqnarray}$

应用(S3),可以得到

(3.26) $\begin{eqnarray}\label{lattice2-3} {\alpha}_n^{(i-1)}(q^{m}, q)={\alpha}_n^{(i)}(q^{2m}, q^2), \end{eqnarray} $

(3.27) $\begin{eqnarray}\label{lattice2-4} {\beta}_{n_{i-1}}^{(i-1)}(q^m, q)=\sum\limits_{{n_{i}\leq n_{i-1}}} \frac{(-1)^{n_{i-1}-n_{i}}q^{2{n^{2}_{i-1}}+2{n^{2}_{i}}-4{n_{i-1}{n_i}}}}{(q^4;q^4)_{n_{i-1}-n_{i}}(-q^{2+2m};q^2)_{2n_{i-1}}} {\beta}_{n_{i}}^{(i)}(q^{2m}, q^2). \end{eqnarray}$

然后应用(S1) $k-i$次,并且令$q \rightarrow q^2$,则有

(3.28) $\begin{eqnarray}\label{lattice2-5} {\alpha}_n^{(i)}(q^{2m}, q^2)=q^{2(k-i)n^2+2(k-i)mn}{\alpha}_n^{(k)}(q^{2m}, q^2), \end{eqnarray} $

(3.29) $\begin{eqnarray}\label{lattice2-6} {\beta}_{n_i}^{(i)}(q^{2m}, q^2)=\sum\limits_{{n_{k}\leq n_{k-1}\leq\cdots\leq n_{i}}} \frac{q^{2(n^{2}_{i+1}+\cdots+n^{2}_{k})+2m(n_{i+1}+\cdots+n_{k})}}{(q^2;q^2)_{n_{i}-n_{i+1}}\cdots(q^2;q^2)_{n_{k-1}-n_{k}}} {\beta}_{n_{k}}^{(k)}(q^{2m}, q^2). \end{eqnarray} $

联立等式(3.24)-(3.29)式,有

(3.30) $\begin{eqnarray}\label{lattice2-7} {\alpha}_n^{(0)}(q^{m}, q)=q^{(2k+2i-4)n^2+(2k+2i-4)mn}{\alpha}_n^{(k)}(q^{2m}, q^2), \end{eqnarray} $

(3.31) $\begin{eqnarray}\label{lattice2-8} &&{\beta}_n^{(0)}(q^m, q)\nonumber\\&=&\sum\limits_{{n_{k}\leq n_{k-1}\leq\cdots\leq n_1\leq n}} \frac{q^{4(n^{2}_1+\cdots+n^{2}_{i-2})+6{n^{2}_{i-1}}+2(n^{2}_{i}+\cdots+n^{2}_{k})+4m(n_1+\cdots+n_{i-1})+2m(n_{i+1}+\cdots+n_{k})}} {(q^4;q^4)_{n-n_{1}}\cdots(q^4;q^4)_{n_{i-1}-n_{i}}(q^2;q^2)_{n_{i}-n_{i+1}}\cdots(q^2;q^2)_{n_{k-1}-n_{k}}}\nonumber\\ &&\times\frac{(-1)^{n_{i-1}-n_{i}}q^{-4{n_{i-1}{n_i}}}}{(-q^{2+2m};q^2)_{2n_{i-1}}}{\beta}_{n_{k}}^{(k)}(q^{2m}, q^2). \end{eqnarray}$

将(1.7)和(1.8)式代入(3.30)-(3.31)式中,然后将得到的移位Bailey对代入等式(1.4),并且将等式(1.4)中的$a$替换成$q^{4m}$, $q$替换成$q^4$,最后令$n\rightarrow\infty$, $q \rightarrow q^{1/2}$.定理得证.

推论3.14  对所有的$k\in{\Bbb N^{*}}$和$m\in{\Bbb N}$,有

(3.32) $\begin{eqnarray}\label{3-16}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_k\leq n_{k-1}\leq\cdots \leq n_1}}\frac{(-1)^{n_1-n_2+n_k}q^{3{n^{2}_{1}}+{{n^{2}_2}}+\cdots+{{n^{2}_k}}+m(2{n_1}+n_2+\cdots+n_k)-2{n_1n_2}}}{(q^2;q^2)_{n_1-n_2}(q)_{n_{2}-n_3}\cdots(q)_{n_{k-1}-n_k}(-q;q)_{2{n_1}+m}}\nonumber\\&&\times q^{n_{k}\choose2}\left[{m+n_{k}\atop m+2n_{k}}\right]_q\nonumber\\&=&\frac{(q^{2k+1}, q^{k(m+1)}, q^{k(1-m)+1};q^{2k+1})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  在定理3.2中令$i=2$.

