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wbridges@uni-koeln.de Department of Mathematics and −1 We prove asymptotic formulas for the complex coeﬃcients of (ζ q; q) ,where ζ is a Computer Science, University of ∞ Cologne, Weyertal 86-90, 50931 root of unity, and apply our results to determine secondary terms in the asymptotics for Cologne, Germany p(a, b, n), the number of integer partitions of n with number of parts congruent a modulo b. Our results imply that, as n →∞, the diﬀerence p(a ,b,n) − p(a ,b,n) for 1 2 a = a oscillates like a cosine when renormalized by elementary functions. Moreover, 1 2 we give asymptotic formulas for arbitrary linear combinations of {p(a, b, n)} . 1≤a≤b Keywords: Circle method, Partitions, Asymptotics, Sign-changes, Secondary terms Mathematics Subject Classiﬁcation: 11P82, 11P83 1 Introduction and statement of results Let n be a positive integer. A partition of n is a weakly decreasing sequence of positive integers that sum to n. The number of partitions of n is denoted by p(n). For example, one has p(4) = 5, since the relevant partitions are 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1. When λ +···+ λ = n,the λ are called the parts of the partition λ = (λ , ... , λ ), and we 1 r j 1 r write λ n to denote that λ is a partition of n. The partition function has no elementary closed formula, nor does it satisfy any ﬁnite order recurrence. However, when deﬁning p(0) := 1, its generating function has the following product expansion 1 q p(n)q = = , (1.1) 1 − q η(τ) n≥0 k≥1 2πiτ where as usual q := e and η(τ) denotes the Dedekind eta function. In [13], Hardy and Ramanujan used (1.1) to show the asymptotic formula 1 2n p(n) ∼ √ exp π ,n →∞. (1.2) 4 3n © The Author(s) 2022. This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. 0123456789().,–: volV 61 Page 2 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 For their proof they introduced the circle method, certainly one of the most useful tools in modern analytic number theory. The now so-called Hardy–Ramanujan circle method uses modular type transformations to obtain a divergent asymptotic expansion whose truncations approximate p(n) up to small errors. A later reﬁnement by Rademacher [20], exploiting the modularity of η(τ), provides a convergent series for p(n). In this work we study the distribution of ordinary partitions whose number of parts are congruent to some residue class a modulo b. The number of such partitions λ n is denoted by p(a, b, n). It is well known that the numbers p(a, b, n) are asymptotically equidistributed; i.e., p(n) p(a, b, n) ∼ ,n →∞. (1.3) The proof of (1.3) begins by writing the generating function for p(a, b, n) in terms of non- modular eta-products twisted with roots of unity modulo b which is a direct consequence of the orthogonality of roots of unity: ⎛ ⎞ −ja 24 ζ 1 q ⎜ ⎟ n b p(a, b, n)q = + , (1.4) ⎝ ⎠ b η(τ) ζ q; q n≥0 b−1≥j≥1 2πi where (a; q) := (1 −aq ) denotes the usual q-Pochhammer symbol and ζ := e . ∞ b n≥0 Since the ﬁrst term does not depend on a and dominates the other summands, (1.3)can be seen as a corollary of (1.4). Similar results have also been proved for related statistics. For example, the rank of a partition is deﬁned to be the largest part minus the number of parts; Males [17] showed that the Dyson rank function N(a, b, n) (the number of partitions with rank congruent to a modulo b) is asymptotically equidistributed. Males’ proof uses Ingham’s Tauberian theorem [15] and exploits the modularity of the generating function of N(a, b, n), which is given in terms of twisted generalizations of one of Ramanujan’s famous mock theta functions. In contrast, the twisted eta-products in (1.4) lack modularity. If we now consider diﬀerences, p(a ,b,n) − p(a ,b,n), the main terms in (1.4) cancel 1 2 and the behavior must be determined by secondary terms. In the following example we look at the diﬀerences of two modulo 5 partition functions (Fig. 1). Example 1.1 Consider the case a = 1, a = 4and b = 5. The q-series of the diﬀerences 1 2 has the following oscillating shape n 2 3 7 8 9 10 11 12 13 14 (p(1, 5,n) − p(4, 5,n)) q =q + q + q − q − 2q − 3q − 4q − 4q − 5q − 6q − 7q n≥0 15 16 22 40 48 49 − 7q − 7q − ··· + 2q + ··· + 109q + ··· + 11q − 24q 75 85 86 120 − ··· − 1998q − ··· − 266q + 163q + ··· + 40511q + ··· 133 134 + 3701q − 3587q − ··· . A similar sounding, but entirely diﬀerent problem is to study the total number of parts that are all congruent to a modulo b. This problem was studied in detail by Beckwith–Mertens [2,3] for ordinary partitions and recently by Craig [12] for distinct parts partitions. Note that the number of partitions λ n with largest part (λ) equals the number of partitions with number of parts (λ) (see also [1], Ch. 1), so the results of this paper can be reformulated taking number of parts into account instead. W. Bridges et al. Res Math Sci (2022) 9:61 Page 3 of 31 61 Fig. 1 For b = 5, a = 1and a = 4, we plot the diﬀerence p(1, 5,n) − p(4, 5,n) with log-scaling. The abrupt 1 2 vertical