Some partition and analytical identities arising from the Alladi, Andrews, Gordon bijection

In the work of Alladi et al. (J Algebra 174:636–658, 1995) the authors provided a generalization of the two Capparelli identities involving certain classes of integer partitions. Inspired by that contribution, in particular as regards the general setting and the tools the authors employed, we obtain new partition identities by identifying further sets of partitions that can be explicitly put into a one-to-one correspondence by the method described in the 1995 paper. As a further result, although of a different nature, we obtain an analytical identity of Rogers–Ramanujan type, involving generating functions, for a class of partition identities already found in that paper and that generalize the first Capparelli identity and include it as a particular case. To achieve this, we apply the same strategy as Kanade and Russell did in a recent paper. This method relies on the use of jagged partitions that can be seen as a more general kind of integer partitions.


Introduction
In a 1969 paper, [2], Andrews characterized the type of partition sets that could be set into a bijection using Euler's classical trick to show that partitions of n into distinct P. Mercuri is supported by the research grant "Ing. Giorgio Schirillo" of the Istituto Nazionale di Alta Matematica "F. Severi", Rome. B A. Vietri andrea.vietri@uniroma1.it 1 Università degli Studi di Roma La Sapienza, Rome, Italy 2 Università degli Studi di Roma "Tor Vergata", Rome, Italy parts are as many as partitions of n into odd parts. In particular, Andrews proved that an identity of Schur ([12]) and one of Göllnitz ([10]) provide examples of "Euler-pairs". Inspired by that paper, here we look at one of the identities given in [6], see also [8].
We study the bijection provided by Alladi et al. in [1] and we find new sets of partitions that can be set into a bijection using the same approach. For further details on this subject and some generalizations we refer the reader to [3][4][5]7,9].
As the starting point of our research, we consider the partition identity which was proved in [1] and in particular the Concluding Remarks (Section 7), according to which the first Capparelli's identity (see [7]) can be generalized from modulo 3 to modulo t by means of suitable dilations. In Sect. 2 we find an analytical identity for the partition identity modulo t. This is done using the same method as in [11] to compute the generating functions of the sum side. In Sect. 3 we again look back at [1], this time by generalizing the machinery which led the authors to build up the partition identity. In particular, we study a different class of partition identities which are indexed by two coprime integers s and t. As better clarified in Sect. 3, these two parameters play different roles and actually generalize the roles of s = 2 and t = 3 in [1]. In the present setting we obtain a new family of partition identities.

An analytical identity for an Alladi-Andrews-Gordon bijection
Let n, t be positive integers with t > 2 and denote by C(n) t the set of partitions of n with distinct parts that are either divisible by t or congruent to t ± 1 (mod 2t). Furthermore, let D(n) t denote the set of partitions of n with distinct parts larger than 1 that are either divisible by t or congruent to ± 1 (mod t) and whose difference is at least t + 1, with the following exception: the difference between two adjacent parts can be smaller than t + 1 if they are both divisible by t or their sum is divisible by 2t. Alladi et al. proved the following proposition (see [1]).

Proposition 1 C(n) t and D(n) t have the same cardinality.
In the present section we find an analytical expression for the above partition identity. The proof of our result is inspired by the argument in [11]. In accordance with that paper, we provide the following definition. Definition 2 For a fixed positive integer k, a k-jagged partition is a finite sequence (a 1 , . . . , a m ) such that a 1 , . . . , a m ∈ Z, a 1 > 0, and a i+1 − a i ≥ −k, for every i = 1, . . . , m − 1.
Clearly, if k = 0, we obtain the classical partitions written in weakly increasing order, as in [11]. The set of classical partitions (b 1 , . . . , b m ) is easily seen to be in bijection with the set of k-jagged partitions (a 1 , . . . , a m ) by associating b i to a i + (i − 1)k; essentially, we add a k-staircase. In the present paper, this bijection plays an important role when passing to generating functions. For our purposes, we now consider a special class of k-jagged partitions. With this definition it is now possible to introduce the key concept we require for our proof.

Definition 4
For each positive integer j and strong k-jagged partition μ, we define the maximal block M j corresponding to j in the following way. If there is an element a i of μ such that a i = j and with the property that every element before a i is smaller than j, then M j is the maximal subsequence of μ starting with a i whose elements belong to the set { j, j − 1, j − 2, . . . , j − k}. If there is no element a i satisfying the above conditions, then M j is the empty set.

Theorem 5 Let t be an integer greater than 2 and
We have n≡0,t−1,t,t+1 (mod 2t) where the left-hand side is the generating function of the partitions with distinct parts congruent to 0, t −1, t, t +1 (mod 2t), and the right-hand side is the generating function of partitions with distinct parts greater than 1 and congruent to 0, ± 1 (mod t) such that the difference between consecutive parts is at least t + 1 unless they are both 0 (mod t) or their sum is 0 (mod 2t).
Proof It is straightforward to see that the infinite product is the generating function of C(n) t . With Proposition 1 in mind, it suffices to show that the quadruple sum is the generating function of D(n) t . For every positive integer j, the configurations that are not allowed in D(n) t are the following: If we subtract a t-staircase, we get Hence, the maximal blocks are the following: It follows that the generating function is . Moreover, since we are not interested in taking account of the number of parts, we set x = 1 and this gives the result.

