ANALYTIC PROOF OF THE PARTITION IDENTITY A_{5,3,3}(n) = B_{5,3,3}^{0}(n)

PADMAVATHAMMA^{†}, CHANDRASHEKARA B.M.^{†}, RAGHAVENDRA R.^{†}, AND C. KRATTENTHALER*^{‡}

Department of Studies in Mathematics University of Mysore, Manasagangotri Mysore – 570 006, Karnataka, India

  E-mail: padma vathamma@yahoo.com, chandru alur@yahoo.com,
                  raghu maths@yahoo.co.in

*Institut Girard Desargues, Université Claude Bernard Lyon-I 21, avenue Claude Bernard, F-69622 Villeurbanne Cedex, France

             E-mail: kratt@euler.univ-lyon1.fr
               WWW: http://igd.univ-lyon1.fr/~kratt

Abstract. In this paper we give an analytic proof of the identity A_{5,3,3}(n) = B_{5,3,3}^{0}(n), where A_{5,3,3}(n) counts the number of partitions of n subject to certain restrictions on their parts, and B_{5,3,3}^{0}(n) counts the number of partitions of n subject to certain other restrictions on their parts, both too long to be stated in the abstract. Our proof establishes actually a refinement of that partition identity. The original identity was first discovered by the first author jointly with M. Ruby Salestina and S. R. Sudarshan in [“A new theorem on partitions,” Proc. Int. Conference on “Analytic Number Theory with Special Emphasis on L-functions,” Ramanujan Math. Soc., Mysore, 2005, pp. 57–70] where it was also given a combinatorial proof, thus answering a question of Andrews.

1. Introduction

For an even integer λ, let A_{λ,k,a}(n) denote the number of partitions of n such that • no part ≢ 0 (mod λ + 1) may be repeated, and

• no part is ≡ 0, ± (a − \frac{λ}{2}) (λ + 1) (mod (2k − λ + 1)(λ + 1)). For an odd integer λ, let A_{λ,k,a}(n) denote the number of partitions of n such that

2000 Mathematics Subject Classification. Primary 11P81, 11P82, 11P83; Secondary 05A15, 05A17, 05A19.

Key words and phrases. Integer partitions, generating functions, analytic proof. †The first author is partially and the second and third authors are fully financially supported by

Department of Science and Tecnology, New Delhi, India, Grant DST\MS\116\2K ”Some Studies in Partition Theory”.

‡Research partially supported by EC’s IHRP Programme, grant HPRN-CT-2001-00272, “Algebraic Combinatorics in Europe,” and by the Austrian Science Foundation FWF, grant S9607-N13, in the framework of the National Research Network “Analytic Combinatorics and Probabilistic Number Theory”.

• no part ≢ 0 (mod \frac{λ+1}{2}) may be repeated,

• no part is ≡ λ + 1 (mod 2λ + 2), and

• no part is ≡ 0, ±(2a − λ) (\frac{λ+1}{2}) (mod (2k − λ + 1)(λ + 1)).

Let B_{λ,k,a}(n) denote the number of partitions of n of the form b_{1} + ⋯ + b_{s} with b_{i} ≥ b_{i+1}, such that

• no part ≢ 0 (mod λ + 1) is repeated,

• b_{i} − b_{i+k−1} ≥ λ + 1, with strict inequality if b_{i} is a multiple of λ + 1, and • ∑_{i=j}^{λ−j+1} f_{i} ≤ a − j for 1 ≤ j ≤ \frac{λ+1}{2} and f_{1} + ⋯ + f_{λ+1} ≤ a − 1, where f_{j} is

the number of appearances of j in the partition.

In 1969, Andrews [1] proved the following theorem.

Theorem 1 ([1, Th. 2]). If λ, k, and a are positive integers with \frac{λ}{2} ≤ a ≤ k and k ≥ 2λ − 1, then, for every positive integer n, we have

A_{λ,k,a}(n) = B_{λ,k,a}(n).

Schur’s theorem [6] addresses the case λ = k = a = 2. Hence, it is not a particular case of Theorem 1 as k ≥ 2λ − 1 is not satisfied. Motivated by this observation, Andrews [1] first conjectured, and later proved in [2], that Theorem 1 is still true for k ≥ λ.

In the same paper [2], Andrews raised the following question: Is it possible to modify the conditions on the partitions enumerated by B_{λ,k,a}(n) so that values of k < λ would be admissible? In fact, Schur [6] had proved that A_{3,2,2}(n) = B_{3,2,2}^{0}(n), where B_{3,2,2}^{0}(n) denotes the number of partitions enumerated by B_{3,2,2}(n) with the added condition that no parts are ≡ 2 (mod 4).