changes in the graph indicate the location of the sign changes in the sequence Our ﬁrst result predicts this oscillation as follows. As usual, we deﬁne the dilogarithm for |z|≤ 1 by the generating series Li (z):= . n≥1 Throughout we use the principle branch of the complex square root and logarithm. Theorem 1.2 Let b ≥ 5 be an integer. Then for any two residue classes a ≡ a (mod b), 1 2 we have p(a ,b,n) − p(a ,b,n) 1 2 = cos β + 2λ n + o(1), Bn exp 2λ n as n →∞,where λ + iλ := Li (ζ ) and B > 0 and 0 ≤ β< 2π are implicitly deﬁned 1 2 2 b by 1 (1 − ζ )(λ + iλ ) −a −a b 1 2 iβ 1 2 Be := (ζ − ζ ) . b b b π A more general version of Theorem 1.2 holds for values b ≥ 2, see Theorem 3.5, where the special cases b ∈{2, 3, 4} have to be treated slightly diﬀerently. Figure 2 shows that this prediction is surprisingly accurate even for small n. The proof makes use of (1.4) and a detailed study of the coeﬃcients Q (ζ )q := n≥0 −1 (ζ q; q) when ζ is a root of unity. Prompted by a question of Stanley, a study of the polynomials Q (ζ ) and their complex zeroes was undertaken in a series of papers by Boyer, −1 Goh, Keith and Parry (see [4–7,18]), and the functions (ζ q; q) have also been studied in recent work of Bringmann, Craig, Males, and Ono [8] in the context of distribution of homology of Hilbert schemes and t-hook lengths. Asymptotics for Q (ζ ) were studied Here and throughout we take the principal branch of the square-root and upcoming logarithms. 61 Page 4 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 Fig. 2 The plot shows the sign changes of p(1, 5,n) − p(4, 5,n). The blue dots depict the function p(1,5,n)−p(4,5,n) and the red line is the asymptotic prediction cos β + 2λ n . The approximate values of the √ 2 − 2λ n 4 1 Bn e constants are B ≈ 0.23268, β ≈ 1.4758, λ ≈ 0.72984 and λ ≈ 0.68327 1 2 by Wright [24] when ζ is any positive real number, and then by Boyer and Goh [4,5] for |ζ | = 1. Our results essentially complete this study, proving asymptotics when ζ is any root of unity. We use the circle method in the form of Parry [18] as a template; however, signiﬁcant technical eﬀort is required to bound minor arcs that one does not encounter when |ζ | = 1. For example, we have to overcome the fact that when ζ is a root of unity, the series representation of the polylogarithm Li (ζ ) does not converge for Re(s) < 0. Asymptotics in the case that b ≥ 5 are as follows, and a general version with the sporadic cases b ∈{2, 3, 4} is stated in Theorem 3.2. Theorem 1.3 Let b ≥ 5.Thenwehave 1 − ζ Li (ζ ) b b Q (ζ ) ∼ exp 2 Li (ζ ) n ,n →∞. n b 2 b 2 πn Remark 1.4 Our techniques for bounding minor arcs seem not to readily apply when |ζ|= 1and ζ is not a root of unity, so we leave it as an open problem to prove asymptotic formulas for Q (ζ)inthiscase. Boyerand Goh[4] prove that the unit circle is part of the zero attractor of the polynomials Q (ζ ), and it follows from their results that asymptotic formulas for Q (ζ ) cannot be uniform on any subset of C which is an open neighborhood of an arc of the unit circle. Furthermore, note that taking ζ → 1inTheorem 1.3 does not recover the Hardy– Ramanujan asymptotic formula (1.2). The above results can be further developed to linear combinations of p(a, b, n) (where b is ﬁxed). Each relevant combination c p(a, b, n) corresponds to a polynomial 0≤a<b a a P(x):= c x via p(a, b, n) → x . Perhaps the most combinatorially interesting 0≤j<b W. Bridges et al. Res Math Sci (2022) 9:61 Page 5 of 31 61 cases are c ∈{0, ±1}; i.e., diﬀerences of partition numbers. This means that for any two nonempty disjoint subsets S ,S ⊂ [0,b − 1] of integers, we consider the diﬀerences 1 2 p(a, b, n) − p(a, b, n). a∈S a∈S 1 2 The asymptotic behavior of these diﬀerences is described in Theorem 3.7. When choos- ing the coeﬃcients c properly we can reduce the growth of the diﬀerence terms by canceling main terms. By doing so, we see that actually all formulas of Theorem 1.3 are required (as well as the sporadic cases described in Theorem 3.2). In contrast, one can deduce formula (1.3) by only elementary means without a thorough analysis of the coef- ﬁcients Q (ζ ). Moreover, for families of partition numbers, one can consider growth and sign changes simultaneously. Of special interest here is the fact that any shift of (S ,S )to(S +r, S +r) 1 2 1 2 changes the “phase” but not the “amplitude” of the asymptotic terms. The following example demonstrates this fact. Example 1.5 Let b = 12. For the sets S :={1, 2, 5} and S :={0, 3, 4},weﬁnd P (x) = 1 2 S ,S 1 2 5 4 3 2 x −x −x +x +x −1 = (x −1) (x), where (x) is the 12th cyclotomic polynomial. 12 12 The corresponding diﬀerence of partition numbers is p(5, 12,n) − p(4, 12,n) − p(3, 12,n) + p(2, 12,n) + p(1, 12,n) − p(0, 12,n). By shifting the sets with integers 0 ≤ r ≤ 6, we ﬁnd that, more generally, all diﬀerences (n):= p(5 + r, 12,n) − p(4 + r, 12,n) − p(3 + r, 12,n) + p(2 + r, 12,n) + p(1 + r, 12,n) − p(r, 12,n) have the same growth in their amplitudes but have diﬀerent phases of sign changes. Since P (x)has zeros {1, ±ζ , ±ζ }, the dominating term in the asymptotic expansion is S ,S 12 1 2 12 induced by the root of unity ζ . In fact, we obtain (n) 2πr √ = cos α − + 2λ n + o(1), (1.5) 3 √ An exp 2λ n with λ + iλ = Li (ζ ), where A > 0and α ∈ [0, 2π) are deﬁned by 1 2 2 6 1 − i 3 iα Ae = √ Li (ζ ) . 