1 Introduction

In [9], Bressoud provides a stunningly elementary proof of the celebrated Rogers–Ramanujan identities [2, Chapter 7]:

$$\begin{aligned} \sum _{n=0}^\infty \frac{q^{n^2}}{(q)_n} = \frac{1}{(q,q^4;q^5)_\infty } \end{aligned}$$
(1.1)

and

$$\begin{aligned} \sum _{n=0}^\infty \frac{q^{n^2+n}}{(q)_n} = \frac{1}{(q^2,q^3;q^5)_\infty }, \end{aligned}$$
(1.2)

where

$$\begin{aligned} (A)_n = (A;q)_n = (1-A)(1-Aq)\cdots (1-Aq^{n-1}) = \prod _{j=0}^\infty \frac{(1-Aq^j)}{(1-Aq^{j+n})} \end{aligned}$$
(1.3)

and

$$\begin{aligned} \left( A_1,A_2,\dots , A_k;q\right) _N = \prod _{j=1}^k(A_j;q)_N. \end{aligned}$$
(1.4)

In fact, the Rogers–Ramanujan identities are the cases \(k=2\), \(i=1,2\) of the following general identity [1]:

$$\begin{aligned} \sum _{s_1\ge \cdots \ge s_{k-1}\ge 0}\frac{q^{s_1^2+s_2^2+\cdots \!+\! s_{k-1}^2\! +\! s_i \!+\! s_{i-1} + \cdots \!+\! s_{k-1}}}{(q)_{s_1-s_k}(q)_{s_i-s_3}\cdots (q)_{s_{k-2}-s_{k-1}}(q)_{s_{k-1}}} \!=\! \frac{(q^i,q^{2k+1-i},q^{2k+1};q^{2k\!+\!1})_\infty }{(q)_\infty }. \end{aligned}$$
(1.5)

Note that it is unnecessary to require that the \(s_j\)’s are nonincreasing because \(\tfrac{1}{(q)_M}=0\) if \(M<0\) by the final expression in (1.3).

Now in Bressoud’s paper [9], after providing his elegant, easy proof, he provides a generalization.

Bressoud’s Theorem. Given positive integers \(k\) and \(N\), we have that

$$\begin{aligned} \sum _{s_1,\dots , s_k} \frac{q^{s_1^2+s_2^2+\cdots +s_k^2}}{(q)_{N-s_1}(q)_{s_1-s_3}\cdots (q)_{s_{k-1}-s_k}(q)_{2s_k}} \prod _1^{s_k}(1 + xq^m)(1+x^{-1}q^{m-1}) \\ = (q)_{2N}^{-1}\sum _m x^m q^{((2k+1)m^2+m)/2} \begin{bmatrix} 2N\\N-m \end{bmatrix},\nonumber \end{aligned}$$
(1.6)

where

$$\begin{aligned} \begin{bmatrix} N\\ m \end{bmatrix} = \frac{(q)_N}{(q)_m(q)_{N-m}}. \end{aligned}$$
(1.7)

Bressoud notes that the case \(k=2\) is the instance from which the Rogers–Ramanujan identities follow, and he points out that the case \(x=-1\) provides a finite version of (1.5) when \(i=k\). Indeed, when \(i=k\), then (1.5) follows with \(x=-1\) and \(N\rightarrow \infty ;\) all that is necessary is to invoke Jacobi’s triple product [2, p. 22, Corollary 2.9].

If we explore (1.6) a little further, something quite surprising is in store. Namely, setting \(x=-q^{k-i}\) and letting \(N\rightarrow \infty \), we find (see Theorems 2 and 3 in Sect. 4)

$$\begin{aligned}&\sum _{s_1\ge \cdots \ge s_{k-1}\ge 0} \frac{q^{s_1^2+s_2^2+\cdots +s_{k-1}^2}B_{k-i}(q^{s_{k-1}})}{(q)_{s_1-s_2}(q)_{s_2-s_3}\cdots (q)_{s_{k-2}-s_{k-1}}(q)_{s_{k-1}}}\\&\quad = \frac{1}{(q)_\infty } \sum _{n=-\infty }^\infty (-1)^n q^{(2k+1){n+1\atopwithdelims ()2}-in}= \frac{(q^i,q^{2k+1-i},q^{2k+1};q^{2k+1})_n}{(q)_\infty },\nonumber \end{aligned}$$
(1.8)

where

$$\begin{aligned} B_n(z,q) := \sum _{j=0}^n \begin{bmatrix} n+j\\2j\end{bmatrix} (-1)^j q^{j(3j-1)/2-nj}(zq^{1-j})_j. \end{aligned}$$
(1.9)

Now the series in (1.8) is identical with the series in (1.5) when \(i=k\) and \(i=k-1\), because \(B_0(z)=1\) and \(B_1(z) = z\). However, while the right-hand sides of (1.5) and (1.8) are identical for \(1\le i\le k\), the left-hand sides differ for \(i< k-1\).

In light of the fact that (1.8) is a special case of (1.5), we shall call the \(B_n(z,q)\) Bressoud polynomials. We shall, in Sect. 2, introduce three further polynomial sequences closely related to \(B_n(z,q)\) and derive some of their basic properties.

In Sect. 3, we put the Bressoud polynomials into the theory of Bailey pairs and Bailey chains. This will be applied in Sects. 5 and 6.

In Sect. 4, we shall reconsider Bressoud’s Theorem from the viewpoint of Bailey chains and derive related results.

Section 5 will look at applications of Bressoud polynomials to Rogers–Ramanujan type identities, and Sect. 6 will consider their relationship to the mock theta functions.

The last two sections provide a discussion of open questions connected with the Bressoud polynomials.