This led Andrews [2] to state the following conjecture.

Conejecture 2. There holds the identity A_{4,3,3}(n) = B_{4,3,3}^{0}(n) for all positive integers n, where B_{4,3,3}^{0}(n) denotes the number of partitions of n enumerated by B_{4,3,3}(n) with the added restrictions:

f_{5j+2} + f_{5j+3} ≤ 1 for j ≥ 0,
f_{5j+4} + f_{5j+6} ≤ 1 for j ≥ 0,
f_{5j−1} + f_{5j} + f_{5j+5} + f_{5j+6} ≤ 3 for j ≥ 1,

where, as before, f_{j} denotes the number of appearances of j in the partition.

In the year 1994, Andrews et al. [3] gave an analytic proof of the above conjecture. The first author and Ruby Salestina, M. gave a combinatorial proof in [4]. In [5], these two authors and Sudarshan, S.R. first conjectured, and then proved combinatorially, the following result, which is analogous to Conjecture 2.

Theorem 3. There holds the identity A_{5,3,3}(n) = B_{5,3,3}^{0}(n) for all positive integers n, where B_{5,3,3}^{0}(n) denotes the number of partitions of n enumerated by B_{5,3,3}(n) with the added restrictions:

f_{6j+3} = 0 for j ≥ 0,
f_{6j+2} + f_{6j+4} ≤ 1 for j ≥ 0,
f_{6j+5} + f_{6j+7} ≤ 1 for j ≥ 0,
f_{6j−1} + f_{6j} + f_{6j+6} + f_{6j+7} ≤ 3 for j ≥ 1.

The object of this paper is to give an analytic proof of the partition identity stated in Theorem 3. Actually, we are going to prove a new refinement of that partition identity, which we state in the next section. The method of our proof in Section 3 is similar to that of Andrews et al. in [3].

2. A refinement of the partition identity in Theorem 3

Before being able to state the announced refinement of Theorem 3, we need to make two definitions.

Definition 1. Let A(μ, ν, N) denote the number of partitions of N into distinct nonmultiples of 6 of which μ are congruent to 1 or 2 mod 6 and ν are congruent to 4 or 5 mod 6.

Clearly, we have

(2.1) ∑_{μ,ν,N≥0} A(μ, ν, N)a^{μ}b^{ν}q^{N} = ∏_{n=0}^{∞}(1 + aq^{6n+1})(1 + aq^{6n+2})(1 + bq^{6n+4})(1 + bq^{6n+5}).

Definition 2. Let B(μ, ν, N) denote the number of partitions λ = b_{1} + ⋯ + b_{s} of N satisfying the following conditions:

(i) Only multiples of 6 may be repeated.

(ii) b_{i} − b_{i+2} ≥ 6 with strict inequality if b_{i} is a multiple of 6.

(iii) The multiplicities f_{i}, 1 ≤ i ≤ s, satisfy

f_{6j+3} = 0 for all j ≥ 0,
f_{6j+2} + f_{6j+4} ≤ 1 for all j ≥ 0,
f_{6j+5} + f_{6j+7} ≤ 1 for all j ≥ 0,
f_{6j−1} + f_{6j} + f_{6j+6} + f_{6j+7} ≤ 3 for all j ≥ 1.

(iv) There are μ parts of the partition ≡ 0, 1 or 2 (mod 6).

(v) There are ν parts of the partition ≡ 0, 4 or 5 (mod 6).

The following theorem is the announced refinement of Theorem 3.

Theorem 4. For each μ, ν, N ≥ 0 we have

(2.2)  A(μ, ν, N) = B(μ, ν, N).

It is obvious that Theorem 3 follows immediately from Theorem 4 by summing both sides of (2.2) over all μ and ν. The proof of Theorem 4 is given in the next section.