2 6 12 2π Note that A, α, λ and λ all do not depend on the choice of r. One can then asymptotically 1 2 describe the regions of n, where p(5, 12,n) + p(2, 12,n) + p(1, 12,n) > p(4, 12,n) + p(3, 12,n) + p(0, 12,n), and vice versa, using formula (1.5). Additionally, since Re( Li (ζ )) < Re( Li (ζ )), we 2 6 2 12 note with the help of Theorem 1.2 that p(a , 12,n) − p(a , 12,n) 1 2 lim sup =∞. 3 √ n→∞ n exp 2Re(Li (ζ )) n 2 6 61 Page 6 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 This implies some form of cancellation within the higher diﬀerences (n), that “exceeds” that of simple diﬀerences modulo 12. In contrast to Example 1.5, when b is prime, one cannot decrease the order of growth below simple diﬀerences using any rational combination of the p(a, b, n). This is shown in Section 3. The paper is organized as follows. In Section 2, we collect some classic analytical tools and prove some key lemmas. This includes a careful study of the dilogarithm function Li (z) for values |z|= 1. In Section 3, we state our main result, Theorem 3.2 and applica- tions. We prove Theorem 3.2 in Sections 4 and 5 using the circle method. Section 5 deals with the primary diﬃculty of bounding the minor arcs. The ﬁrst author is partially supported by the SFB/TRR 191 “Symplectic Structures in Geometry, Algebra and Dynamics”, funded by the DFG (Projektnummer 281071066 TRR 191). The second author is partially supported by the Alfried Krupp prize. 2 Preliminaries We will need several analytical results in this work, which we collect in this section. Much of the items here are discussed in works such as [9–11,16,19,22]. The reader can skip this section and refer back to it as needed as we work through the proofs of our main theorems. 2.1 Classical asymptotic analysis and integration formulas A ﬁrst tool is the well-known Laplace’s method for studying limits of deﬁnite integrals with oscillation, which we will use for evaluating Cauchy-type integrals. Theorem 2.1 (Laplace’s method, see Section 1.1.5 of [19]) Let A, B :[a, b] → C be continuous functions. Suppose x = x ∈ [a, b] such that Re(B(x)) < Re(B(x )),and that 0 0 B(x) − B(x ) lim =−k ∈ C, x→x 0 (x − x ) with Re(k) > 0.Thenast →∞ π 1 tB(x) tB(x ) A(x)e dx = e A(x ) + o √ . tk We are ultimately interested in how the coeﬃcients of a series S(q):= a(n)q n≥0 grow as n →∞. The classical Euler–Maclaurin summation formulas can be applied in many cases to link the growth of a(n) to the growth of S(q)as q approaches the unit circle. We state the Euler–Maclaurin summation formulas in two diﬀerent forms. Theorem 2.2 (classical Euler–Maclaurin summation, see p. 66 of [16]) Let {x} := x −x denote the fractional part of x. For N ∈ N and f :[1, ∞) → C a continuously diﬀerentiable function, we have N N 1 1 f (n) = f (x)dx + (f (N) + f (1)) + f (x) {x}− dx. 2 2 1 1 1≤n≤N We use a second formulation, due to Bringmann–Mahlburg–Jennings-Shaﬀer, to study the asymptotics of series with a complex variable approaching 0 within a ﬁxed cone in W. Bridges et al. Res Math Sci (2022) 9:61 Page 7 of 31 61 the right-half plane. It is not stated in [10], but one can conclude from the proof that Theorem 2.3 is uniform in 0 ≤ a ≤ 1. Theorem 2.3 (uniform complex Euler–Maclaurin summation, [10], Theorem 1.2) Sup- pose 0 ≤ θ< and suppose that f : C → C is holomorphic in a domain containing iα {re : |α|≤ θ } with derivatives of suﬃcient decay; i.e., there is an ε> 0 such that for all m (m) −1−ε f (z) z as z →∞. Then uniformly for a ∈ [0, 1],wehave N −1 ∞ (n) 1 f (0)B (a) n+1 n N f (w( + a)) = f (w)dw − w + O w , w (n + 1)! ≥0 n=0 uniformly as w → 0 in Arg(w) ≤ θ. To identify a constant term in our asymptotic formula, we cite the following integral calculation of Bringmann, Craig, Males and Ono. Lemma 2.4 ([8], Lemma 2.3) For a ∈ R , ∞ −ax −ax e 1 1 e 1 1 − − − a dx = log a + − a log a − log(2π). ( ( )) ( ) −x 2 x(1 − e ) x 2 x 2 2 Finally, we will use Abel partial summation extensively when bounding the twisted eta-products on the minor arcs. Proposition 2.5 (Abel partial summation, see p. 3 of [23]) Let N ∈ N and M ∈ N.For sequences {a } , {b } of complex numbers, if A := a ,then n n≥N n n≥N n m N <m≤n a b = A b + A (b − b ). n n N +M N +M n n n+1 N <n≤N +M N <n<N +M 2.2 Elementary bounds The following bound for diﬀerences of holomorphic functions will be used in conjunction with Abel partial summation during the course of the circle method. The proof of the following is a straightforward application of the fundamental theorem of calculus and the maximum modulus principle. Lemma 2.6 Let f : U → C be a holomorphic function and B (c) ⊂ U a compact disk. Then, for all a, b ∈ B (c) with a = b, we have f (b) − f (a) ≤ max |f (z)||b − a|. |z−c|=r We also need the following elementary maximum; for a proof see [18], Lemma 5.2. π π Lemma 2.7 For a ∈ − , and b ∈ R,wehave 2 2 ia e a Re ≤ cos , 1 + ib 2 with equality if and only if b satisﬁes Arg(1 + ib) = . 