2 Recurrences: the four families of Bressoud polynomials

In addition to the \(B_n(z,q)\) defined by (1.9), we shall also consider

$$\begin{aligned} \overline{B}_n(z,q)&= B_n(z,\tfrac{1}{q}) = \sum _{j=0}^n \begin{bmatrix} n+j\\2j\end{bmatrix} (-1)^jq^{j(j+1)/2-nj}(z)_j, \end{aligned}$$
(2.1)
$$\begin{aligned} D_n(z,q)&\sum _{j=0}^n \begin{bmatrix} n+j+1\\2j+1\end{bmatrix} (-1)^jq^{j(3j+1)/2-n(j+1)}(zq^{1-j})_j, \end{aligned}$$
(2.2)

and

$$\begin{aligned} \overline{D}_n(z,q)&= D_n(z,\tfrac{1}{q}) = \sum _{j=0}^n\begin{bmatrix}n+j+1\\2j+1\end{bmatrix}(-1)^jq^{j(j+1)/2-nj}(z)_j. \end{aligned}$$
(2.3)

The following theorem is fundamental to each application of the Bressoud polynomials:

Theorem 1

If we define for \(n<0\)

$$\begin{aligned} B_n(z,q) = \overline{B}_n(z,q) = D_n(z,q) = \overline{D}_(z,q) = 0 \end{aligned}$$
(2.4)

and

$$\begin{aligned} B_0(z,q) = \overline{B}_0(z,q) = D_0(z,q) = \overline{D}_0(z,q) = 1, \end{aligned}$$
(2.5)

then for \(n\ge 3\),

$$\begin{aligned} B_n(z,q)&= zB_{n-1}(z,q) + zq^{1-n}B_{n-2}(z,q)-q^{1-n}B_{n-3}(z,q),\end{aligned}$$
(2.6)
$$\begin{aligned} \overline{B}_n(z,q)&= z\overline{B}_{n-1}(z,q) ? zq^{n-1}\overline{B}_{n-2}(z,q)-q^{n-1}\overline{B}_{n-3}(z,q),\end{aligned}$$
(2.7)
$$\begin{aligned} D_n(z,q)&= zD_{n-1}(z,q) + zq^{-n}D_{n-2}(z,q) - q^{-n-1}D_{n-3}(z,q),\end{aligned}$$
(2.8)
$$\begin{aligned} \overline{D}_n(z,q)&= z\overline{D}_{n-1}(z,q) + zq^n\overline{D}_{n-2}(z,q)-q^{n+1}\overline{D}_{n-3}(z,q). \end{aligned}$$
(2.9)

Proof

We shall prove (2.7) and (2.9). Recurrence (2.6) follows from (2.7) by (2.1), and (2.8) follows from (2.9) by (2.3).

First we note that

$$\begin{aligned} z(z)_j = q^{-j}(1-(1-zq^j))(z)_j = q^{-j}(z)_j -q^{-j}(z)_{j+1}. \end{aligned}$$
(2.10)

So if we compare coefficients of \((z)_j\) on each side of (2.7), we see that we need to prove

$$\begin{aligned} t(n,j)&= q^{-j}t(n-1,j) - q^{1-j}t(n-1,j-1)\\&\quad + q^{n-1-j}t(n-2,j) - q^{n-j}t(n-2,j-1)\nonumber \\&\quad -q^{n-1}t(n-3,j),\nonumber \end{aligned}$$
(2.11)

where

$$\begin{aligned} t(n,j) = \begin{bmatrix}n+j\\2j \end{bmatrix} (-1)^jq^{j(j+1)/2-nj}. \end{aligned}$$
(2.12)

Now multiplying both sides of (2.11) by \((q)_{2j}(q)_{n-j}/(q)_{n+j-3}\) reduces the expression to an easily verified polynomial identity.

Similarly if we compare coefficients of \((z)_j\) on each side of (2.9), we see that we need to prove

$$\begin{aligned} u(n,j)&= q^{-j}u(n-1,j)-q^{1-j}u(n-1,j-1)-q^{n-j}u(n-2,j)\\&\quad + q^{n-j+1}u(n-2,j-1) + q^{n+1}u(n-3,j),\nonumber \end{aligned}$$
(2.13)

where

$$\begin{aligned} u(n,j) = \begin{bmatrix}n+j+1\\2j+1\end{bmatrix} (-1)^jq^{j(j+1)/2-nj}. \end{aligned}$$
(2.14)

Finally, multiplying both sides of (2.13) by \((q)_{2j+1}(q)_{n-j}/(q)_{n+j-2}\) reduces this expression to an easily verified polynomials identity. \(\square \)

3 Bressoud polynomials in Bailey pairs

Let us recall fundamental aspects of the theory of Bailey chains and Bailey pairs [3]. We start with a pair of sequences of rational functions \((\alpha _n,\beta _n)\) defined to be a Bailey pair provided [3, p. 278]

$$\begin{aligned} \alpha _n = \frac{(1-aq^{2n})(-1)^nq^{n(n-1)/2}(a)_n}{(1-a)(q)_n} \sum _{j=0}^n(q^{-n})_j(aq^n)_jq^j \beta _j, \end{aligned}$$
(3.1)

or equivalently

$$\begin{aligned} \beta _n = \sum _{r=0}^n \frac{\alpha _r}{(q)_{n-r}(aq)_{n+r}}. \end{aligned}$$
(3.2)

Whenever you have a Bailey pair, the following holds for integers \(K\ge 1\) [3, p. 273, Eq. (3.1)]:

$$\begin{aligned} \frac{1}{(aq)_\infty }&\sum _{n\ge 0}q^{Kn^2}a^{Kn}\alpha _n\\&= \sum _{s_1\ge s_2 \ge \cdots \ge s_K\ge 0}\frac{a^{s_1+\cdots +s_K}q^{s_1^2+\cdots +s_K^2}\beta _{s_K}}{(q)_{s_1-s_2}(q)_{s_2-s_3}\cdots (q)_{s_{K-1}-s_K}},\nonumber \end{aligned}$$
(3.3)

and in particular when \(K=1\),

$$\begin{aligned} \frac{1}{(aq)_\infty } \sum _{n=0}^\infty q^{n^2}a^n \alpha _n = \sum _{s=0}^\infty a^s q^{s^2}\beta _n. \end{aligned}$$
(3.4)

Now let us consider four Bailey pairs associated with the Bressoud polynomials. We begin with \(B_n(z,q)\):

$$\begin{aligned} B_n(z,q)&= \sum _{j=0}^n \begin{bmatrix} n+j\\2j\end{bmatrix}(-1)^jq^{j^2+j(j-1)/2-nj}(zq^{1-j})_j \\&= \sum _{j=0}^n (q^{-n})_j(q^{n+1})_jq^j \frac{q^{j(j+1)/2}z^j(z^{-1})_j}{(q)_{2j}}. \end{aligned}$$

Hence with

$$\begin{aligned} \alpha 1_n = \frac{(1-q^{2n+1})(-1)^nq^{n(n-1)/2}}{(1-q)}B_n(z,q), \end{aligned}$$
(3.5)

we see that

$$\begin{aligned} \left( \alpha 1_n, \frac{(-1)^nz^nq^{n(n-1)/2}(z^{-1})_n}{(q)_{2n}}\right) = \left( \alpha 1_n,\frac{q^{n^2-n}(zq^{1-n})_n}{(q)_{2n}}\right) \end{aligned}$$
(3.6)

is a Bailey pair for \(a=q\).

