Overpartitions with Restricted Odd Differences

We use q-difference equations to compute a two-variable q-hypergeometric generating function for overpartitions where the difference between two successive parts may be odd only if the larger part is overlined. This generating function specializes in one case to a modular form, and in another to a mixed mock modular form. We also establish a two-variable generating function for the same overpartitions with odd smallest part, and again find modular and mixed mock modular specializations. Applications include linear congruences arising from eigenforms for 3-adic Hecke operators , as well as asymptotic formulas for the enumeration functions. The latter are proven using Wright's variation of the circle method.

The second identity follows from a straightforward combinatorial argument, but the first is more subtle and our proof depends on showing that both sides satisfy a certain qdifference equation.Note that the positivity of the coefficients on the right-hand side of (1) is not immediately apparent.When x = 1 in (1) or −1 in (2), then we have a modular form, and when x = −1 in (1) or 1 in (2), then we have the product of a modular form and a mock theta function, a so-called mixed mock modular form (see [11]).More precisely, define the mock theta functions γ(q) and χ(q) by γ(q) := n 0 the electronic journal of combinatorics 22(3) (2015), #P3.17 and (That these are mock theta functions follows from work of the first and the third authors [5].)Let t ± (n) (resp.s ± (n)) denote the number of overparititions counted by t(n) (resp.s(n)) with largest part even (resp.odd).
Corollary 2. We have 1 + 3 Equation ( 6) combined with work of the first and the third author on overpartitions and class numbers [6,7] implies that the generating function for t + (n) − t − (n) is an eigenform modulo 3 for the weight 3/2 Hecke operators.This is recorded below along with a congruence for t(n) modulo 3.Even though the congruence for t(n) is immediate from (5), it would be interesting to see if it can be deduced from an explicit 3-fold symmetry for the overpartitions counted by t(n).

Corollary 3.
1.For a prime = 2, 3, and n 0, we have 2. For n 1, we have Our final results give asymptotic formulas for s(n) and t + (n) − t − (n), which are the cases from Corollary 2 in which the generating functions are mixed mock modular forms.Theorem 4. As n → ∞, we have Remark.Since ( 5) and ( 7) are (up to rational q-powers) weakly holomorphic modular forms of non-positive weight, Rademacher and Zuckerman's famous refinement of the Hardy-Ramanujan Circle Method applies [9,12,13].These results allow one to use the cuspidal principal parts in order to calculate exact formulas for the coefficients.In fact, one finds that t(n) is asymptotically equivalent to (11) and s + (n)−s − (n) is asymptotically equivalent to (12).
We have highlighted the two cases in Theorem 4 because the Hardy-Ramanujan-Rademacher Circle Method does not generally apply to mixed mock modular forms.Instead we use Wright's Circle Method [15], which uses a single "Major Arc" centered around a dominant cusp to derive an asymptotic expansion for the coefficients.In most examples in the literature this cusp is q = 1, but for (12) the Major Arc is instead centered at q = −1.
The rest of the paper is organized as follows.In the next section we prove Theorem 1 using analytic and combinatorial arguments.In Section 3 we then apply q-series identities to establish Corollaries 2 and 3. Section 4 outlines the proof of Theorem 4 using Wright's Circle Method.We conclude in Section 5 by briefly tying the present work to other recent studies of mixed mock modular q-series and q-difference equations.

Combinatorial recurrences and the proof of Theorem 1
In this section we use the analytic and combinatorial theory of q-difference equations to prove Theorem 1.We begin by considering (1).Define H(x; q) by Now, if λ is an overpartition counted by t(m, n), then its smallest part is either 1 or at least 2. In the first case, we may remove the 1 along with any other 1s in λ and then subtract 1 from each remaining part to obtain an overpartition counted by t(m−a, n−m), where a 1 is the number of 1s occurring in λ.In the second case, we may remove any 2s in λ and then subtract 2 from each remaining part to obtain an overpartition counted by t(m − b, n − 2m), where b 0 is the number of 2s occurring in λ.Note that if the smallest part in this new partition is odd, then it must have correspondingly occurred as an overlined part in λ.In other words, we have the electronic journal of combinatorics 22(3) (2015), #P3.17 Together with the fact that H(0; q) = 1, this uniquely defines H(x; q).We claim that the right-hand side of (1) also satisfies (13) with the same initial condition.To see this, first define We have that x n q n n−1 j=1 1 − q j + q 2j (q 2 ; q x n q n n−1 j=1 1 − q j + q 2j (q 2 ; q 2 ) n−1 Now, the right-hand side of (1) is G(x; q), where and applying (14) to G(x; q) gives Comparing ( 13) and ( 15) and noting the initial condition G(0; q) = 1, yields (1).Next we consider (2).Here we require the notion of the conjugate of an overpartition, which is obtained by reading the columns of the Ferrers diagram (an overlined part is designated by a mark at the end of a row).The right-hand side of (2) is n 1 Here the mth summand is the generating function for overpartitions where the largest part m occurs overlined and an odd number of times, while each part less than m occurs an even number of times if it does not occur overlined.Conjugating we obtain an overpartition counted by s(m, n).This is illustrated in Figure 1.This establishes (2) and completes the proof of Theorem 1.The overpartition (11,11,11,8,8,7,6,6,3,3,3,3) and its conjugate (12,12,12,8,8,8,6,5,3,3,3).

