A Class of New m-Multisum Rogers-Ramanujan Identities and Applications

Zhang Zhizheng,1,2, Li Xiaoqian,2

 基金资助: 国家自然科学基金.  11871258

 Fund supported: the NSFC.  11871258

Rogers-Ramanujan恒等式是分拆理论和组合学中著名的恒等式，被广泛的证明和推广.该文应用双边Bailey引理和迭代技巧建立一类新的多重和Rogers-Ramanujan恒等式.

Abstract

Rogers-Ramanujan identities are among the most famous q-series in partition theory and combinatorics, they have been proved and generalized widely. The purpose of this paper is to establish a class of new multisum Rogers-Ramanujan identities by applying the bilateral Bailey lemma and iterating technique.

Keywords： Bilateral Bailey lemma ; Shifted Bailey pair ; Multisum Rogers-Ramanujan identities

Zhang Zhizheng, Li Xiaoqian. A Class of New m-Multisum Rogers-Ramanujan Identities and Applications. Acta Mathematica Scientia[J], 2019, 39(4): 851-864 doi:

1 引言

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.

$q$ -二项式系数定义为

Jacobi三重积恒等式(参见文献[10, (1.6.1)])

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

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

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

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

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

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

在定理3.1中令$i=1, n_0 \rightarrow\infty$.

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

在推论3.1中令$k=1$.

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

在推论3.1中令$k=2$.

$m$-等价形式.

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

在推论3.1中令$k=3$.

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

在推论3.1中令$k=4$.

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

在定理3.1中令$i=3$.

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

在推论3.10中令$k=3$.

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

在推论3.10中令$k=4$.

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

在定理3.1中令$k=4, i=4$.

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

在定理3.2中令$i=2$.

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

在推论3.14中令$k=2$.

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

在推论3.14中令$k=3$.

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

在推论3.14中令$k=4$.

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

在定理3.2中令$k=4, i=4$.

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

在定理3.3中令$i=3$.

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

在推论3.25中,令$k=3$.

参考文献 原文顺序 文献年度倒序 文中引用次数倒序 被引期刊影响因子

Agarwal A , Andrew G E , Bressoud D .

The Bailey lattice

J Indian Math Soc, 1987, 51: 57- 73

Andrews G E .

An analytic generalization of the Rogers-Ramanujan identities for odd moduli

Proc National Academy of Science, USA, 1974, 71 (10): 4082- 4085

Andrews G E .

Multiple series Rogers-Ramanujan type identities

Pac J Math, 1984, 114: 267- 283

Andrews G E. q-Series:Their Development and Application in Analysis, Number Theory, Combinatorics, Physics and Computer Algebra. Providence, RI:Amer Math Soc, 1986

Andrews G E , Schilling A , Warnaar S O .

An A2 Bailey lemma and Rogers-Ramanujan-type identities

J Amer Math Soc, 1999, 12 (3): 677- 702

Bailey W N .

Identities of the Rogers-Ramanujan type

Proc London Math Soc, 2002, 66: 529- 549

Bressoud D , Ismail M , Stanton D .

Change of base in Bailey pairs

Ramanujan J, 2000, 4: 435- 453

Berkovich A , McCoy B M , Schilling A .

N=2 supersymmetry and Bailey pairs

Physical A, 1996, 228: 33- 62

Frederic J .

Shifted versions of the Bailey and WP-Poised Bailey lemmas

Ramnujan J, 2010, 23: 315- 333

Gasper G , Rahman M . Basic Hypergeometric Series. Cambridge: Cambridge University Press, 1990

Andrew V Sills .

On identities of the Rogers-Ramanujan type

Ramanujan J, 2006, 11: 403- 429

Slater L J .

A new proof of Rogers's transformations of infinite series

Proc London Math Soc, 1951, 53 (2): 460- 475

Slater L J .

Further identities of the Rogers-Ramanujan type

Proc London Math Soc, 1952, 54: 147- 167

Watson G N .

A new proof of the Rogers-Ramanujan identities

J London Math Soc, 1929, 4: 4- 9

Chu W , Yan Q .

Jacobi's triple product identity and theta functions

Math Commun, 2011, 16: 191- 203

Zhu J M , Zhang Z Z .

Some series-product identities and expansions on Dedekind's Eta function

Acta Math Sci, 2015, 35A (1): 110- 117

/

 〈 〉