By exactly the same reasoning, with

$$\begin{aligned} \alpha 2_n = \frac{(1-q^{2n+1})}{(1-q)}(-1)^nq^{n(n-1)/2}\overline{B}_n(z,q), \end{aligned}$$
(3.7)

we see that

$$\begin{aligned} \left( \alpha 2_n,\frac{(z)_n}{(q)_{2n}}\right) \end{aligned}$$
(3.8)

is a Bailey pair for \(a=q\).

Similarly,

$$\begin{aligned} \alpha 3_n = \frac{(1-q^{2n+2})(-1)^nq^{n(n-1)/2}}{(1-q)(1-q^2)}D_n(z,q), \end{aligned}$$
(3.9)

we see that

$$\begin{aligned} \left( \alpha 3_n,\frac{(-1)^nq^{n(n+1)/2}z^n(z^{-1})_n}{(q)_{2n+1}}\right) = \left( \alpha 3_n, \frac{q^{n^2}(zq^{1-n})_n}{(q)_{2n+1}}\right) \end{aligned}$$
(3.10)

is a Bailey pair for \(a=q^2\).

Finally,

$$\begin{aligned} \alpha 4_n = \frac{(1-q^{2n+2})(-1)^nq^{n(n-1)/2}}{(1-q)(1-q^2)}\overline{D}_n(z,q), \end{aligned}$$
(3.11)

we see that

$$\begin{aligned} \left( \alpha 4_n,\frac{(z)_n}{(q)_{2n+1}}\right) \end{aligned}$$
(3.12)

is a Bailey pair for \(a=q^2\).

4 Bressoud polynomials and the generalized Rogers–Ramanujan series

The object of this section is first to show how Bressoud’s theorem relates to the generalized Rogers–Ramanujan series [1] via Bailey chains, and second to provide a companion result in which the \(D_n(z)\) appears.

Theorem 2

For \(1\le i\le k\),

$$\begin{aligned} \sum _{s_1\ge \cdots \ge s_{k-1}\ge 0}&\frac{q^{s_1^2+\cdots +s_{k-1}^2+s_i+s_{i+1}+\cdots +s_{k-1}}}{(q)_{s_1-s_2}(q)_{s_2-s_3}\cdots (q)_{s_{k-2}-s_{k-1}}(q)_{s_k}}\\&= \sum _{s_1\ge \cdots s_{k-1}\ge 0} \frac{q^{s_1^2+\cdots +s_{k-1}^2}B_{k-i}(q^{s_{k-1}})}{(q)_{s_1-s_2}(q)_{s_2-s_3}\cdots (q)_{s_{k-2}-s_{k-1}}(q)_{s_k}} \nonumber \\&= \frac{(q^i,q^{2k+1-i},q^{2k+1};q^{2k+1})_\infty }{(q)_\infty }.\nonumber \end{aligned}$$
(4.1)

Remark

This result may also be derived from Bressoud’s original proof of Bressoud’s Theorem. Our goal here is to put the proof into the Bailey chain scenario.

Proof

The \(q\)-binomial theorem [1, p. 36, Eq. (3.3.6)] asserts that

$$\begin{aligned} (xq^{1-n})_{2n} = \sum _{j=0}^{2n} \begin{bmatrix}2n\\ j\end{bmatrix}(-1)^jx^jq^{{j+1\atopwithdelims ()2}-nj}, \end{aligned}$$
(4.2)

and this result may be rewritten as

$$\begin{aligned} \frac{(x^{-1})_n(xq)_n}{(q)_{2n}} = \frac{1}{(q)_n^2} + \sum _{r=1}^n \frac{(-1)^r(q^{r+1\atopwithdelims ()2}x^r+q^{r\atopwithdelims ()2}x^{-r})}{(q)_{n-r}(q)_{n+r}}. \end{aligned}$$
(4.3)

Thus, \((\alpha _n,\beta _n)\) is a Bailey pair for \(a=1\), where

$$\begin{aligned} \beta _n = \frac{(x^{-1})_n(xq)_n}{(q)_{2n}} \end{aligned}$$
(4.4)

and

$$\begin{aligned} \alpha _n = {\left\{ \begin{array}{ll} 1 &{} \text {if } n=0,\\ (-1)^n\left( q^{n+1\atopwithdelims ()2}x^n+q^{n \atopwithdelims ()2}x^{-n}\right) &{} \text {if } n>0. \end{array}\right. } \end{aligned}$$
(4.5)

We now insert this pair into (3.3) with \(a=1\) and \(K=k\):

$$\begin{aligned} \frac{1}{(q)_\infty }&\left( 1+\sum _{n=1}^\infty q^{kn^2}(-1)^n\left( q^{-n\atopwithdelims ()2}x^n + q^{n\atopwithdelims ()2}x^{-n}\right) \right) \\&= \sum _{s_1\ge \cdots \ge s_{k-1}\ge 0}\frac{q^{s_1^2+\cdots s_{k-1}^2}}{(q)_{s_1-s_2}\cdots (q)_{s_{k-2}-s_{k-1}}(q)_{s_{k-1}}}\sum _{s_k=0}^{s_{k-1}} \frac{(q)_{s_{k-1}}(x^{-1})_{s_k}(xq)_{s_k}q^{s_k^2}}{(q)_{2s_k}(q)_{s_{k-1}-s_k}}.\nonumber \end{aligned}$$
(4.6)

Now set \(x=q^{k-i}\). The left-hand side of (4.6) becomes

$$\begin{aligned}&\frac{1}{(q)_\infty } \sum _{n=-\infty }^\infty (-1)^n q^{kn^2+n(n+1)/2 + (k-i)n}\nonumber \\&\quad = \frac{1}{(q)_\infty } \left( q^i, q^{2k+1-i}, q^{2k+1};q^{2k+1}\right) _\infty \nonumber \\&(\hbox {by Jacobi's Triple product }\,[2, \hbox {p. }\,22, \hbox {Corollary}\,2.9]), \end{aligned}$$
(4.7)

which is the third expression in (4.1), and is thus equal to the first expression in (4.1) by (1.5).