2 61 Page 8 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 2.3 Bounds for trigonometric series and the polylogarithm We recall that for complex numbers s and z with |z| < 1 the polylogarithm Li (z)isdeﬁned by the series Li (z):= . n≥1 We are especially interested in the case s = 2, where Li (z) is called the dilogarithm. The appendix of a recent preprint of Boyer and Parry [7] contains many useful results on the dilogarithm, including the following key lemma. 2πiθ Lemma 2.8 (Proposition 4, [7]) The function θ → Re Li (e ) is decreasing on the interval [0, ]. 2πiθ We also need to consider the derivative of the function θ → Li (e ) and its partial sums. Let 2πiθm G (θ):= . 1≤m≤M Then the following bound holds. Lemma 2.9 We have, uniformly for 0 <θ < 1, 1 1 G (θ) log + log ,M →∞. θ 1 − θ Proof Note that we have cos(2πθm) sin(2πθm) G (θ) = + i , m m 1≤m≤M 1≤m≤M sin(2πθm) and as it is well known that is uniformly bounded in R (see [21]onp. 1≤m≤M 94) we are left with the cosine sum. Let 0 <θ ≤ and M ∈ N. Consider the meromorphic function cos(2πθz)(cot(πz) − i) h (z):= , 2iz 1 1 i i 1 1 together with the rectangle R with vertices −iM, + , +M + ,and −iM +M + . 2 2 θ θ 2 2 Notice that in a punctured disk of radius r < 1 centered at k ≥ 1, we have cos(2πθk) 1 h (z) = − i + O(z − k) . 2ik π(z − k) With the residue theorem we ﬁnd cos(2πθm) h (z)dz = 2πi Res h (z) = , θ z=m θ ∂R 1≤m≤M 1≤m≤M where the contour is taken once in positive direction. By the invariance of the function under the reﬂection θ → 1 − θ, it suﬃces to consider values 0 <θ ≤ . A straightforward calculation shows that the bottom side integrals are bounded uniformly in 0 <θ ≤ . 2 W. Bridges et al. Res Math Sci (2022) 9:61 Page 9 of 31 61 Similarly, also using that cot(πz) is uniformly bounded on lines {Re(z) ∈ Z \ Z} for the second part, one argues 1 i 1 i 1 i M+ + M+ + + 2 θ 2 θ 2 θ dz 1 h (z)dz = O(1) + h (z)dz log . θ θ 1 1 1 z + M θ M+ −Mi M+ 2 2 2 A similar argument yields −Mi h (z)dz log . 1 i 2 θ For the rectangle top, note with the use of the substitution z → we have 1 i θ θ πz + +i +i 2 θ 2 2 cos(2πz)(cot + i) cos(2πz) h (z)dz = dz − 2i dz. (2.1) 1 i 1 1 z z M+ + θ M+ +i θ M+ +i 2 θ 2 2 With the Residue Theorem we obtain 1 1 θ(M+ )+i i∞ θ(M+ )+i 2 2 = + . θ θ 1 +i +i θ(M+ )+i∞ 2 2 2 πz −2πIm(z) 1 and the bound cos(2πz)(cot + i) = O(e )asIm(z) →∞ (since 0 <θ ≤ ) θ 2 yields the uniform boundedness of the ﬁrst integral in (2.1). The integral on the right-hand side of (2.1) can be bounded via standard contour integration, for instance, by using that cos(x) dx 1 for all α ≥ 1. This completes the proof. α x Lemma 2.9 has the following important consequence. Lemma 2.10 Let a, b, h and k be positive integers, such that gcd(a, b) = gcd(h, k) = 1. We assume that a and b are ﬁxed. Then we have a hj max G + = O(k), m≥1 b k 1≤j≤k ak+bjh≡0(mod bk) as k →∞, uniformly in h. Proof The assumption gcd(h, k) = 1 implies that the map j → hj is a bijection modulo k. One can show by elementary means that there is at most one j such that bk divides ak +bj h. First, we assume that such a j exists. Using G (x +1) = G (x), we ﬁrst reorder 0 0 m m the sum, then apply Lemma 2.9 to obtain a hj j k k max G + = max G log + log m m m≥1 b k m≥1 k j k − j 1≤j≤k 1≤j≤k−1 1≤j≤k−1 j=j = O(k) by Stirling’s formula. Now, assume that there is no such j .Inthiscaseweﬁnd 0 <α < 0 a,b,k b, such that a hj α α 1 a,b,k a,b,k max G + = max G + max G + 1 − m m m m≥1 b k m≥1 bk m≥1 bk k 1≤j≤k 61 Page 10 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 α j a,b,k + max G + . m≥1 bk k 1≤j≤k−2 Using again Lemma 2.9 and Stirling’s formula, we obtain α j k k a,b,k max G + log + log = O(k). m≥1 bk k j k − j 1≤j≤k−2 1≤j≤k−2 Since clearly α α 1 a,b,k a,b,k max G + max G + 1 − = O(log(k)), m m m≥1 bk m≥1 bk k the lemma follows. 3 Main results and applications 3.1 Twisted eta-products In this section, we record the general asymptotic formula for Q (ζ ) where ζ is any root of unity. Note that Q (ζ ) = Q (ζ ), so it suﬃces to ﬁnd asymptotic formulas for ζ in the n n upper-half plane. The following theorem of Boyer and Goh [4] regarding the dilogarithm distinguishes several cases in our main theorem. Following [4], we deﬁne ikθ Li (e ) (θ):= Re , for 0 ≤ θ ≤ π, and 2π 0 <θ < <θ <π 13 23 where each θ is a solution to θ = θ . Here, ( ) ( ) jk j k θ = 2.06672 ... , θ = 2.36170 ... . 13 23 Since the values θ and θ arise as solutions to a non-algebraic equation, it is very unlikely 13 23 that they are rational multiples of π. Therefore, we will no longer consider them in future investigations. Theorem 3.1 ([4], discussion prior to Theorem 2) For 0 ≤ θ ≤ π,wehave (θ) if θ ∈ [0, θ ], ⎪ 1 13 max (θ) = (θ) if θ ∈ [θ , π], k 2 23 k≥1 (θ) if θ ∈ [θ , θ ]. 