3. Proof of Theorem 4

We first observe that for any partition which satisfies (i)–(iii) of Definition 2 there are exactly 16 possibilities (numbered 0–15 in Table 1) for the subset of summands of the partition that lie in the interval [6i + 1, 6i + 6].

\begin{matrix}0 : & ∅ : empty \\ 1 : & 6i + 1 \\ 2 : & 6i + 2 \\ 3 : & 6i + 2, 6i + 1 \\ 4 : & 6i + 4 \\ 5 : & 6i + 4, 6i + 1 \\ 6 : & 6i + 5 \\ 7 : & 6i + 5, 6i + 1 \\ 8 : & 6i + 5, 6i + 2 \\ 9 : & 6i + 5, 6i + 4 \\ 10 : & 6i + 6 \\ 11 : & 6i + 6, 6i + 1 \\ 12 : & 6i + 6, 6i + 2 \\ 13 : & 6i + 6, 6i + 4 \\ 14 : & 6i + 6, 6i + 5 \\ 15 : & 6i + 6, 6i + 6 \end{matrix}

Table 1

We now refine the partitions from Definition 2 further, using the classification given in Table 1.

Definition 3. Let S_{n}(j, a, b, q) denote the generating function

∑ B_{n}(μ, ν, N)a^{μ}b^{ν}q^{N},

where B_{n}(μ, ν, N) is the number of all partitions considered in Definition 2, which in addition satisfy the two conditions

(vi) all parts are ≤ 6n + 6, and (vii) the subset of summands that lie in the interval [6n + 1, 6n + 6] must have

number ≤ j in Table 1. When n = −1, we define S_{−1}(j, a, b, q) = 1 and for n < −1, we define S_{n}(j, a, b, q) = 0.

For example,

S_{0}(9, a, b, q) = 1 + aq + aq^{2} + a^{2}q^{3} + bq^{4} + abq^{5} + bq^{5} + abq^{6} + abq^{7}

and

S_{0}(15, a, b, q) = 1 + aq + aq^{2} + a^{2}q^{3} + bq^{4} + abq^{5} + bq^{5} + 2abq^{6} + a^{2}bq^{7} + abq^{7}

+ a^{2}bq^{8} + b^{2}q^{9} + ab^{2}q^{10} + ab^{2}q^{11} + a^{2}b^{2}q^{12}.

It is easy to verify that

S_{0}(15, a, b, q) = (1 + aq)(1 + aq^{2})(1 + bq^{4})(1 + bq^{5}).

For convenience, we write S_{n}(j) for S_{n}(j, a, b, q). Along the lines of [3], we obtain the following recurrence relations for S_{n}(j):

(3.1)  S_{n}(0) = S_{n−1}(15),
(3.2)  S_{n}(1) = S_{n}(0) + aq^{6n+1}[S_{n−1}(11) − S_{n−1}(9) + S_{n−1}(5)]
− a^{3}b^{3}q^{24n−12}S_{n−3}(9),
(3.3)  S_{n}(2) = S_{n}(1) + aq^{6n+2}[S_{n−1}(12) − S_{n−1}(9) + S_{n−1}(8)],
(3.4)  S_{n}(3) = S_{n}(2) + a^{2}q^{12n+3}S_{n−1}(3),
(3.5)  S_{n}(4) = S_{n}(3) + bq^{6n+4}S_{n−1}(13),
(3.6)  S_{n}(5) = S_{n}(4) + abq^{12n+5}S_{n−1}(5),
(3.7)  S_{n}(6) = S_{n}(5) + bq^{6n+5}S_{n−1}(14),
(3.8)  S_{n}(7) = S_{n}(6) + abq^{12n+6}S_{n−1}(5),
(3.9)  S_{n}(8) = S_{n}(7) + abq^{12n+7}S_{n−1}(8),
(3.10)  S_{n}(9) = S_{n}(8) + b^{2}q^{12n+9}S_{n−1}(9),
(3.11)  S_{n}(10) = S_{n}(9) + abq^{6n+6}S_{n−1}(14),
(3.12)  S_{n}(11) = S_{n}(10) + a^{2}bq^{12n+7}S_{n−1}(5),
(3.13)  S_{n}(12) = S_{n}(11) + a^{2}bq^{12n+8}S_{n−1}(8),
(3.14)  S_{n}(13) = S_{n}(12) + ab^{2}q^{12n+10}S_{n−1}(9),
(3.15)  S_{n}(14) = S_{n}(13) + ab^{2}q^{12n+11}S_{n−1}(9),
(3.16)  S_{n}(15) = S_{n}(14) + a^{2}b^{2}q^{12n+12}S_{n−1}(9).