On the right-hand side of (4.6), the inner sum on \(s_k\) becomes

$$\begin{aligned} \sum _{s_k=0}^{s_{k-1}}&\frac{\left( q^{-(k-i)}\right) _{s_k}\left( q^{k-i+1}\right) _{s_k}q^{s_k}}{(q)_{2s_k}}\left( 1-q^{s_{k-1}}\right) \cdots \left( 1-q^{s_{k-1}-s_k+1}\right) \\&= \sum _{S=0}^{k-i} \begin{bmatrix} S + k-i \\ 2S\end{bmatrix} (-1)^S q^{S(3S-1)/2-(k-i)S}\left( q^{s_k+1-S}\right) _S\\&= B_{k-i}\left( q^{s_k}\right) , \end{aligned}$$

and this confirms that the resulting right-hand side of (4.6) equals the expression in (4.7) and the theorem is proved. \(\square \)

Theorem 3

For \(1\le i \le k\),

$$\begin{aligned} \sum _{s_1\ge \cdots s_{k-1}\ge 0}&\frac{q^{s_1^2+\cdots s_{k-1}^2 + s_i + s_{i+1}+ \cdots s_{k-1}}}{(q)_{s_1-s_2}(q)_{s_2-s_3}\cdots (q)_{s_{k-2}-s_{k-1}}(q)_{s_k}}\\&= \sum _{s_1\ge \cdots s_{k-1}\ge 0} \frac{q^{s_1^2+\cdots s_{k-1}^2 + s_1 + s_2+ \cdots s_{k-1}}D_{i-1}\left( q^{s_{k-1}}\right) }{(q)_{s_1-s_2}(q)_{s_2-s_3}\cdots (q)_{s_{k-2}-s_{k-1}}(q)_{s_k}}\nonumber \\&= \frac{\left( q^i, q^{2k+1-i}, q^{2k+1};q^{2k+1}\right) _\infty }{(q)_\infty }. \nonumber \end{aligned}$$
(4.8)

Proof

The \(q\)-binomial theorem [1, p. 36, Eq. (3.3.6)] asserts that

$$\begin{aligned} (xq^{1-n})_{2n+1} = \sum _{j=0}^{2n+1} \begin{bmatrix}2n+1\\ j\end{bmatrix}(-1)^jx^jq^{{j+1\atopwithdelims ()2}-nj}, \end{aligned}$$
(4.9)

and the result may be rewritten as

$$\begin{aligned} \frac{(x^{-1})_n(xq)_{n+1}}{(q^2;q)_{2n}} = \sum _{r=0}^n \frac{(-1)^r(x^{-r}q^{r\atopwithdelims ()2}-x^{r+1}q^{-r-1\atopwithdelims ()2})}{(q)_{n-r}(q^2;q)_{n+r}}. \end{aligned}$$
(4.10)

Thus, \((\overline{\alpha }_n, \overline{\beta }_n)\) is a Bailey pair for \(a=q\), where

$$\begin{aligned} \overline{\beta }_n = \frac{(x^{-1})_n (xq)_{n+1}}{(q^2;q)_{2n}} \end{aligned}$$
(4.11)

and

$$\begin{aligned} \overline{\alpha }_n = (-1)^n \left( x^{-n}q^{n\atopwithdelims ()2}-x^{n+1}q^{-n-1\atopwithdelims ()2}\right) . \end{aligned}$$
(4.12)

We now insert this pair into (3.3) with \(a=q\) and \(K=k\) (and we multiply both sides by \(\tfrac{1}{(1-q)}\) ):

$$\begin{aligned} \frac{1}{(q)_\infty }&\sum _{n=0}^\infty q^{kn^2+kn}(-1)^n\left( x^{-n}q^{n\atopwithdelims ()2}-x^{n+1}q^{-n-1\atopwithdelims ()2}\right) \\&= \sum _{s_1\ge \cdots \ge s_{k-1}\ge 0}\frac{q^{s_1^2+\cdots +s_{k-1}^2}}{(q)_{s_1-s_2}(q)_{s_2-s_3}\cdots (q)_{s_{k-2}-s_{k-1}}(q)_{s_{k-1}}}\nonumber \\&\quad \times \sum _{s_k=0}^{s_{k-1}}\frac{(q)_{s_{k-1}}\left( x^{-1}\right) _{s_k}(xq)_{s_k+1}q^{s_k^2}}{(q)_{2s_k+1}(q)_{s_{k-1}-s_k}}.\nonumber \end{aligned}$$
(4.13)

Now set \(x=q^{i-1}\). The left-hand side of (4.13) becomes

$$\begin{aligned} \frac{1}{(q)_\infty }\sum _{n=-\infty }^{\infty } (-1)^n q^{kn^2+n(n-1)/2+(k-i+1)n}\nonumber \\ = \frac{1}{(q)_\infty }\left( q^i,q^{2k+1-i},q^{2k+1};q^{2k+1}\right) _\infty \nonumber \\ (\hbox {by Jacobi's Triple product} [1, \hbox {p. }\,22, \hbox {Corollary}\,2.9]), \end{aligned}$$
(4.14)

which is the third expression in (4.8), and is thus equal to the first expression in (4.8) by (1.5).

On the right-hand side of (4.13), the inner sum on \(s_k\) becomes

$$\begin{aligned} \sum _{s_k=0}^{s_{k-1}}&\frac{\left( q^{-(i-1)}\right) _{s_k}\left( q^i\right) _{s_k+1}q^{s_k^2}}{(q)_{2s_k+1}(q)_{s_{k-1}-s_k}}\\&= \sum _{S=0}^{i-1} \begin{bmatrix} S+i\\2S+1 \end{bmatrix} (-1)^S q^{S(3S+1)/2-iS}\left( q^{s_{k-1}+1-S}\right) _S\\&= D_{i-1}\left( q^{s_{k-1}}\right) . \end{aligned}$$

This confirms that the resulting right-hand side of (4.13) equals the expression in (4.14), and the theorem is proved. \(\square \)

To close this section, we note that the first two expressions in Theorem 3 are identical at \(i=1\) because \(D_0(x)=1\).

5 Rogers–Ramanujan–Slater identities

In this short section, we note that eight identities from [11] are direct corollaries of the results in Sect. 2.