3 13 23 Following [18], deﬁne k j −jh k 2 ω (z):= 1 − zζ . h,k j=1 We have the following asymptotic formulas. Theorem 3.2 (1) If 2π ∈ (0, θ ),then a a 1 − ζ Li ζ √ a b b Q ζ ∼ exp 2 Li ζ n ,n →∞. n 2 b √ 3 2 πn W. Bridges et al. Res Math Sci (2022) 9:61 Page 11 of 31 61 (2) If 2π ∈ (θ , π),then n a 2a (−1) 1 − ζ Li ζ √ a b b 2a Q ζ ∼ exp Li ζ n ,n →∞. n √ 2 b 3 2 2πn a 2π (3) If 2π ∈ (θ , θ ) \ ,then 13 23 b 3 3a Li (ζ ) 2 √ a −n a −2n a b 3a Q ζ ∼ (ζ ω (ζ ) + ζ ω (ζ )) exp Li (ζ ) n ,n →∞. n 1,3 2,3 √ 2 b 3 b 3 b 3 2 3πn (4) We have 1 1 −2n 2 1 6 2 ζ (1 − ζ ) (1 − ζ ) 2π n 3 3 3 Q ζ ∼ exp ,n →∞. ( ) n 3 3 6 2(6πn) We prove Theorem 3.2 in Sections 4 and 5. Remark 3.3 Recall that we have the asymptotic formula (1.2) for Q (1) = p(n), whereas for Q (−1), standard combinatorial methods give 1 (q; q) n ∞ 2 n n Q (−1)q = = = q; q = (−1) p (n)q , n DO 2 2 (−q; q) (q ; q ) ∞ ∞ n≥0 n≥0 where p (n) counts the number of partitions of n into distinct odd parts. An asymptotic DO formula for p (n) can be worked out using standard techniques. For example, Ing- DO ham’s Tauberian theorem (see [10], Theorem 1.1) with the modularity of the Dedekind η-function yields (−1) n Q (−1) = (−1) p (n) ∼ exp π . n DO 1 3 4 4 2(24) n Note the lack of uniformity in the asymptotic formulas for ζ near ±1; in particular, the asymptotic formulas for Q (1) and Q (−1) cannot be obtained by taking →±1incases n n (1) and (2). 3.2 Applications to diﬀerences of partition functions In this section, we apply Theorem 3.2 to diﬀerences of the partition functions p(a, b, n), partitions of n with number of parts congruent to a modulo b. The following elementary proposition relates the numbers p(a, b, n) to the coeﬃcients Q (ζ ). Proposition 3.4 We have −ja j p(a, b, n) = ζ Q ζ . b b 0≤j≤b−1 Proof Using orthogonality, we can write the indicator functions of congruence classes as b−1 −ja jx 1 = ζ ζ , x≡a (mod b) b b j=0 and by standard combinatorial techniques (see [1], Ch. 1) one has (λ) Q (z) = z . λn 61 Page 12 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 Hence, (λ) 1 1 1 −ja j(λ) −ja j −ja j p(a, b, n) = ζ ζ = ζ ζ = ζ Q ζ , b b b b b b b b b λn 0≤j≤b−1 0≤j≤b−1 λn 0≤j≤b−1 which completes the proof. Thus, asymptotics for diﬀerences of p(a, b, n) can be identiﬁed using the asymptotic for- mulas found for Q (ζ)inTheorem 3.2, in particular the equidistribution of the largest part p(a,b,n) in congruence classes follows immediately: ∼ . Furthermore, using the Hardy– p(n) b Ramanujan–Rademacher exact formula for p(n)([1], Theorem 5.1) with Theorem 3.2, one can improve this in the various cases. For example, for b ≥ 5, we have ⎛ ⎞ 2 3 π 24n−1 6 b · p(a, b, n) − exp 1 − √ 24n−1 6 2πa √ π 24n−1 ⎝ ⎠ lim − cos α − + 2λ n = 0, 3 √ n→∞ − An exp(2λ n) where λ + iλ = Li (ζ ),A ≥ 0, and α ∈ [0, 2π) are deﬁned by 1 2 2 (λ + iλ )(1 − ζ ) 1 2 iα b Ae = . We record a number of results below in the same vein. Considering simple diﬀerences, p(a ,b,n) − p(a ,b,n), it follows from Proposition 3.4 that when rewriting each p(a, b, n) 1 2 in terms of Q the two summands Q (1) = p(n) cancel. With a few exceptions for b = 4, n n b−1 the dominant terms are always Q (ζ )and Q (ζ ). n b n Theorem 3.5 Let 0 ≤ a < a ≤ b − 1. 1 2 (1) For b = 2,wehavep(0, 2,n) − p(1, 2,n) = Q (−1). (2) For b = 3,wehave p(a , 3,n) − p(a , 3,n) 4πn 1 2 = cos α − + o(1), a ,a 1 2 2π n − 3 A n exp a ,a 1 2 3 6 where A ≥ 0 and α ∈ [0, 2π) are deﬁned by a ,a a ,a 1 2 1 2 1 1 −a −a iα 1 2 2 3 a ,a 1 2 6 2 A e = (ζ − ζ )(1 − ζ ) (1 − ζ ) . a ,a 3 1 2 3 3 3 2 3(6π) (3) For b = 4,and a ,a of opposite parity, we have 1 2 a a 1 2 (−1) − (−1) p(a , 4,n) − p(a , 4,n) ∼ Q (−1). 1 2 n (4) For b ≥ 5,orfor b = 4 and a ,a of the same parity, we have 1 2 p(a ,b,n) − p(a ,b,n) 1 2 = cos β + 2λ n + o(1), a ,a ,b 2 √ 1 2 B n exp(2λ n) a ,a ,b 1 1 2 where λ + iλ = Li (ζ ),B ≥ 0 and β ∈ [0, 2π) are deﬁned by 1 2 2 b a ,a ,b a ,a ,b 1 2 1 2 −a −a 1 2 ζ − ζ (1 − ζ )(λ + iλ ) b b b 1 2 iβ a ,a ,b 1 2 B e = . a ,a ,b 1 2 b π W. Bridges et al. Res Math Sci (2022) 9:61 Page 13 of 31 61 Proof When b = 2, Theorem 3.5 follows from Proposition 3.4.Let b ≥ 5. By Proposi- tion 3.4, it follows that 1 1 −ja −ja j −a −a 1 2 1 2 p(a ,b,n) − p(a ,b,n) = Re (ζ − ζ )Q (ζ ) + (ζ − ζ )Q ζ . 1 2 n n b b b b b b b 1≤j≤b−2 3 √ Upon dividing both sides by B n exp(2λ n), Lemma 2.8 implies that the sum on a ,a ,b 1 2 −c n the right is O(e ), for some c > 0, and it then follows from Theorem 3.2 that p(a ,b,n) − p(a ,b,n) i β +2λ n 1 2 a ,a ,b 2 −c n 1 2 = Re e + o(1) + o e , (3.1) 3 √ B n exp 2λ n a ,a ,b 1 1 2 which gives the claim of Theorem 3.5. The other cases are proved similarly by noting that Lemma 2.8 and Theorem 3.2 imply −1 that the Q (ζ )and Q (ζ ) terms always dominate Q (ζ ) for j =±1, except in the n b n n b b case that b = 4, where Q (−1) dominates when a ,a have opposite parity. But Q (−1) n 1 2 n vanishes from the sum in Proposition 3.4 when b = 4and a ,a have thesameparity, 1 2 which leads to the third case in Theorem 3.5. a b More generally, if P (x):= v x ∈ R[x]with v := (v , ... ,v ) ∈ R , then v a 0 b−1 0≤a≤b−1 b−1 Theorem 3.2 implies asymptotic formulas for any weighted count v p(a, b, n). To a=0 state our general theorem, we let iθ Li (e )if0 ≤ θ< θ , 2 13 3iθ iθ Li (e ) L(e ):= if θ <θ <θ , 13 23 ⎪ 2iθ Li (e ) ⎩ 2 if θ <θ ≤ π. Let Z(P ) be the roots of P ,and let v v 0 b λ + iλ = L(ζ ) for a ≤ , whenever 1 2 0 b 2 a a Re L(ζ ) = max Re L(ζ ) . 