Remark 6
For t = 3 the sum of the previous proposition gives the same sum obtained in [11] using the substitution x → 1, i → b, j → c, k → d, l → a and also the same sum obtained in [9, Equation (2.6)] using the substitution

Partition identities for the "s-rate , t-stack" case
As mentioned in the Introduction, in the present section we construct an original class of partition identities which are indexed by two coprime integers s and t, with t odd. Given a partition of a positive integer n into distinct parts, we list its parts in decreasing order, as in [1]. Fix three positive integers n, s, t, with s, t coprime and greater than 1; denote by C(n) t s the set of partitions of n with distinct parts multiple of s or multiple of t.
and with the following conditions which all parts d i must fulfill.
we require that f p be congruent to 0 or t (mod s), and the same must hold for for some positive integer r , then at least one of the following conditions must be satisfied: Notice that C(n) t 2 has a different meaning from C(n) t ; a similar remark concerns D(n) t 2 and D(n) t . We are going to prove the following result. Proposition 7 C(n) t s and D(n) t s have the same cardinality.
In order to establish the above proposition we prove a stronger result, namely, Proposition 8, for which some more terminology is needed. We denote by C(n; i 1 , . . . , i t−1 , k) t s the subset of C(n) t s whose parts are grouped according to their congruence class as follows: i 1 parts congruent to s (mod st), . . . , i t−1 parts congruent to (t − 1)s (mod st), k parts larger than t( t−1 j=1 i j ) and congruent to 0 (mod t). Similarly, set D(n; i 1 , . . . , i t−1 , k) t s to be the subset of D(n) t s having i h parts congruent to hs (mod t), with 1 ≤ h ≤ t − 1, and k further parts congruent to 0 (mod t). Now we proceed with the proof of the stronger result.
Proof Given a partition in C(n; i 1 , . . . , i t−1 , k) t s , we associate to it a partition in D(n; i 1 , . . . , i t−1 , k) t s , with an algorithm that generalizes the classical case t = 3 in [1]; also our terminology traces back to that paper. Later we show that such a procedure is reversible.
Step 1 Setting p = t−1 j=1 i j , split π into the subpartition π 1 = (a 1 , . . . , a p ) of those elements not congruent to 0 (mod t) and the subpartition π 2 made up of those elements congruent to 0 (mod t). Notice that two elements in π 1 have difference s j only if they have sum not congruent to ±s j (mod st), i.e., they satisfy D3-II. Indeed, suppose we have two parts sα, sβ ∈ π 1 such that sα − sβ = s j and s j. Assuming, by contradiction, that sα + sβ ≡ ±s j (mod st), since α = β + j we have either 2sβ + s j ≡ s j (mod st) or 2sβ + s j ≡ −s j (mod st). In the first case 2β ≡ 0 (mod t). In the second, likewise, we have 2α ≡ 0 (mod t). Since t is odd, it follows that α or β are congruent to 0 (mod t), in both cases contradicting the assumption.
Step 4 Let α 1 ≥ α 2 ≥ · · · ≥ α u 1 where α i is the cardinality of the ith row of the diagram associated to π * 4 . Add the partitions π 1 and π * 4 by adding the corresponding parts a i + α i for 1 ≤ i ≤ u 1 and leaving alone the elements a u 1 +1 . . . a p . Notice that this is possible as u 1 ≤ p, since b 1 = u 1 t ≤ t p. We thus get a new partition π 6 = (a 1 , . . . , a p ), with a i > a i+1 , for i = 1, . . . , p − 1. This operation either leaves the differences between the parts unchanged or increases them by multiples of t so that condition D3-II still holds. Moreover the condition D0 for the symbols a i in place of f i holds; in fact the difference between a i and a i+1 can be either 0 or t (mod s) and no larger multiples are possible, because of the structure of the diagram.
Step 5 Construct a string π 5 /π 6 by juxtaposing, left to right, first the elements of π 5 , then those of π 6 . In this string i h parts are congruent to hs (mod t), for h = 1, . . . , t − 1, and k parts congruent to 0 (mod t), the latter being larger than t p.
Step 6 Subtract multiples of t to the elements of π 5 /π 6 , by obtaining the following new elements:b Notice that while the elementsb i remain in a nonincreasing order, this does not necessarily happen for the elementsā i . To be more precise, the nondecreasing order fails whenever a i − a i+1 = j < t. In such a caseā i+1 =ā i + t − j >ā i . Moreover the elementsb i are all strictly positive since they are above the threshold value t p, while the elementsā i may be negative.
Step 7 Starting from the string S 0 = (b 1 , . . . ,b k ,ā 1 , . . . ,ā p ), we define a recursive algorithm which will lead us to a final string S f in k steps. Define the generic i-th step, 1 ≤ i ≤ k. Placeb i in the string S i−1 in the rightmost possible position so that all the elements to its left are larger than itself. At the end of the process we get the desired string S f .
Step 8 Denote by c 1 , . . . , c p+k the elements of S f . We construct the following elements d i , for i = 1, . . . , p + k: . . .
The difference between two elements d i and d i+1 that are not congruent to 0 (mod t) may be less than t + 1 only if is a multiple of s and the sum is not congruent to ± (mod t).
Notice that D0 holds. Indeed, the elements f i obtained from the elements d i not congruent to 0 (mod t), by the formulas (1), coincide with a i for which we already observed that they satisfy the requirements in Step 4. Therefore this algorithm transforms a partition π in C(n; i 1 , . . . , i t−1 , k) t s into a partition in D(n; i 1 , . . . , i t−1 , k) t s that we denote by π 3 again in accordance with [1]. Finally, we only need to show that this procedure is completely reversible. Given a partitionπ of n in D(n; i 1 , . . . , i t−1 , k) t s it is trivial to trace back the steps up to Step 5. We thus get a partition of n made of some elements not congruent to 0 (mod t) and of some elements congruent to 0 (mod t). The set of the first type of elements, consistently with previous notation, we denote by π 6 = (a 1 , . . . , a p ), with a i > a i+1 , for i = 1, . . . , p − 1. The set of the second type of elements we denote by Now, check whether a p is congruent to 0 or t modulo s. In the first case do nothing. In the second case create a diagram with one row of t squares. Proceed inductively by creating, in corresponding with each a i , a row of squares to be stacked on top of the row corresponding to a i+1 , with the same number of squares of the row corresponding to a i+1 if a i ≡ a i+1 (mod s); otherwise add t new squares to the row. We thus form a diagram that gives, by using stacks of t columns, the elements of π 4 and, using the rows (corresponding to π * 4 ), the quantities to be subtracted from the elements of π 6 , in order to obtain the elements of π 1 . In this fashion, the partition π in C(n, i 1 , . . . , i t−1 , k) t s is completely reconstructed.