Theorem 4

$$\begin{aligned} B_n(0,q) = {\left\{ \begin{array}{ll} 0 &{} \text {if } n = 3\nu +1,\\ (-1)^\nu q^{-\nu (3\nu +1)/2} &{} \text {if } n = 3\nu , \\ (-1)^\nu q^{-\nu (3\nu -1)/2} &{} \text {if } n = 3\nu -1, \end{array}\right. } \end{aligned}$$
(5.1)
$$\begin{aligned} B_n\left( -\frac{1}{q},q\right) = {\left\{ \begin{array}{ll} (-1)^\nu q^{-\nu ^2} &{} \text {if } n = 2\nu ,\\ (-1)^\nu q^{-\nu ^2} &{} \text {if } n = 2\nu -1, \end{array}\right. } \end{aligned}$$
(5.2)
$$\begin{aligned} D_n(0,q) = {\left\{ \begin{array}{ll} 0 &{} \text {if } n = 3\nu -1\\ (-1)^\nu q^{-(\nu +1)(3\nu +2)/2+1} &{} \text {if } n = 3\nu ,\\ (-1)^\nu q^{-(\nu +1)(3\nu +4)/2+1} &{} \text {if } n = 3\nu +1, \end{array}\right. } \end{aligned}$$
(5.3)
$$\begin{aligned} D_n\left( -\frac{1}{q},q\right) = {\left\{ \begin{array}{ll} 0 &{} \text {if } n = 2\nu +1,\\ (-1)^\nu q^{-(\nu +1)^2+1} &{} \text {if } n = 2\nu , \end{array}\right. } \end{aligned}$$
(5.4)
$$\begin{aligned} \overline{B}_n(0,q) = B_{n}\left( 0,\tfrac{1}{q}\right) = {\left\{ \begin{array}{ll} 0 &{} \text {if } n = 3\nu +1,\\ (-1)^\nu q^{\nu (3\nu +1)/2} &{} \text {if } n = 3\nu ,\\ (-1)^\nu q^{\nu (3\nu -1)/2} &{} \text {if } n = 3\nu -1, \end{array}\right. } \end{aligned}$$
(5.5)
$$\begin{aligned} \overline{B}_n(-q,q) = B_n\left( -q,\tfrac{1}{q}\right) = {\left\{ \begin{array}{ll} (-1)^\nu q^{\nu ^2} &{} \text {if } n = 2\nu ,\\ (-1)^\nu q^{\nu ^2} &{} \text {if } n = 2\nu -1, \end{array}\right. } \end{aligned}$$
(5.6)
$$\begin{aligned} \overline{D}_n(0,q) = D_n\left( 0,\tfrac{1}{q}\right) = {\left\{ \begin{array}{ll} 0 &{} \text {if } n = 3\nu -1,\\ (-1)^\nu q^{(\nu +1)(3\nu +2)/2-1} &{} \text {if } n = 3\nu ,\\ (-1)^\nu q^{(\nu +1)(3\nu +4)/2-1} &{} \text {if } n = 3\nu +1, \end{array}\right. } \end{aligned}$$
(5.7)
$$\begin{aligned} \overline{D}_n(-q,q) = D\left( -q,\tfrac{1}{q}\right) = {\left\{ \begin{array}{ll} 0 &{} \text {if } n = 2\nu +1,\\ (-1)^\nu q^{(\nu +1)^2-1} &{} \text {if } n = 2\nu ,\\ \end{array}\right. } \end{aligned}$$
(5.8)
$$\begin{aligned} \overline{B}_n(q,q^2) = B_n\left( q,\tfrac{1}{q^2}\right) = q^{n(n+1)/2}. \end{aligned}$$
(5.9)

Proof

We only need to prove (5.1)–(5.4) in light of the fact that (5.5)–(5.8) follow by replacing \(q\) by \(\tfrac{1}{q}\).

As for (5.1)–(5.4), these follow directly by mathematical induction using the recurrences in Theorem 1. For example, by (2.6)

$$\begin{aligned} B_n(0,q) = -q^{1-n} B_{n-3}(0,q), \end{aligned}$$
(5.10)

and one verifies directly that the right-hand side of (5.1) satisfies the same recurrence.

Equations (5.2)–(5.4) follow in the same manner. \(\square \)

Seven of these eight identities imply identities listed in Slater’s compendium [11]. In each case, one combines one of the above with the corresponding Bailey pair from Sect. 3 and inserts the result in (3.4). Our (5.1) implies Eq. (83) of [11].

  • Our (5.2) implies Eq. (46) of [11].

  • Our (5.3) implies Eq. (86) of [11].

  • Our (5.4) implies Eq. (44) of [11].

  • Our (5.5) implies Eq. (99) of [11].

  • Our (5.7) implies Eq. (96) of [11].

  • Our (5.8) implies Eq. (59) of [11].

The first entry in (5.9) implies Eq. (32) of [11], and the second entry in (5.9) implies Eq. (19) of [11].

The missing case (which turns out to be a linear combination of two of Slater’s identities) can be written out in full to illustrate the method.

Corollary 5

$$\begin{aligned} \sum _{n=0}^\infty&\frac{q^{n^2+n}(-q;q)_n}{(q)_{2n}} \\&= \frac{\left( q^6, q^8, q^{14};q^{14}\right) _\infty }{(q)_\infty } - \frac{q\left( q^2,q^{12},q^{14};q^{14}\right) _\infty }{(q)_\infty }. \nonumber \end{aligned}$$
(5.11)

Proof

Substituting (5.6) into (3.7), we see that \(\left( \alpha _n,\beta _n\right) \) is a Bailey pair for \(a=q\), with

$$\begin{aligned} \alpha _n&= \frac{(1-q^{2n+1})}{(1-q)}(-1)^{n+\lfloor \frac{n+1}{2}\rfloor }q^{n(n-1)+\lfloor \frac{n+1}{2}\rfloor ^2} \\ \beta _n&= \frac{(-q)_n}{(q)_{2n}}. \end{aligned}$$

Inserting this pair into (3.4) with \(a=q\), we find

$$\begin{aligned} \frac{1}{(q)_\infty }&\left( \sum _{n=-\infty }^\infty (-1)^n q^{7n^2-n} - \sum _{n=-\infty }^\infty (-1)^n q^{7n^2-5n+1}\right) \\&= \sum _{n=0}^\infty \frac{q^{n^2+n}(-q)_n}{(q)_{2n}},\nonumber \end{aligned}$$
(5.12)

and if we apply Jacobi’s triple product to the two series on the left-hand side, we obtain the desired result. \(\square \)