推论3.15  对所有的$m\in{\Bbb N}$,有

(3.33) $\begin{eqnarray}\label{3-18}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_2\leq n_1}}\frac{(-1)^{n_1}q^{3{n^{2}_{1}}+{n^{2}_2}+m(2{n_1}+n_2)-2{n_1n_2}}}{(q^2;q^2)_{n_1-n_2}(-q;q)_{2{n_1}+m}}q^{n_{2}\choose2}\left[{m+n_{2}\atop m+2n_{2}}\right]_q\nonumber\\&=&\frac{(q^{5}, q^{2m+2}, q^{3-2m};q^{5})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  在推论3.14中令$k=2$.

在(3.33)式中,当$m=0$时,令$n_1\rightarrow m$,则有

$\begin{eqnarray*} \sum\limits_{{m\geq 0}}\frac{(-1)^{m}q^{3{m^2}}}{(q^2;q^2)_m(-q;q)_{2m}} =\frac{(q^{5}, q^{2}, q^{3};q^{5})_\infty}{(q^2;q^2)_\infty}. \end{eqnarray*}$

推论3.16  对所有的$m\in{\Bbb N}$,有

(3.34) $\begin{eqnarray}\label{3-19}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_3\leq n_2\leq n_1}}\frac{(-1)^{n_1-n_2+n_3}q^{3{n^{2}_{1}}+{n^{2}_2}+{n^{2}_3}+m(2{n_1}+n_2+n_3)-2{n_1n_2}}}{(q^2;q^2)_{n_1-n_2}(q)_{n_{2}-n_3}(-q;q)_{2{n_1}+m}}q^{n_{3}\choose2}\left[{m+n_{3}\atop m+2n_{3}}\right]_q\nonumber\\&=&\frac{(q^{7}, q^{3m+3}, q^{4-3m};q^{7})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  在推论3.14中令$k=3$.

在(3.34)式中,当$m=0$时,令$n_1\rightarrow m$, $n_2\rightarrow n$,则有

$\begin{eqnarray*} \sum\limits_{{m, n\geq 0}}\frac{(-1)^{m}q^{3{m^2}+2n^2+4mn}}{(q^2;q^2)_m(q;q)_n(-q;q)_{2m+2n}} =\frac{(q^{7}, q^{3}, q^{4};q^{7})_\infty}{(q^2;q^2)_\infty} .\end{eqnarray*}$

推论3.17  对所有的$m\in{\Bbb N}$,有

(3.35) $\begin{eqnarray}\label{3-21}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_4\leq n_3\leq n_2\leq n_1}}\frac{(-1)^{n_1-n_2+n_4}q^{3{n^{2}_{1}}+{{n^{2}_2}}+{n^{2}_3}+{n^{2}_4}+m(2{n_1}+n_2+n_3+n_4)-2{n_1n_2}}}{(q^2;q^2)_{n_1-n_2}(q)_{n_{2}-n_3}(q)_{n_3-n_4}(-q;q)_{2{n_1}+m}}q^{n_{4}\choose2}\left[{m+n_{4}\atop m+2n_{4}}\right]_q\nonumber\\&=&\frac{(q^{9}, q^{4m+4}, q^{5-4m};q^{9})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  在推论3.14中令$k=4$.

在(3.35)式中,当$m=0$时,令$n_1\rightarrow m$, $n_2\rightarrow n$, $n_3\rightarrow k$,则有

$\begin{eqnarray*} \sum\limits_{{m, n, k\geq 0}} \frac{(-1)^{m}q^{{m^2}+4n^2+5k^2+4mn+4mk+8nk}}{(q^2;q^2)_m(q;q)_n(q;q)_k(-q;q)_{2m+2n}} =\frac{(q^{9}, q^{5}, q^{4};q^{9})_\infty}{(q^2;q^2)_\infty}. \end{eqnarray*}$