Remark 9
If t is even we can state an analogous proposition, but we must substitute the D3 condition with the following one: D3 . If d i − d i+r < t + 1 for some positive integer r , then at least one of the following conditions must be satisfied: II. either at least one between d i and d i+r is equal to sk t 2 +ht for some nonnegative integers h and k or d i + d i+r ≡ ±s j (mod st).

Remark 10
We observe that D(n) t s and D(n) s t have the same cardinality, because C(n) t s = C(n) s t .
We conclude this section with some remarks on the particular case s = 2.

Remark 11
If s = 2, the condition D0 is trivially satisfied.

Proposition 12
Condition D2 is redundant if s = 2 and t = 3.
Proof By contradiction, assuming that D2 does not hold, let i 0 be the largest integer such that 3 | d i 0 ≤ 3(m − i 0 ) and let d i 0 +h be the next multiple of 3 from left to right -if there is no such multiple, set h = m − i 0 + 1 and define d m+1 = 0. Clearly, Notice that h must be larger than 1. The h − 1 parts between d i 0 and d i 0 +h are all congruent to 1 or 2 (mod 3). Now condition D3 forces the leftmost part to be not larger than d i 0 − 4 and the rightmost to be not smaller than d i 0 +h + 4. Furthermore, using condition D3, it is easy to see that every interval of the form [α, α + 5] contains at most 2 this h − 1 parts. It follows that there are no more than , which is less than h − 1, a contradiction. In the remaining case, namely if h = m − i 0 + 1, the element d m might be equal to 2 but the above argument is still valid, using similar calculations, as long as h is even. Instead, if h is odd (necessarily h ≥ 3), we slightly improve the argument. Since d m−1 ≥ 4, counting the h − 2 parts from d i 0 +1 to d m−1 leads to the following contradiction: Notice that for t odd and larger than 3 Condition D2 is necessary.

Examples
In this section we give three examples of the bijection described in the previous section.
So p = 8 and the threshold is t p = 56.
Step 6 Subtract from the string just obtained multiples of 7 as follows:  7 2 as desired. We now check that this process is reversible. Given the partition π 3 = (101, 77,67,56,47,45,17,8,4,2), the steps 8,7,6,5 are easily reversible and lead to π 5 and π 6 = (84, 70,87,60,47,45,17,8,4,2). To recover π 4 proceed as follows. Note the position where, starting from the right, we find the first odd part. In our example it is 17 and, correspondingly, we draw a row of 7 squares. Going leftward and ignoring the multiple of 7, we have the sequence (17, 45, 47, 60, 87) that is congruent to (1, 1, 1, 0, 1) modulo 2. For each element with the same parity of the previous one, we draw a row upon the others with the same number of squares than the row below it. For each element with a different parity than the previous one, we draw a row upon the others with 7 squares more than the row below it. This generates the diagram that coincides with the diagram in Step 3. In this diagram there are 3 blocks of 7 columns containing 35, 14 and 7 squares, respectively. Hence π 4 = (35, 14, 7). The rows of the diagram give the string (21, 14, 7, 7, 7) which must be subtracted to (87, 60, 47, 45, 17) to get π 1 .

Example 14
We illustrate an example of the bijection when s = 3 and t = 7.
So p = 7 and the threshold is t p = 49.
Also in this case it is easy to invert the process.