Turning to Slater’s list, we find that the two infinite products in Corollary 5 correspond to Eqs. (51) and (59) of [11]. Not surprisingly, Slater’s identities imply that the left-hand side of (5.11) is

$$\begin{aligned} \sum _{n=0}^\infty&\frac{q^{n^2}(-q)_n}{(q)_{2n}} - q \sum _{n=0}^\infty \frac{q^{n^2+2n}(-q)_n}{(q)_{2n+1}} \\&= 1 + \sum _{n=1}^\infty \frac{q^{n^2}(-q)_n}{(q)_{2n}} - \sum _{n=1}^\infty \frac{q^{n^2}(-q)_{n-1}}{(q)_{2n-1}} \nonumber \\&= 1 + \sum _{n=1}^\infty \frac{q^{n^2}(-q)_{n-1}}{(q)_{2n}}\left( (1+q^n)-(1-q^{2n})\right) \nonumber \\&= 1 + \sum _{n=1}^\infty \frac{q^{n^2+n}(-q)_n}{(q)_{2n}}. \end{aligned}$$

6 Mock theta functions

It should be noted that the Bailey pair (3.8) was suggested in [5] and studied extensively in [6]. Thus, it is natural that it will arise here among the five instances we shall exhibit that yield the Hecke-type expansions involving indefinite quadratic forms. The cases we have chosen illustrate how often Bressoud polynomials arise in this context. We emphasize that this is only a sample.

Theorem 6

$$\begin{aligned} B_n(-1,q) = (-1)^n \sum _{j=-\lfloor \frac{n}{2}\rfloor }^{\lfloor \frac{n}{2}\rfloor } (-1)^jq^{-j^2}, \end{aligned}$$
(6.1)
$$\begin{aligned} \overline{B}_n(-1,q) =B_n(-1,\tfrac{1}{q})= (-1)^n \sum _{j=-\lfloor \frac{n}{2}\rfloor }^{\lfloor \frac{n}{2}\rfloor } (-1)^jq^{j^2}, \end{aligned}$$
(6.2)
$$\begin{aligned} \overline{D}_n(q,q) = {\left\{ \begin{array}{ll} q^{\nu ^2+2\nu } \sum _{j=-\nu }^\nu q^{-j^2} &{} \text {if } n =2\nu ,\\ 2q^{\nu (\nu +1)-1}\sum _{j=0}^{\nu -1} q^{-j^2-j} &{} \text {if } n = 2\nu -1,\\ \end{array}\right. } \end{aligned}$$
(6.3)
$$\begin{aligned} D_n(\tfrac{1}{q},q) = \overline{D}_n(\tfrac{1}{q},\tfrac{1}{q}) = {\left\{ \begin{array}{ll} q^{-\nu ^2-2\nu } \sum _{j=-\nu }^\nu q^{j^2} &{} \text {if } n =2\nu ,\\ 2q^{-\nu ^2-\nu -1}\sum _{j=0}^{\nu -1} q^{j^2+j} &{} \text {if } n = 2\nu -1,\\ \end{array}\right. } \end{aligned}$$
(6.4)
$$\begin{aligned} \overline{D}_n(q,q^2) = q^n\sum _{j=0}^n q^{j(j+1)/2}, \end{aligned}$$
(6.5)
$$\begin{aligned} D_n(\tfrac{1}{q},q^2) = \overline{D}_n(\tfrac{1}{q},\tfrac{1}{q^2})= q^{-n} \sum _{j=0}^n q^{-j(j+1)/2}. \end{aligned}$$
(6.6)

Proof

As in Sect. 5, these results are easily deduced from the recurrences in Theorem 1. As an example, we treat (6.5). It is immediate by inspection that (6.5) is valid for \(n=0,1,2.\) Now let \(\Delta _n\) denote the right-hand side of (6.5). Clearly

$$\begin{aligned} \Delta _n - q \Delta _{n-1}&= q^n \sum _{j=0}^n q^{j(j+1)/2}-q^n\sum _{j=0}^{n-1}q^{j(j+1)/2}\\&= q^{n(n+3)/2}\nonumber \end{aligned}$$

while

$$\begin{aligned} q^{1+2n}&\Delta _{n-2} - q^{2n+2}\Delta _{n-3} \\&= q^{1+2n} \left( \Delta _{n-2} -q\Delta _{n-3}\right) \\&= q^{1+2n+(n-2)(n+1)/2}\\&= q^{\frac{n^2}{2}+\frac{3n}{2}}, \end{aligned}$$

so

$$\begin{aligned} \Delta _n - q\Delta _{n-1} = q^{1+2n}\Delta _{n-2}-q^{2n+2}\Delta _{n-3}, \end{aligned}$$

and the relevant recurrence is established. \(\square \)

We may immediately use these results in the Bailey pairs in Sect. 3, and thus obtain the following identities of mock theta or false theta type:

Theorem 7

$$\begin{aligned} \sum _{n=0}^\infty \frac{q^{n(3n+1)/2}(-1;q)_n}{(q)_{2n}} = \frac{1}{(q)_\infty }\sum _{n=0}^\infty q^{n(3n+1)/2}\left( 1-q^{2n+1}\right) \sum _{|j |\le \lfloor \frac{n}{2}\rfloor }(-1)^jq^{-j^2}, \end{aligned}$$
(6.7)
$$\begin{aligned} \sum _{n=0}^\infty \frac{q^{n^2+n}(-1;q)_n}{(q)_{2n}} = \frac{1}{(q)_\infty }\sum _{n=0}^\infty q^{n(3n+1)/2}(-1)^n\left( 1-q^{2n+1}\right) \sum _{|j |\le \lfloor \frac{n}{2}\rfloor } (-1)^jq^{j^2}, \end{aligned}$$
(6.8)
$$\begin{aligned} \sum _{n=0}^\infty&\frac{q^{n^2+2n}}{\left( q^{n+1};q\right) _{n+1}}\\&= \frac{1}{(q)_\infty }\Bigg \{\sum _{n=0}^\infty q^{7n^2+4n}\left( 1-q^{4n+2}\right) \sum _{|j |\le n} q^{-j^2}\nonumber \\&\quad - 2\sum _{n=0}^\infty q^{7n^2+12n+4}\left( 1-q^{4n+4}\right) \sum _{j=0}^n j^{-j^2-j}\Bigg \},\nonumber \end{aligned}$$
(6.9)
$$\begin{aligned} \sum _{n=0}^\infty&\frac{(-1)^n q^{3n(n+1)/2}}{\left( q^{n+1};q\right) _{n+1}} \\&= \frac{1}{(q)_\infty } \Bigg \{ \sum _{n=0}^\infty q^{5n^2+3n} \left( 1-q^{4n+2}\right) \sum _{|j |\le n} q^{j^2}\nonumber \\&\quad -2\sum _{n=0}^\infty q^{5n^2+8n+1} \left( 1-q^{4n+4}\right) \sum _{j=0}^\infty q^{j^2+j} \Bigg \}\nonumber \\&= \sum _{n=0}^\infty (-1)^n q^{n(5n+7)/2} \left( 1+q^{3n+1}+q^{5n+2}+q^{8n+4}\right) ,\nonumber \end{aligned}$$
(6.10)
$$\begin{aligned} \sum _{n=0}^\infty&\frac{q^{2n^2+4n}\left( q;q^2\right) _n}{\left( q^2;q^2\right) _{2n+1}}\\&= \frac{1}{\left( q^2;q^2\right) _\infty } \sum _{n=0}^\infty (-1)^n q^{3n^2+4n}\left( 1-q^{4n+4}\right) \sum _{j=0}^n q^{j(j+1)/2},\nonumber \end{aligned}$$
(6.11)
$$\begin{aligned} \sum _{n=0}^\infty&\frac{(-1)^nq^{3n^2+4n}\left( q;q^2\right) _n}{\left( q^2;q^2\right) _{2n+1}}\end{aligned}$$
(6.12)
$$\begin{aligned}&= \frac{1}{\left( q^2;q^2\right) _\infty } \sum _{n=0}^\infty (-1)^nq^{3n^2+4n} \left( 1-q^{4n+4}\right) \sum _{j=0}^n q^{-j(j+1)/2}. \end{aligned}$$
(6.13)

Proof

In each of the six identities, we take the corresponding evaluation of the Bressoud polynomial in Theorem 6 and then apply it to the corresponding Bailey pair and the Bailey Lemma from Sect. 3.

We should remark that (6.9) is a small variation of the seventh-order mock theta function identity in [5, p. 132, Eq. (7.23)]. Also the last line of (6.10) is due to L. J. Rogers [10, Sect. 9]. \(\square \)

7 Bressoud’s theorem and the generalized Rogers–Ramanujan series

The object of this section is to reconsider the left-hand sides of (1.5) and (1.8).

Theorem 8

For \(1\le i\le k\),

$$\begin{aligned} \sum _{s_1\ge \cdots \ge s_{k-1}\ge 0}&\frac{q^{s_1^2 + \cdots +s_{k-1}^2 + s_i + s_{i+1}+ \cdots + s_{k-1}}}{(q)_{s_1-s_2}(q)_{s_2-s_3}\cdots (q)_{s_{k-2}-s_{k-1}}(q)_{s_k}}\\&= \sum _{s_1\ge \cdots \ge s_{k-1}\ge 0} \frac{q^{s_1^2 + \cdots +s_{k-1}^2}B_{k-i}\left( q^{s_{k-1}}\right) }{(q)_{s_1-s_2}(q)_{s_2-s_3}\cdots (q)_{s_{k-2}-s_{k-1}}(q)_{s_{k-1}}}.\nonumber \end{aligned}$$
(7.1)

Proof

The above assertion follows immediately from the fact that the right-hand sides of (1.5) and (1.8) are identical. \(\square \)

The object here is to consider whether (7.1) can be proved directly without using either (1.5) or (1.8).

The cases \(i=k\) and \(i=k-1\) are, in fact, tautologies because \(B_0(z,q) =1\) and \(B_1(z,q) =z\).

We go through the \(i=k-2\) case step by step because it illustrates the general case.

For simplicity, we write

$$\begin{aligned} T\left( s_1,\dots ,s_{k-3}\right) = \frac{q^{s_1^2+\cdots + s_{k-3}^2}}{(q)_{s_1-s_2}\cdots (q)_{s_{k-1}-s_{k-3}}}. \end{aligned}$$

Then

$$\begin{aligned}&\sum _{s_1\ge \cdots \ge s_{k-1}\ge 0} \frac{q^{s_1^2 + \cdots +s_{k-1}^2 + s_{k-2}+ s_{k-1}}}{(q)_{s_1-s_2}(q)_{s_2-s_3}\cdots (q)_{s_{k-2}-s_{k-1}}(q)_{s_{k-1}}}\\&= \sum _{\underline{s}} T\left( s_1,\dots ,s_{k-3}\right) \frac{q^{s_{k-2}^2+s_{k-1}^2+s_{k-2}+s_{k-1}}}{(q)_{s_{k-3}-s_{k-2}}(q)_{s_{k-2}-s_{k-1}}(q)_{s_{k-1}}}\\&= \sum _{\underline{s}}T\left( s_1,\dots , s_{k-3}\right) \frac{q^{s_{k-2}^2+s_{k-1}^2+ 2s_{k-1}}\left( 1-\left( 1-q^{s_{k-2}-s_{k-1}}\right) \right) }{(q)_{s_{k-3}-s_{k-2}}(q)_{s_{k-2}-s_{k-1}}(q)_{s_{k-1}}} \\&= \sum _{\underline{s}}T\left( s_1,\dots , s_{k-3}\right) \frac{q^{s_{k-2}^2+s_{k-1}^2+ 2s_{k-1}}}{(q)_{s_{k-3}-s_{k-2}}(q)_{s_{k-2}-s_{k-1}}(q)_{s_{k-1}}}\\&\quad - \sum _{\underline{s}}T\left( s_1,\dots , s_{k-3}\right) \frac{q^{s_{k-2}^2+(s_{k-1}-1)^2+2(s_{k-1}-1)}\left( 1-q^{s_{k-1}}\right) }{(q)_{s_{k-3}-s_{k-2}}(q)_{s_{k-2}-s_{k-1}}(q)_{s_{k-1}}}\\&{(\mathrm{where}\,s_{k-1}\rightarrow s_{k-1}-1\,\mathrm{in\,second\,sum})}\\&= \sum _{\underline{s}}\frac{T\left( s_1,\dots , s_{k-3}\right) q^{s_{k-2}^2+s_{k-1}^2}}{(q)_{s_{k-3}-s_{k-2}}(q)_{s_{k-2}-s_{k-1}}(q)_{s_{k-1}}}\left( q^{2s_{k-1}}+q^{s_{k-1}-1}-q^{-1}\right) \\&= \sum _{\underline{s}}\frac{q^{s_1^2+\cdots +s_{k-1}^2}B_2\left( q^{s_{k-1}}\right) }{(q)_{s_1-s_2}\cdots (q)_{s_{k-2}-s_{k-1}}(q)_{s_{k-1}}}, \end{aligned}$$

because

$$\begin{aligned} B_2(z,q) = z^2+\frac{z}{q} - \frac{1}{q}. \end{aligned}$$

The simplification applied in the \(i=k-2\) case can be made applicable to the general problem of reducing the left-hand side of (7.1) to the right-hand side.