推论3.18  对所有的$k\in{\Bbb N^{*}}$和$m\in{\Bbb N}$,有

(3.36) $\begin{eqnarray}\label{3-17}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_k\leq n_{k-1}\leq\cdots \leq n_1}}\frac{(-1)^{n_2-n_3}q^{2{n^{2}_1}+3{n^{2}_{2}}+{{n^{2}_3}}+\cdots+{{n^{2}_k}}+m(2{n_1}+2{n_2}+n_4+\cdots+n_k)-2{n_2n_3}}}{(q^2;q^2)_{n_1-n_2}(q^2;q^2)_{n_2-n_3}(q)_{n_{3}-n_{4}}\cdots(q)_{n_{k-1}-n_k}(-q;q)_{2{n_2}+m}}\nonumber\\&&\times(-1)^{n_{k}}q^{n_{k}\choose2}\left[{m+n_{k}\atop m+2n_{k}}\right]_q\nonumber\\&=&\frac{(q^{2k+3}, q^{(k+1)(m+1)}, q^{(k+1)(1-m)+1};q^{2k+3})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  在定理3.2中令$i=3$.

推论3.19  对所有的$m\in{\Bbb N}$,有

(3.37) $\begin{eqnarray}\label{3-20}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_3\leq n_2\leq n_1}}\frac{(-1)^{n_2}q^{2{n^{2}_1}+3{n^{2}_{2}}+{n^{2}_3}+m(2{n_1}+2{n_2})-2{n_2n_3}}}{(q^2;q^2)_{n_1-n_2}(q^2;q^2)_{n_2-n_3}(-q;q)_{2{n_2}+m}}q^{n_{3}\choose2}\left[{m+n_{3}\atop m+2n_{3}}\right]_q\nonumber\\&=&\frac{(q^{9}, q^{4m+4}, q^{5-4m};q^{9})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  在推论3.18中令$k=3$.

在(3.37)式中,当$m=0$时,令$n_1\rightarrow m$, $n_2\rightarrow n$,则有

$\begin{eqnarray*} \sum\limits_{{m, n\geq 0}}\frac{(-1)^{n}q^{2{m^2}+5n^2+4mn}}{(q^2;q^2)_m(q^2;q^2)_n(-q;q)_{2n}} =\frac{(q^{9}, q^{5}, q^{4};q^{9})_\infty}{(q^2;q^2)_\infty}. \end{eqnarray*}$

推论3.20  对所有的$m\in{\Bbb N}$,有

(3.38) $\begin{eqnarray}\label{3-22}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_4\leq n_3\leq\ n_2\leq n_1}}\frac{(-1)^{n_2-n_3+n_4}q^{2{n^{2}_1}+3{n^{2}_{2}}+{{n^{2}_3}}+{n^{2}_k}+m(2{n_1}+2{n_2}+n_4)-2{n_2n_3}}}{(q^2;q^2)_{n_1-n_2}(q^2;q^2)_{n_2-n_3}(q)_{n_{3}-n_{4}}(-q;q)_{2{n_2}+m}}q^{n_{4}\choose2}\left[{m+n_{4}\atop m+2n_{4}}\right]_q\nonumber\\&=&\frac{(q^{11}, q^{5(m+1)}, q^{6-6m};q^{11})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  在推论3.18中令$k=4$.

在(3.38)式中,当$m=0$时,令$n_1\rightarrow m$, $n_2\rightarrow n$, $n_3\rightarrow k$,则有

$\begin{eqnarray*} \sum\limits_{{m, n, k\geq 0}} \frac{(-1)^{n}q^{2{m^2}+3n^2+6k^2+4mn+4mk+8nk}}{(q^2;q^2)_m(q^2;q^2)_n(q;q)_k(-q;q)_{2n+2k}} =\frac{(q^{11}, q^{5}, q^{6};q^{11})_\infty}{(q^2;q^2)_\infty}. \end{eqnarray*}$

推论3.21  对所有的$k\in{\Bbb N^{*}}$, $m\in{\Bbb N}$和$1\leq i\leq k$,我们有

(3.39) $\begin{eqnarray}\label{3-23}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_4\leq n_3\leq n_2\leq n_1}}\frac{(-1)^{n_3}q^{2{n^{2}_1}+2{n^{2}_2}+3{n^{2}_3}+{n^{2}_4}+m(2{n_1}+2{n_2}+2{n_3})-2{n_3n_4}}}{(q^2;q^2)_{n_1-n_2}(q^2;q^2)_{n_2-n_3}(q^2;q^2)_{n_3-n_4}(-q;q)_{2{n_3}+m}}q^{n_{4}\choose2}\left[{m+n_{4}\atop m+2n_{4}}\right]_q\nonumber\\&=&\frac{(q^{13}, q^{6m+6}, q^{7-6m};q^{13})_\infty}{(q^2;q^2)_{\infty}}.\end{eqnarray}$

证  在定理3.2中令$k=4, i=4$.

