On the Positive Moments of Ranks of Partitions

By introducing $k$-marked Durfee symbols, Andrews found a combinatorial interpretation of $2k$-th symmetrized moment $\eta_{2k}(n)$ of ranks of partitions of $n$ in terms of $(k+1)$-marked Durfee symbols of $n$. In this paper, we consider the $k$-th symmetrized positive moment $\bar{\eta}_k(n)$ of ranks of partitions of $n$ which is defined as the truncated sum over positive ranks of partitions of $n$. As combintorial interpretations of $\bar{\eta}_{2k}(n)$ and $\bar{\eta}_{2k-1}(n)$, we show that for fixed $k$ and $i$ with $1\leq i\leq k+1$, $\bar{\eta}_{2k-1}(n)$ equals the number of $(k+1)$-marked Durfee symbols of $n$ with the $i$-th rank being zero and $\bar{\eta}_{2k}(n)$ equals the number of $(k+1)$-marked Durfee symbols of $n$ with the $i$-th rank being positive. The interpretations of $\bar{\eta}_{2k-1}(n)$ and $\bar{\eta}_{2k}(n)$ also imply the interpretation of $\eta_{2k}(n)$ given by Andrews since $\eta_{2k}(n)$ equals $\bar{\eta}_{2k-1}(n)$ plus twice of $\bar{\eta}_{2k}(n)$. Moreover, we obtain the generating functions of $\bar{\eta}_{2k}(n)$ and $\bar{\eta}_{2k-1}(n)$.


Introduction
This paper is concerned with a combinatorial study of the symmetrized positive moments of ranks of partitions. The notion of symmetrized moments was introduced by Andrews [1]. The odd symmetrized moments are zero due to the symmetry of ranks. For even symmetrized moments, Andrews found a combinatorial interpretation by introducing kmarked Durfee symbols. It is natural to investigate the combinatorial interpretation of the odd symmetrized moments which are truncated sum over positive ranks of partitions of n. We give combinatorial interpretations of the even and odd positive moments in terms of k-marked Durfee symbols, which also lead to the combinatorial interpretation of the even symmetrized moments of ranks given by Andrews.
The rank of a partition λ introduced by Dyson [6] is defined as the largest part minus the number of parts. Let N(m, n) denote the number of partitions of n with rank m. The generating function of N(m, n) is given by Theorem 1.1 (Dyson-Atkin-Swinnerton-Dyer [3]). For fixed integer m, we have +∞ n=0 N(m, n)q n = 1 (1.1) Recently, Andrews [1] introduced the k-th symmetrized moment η k (n) of ranks of partitions of n as given by It can be easily seen that for given k, η k (n) is a linear combination of the moments N j (n) of ranks given by Atkin and Garvan [4] For example, In view of the symmetry N(−m, n) = N(m, n), we have η 2k+1 (n) = 0. As for the even symmetrized moments η 2k (n), Andrews gave the following combinatorial interpretation by introducing k-marked Durfee symbols. For the definition of k-marked Durfee symbols, see Section 2. Theorem 1.2 (Andrews [1]). For fixed k ≥ 1, η 2k (n) is equal to the number of (k + 1)marked Durfee symbols of n.
Andrews [1] proved the above theorem by using the k-fold generalization of Watson's q-analog of Whipple's theorem. Ji [8] gave a combinatorial proof of Theorem 1.2 by establishing a map from k-marked Durfee symbols to ordinary partitions. Kursungoz [9] provided another proof of Theorem 1.2 by using an alternative representation of k-marked Durfee symbols.
In this paper, we introduce the k-th symmetrized positive momentη k (n) of ranks as given by or equivalently, Furthermore, it is easy to see that for given k,η k (n) is a linear combination of the positive moments N j (n) of ranks introduced by Andrews, Chan and Kim [2] as given by For example,η By the symmetry N(−m, n) = N(m, n), it is readily seen that The main objective of this paper is to give combintorial interpretations ofη 2k (n) and η 2k−1 (n). We show that for given k and i with 1 ≤ i ≤ k + 1,η 2k−1 (n) equals the number of (k + 1)-marked Durfee symbols of n with the i-th rank being zero andη 2k (n) equals the number of (k + 1)-marked Durfee symbols of n with the i-th rank being positive. It should be noted thatη 2k−1 (n) andη 2k (n) are independent of i since the ranks of k-marked Durfee symbols are symmetric, see Andrews [1,Corollary 12].
With the aid of Theorem 2.1 and Theorem 2.2 together with the generating function (1.1) of N(m, n), we obtain the generating functions ofη 2k (n) andη 2k−1 (n).

Combinatorial interpretations
In this section, we give combinatorial interpretations ofη 2k−1 (n) andη 2k (n) in terms of the k-marked Durfee symbols. For a partition λ, we write λ = (λ 1 , λ 2 , . . . , λ s ), so that λ 1 is the largest part and λ s is the smallest part of λ. Recall that a k-marked Durfee symbol of n introduced by Andrews [1] is a two-line array composed of k pairs (α i , β i ) of partitions along with a positive integer D which is represented in the following form: where the partitions α i and β i satisfy the following four conditions: (1) The partitions α i (1 ≤ i < k) are nonempty, while α k and β i (1 ≤ i ≤ k) are allowed to be empty; For example, consider the following 3-marked Durfee symbol τ .
Theorem 2.1. For fixed positive integers k and i with 1 ≤ i ≤ k + 1,η 2k−1 (n) is equal to the number of (k + 1)-marked Durfee symbols of n with the i-th rank equal to zero.
For the even case, we have the following interpretation.
Theorem 2.2. For fixed positive integers k and i with 1 ≤ i ≤ k + 1,η 2k (n) is equal to the number of (k + 1)-marked Durfee symbols of n with the i-th rank being positive.
The proofs of the above two interpretations are based on the following partition identity given by Ji [8]. We shall adopt the notation D k (m 1 , m 2 , . . . , m k ; n) as used by Andrews [1] to denote the number of k-marked Durfee symbols of n with i-th rank equal to m i . Theorem 2.3. Given k ≥ 2 and n ≥ 1, we have (2.1) To prove the above two interpretations, we also need the following symmetric property given by Andrews [1]. Boulet and Kursungoz [5] found a combinatorial proof of this fact.
Given k and n, let c k (n) denote the number of integer solutions to the equation where the variables m i are integers and the variables t i are nonnegative integers. It is easy to see that the generating function of c k (n) is equal to ∞ n=0 c k (n)q n = (1 + 2q + 2q 2 + 2q 3 + · · · ) k (1 + q 2 + q 4 + q 6 + · · · ) k Equating the coefficients of q n on the both sides of (2.4), we get which is equal toη 2k−1 (n). This completes the proof.  Given k and n, letc k (n) denote the number of integer solutions to the equation where the variable m 1 is a positive integer, the variables m i (2 ≤ i ≤ k + 1) are integers and the variables t i are nonnegative integers. An easy computation shows that so thatc We writec

It follows that
which equalsη 2k (n), as required.
Note that the number D k (m 1 , . . . , m k ; n) has the mirror symmetry with respect to each m i , that is, for 1 ≤ i ≤ k, we have Using this mirror symmetry, Theorem 2.2 can be restated as follows.
By Theorem 2.3, we have