Lemma 9

Let

$$\begin{aligned} T_{R,S}(u,t,n) = \sum _{i_1,i_2,i_3\ge 0} \frac{q^{i_1^2+i_2^2+i_3^2+ui_1+ti_2+ni_3}}{(q)_{R-i_1}(q)_{i_1-i_2}(q)_{i_2-i_3}(q)_{i_3-S}}. \end{aligned}$$
(7.2)

Then

$$\begin{aligned} T_{R,S}(u,t,n) \!= \!T_{R,S}(u\!-\!1,t\!+\!1,n)-q^{-t}T_{R,S}(u-1,t-1,n) - q^{-t}T_{R,S}(u-1,t,n-1). \end{aligned}$$
(7.3)

Proof

We note that

$$\begin{aligned} 1 = q^{-i_1+i_2}\left( 1-\left( 1-q^{i_1-i_2}\right) \right) . \end{aligned}$$

Hence

$$\begin{aligned} T_{R,S}(u,t,n)&= \sum _{i_1,i_2,i_3\ge 0} \frac{q^{i_1^2+i_2^2+i_3^2 + ui_1+ ti_2 + ni_3-i_1-i_2}\left( 1-\left( 1-q^{i_1-i_2}\right) \right) }{(q)_{R-i_1}(q)_{i_1-i_2}(q)_{i_2-i_3}(q)_{i_3-S}}\\&= T_{R,S}(u-1,t+1,n) - \sum _{i_1,i_2,i_3\ge 0} \frac{q^{i_1^2+i_2^2+i_3^2 + (u-1)i_1+ (t+1)i_2 + ni_3}}{(q)_{R-i_1}(q)_{i_1-i_2-1}(q)_{i_2-i_3}(q)_{i_3-S}}\\&= T_{R,S}(u-1,t+1,n) \\&\quad - \sum _{i_1,i_2,i_3\ge 0} \frac{q^{i_1^2+(i_2-1)^2+i_3^2 + (u-1)i_1+ (t+1)(i_2-1) + ni_3}\left( 1-\left( 1-q^{i_2-i_3}\right) \right) }{(q)_{R-i_1}(q)_{i_1-i_2}(q)_{i_2-i_3}(q)_{i_3-S}}\\ {(\mathrm{where}\, i_2 \rightarrow i_2-1)}\\&= T_{R,S} (u-1,t+1,n) -q^{-t}T_{R,S}(u-1,t-1,n) \\ {}&- q^{-t}T_{R,S}(u-1,t,n-1). \end{aligned}$$

\(\square \)

I believe that successive applications of Lemma 9 to the left-hand side of (7.1) will yield the right-hand side of (7.1) just as we did in the case \(i=k-2\) with effectively one application of Lemma 9.

In particular, let us define

$$\begin{aligned} \sum _{s_1\ge \cdots \ge s_k\ge 0}&\frac{q^{s_1^2+s_2^2+\cdots + s_k^2 + s_{j+1}+ s_{j+2}+\cdots + s_{k-1} + hs_k}}{(q)_{s_1-s_2}(q)_{s_2-s_3}\cdots (q)_{s_{k-1}-s_k}(q)_{s_k}}\\&= \sum _{s_1\ge \cdots \ge s_k\ge 0} \frac{q^{s_1^2+s_2^2+\cdots + s_k^2 }f_k(k-j,h;q^{s_k})}{(q)_{s_1-s_2}(q)_{s_2-s_3}\cdots (q)_{s_{k-1}-s_k}(q)_{s_k}}.\\ \end{aligned}$$

Then successive application of Lemma 9 reveals that

$$\begin{aligned} f_k(k-1,h;z)&= z^h,\\ f_k(k-2,h;z)&= z^{h+1}-\frac{z^{h-1}}{qh} + \frac{1}{q^h}, \end{aligned}$$

and for \(j>2\)

$$\begin{aligned} f_k(k-j,h;z)&= f_k(k-j+1,h+1;z) \\&\quad + (q^{-1}-q^{-h})f_k(k-j+1,h-1;z)\\&\quad - q^{-1}f_k(k-j+2,h;z) + q^{-h}f_k(k-j+1,h;z). \end{aligned}$$

Conjecture 1

$$\begin{aligned} f_k(k-j,1;z) = B_j(z,q). \end{aligned}$$

We have already shown that this is true for \(j=0,1,2\). The assertion has been verified for \(0\le j\le 10\).

8 Conclusion

There are many questions yet to be examined in connection with the topics explored in this paper. The fact that \(B_n(z,q)\) (actually a normalized version) played a central role in the recent study of the seventh-order mock theta functions [3] suggests that the other three Bressoud polynomials might well reveal interesting results from a similar study.

The work in Sect. 7 is effectively dual to the work of Berkovich and Paule [7, 8]. In their work, they were able to represent

$$\begin{aligned} \sum _{s_1\ge \cdots \ge s_{k-1}\ge 0} \frac{q^{s_1^2+s_2^2+\cdots +s_{k-1}^2 + M_1s_1+M_2s_2+\cdots +M_{k-1}s_{k-1}}}{(q)_{s_1-s_2}(q)_{s_2-s_3}\cdots (q)_{s_{k-2}-s_{k-1}}(q)_{s_{k-1}}} \end{aligned}$$

as linear combinations of the series given in (1.5). In section 8, we examine when various linear combinations are identical. Presumably, identical combinations could be identified by means of Lemma 9.

Finally, the fact that the Bressoud polynomials have such diverse applications as those considered here suggests that they merit further study in their own right.