Proof of Corollaries 2 and 3
In this section we use q-series identities to obtain the generating functions for overpartitions with restricted odd differences in Corollary 2. We require the q-Gauss summation formula [8, Equation (1.5.1)], ) and c = −q, we obtain which proves (5).
The final two equations in the statement of the corollary are proven similarly.For (7) we set x = −1 in (2) and apply (16) with a = 1/b = ζ 3 and c = −q, whereas for ( 8 We remark that (5) can also be established by a simple combinatorial argument, as in the proof of (2).Indeed, conjugating an overpartition λ counted by t(n) gives an overpartition with the property that if m doesn't occur, then m occurs an even number of times.Hence This argument easily generalizes.Let t (k) (n) denote the number of overpartitions of n where (i) consecutive parts differ by a multiple of (k + 1) unless the larger of the two is overlined, and (ii) the smallest part is overlined unless it is divisible by k + 1.Then we have We close this section by determining the behavior of these generating functions modulo 3, which is achieved by relating them to previously studied Hecke eigenforms.
Proof of Corollary 3. Reading (6) modulo 3, we have We recall from Proposition 5.1 of [7] that if α(n) is defined by the electronic journal of combinatorics 22(3) (2015), #P3.17 This proves (9).To show (10), we read (5) modulo 3, obtaining The equality is Gauss' identity [1, Equation (2.2.12)].4 Wright's Circle Method and the proof of Theorem 4 In this section we apply Wright's Circle Method [15] to prove Theorem 4. As is typical of the Circle Method, Wright's approach recovers the coefficients of a generating series by applying Cauchy's Theorem, with the main asymptotic contribution coming from a single Major Arc (for our two results, this is centered around q = 1 and q = −1, respectively).
For notational convenience we recall ( 6) and ( 8) and write

Analytic behavior of f 1 (q)
We begin with s(n), as the asymptotic analysis is analogous to Wright's original examples, which were also centered around q = 1.Throughout the section we therefore use the standard parameterization q = e 2πiτ , where τ = x + iy and y > 0. Recalling (3), we use an alternative expression for the mock theta function, which follows from equation (1.3) in [5], namely with ζ 3 := e 2πi 3 .We show that the overall asymptotic behavior of f 1 (q) is largely controlled by the singularities of the infinite product from (18), so the dominant pole is at q = 1.The following result describes the behavior of f 1 (q) in a neighborhood of q = 1 and gives a uniform bound away from this point.
the electronic journal of combinatorics 22(3) (2015), #P3.17 A short calculation shows that the bound in part 2 is indeed an error term as long as = 5.543 . . . .Corollary 6.If M 6 and M y < |x| 1  2 , then for some ε > 0 we have Proof.In order to determine the behavior of f 1 (q) near q = 1, we begin by studying γ(q) near this point.By Taylor's Theorem, we have and by (3), we directly calculate Since |x| M y, we therefore have the following expansion as n → ∞: We next determine the asymptotic behavior of the infinite product.The modular inversion formula for Dedekind's eta-function (page 121, Proposition 14 of [10]) implies that, as τ → 0, This formula directly gives that, as n → ∞, Combining ( 22) and (24) completes the proof of (20).We next consider f 1 (q) away from q = 1.We begin by considering the sum in (19).Bounding each term in the sum absolutely, we get where the final bound follows either by an integral comparison or using the transformation law of the theta function.
For convenience denote the above product by g 1 (q).We calculate log (g .
This implies that Using the transformation (23), we find that Converting ( 28) and (29) to Laurent series and combining with (26) and ( 27), we obtain This completes the proof of (21) and Theorem 5.
This gives (34) Next, using a similar argument as in (25), we conclude that Combining the products from ( 6) and (33), we therefore write After some simplification, we obtain log (g 2 (q)) = ) .
We therefore have the bound Once again (23) implies that  The remainder of the proof of (35) proceeds analogously to the arguments following equation (28).
Proof of (12).We proceed as in Section 4.2, with the main differences due to the different parameterization around q = −1.By Cauchy's Theorem, we again have that for all n 1, After pulling the (−1) n out of the integral, the remainder of the argument follows as in (32) (note that Lemma 7 is now applied with u = π √ n 3 and s = 0).