0≤a≤ ζ ∈ /Z(P ) Theorem 3.6 With notation as above, we have the following asymptotic formulas. P (1) (1) If a = 0,then v p(a, b, n) ∼ p(n). 0 a 0≤a≤b−1 b (2) If 0 < 2π <θ ,then v p(a, b, n) √ 0≤a≤b−1 = cos α + 2λ n + o(1), √ a ,b,v 3 0 A n exp(2λ n) a ,b,v 1 where A ≥ 0 and α ∈ [0, 2π) are deﬁned by a ,b,v a ,b,v 0 0 −a a 0 0 P (ζ ) (λ + iλ )(1 − ζ ) v 1 2 iα b b a ,b,v A · e = . a ,b,v b π a b (3) If θ < 2π <θ and a = ,then 13 23 0 b 3 v p(a, b, n) 2πn √ 0≤a≤b−1 = B cos α + β − + 2λ n a ,b a ,b,v a ,b 2 3 √ 0 0 0 A n exp(2λ n) a ,b,v 4πn + C cos α + γ − + 2λ n + o(1), a ,b a ,b,v a ,b 2 0 0 0 3 61 Page 14 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 iθ Fig. 3 Re(L(e )) for 0 ≤ θ ≤ π where B ,C ≥ 0 and β , γ ∈ [0, 2π) are deﬁned by a ,b a ,b a ,b a ,b 0 0 0 0 iβ a a ,b B e = ω (ζ ), 1,3 a ,b 0 b iγ a a ,b C e = ω (ζ ). 2,3 a ,b 0 b (4) If a = ,then b v p(a, b, n) a 4πn 0≤a≤b−1 = cos δ − + o(1), 2π n − 3 D n exp 3 6 where D ≥ 0 and δ ∈ [0, 2π) are deﬁned by v v 1 1 2 1 6 2 (1 − ζ ) (1 − ζ ) ( ) iδ 3 3 −1 D · e = P ζ . v v 2(6π) (5) If θ < 2π <π,then c p(a, b, n) √ 0≤a≤b−1 = cos α + πn + 2λ n + o(1). a ,b,v 2 3 √ 0 A n exp 2λ n a ,b,v 1 P (−1) b a v (6) If a = ,then (−1) v p(a, b, n) ∼ Q (−1). 0 a n 0≤a≤b−1 2 b Proof Theorem 3.6 is proved similarly to Theorem 3.5. For Cases (1) and (6), the asymp- totic formula for Q (1) ∼ p(n) and Proposition 3.4 directly implies 1 P (1)p(n) v p(a, b, n) ∼ (v Q (1) + v Q (1) + ··· + v Q (1)) = , a 0 n 1 n b−1 n b b 0≤a≤b−1 a b−1 (−1) v p(a, b, n) ∼ (v Q (−1) − v Q (−1) + ··· + (−1) v Q (−1)) a 0 n 1 n b−1 n 0≤a≤b−1 P (−1)Q (−1) v n = , for Cases (1) and (6), respectively. W. Bridges et al. Res Math Sci (2022) 9:61 Page 15 of 31 61 For Cases (2), (3) and (5), the asymptotic main term (using that Q (ζ ) = Q (ζ )) is n n 1 2 −a a a −a a a −a a a 0 0 o 0 0 0 p(a, b, n) ∼ ζ Q (ζ ) + ζ Q (ζ ) = Re ζ Q (ζ ) . n n n b b b b b b b b The proof then follows by applying Theorem 3.2 and dividing by the appropriate normal- izing factors in analog to Equation (3.1). For case (4), there is only one main term in the sum for a 1 3 3 p(a, b, n) ∼ ζ Q ζ . Applying Theorem 3.2 and the analog to Equation (3.1) again proves the claim. One application of the above theorem generalizes Theorem 3.5 to diﬀerences of par- titions with number of parts modulo b in one of two disjoint sets of residue classes S ,S ⊂ [0,b − 1]. That is, we consider 1 2 a a P (x):= x − x , S ,S 1 2 a∈S a∈S 1 2 and prove a more explicit version of Theorem 3.6 in this case. Since P (x) is a polynomial S ,S 1 2 of degree at most b − 1 with integer coeﬃcients, there must exist some d | b such that ζ ∈ / Z(P ); otherwise, P would be divisible by (x) = x − 1, where is S ,S S ,S d 1 2 1 2 d d d|b the d-th cyclotomic polynomial, a contradiction. Theorem 3.7 Let b ≥ 2 and let S ,S ⊂ [0,b − 1] be disjoint subsets of integers. If 1 2 |S | =|S |,then 1 2 (|S |−|S |)p(n) 1 2 p(a, b, n) − p(a, b, n) ∼ . a∈S a∈S 1 2 Otherwise, if |S |=|S |, then we have the following cases. Let d be the largest integer such 1 2 0 that d | band ζ ∈ / Z(P ). 0 d S ,S 0 1 2 (1) If d ≥ 5 or if d = 4 and −1 ∈ Z(P ),then 0 0 S ,S 1 2 p(a, b, n) − p(a, b, n) a∈S a∈S 1 2 = cos α + 2λ n + o(1), d ,S ,S 2 3 √ 0 1 2 A n exp(2λ n) d ,S ,S 1 0 1 2 where λ + iλ = Li (ζ ),A ≥ 0 and α ∈ [0, 2π) are deﬁned by 1 2 2 d d ,S ,S d ,S ,S 0 0 1 2 0 1 2 −1 P ζ S ,S 1 2 d (λ + iλ )(1 − ζ ) 1 2 d iα 0 d ,S ,S 0 1 2 A · e = . d ,S ,S 0 1 2 b π (2) If d = 4 or 3 with b even, and −1 ∈ / Z(P ), we have the following allowable sets: 0 S ,S 1 2 • If d = 4 and (k + k ) ≡ 0 (mod 2), 0 1 2 S × S ={a : a ≤ b − 1,a ≡ k (mod 4)}×{a : a ≤ b − 1,a ≡ k (mod 4)}. 1 2 1 2 • If d = 3, b is even and k ≡ k (mod 2), 0 1 2 S × S ={a : a ≤ b − 1,a ≡ k (mod 2)}×{a : a ≤ b − 1,a ≡ k (mod 2)}. 1 2 1 2 The asymptotic formulas are then given by S ,S 1 2 p(a, b, n) − p(a, b, n) ∼ Q (−1), a∈S a∈S 1 2 61 Page 16 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 where ⎨ k (−1) b if d = 3,b even, N = S ,S 1 2 k b ⎩ 1 (−1) if d = 4. (3) If d = 3, b is even and −1 ∈ Z(P ),ord = 3 and b is odd, we have the following 0 S ,S 0 1 2 sets S and S : 1 2 • If b is odd, S and S must contain distinct residue classes modulo 3. 1 2 • If b is even, S × S ={a : a ≤ b − 1,a ≡ k or k (mod 6)}×{a : a ≤ b − 2,a ≡ k or k (mod 6)}, 1 2 1 2 3 4 where (k ,k ,k ,k ) = (0, 3, 2, 5), (2, 5, 0, 3), (1, 4, 2, 5), (2, 5, 1, 4), (1, 3, 1, 4), or (1, 4, 1, 3). 