在(3.39)式中,当$m=0$时,令$n_1\rightarrow m$, $n_2\rightarrow n$, $n_3\rightarrow k$,则有

$\begin{eqnarray*}\sum\limits_{{m, n, k\geq 0}}\frac{(-1)^{k}q^{2{m^2}+4n^2+7k^2+4mn+4mk+8nk}}{(q^2;q^2)_m(q^2;q^2)_n(q^2;q^2)_k(-q;q)_{2k}}=\frac{(q^{13}, q^{6}, q^{7};q^{13})_\infty}{(q^2;q^2)_\infty}.\end{eqnarray*}$

定理3.3  对所有的$k\in{\Bbb N^{*}}$, $m\in{\Bbb N}$和$1\leq i\leq k$,我们有

(3.40) $\begin{eqnarray}\label{3-24}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_k\leq n_{k-1}\leq\cdots \leq n_1}}\frac{q^{{3{n^{2}_1}}+\cdots+{3{n^{2}_{i-1}}}+{{n^{2}_i}}+\cdots+{{n^{2}_k}}+m(3{n_1}+\cdots+3{n_{i-1}}+n_i+\cdots+n_k)}}{(q^3;q^3)_{n_1-n_2}\cdots(q^3;q^3)_{n_{i-1}-n_{i}}(q)_{n_{i}-n_{i+1}}\cdots(q)_{n_{k-1}-n_k}}\nonumber\\&&\times\frac{(q;q)_{3{n_{i-1}}-n_i+m}}{(q^3;q^3)_{2{n_{i-1}}+m}}(-1)^{n_{k}}q^{n_{k}\choose2}\left[{m+n_{k}\atop m+2n_{k}}\right]_q\nonumber\\&=&\frac{(q^{2k+4i-3}, q^{(k+2i-2)(m+1)}, q^{(k+2i-2)(1-m)+1};q^{2k+4i-3})_\infty}{(q^3;q^3)_{\infty}}.\end{eqnarray}$

证  若Bailey链表示如下

$({\alpha}^{(k)}, {\beta}^{(k)})\rightarrow\cdots\rightarrow({\alpha}^{(i)}, {\beta}^{(i)})\rightarrow({\alpha}^{(i-1)}, {\beta}^{(i-1)})\rightarrow\ldots\rightarrow({\alpha}^{(0)}, {\beta}^{(0)}), $

其中${\alpha}^{(j)}={\alpha}_n^{(j)}, {\beta}^{(j)}={\beta}_n^{(j)}$.迭代(S1) $i-1$($i\geq1$)次,并且将$a$替换成$q^{4m}$,则有

(3.41) $\begin{eqnarray}\label{lattice3-1}{\alpha}_n^{(0)}(q^{m}, q)=q^{3(i-1)n^2+3(i-1)mn}{\alpha}_n^{(i-1)}(q^m, q), \end{eqnarray} $

(3.42) $\begin{eqnarray}\label{lattice3-2} {\beta}_n^{(0)}(q^m, q)=\sum\limits_{{n_{i-1}\leq n_{i-2}\leq\cdots\leq n_1\leq n}} \frac{q^{3(n^{2}_1+\cdots+n^{2}_{i-1})+3m(n_1+\cdots+n_{i-1})}}{(q^3;q^3)_{n-n_{1}}\cdots(q^3;q^3)_{n_{i-2}-n_{i-1}}} {\beta}_{n_{i-1}}^{(i-1)}(q^m, q). \end{eqnarray} $

应用定理2.2,可以得到

(3.43) $ \begin{eqnarray}\label{lattice3-3} {\alpha}_n^{(i-1)}(q^{m}, q)=q^{n^2+mn}{\alpha}_n^{(i)}(q^{m}, q), \end{eqnarray} $

(3.44) $\begin{eqnarray}\label{lattice3-4} {\beta}_{n_{i-1}}^{(i-1)}(q^m, q)=\sum\limits_{{n_{i}\leq n_{i-1}}} \frac{q^{n^{2}_{i}+mn_{i}}}{(q^3;q^3)_{n_{i-1}-n_{i}}(q^{3+3m};q^3)_{2n_{i-1}}} {\beta}_{n_{i}}^{(i)}(q^{m}, q). \end{eqnarray}$

然后,应用(S1)$k-i$次,则有

(3.45) $\begin{eqnarray}\label{lattice3-5} {\alpha}_n^{(i)}(q^{m}, q)=q^{(k-i)n^2+(k-i)mn}{\alpha}_n^{(k)}(q^{m}, q), \end{eqnarray} $