We now define two linear combinations of the S_{n}(9)’s and the S_{n}(15)’s,

J(n) := S_{n}(9) − (1 − q^{6n})(1 + aq^{6n+1} + aq^{6n+2} + bq^{6n+4} + bq^{6n+5})S_{n−1}(15) (3.17)
− q^{6n}(1 + aq^{6n+1} + aq^{6n+2} + a^{2}q^{6n+3} + bq^{6n+4} + bq^{6n+5} + abq^{6n+5}
+ abq^{6n+6} + abq^{6n+7} + b^{2}q^{6n+9})S_{n−1}(9)
+ (1 − q^{6n})abq^{18n−3}(a^{2} + abq^{2} + abq^{3} + abq^{4} + a^{2}bq^{4} + a^{2}bq^{5} + b^{2}q^{6}
+ ab^{2}q^{7} + ab^{2}q^{8})S_{n−2}(9)
+ a^{3}b^{3}q^{24n−12}(1 − q^{6n})(1 − q^{6n−6})S_{n−3}(9),

and

(3.18)  K(n) := S_{n}(9)−S_{n}(15) + abq^{6n+6}(1 − q^{6n})S_{n−1}(15)
+ abq^{12n+6}(1 + aq + aq^{2} + bq^{4} + bq^{5} + abq^{6})S_{n−1}(9)
− a^{3}b^{3}q^{18n+6}(1 − q^{6n})S_{n−2}(9).

Along lines similar to those in [3], we are able to obtain a recurrence for S_{n}(9) (see Lemma 6).

Lemma 5. For n ≥ 0, J(n) = 0 = K(n).

Sketch of proof. We prove the lemma by using the identities (3.1)–(3.16). The 14 sequences S_{n}(j), where j is different from 9 and 15, can be expressed as linear combinations of the S_{m}(9)’s and the S_{m}(15)’s in the following way: from (3.16), we find that S_{n}(14) is given by

S_{n}(14) = S_{n}(15) − a^{2}b^{2}q^{12n+12}S_{n−1}(9).

Using the above equation in (3.15), S_{n}(13) becomes such a linear combination. Similarly for S_{n}(12) if we use (3.14). Equation (3.10) yields such a linear combination for S_{n}(8). Subsequently, (3.13) yields a linear combination for S_{n}(11), and (3.11) yields a linear combination for S_{n}(10). Replacing n by n + 1 in (3.12), we get

S_{n}(5) = a^{−2}b^{−1}q^{−12n−19}[S_{n+1}(11) − S_{n+1}(10)],

which in turn yields an expression for S_{n}(5) in terms of the S_{m}(9)’s and the S_{m}(15)’s. Equations (3.9), (3.8), (3.6), (3.5), (3.4) and (3.3) yield respectively linear combinations in terms of the S_{m}(9)’s and the S_{m}(15)’s for S_{n}(7), S_{n}(6), S_{n}(4), S_{n}(3), S_{n}(2) and S_{n}(1). Finally, we know already from (3.1) that S_{n}(0) = S_{n−1}(15).

We are now in the position to prove K(n) = 0. Let us consider (3.7), that is

S_{n}(6) = S_{n}(5) + bq^{6n+5}S_{n−10}(14).

Substituting the expression in terms of the S_{m}(9)’s and S_{m}(15)’s obtained earlier for S_{n}(5) and the respective one for S_{n−1}(14) in the equation above, we get a certain identity, (A) say.

On the other hand, from (3.8), we have

S_{n}(6) = S_{n}(7) − abq^{12n+6}S_{n−1}(5).

Substituting the expression in terms of the S_{m}(9)’s and S_{m}(15)’s obtained earlier for S_{n}(7) and the respective one for S_{n−1}(5), we obtain another identity, (B) say. It can now be verified that (A)−(B), when multiplied by a^{2}bq^{12n+19}, is exactly the equation K(n + 1) = 0.

Now we prove J(n) = 0. Substituting the expressions obtained earlier for S_{n}(1), S_{n}(0), S_{n−1}(11) and S_{n−1}(5) into (3.2), we obtain

0 = a^{2}bq^{12n+19}J(n) − K(n + 1) + aq^{6n+13}(1 + aq^{6n+2} + bq^{6n+4} + bq^{6n+5})K(n).