1 2 3 4 The asymptotic formula is then given by −1 p(a, b, n) − p(a, b, n) ∼ Re Q (ζ )P ζ . n 3 S ,S 1 2 3 a∈S a∈S 1 2 (4) If d = 2,thenfor some a ,a of opposite parity we have 0 1 2 (S ,S ) ={a ≡ a (mod 2)}×{a ≡ a (mod 2)} 1 2 1 2 and p(a, b, n) − p(a, b, n) ∼ (−1) Q (−1). a∈S a∈S 1 2 Proof sketch of Theorem 3.7 The asymptotic analysis is similar to the proof of Theo- rem 3.5, since Lemma 2.8 and Theorem 3.2 imply that the sequence Q (ζ ), ranked from n d in asymptotic order from least to greatest, is Q (ζ ),Q (i),Q (−1),Q (ζ ),Q (ζ ), ... ,Q (1) = p(n). n 3 n n n 5 n 6 n Thus,forexampleincase(3),theasymptoticbehaviorof p(a, b, n)− p(a, b, n) a∈S a∈S 1 2 −j j −1 is determined by Q (ζ )and Q (ζ ) since all other Q (ζ )vanishin P (ζ )Q (ζ ). n 3 n n S ,S n 3 1 2 b b b Furthermore, in this case we must have x − 1 | P (x), S ,S 1 2 (x) and deg(P (x)) ≤ b − 1. Since S ,S 1 2 x − 1 b−2 b−3 b−5 b−6 4 3 = x − x + x − x + ··· + x − x + x − 1, (x) this leads directly to the possible sets S and S described in case 3. Finally, note that, given 1 2 S and S deﬁned in terms of a and a modulo b,wehave 1 2 1 2 −a −a −1 −a −a 1 2 P ζ = ζ − ζ = (ζ − ζ ). S ,S 1 2 3 3 3 3 3 0≤a≤b 0≤a≤b a≡a (mod 3) a≡a (mod 3) 1 2 This is taken into account in the deﬁnition of the constant B . The other cases are S ,S 1 2 proved similarly. W. Bridges et al. Res Math Sci (2022) 9:61 Page 17 of 31 61 We make a few remarks. Remark 3.8 (1) Theorem 3.7 case (1), in combination with Lemma 2.8, shows that one can reduce the growth of the amplitudes in the diﬀerences exponentially, as long as the corresponding polynomial P vanishes at the crucial roots of unity. But the options S ,S 1 2 for such types of cancellation strongly depend on b. For example, if b is a prime num- ber, there is not even a rational combination (except for the trivial combination) such that the amplitudes of v p(a, b, n) grow exponentially less than any simple diﬀer- 0≤a<b ence p(a ,b,n) − p(a ,b,n). The simple algebraic reason behind this is that the minimal 1 2 polynomial of ζ has degree b − 1 in this case. It would be interesting to ﬁnd a purely combinatorial interpretation for this fact. (2) Note that if we shift the residue classes in S and S by some integer r and then take 1 2 least residues modulo b to compute P (x), then this polynomial has the same roots S +r,S +r 1 2 of unity as P (x), and so S ,S 1 2 p(a, b, n) − p(a, b, n) a∈S +r a∈S +r 1 2 always has the same asymptotic behavior in Theorem 3.7 as p(a, b, n) − a∈S p(a, b, n). At the same time, the phase in the cosine changes. Indeed, we obtain a∈S p(a, b, n) − p(a, b, n) 2πr √ a∈S +r a∈S +r 1 2 = cos α − + 2λ n + o(1), d ,S ,S 2 3 √ 0 1 2 4 0 A n exp 2λ n d ,S ,S 0 1 2 (3.2) where all the constants are the same as in Theorem 3.7 (1). One can further use trigono- metric identities to obtain a wider class of more classical asymptotic formulas, for instance regarding squared partition diﬀerences. Indeed, if 4|d in (3.2), one ﬁnds using 2 2 sin (x) + cos (x) = 1, as n →∞, ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ p(a, b, n) − p(a, b, n) + p(a, b, n) − p(a, b, n) ⎝ ⎠ d d a∈S a∈S 1 2 0 0 a∈S + a∈S + 1 2 4 4 2 − ∼ A n exp 4λ n . d ,S ,S 0 1 2 (3)Cases 3and 4inTheorem 3.7 show that for d ∈{2, 3}, there are ﬁnitely many sets S and S such that the asymptotic behavior of p(a, b, n) − p(a, b, n)is 1 2 a∈S a∈S 1 2 determined by Q (ζ ), and that the number of such sets is independent of b.Infactthis is true for any d ,and we leaveitasanopenproblem to describe thesets S ,S in general. 0 1 2 The sets S and S need not each consist of one congruence class each modulo d , for 1 2 0 example with b = 10, the polynomial P (x)has d = 5. {1,3,6,8},{0,2,5,7} 0 3.3 Examples In this section, we provide some examples. 61 Page 18 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 Fig. 4 The plot shows the sign changes of p(1, 6,n) − p(5, 6,n) (in blue dots) and the estimated sign changes (the red line) Example 3.9 Let b = 6, and consider the diﬀerence p(1, 6,n) − p(5, 6,n). Then according to Theorem 3.5,weobtain p(1, 6,n) − p(5, 6,n) = cos β + 2λ n + o(1), (3.3) 3 √ Bn exp 2λ n where λ + iλ = Li (ζ ), and B > 0and β ∈ [0, 2π) are given implicitly by 1 2 2 6 # √ −1 −5 ζ − ζ (1 − ζ )(λ + iλ ) i 1 3 6 1 2 iβ 6 6 Be = = − i Li (ζ ). 2 6 6 π 2 2 12π Note that this implies (choosing B to be the absolute value of the right-hand side of the equation above) λ = 0.81408 ... , λ = 0.62336 ... , 1 2 B = 0.23268 ... , β = 1.37394 ... . (3.4) Considering the ﬁrst 900 coeﬃcients numerically yields M :={7, 26, 59, 104, 162, 233, 316, 412, 521, 642, 776, ...}, which are the highest indices until a change of signs in the sequence p(1, 6,n) − p(5, 6,n). We can compare this exact result to the prediction of formula (3.3). By considering the roots of the cosine, we ﬁnd that it changes signs approximately at M :={7, 27, 59, 104, 162, 233, 316, 412, 521, 642, 777, ...}. Note that in the ﬁrst eleven cases, only case two with 27 and case eleven with 777 give slightly wrong predictions (Figs. 3 and 4). W. Bridges et al. Res Math Sci (2022) 9:61 Page 19 of 31 61 The next example refers to higher diﬀerences of partition functions. Example 3.10 Again, we consider the case b = 6. In the spirit of Theorem 3.7 we want to consider multi-termed diﬀerences. To do so, we need two subsets S ,S ⊂ Z/6Z, such 1 2 that the corresponding nontrivial polynomial P (x) of degree at most 5 vanishes at as S ,S 1 2 many roots of unity around x = 1 as possible. We ﬁrst note that 2 3 2 (x − 1) (x) = (x − 1) x − x + 1 = x − 2x + 2x − 1. In this case, we obtain a weighted diﬀerence and no subsets can be found for a growth reduction to an exponent induced by Li (ζ ). We continue by trying to eliminate also the 2 3 3rd roots of unity; i.e., with 5 4 3 2 (x − 1) (x) (x) = x − x + x − x + x − 1. 