(3.46) $\begin{eqnarray}\label{lattice3-6} {\beta}_{n_i}^{(i)}(q^{m}, q)=\sum\limits_{{n_{k}\leq n_{k-1}\leq\cdots\leq n_{i}}} \frac{q^{(n^{2}_{i+1}+\cdots+n^{2}_{k})+m(n_{i+1}+\cdots+n_{k})}}{(q;q)_{n_{i}-n_{i+1}}\cdots(q;q)_{n_{k-1}-n_{k}}} {\beta}_{n_{k}}^{(k)}(q^{m}, q). \end{eqnarray} $

联立等式(3.41)-(3.46),有

(3.47) $\begin{eqnarray}\label{lattice3-7} {\alpha}_n^{(0)}(q^m, q)=q^{(k+2i-2)n^2+(k+2i-2)mn}{\alpha}_n(q^m, q), \end{eqnarray}$

(3.48) $ \begin{eqnarray} &&{\beta}_n^{(0)}(q^m, q)\nonumber\\&=&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_k\leq n_{k-1}\leq\cdots \leq n_1\leq n}} \frac{q^{{3({n} ^{2}_1}+\cdots+{{n^{2}_{i-1}})}+({{n^{2}_i}+\cdots+{{n^{2}_k}})+3m(n_1+\cdots+n_{i-1})+m(n_i+\cdots+n_k)}}} {(q^3;q^3)_{n-n_1}\cdots(q^3;q^3)_{n_{i-1}-n_{i}}(q)_{n_{i}-n_{i+1}}\cdots(q)_{n_{k-1}-n_k}} \nonumber\\\label{lattice3-8} &&\times\frac{1}{(q^{3+3m};q^3)_{2n_{i-1}}}{\beta}_{n_{k}}^{(k)}(q^{m}, q). \end{eqnarray}$

将(1.7)和(1.8)式代入$(3.47)$-$(3.48)$式中,然后将得到的Bailey对代入等式(1.4),并将(1.4)式中的$a$替换成$q^{3m}$, $q$替换成$q^3$,最后令$n\rightarrow\infty$.定理得证.

注3.3  在定理3.3中令$i=1, n\rightarrow\infty$,可以得到推论3.2;在定理3.3中分别令$k=1, $$i=1$; $k=2, $$i=1$和$k=3, i=1$,可以得到恒等式(3.11), (3.12)和(3.13).

推论3.22  对所有的$k\in{\Bbb N^{*}}$和$m\in{\Bbb N}$,我们有

(3.49) $\begin{eqnarray}\label{3-25}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_k\leq n_{k-1}\leq\cdots \leq n_1}}\frac{q^{3{n^{2}_{1}}+n^{2}_{2}+\cdots+{n^{2}_k}+m(3{n_1}+n_2+\cdots+n_k)}(q;q)_{3{n_1}-n_2+m}}{(q^3;q^3)_{n_1-n_2}(q)_{n_2-n_3}\cdots(q)_{n_{k-1}-n_k}(q^3;q^3)_{2{n_1}+m}}\nonumber\\&&\times (-1)^{n_{k}}q^{n_{k}\choose2}\left[{m+n_{k}\atop m+2n_{k}}\right]_q\nonumber\\&=&\frac{(q^{2k+5}, q^{(k+2)(m+1)}, q^{(k+2)(1-m)+1};q^{2k+5})_\infty}{(q^3;q^3)_{\infty}}.\end{eqnarray}$

证  在定理3.3中令$i=2$.

推论3.23  对所有的$m\in{\Bbb N}$,有

(3.50) $\begin{eqnarray}\label{3-27}&&\sum\limits_{{j\geq 0}}\frac{(-1)^{j}q^{8{j\choose2}-(4m-4)j}(q;q)_{m-2j}}{(q^3;q^3)_{m-2j}}\left[{m-j\atop j}\right]_{q}\sum\limits_{{k\geq 0}}\frac{q^{3{k^2}+3(m-2j)k}(q^{m-2j+1};q)_{3k}}{(q^3;q^3)_k(q^{3+3m-6j};q^3)_{2k}}\nonumber\\&=&\frac{(q^9, q^{4+4m}, q^{5-4m};q^9)_\infty}{(q^3;q^3)_\infty}.\end{eqnarray}$

证  在推论3.22中令$k=2$.