Since K(n) = 0 for all n ≥ 0, we conclude that J(n) = 0 for all n ≥ 0. This proves the lemma. □

Lemma 6. For n ≥ 0,

(3.19)  (1 + aq^{6n−5} + aq^{6n−4} + bq^{6n−2} + bq^{6n−1})S_{n}(9)
= p_{1}(n, a, b, q)S_{n−1}(9) + (1 − q^{6n})p_{2}(n, a, b, q)S_{n−2}(9)
+ p_{3}(n, a, b, q)(1 − q^{6n})(1 − q^{6n−6})S_{n−3}(9)
+ a^{4}b^{4}q^{30n−36}(1 − q^{6n})(1 − q^{6n−6})(1 − q^{6n−12})
× [(1 + aq^{6n+1} + aq^{6n+2} + bq^{6n+4} + bq^{6n+5})S_{n−4}(9)],

where

p_{1}(n, a, b, q) = 1 + aq^{6n−5} + aq^{6n−4} + bq^{6n−2} + bq^{6n−1} + abq^{6n} + 2abq^{12n−1} + 3abq^{12n} (3.20)
+ aq^{6n+2} + 2a^{2}q^{12n−3} + a^{2}q^{12n−2} + 2abq^{12n+1} + a^{2}bq^{12n+2} + bq^{6n+4}
+ b^{2}q^{12n+2} + 2b^{2}q^{12n+3} + ab^{2}q^{12n+4} + bq^{6n+5} + b^{2}q^{12n+4} + ab^{2}q^{12n+5}
+ a^{2}q^{12n+3} + 2a^{2}bq^{18n+1} + 2a^{2}bq^{18n+2} + abq^{12n+5} + a^{2}bq^{18n}
+ ab^{2}q^{18n+3} + a^{2}q^{12n−4} + 2ab^{2}q^{18n+4} + abq^{12n+6} + 2ab^{2}q^{18n+5}
+ abq^{12n+7} + a^{2}bq^{18n+3} + ab^{2}q^{18n+6} + b^{2}q^{12n+9} + a^{3}q^{18n−2}
+ a^{3}q^{18n−1} + b^{3}q^{18n+7} + b^{3}q^{18n+8} + a^{2}bq^{12n+1} + aq^{6n+1},

p_{2}(n, a, b, q) = a^{2}bq^{12n−5} + a^{2}bq^{12n−4} + ab^{2}q^{12n−2} + ab^{2}q^{12n−1} + a^{2}b^{2}q^{12n} + a^{3}bq^{18n−4} (3.21)
+ a^{3}bq^{18n−3} + a^{3}bq^{18n−2} + 3a^{2}b^{2}q^{18n} + a^{2}b^{2}q^{18n−1} + ab^{3}q^{18n+2}
+ ab^{3}q^{18n+3} + a^{2}b^{2}q^{18n+1} + ab^{3}q^{18n+4} + a^{3}bq^{18n−10} + a^{2}b^{2}q^{18n−7}
+ 3a^{2}b^{2}q^{18n−6} + a^{3}b^{2}q^{18n−5} + a^{4}bq^{24n−9} + a^{4}bq^{24n−8} + a^{4}bq^{24n−7}
+ 3a^{3}b^{2}q^{24n−5} + a^{3}b^{2}q^{24n−6} + 3a^{2}b^{3}q^{24n−2} + 3a^{3}b^{2}q^{24n−4} + 3a^{2}b^{3}q^{24n−1}
+ a^{3}bq^{18n−9} + a^{3}bq^{18n−8} + a^{2}b^{2}q^{18n−5} + a^{3}b^{2}q^{18n−4} + a^{2}b^{3}q^{24n}
+ ab^{3}q^{18n−4} + ab^{4}q^{24n} + a^{3}b^{2}q^{24n−3} + ab^{4}q^{24n+2} + ab^{3}q^{18n−2}
+ ab^{4}q^{24n+1} + ab^{4}q^{24n+3} + a^{4}bq^{24n−6} + ab^{3}q^{18n−3} + a^{2}b^{3}q^{18n−2}
+ a^{2}b^{3}q^{18n−1} + a^{2}b^{3}q^{24n−3},

and

p_{3}(n, a, b, q) = − a^{3}b^{3}q^{24n−12} − a^{3}b^{3}q^{18n−12} − a^{4}b^{3}q^{24n−11} − a^{4}b^{3}q^{24n−10} − a^{3}b^{4}q^{24n−8} (3.22)
− a^{3}b^{4}q^{24n−7} + 2a^{4}b^{3}q^{30n−17} + 2a^{4}b^{3}q^{30n−16} + 2a^{3}b^{4}q^{30n−14}
+ 2a^{3}b^{4}q^{30n−13} + a^{4}b^{2}q^{24n−21} + a^{5}b^{2}q^{30n−20} + a^{3}b^{3}q^{24n−19}
+ a^{5}b^{2}q^{30n−19} + a^{4}b^{3}q^{30n−18} + a^{3}b^{4}q^{30n−15} + a^{3}b^{3}q^{24n−18} + a^{3}b^{3}q^{24n−17}
+ a^{4}b^{3}q^{30n−15} + a^{3}b^{4}q^{30n−12} + a^{2}b^{5}q^{30n−11} + a^{2}b^{5}q^{30n−10} + a^{2}b^{4}q^{24n−15}.