在(3.50)式中,当$m=0$时,令$n_1\rightarrow k$,则有

$\begin{eqnarray*}\sum\limits_{{k\geq 0}}\frac{q^{3{k^2}}(q;q)_{3k}}{(q^3;q^3)_k(q^{3};q^3)_{2k}}=\frac{(q^{9}, q^{4}, q^{5};q^{9})_\infty}{(q^3;q^3)_\infty}.\end{eqnarray*}$

推论3.24  对所有的$m\in{\Bbb N}$,有

(3.51) $\begin{eqnarray}\label{3-28}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_3\leq n_{2}\leq n_1}}\frac{q^{3{n^{2}_{1}}+n^{2}_{2}+{n^{2}_3}+m(3{n_1}+n_2+n_3)}(q;q)_{3{n_1}-n_2+m}}{(q^3;q^3)_{n_1-n_2}(q)_{n_2-n_3}(q^3;q^3)_{2{n_1}+m}}(-1)^{n_{3}}q^{n_{3}\choose2}\left[{m+n_{3}\atop m+2n_{3}}\right]_q\nonumber\\&=&\frac{(q^{11}, q^{5+5m}, q^{6-5m};q^{11})_\infty}{(q^3;q^3)_{\infty}}.\end{eqnarray}$

证  在推论3.22中令$k=3$.

在(3.51)式中,当$m=0$时,令$n_1\rightarrow m$, $n_2\rightarrow n$,则有

$\begin{eqnarray*}\sum\limits_{{m, n\geq 0}}\frac{q^{3{m^2}+4{n^2}+6mn}(q;q)_{{3m+2n}}}{(q^3;q^3)_{m}(q;q)_n(q^3;q^3)_{2m+2n}}=\frac{(q^{11}, q^{5}, q^{6};q^{11})_\infty}{(q^3;q^3)_\infty}.\end{eqnarray*}$

推论3.25  对所有的$k\in{\Bbb N^{*}}$和$m\in{\Bbb N}$,我们有

(3.52) $\begin{eqnarray}\label{3-26}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_k\leq n_{k-1}\leq\cdots \leq n_1}}\frac{q^{3{n^{2}_{1}+3n^{2}_{2}}+n_3+\cdots+{n^{2}_k}+m(3{n_1}+3{n_2}+n_3+\cdots+n_k)}(q;q)_{3{n_2}-n_3+m}}{(q^3;q^3)_{n_1-n_2}(q^3;q^3)_{n_2-n_3}(q)_{n_3-n_4}\cdots(q)_{n_{k-1}-n_k}(q^3;q^3)_{2{n_2}+m}}\nonumber\\&&\times(-1)^{n_k}q^{n_{k}\choose2}\left[{m+n_{k}\atop m+2n_{k}}\right]_q\nonumber\\&=&\frac{(q^{2k+9}, q^{(k+4)(m+1)}, q^{(k+4)(1-m)+1};q^{2k+9})_\infty}{(q^3;q^3)_{\infty}}.\end{eqnarray}$

证  在定理3.3中令$i=3$.

推论3.26  对所有的$m\in{\Bbb N}$,有

(3.53) $\begin{eqnarray}\label{3-29}&&\sum\limits_{{-\lfloor m/2 \rfloor\leq n_3\leq n_2\leq n_1}}\frac{q^{3{n^{2}_{1}+3n^{2}_{2}}+n_3+m(3{n_1}+3{n_2}+n_3})(q;q)_{3{n_2}-n_3+m}}{(q^3;q^3)_{n_1-n_2}(q^3;q^3)_{n_2-n_3}(q^3;q^3)_{2{n_2}+m}}(-1)^{n_3}q^{n_{3}\choose2}\left[{m+n_{3}\atop m+2n_{3}}\right]_q\nonumber\\&=&\frac{(q^{15}, q^{7+7m}, q^{8-7m};q^{15})_\infty}{(q^3;q^3)_{\infty}}.\end{eqnarray}$

证  在推论3.25中,令$k=3$.

在(3.53)式中,当$m=0$时,令$n_1\rightarrow m$, $n_2\rightarrow n$,则有^{[6, (4.27)]}

$ \begin{eqnarray*} \sum\limits_{{m, n\geq 0}}\frac{q^{3{m^2}+6{n^2}+6mn}(q;q)_{{3n}}}{(q^3;q^3)_{m}(q^3;q^3)_{n}(q^3;q^3)_{2n}} =\frac{(q^{15}, q^{7}, q^{8};q^{15})_\infty}{(q^3;q^3)_\infty}. \end{eqnarray*}$