Sketch of proof. Using the identity J(n) = 0, we find that S_{n−1}(15) is a linear combination of S_{n}(9), S_{n−1}(9), S_{n−2}(9) and S_{n−3}(9). By Lemma 5, we have K(n) = 0. We substitute that linear combination for S_{n−1}(15) and the corresponding one for S_{n}(15) in (3.18). After some simplification, and after replacing n by n − 1, we arrive exactly at (3.19). □

We are now able to prove a recurrence for S_{n}(15).

Lemma 7. For n ≥ 0, we have

(1 + aq^{6n−5} + aq^{6n−4} + bq^{6n−2} + bq^{6n−1})S_{n}(15) (3.23)

= p_{1}(n − 1, aq^{6}, bq^{6}, q)S_{n−1}(15) + (1 − q^{6n−6})p_{2}(n − 1, aq^{6}, bq^{6}, q)S_{n−2}(15)

+ (1 − q^{6n−6})(1 − q^{6n−12})p_{3}(n − 1, aq^{6}, bq^{6}, q)S_{n−3}(15)
+ a^{4}b^{4}q^{30n−18}(1 + aq^{6n+1} + aq^{6n+2} + bq^{6n+4} + bq^{6n+5})
× (1 − q^{6n−6})(1 − q^{6n−12})(1 − q^{6n−18})S_{n−4}(15).

Proof. Since J(n) = 0, we can express S_{n}(15) in terms of the S_{n}(9)′s. Substituting these expressions in (3.23), we get an equation involving S_{n+1}(9), S_{n}(9), …, S_{n−6}(9). In that equation we apply Lemma 6 to S_{n+1}(9). In the result thus obtained, we apply Lemma 6 to S_{n}(9). In the subsequent result obtained, we apply Lemma 6 to S_{n−1}(9). In the result obtained, we again apply Lemma 6, this time to S_{n−2}(9). The result is zero. All these calculations have been performed using Mathematica. □

Lemma 8. For n ≥ 0,

S_{n}(15, a, b, q) = (1 + aq)(1 + aq^{2})(1 + bq^{4})(1 + bq^{5})S_{n−1}(9, aq^{6}, bq^{6}, q).

Proof. Comparing Lemma 7 and Lemma 6, we find that both sides of Lemma 8 satisfy the same fourth order recurrence valid for n ≥ 1. Hence we have only to verify Lemma 8 for n = 0, 1, 2, 3. This is a routine verification, and can therefore be left to the reader. □

Proof of Theorem 4. For 0 ≤ j ≤ 15, we have

lim_{n→∞} S_{n}(j, a, b, q) = ∑_{μ,ν,N≥0} B(μ, ν, N)a^{μ}b^{ν}q^{N} ≡ S(a, b, q).

Letting n → ∞ in Lemma 8, we find that

S(a, b, q) = (1 + aq)(1 + aq^{2})(1 + bq^{4})(1 + bq^{5})S(aq^{6}, bq^{6}, q).

Iterating the above equation, we obtain

S(a, b, q) = ∏_{n=0}^{∞}(1 + aq^{6n+1})(1 + aq^{6n+2})(1 + bq^{6n+4})(1 + bq^{6n+5})

= ∑_{μ,ν,N≥0} A(μ, ν, N)a^{μ}b^{ν}q^{N},

the latter equality being due to (2.1). Thus we get

∑_{μ,ν,N≥0} B(μ, ν, N)a^{μ}b^{ν}q^{N} = ∑_{μ,ν,N≥0} A(μ, ν, N)a^{μ}b^{ν}q^{N},

which, upon comparison of coefficients of a^{μ}b^{ν}q^{N}, implies

A(μ, ν, N) = B(μ, ν, N)

for all non-negative μ, ν and N. This is exactly the claim in Theorem 4. □

References

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[2] Andrews, G.E., On the general Rogers–Ramanujan theorem, Mem. Amer. Math. Soc., No. 152, 1974, pp. 1–86.

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Department of Studies in Mathematics, University of Mysore, Manasagangotri, Mysore – 570 006, Karnataka, India

Institut Girard Desargues, Université Claude Bernard Lyon-I, 21, avenue Claude Bernard, F-69622 Villeurbanne Cedex, France