6 3 This is exactly the case (4) of Theorem 3.7. As a result, setting S :={1, 3, 5} and S := 1 2 {0, 2, 4},weﬁnd,as n →∞, n+1 (−1) n a+1 (−1) p(a, 6,n) ∼−Q (−1) ∼ exp π . 1 3 4 4 2(24) n 0≤a≤5 Also note that = 1.2825 ... is much smaller than the exponent 2λ = 1.6281 ... in (3.3), compare also (3.4). Finally, we give an application of Remark 3.8. Example 3.11 In light of Equation (3.2), we can choose r such that the cosine becomes a sine on the right-hand side. Using Pythagoras’ Theorem, and considering the case b = 8 and r = 2, we obtain an asymptotic formula without oscillating terms. Indeed, according to Theorem 3.5 (4) and Remark 3.8, respectively, we obtain for residue classes a = a 1 2 that 2 2 2 − (p(a , 8,n) − p(a , 8,n)) + (p(a + 2, 8,n) − p(a + 2, 8,n)) ∼ B n exp 4λ n . 1 2 1 2 1 a ,a ,8 1 2 There is no diﬃculty to extend this type of asymptotic formula for higher diﬀerences in the spirit of Theorem 3.7 (1). 4 Proof of Theorem 3.2 Since the asymptotic formulas for Q (1) and Q (−1) are well-known, we assume through- n n out that 1 ≤ a < with gcd(a, b) = 1and b ≥ 3, since Q (ζ ) = Q (ζ ). n n The setup follows the standard Hardy–Ramanujan circle method, expressing Q (ζ)as a contour integral about 0 and breaking the contour apart with the sequence of Farey fractions of order N. For facts about the Farey sequence, we refer the reader to Chapter 3 of [14]. Much of the analysis in this section closely follows [18] after assuming the technical Lemmas 4.2 and 4.3 which we prove in the next section. Let N =δ n, for some δ> 0 to be chosen independently of n and small enough during the course of the proof. Let F be the sequence of Farey fractions of order N with N 61 Page 20 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 mediants θ and θ at .Wewrite h,k h,k k ak Li ζ t := t − 2πiθ, where t n := , (4.1) θ n n with k ∈{1, 2, 3} according to whether we are in case (1), (2) or (3). By Cauchy’s integral formula, we have −1 a a −t +2πiθ −t +2πiθ nt −2πinθ n n n Q ζ = ζ e ; e e dθ b b ⎛ ⎞ ak θ Li ζ h,k b −hn a ⎝ ⎠ = ζ exp + nt + E (ζ ,t ) dθ, θ h,k θ k 2 k t −θ θ h,k ∈F where −1 Li z h −t h −t E (z, t):= Log zζ e ; ζ e − . h,k k k ∞ k t We will show that the integral(s) where k = k dominate, that is they are the major arcs, and all the other integrals are exponentially smaller, that is they are minor arcs.The E h,k will be shown to be error terms on all arcs; the following gives the growth of the E up h,k to o(1) on each of the possible major arcs. Lemma 4.1 (1) For 2π ∈ (0, θ ) and −θ ≤ θ ≤ θ ,wehave b 0,1 0,1 a a E (ζ ,t ) = Log ω ζ + o(1). 0,1 θ 0,1 b b (2) For 2π ∈ (θ , π) and −θ ≤ θ ≤ θ ,wehave 1,2 1,2 a a E (ζ ,t ) = Log ω ζ + o(1). 1,2 θ 1,2 b b a 2π (3) For 2π ∈ (θ , θ ) \ and −θ ≤ θ ≤ θ ,wehave 13 23 1,3 1,3 b 3 a a E (ζ ,t ) = Log ω (ζ ) + o(1), 1,3 θ 1,3 b b and for −θ ≤ θ ≤ θ ,wehave 2,3 2,3 a a E (ζ ,t ) = Log ω (ζ ) + o(1). 2,3 θ 2,3 b b a 1 (4) For = and −θ ≤ θ ≤ θ ,wehave b 3 1,3 1,3 1 1 1 2 6 6 E (ζ ,t ) = Log t + Log 1 − ζ 1 − ζ + log ( ) 1,3 3 θ 3 θ 3 1 1 + log (3) − log(2π) + o(1), 6 2 and for −θ ≤ θ ≤ θ ,wehave 2,3 2,3 1 1 − 1 1 6 2 6 E (ζ ,t ) = Log t + Log 1 − ζ (1 − ζ ) + log 2,3 3 θ 3 1 1 1 + log − log(2π) + o(1). 6 3 2 W. Bridges et al. Res Math Sci (2022) 9:61 Page 21 of 31 61 The next two lemmas give uniform bounds on E to be applied on the minor arcs. The h,k proofs are quite intricate and are provided in the next section. We assume throughout that 0 <ε ≤ . Lemma 4.2 Uniformly for k ≤ n and −θ ≤ θ ≤ θ ,wehave h,k h,k a 3ε− ε E (ζ ,t ) = O n + O (n ) . h,k θ The next lemma treats the case of large denominators. Lemma 4.3 Uniformly for k ≥ n and −θ ≤ θ ≤ θ ,wehave h,k h,k E (ζ ,t ) = O (N) . h,k We postpone the proofs of Lemmas 4.1–4.3 until Section 5. With these key lemmas in hand, the proof of Theorem 3.2 follows [18] closely. a 1 Proof of Theorem 3.2 Cases (1), (2) and (3). Assume = . We write b 3 ak Li (ζ ) λ + iλ := . 1 2 1 a It follows from the choice 0 <ε ≤ , Lemmas 4.2 and 4.3 that E (ζ ,t ) = δO( n) h,k θ 4 b uniformly. Using this and Lemma 4.1,wehave −2λ n a e Q ζ θ 2 h,k √ (λ + iλ ) 0 1 2 −hn a = ζ ω (ζ ) exp −2λ n + + nt + o(1) dθ h,k 1 θ k b −θ θ h,k ∈F ⎛ ⎞ ak θ Li ζ h,k √ √ −hn ⎝ ⎠ + ζ exp −2λ n + + nt + δO n dθ. (4.2) 1 θ k t −θ θ h h,k ∈F k=k Recalling (4.1), we rewrite the ﬁrst term in (4.2)as ⎛ ⎛ ⎞ ⎞ √ h,k √ λ + iλ 1 2 −hn a ⎝ ⎝ ⎠ ⎠ ζ ω ζ exp i2λ n exp n √ − (λ + iλ ) − 2πinθ dθ. h,k 2 1 2 k b 0 2πiθ n −θ 1 − 1≤h<k h,k 0 0 λ +iλ 1 2 (h,k)=1 1 1 − √ √ 2 We can estimate the mediants as θ , θ and setting θ → θn ,the h,k h,k n 0 0 n above integral is asymptotic to c c 1 λ + iλ 1 1 2 nB(θ) exp n − (λ + iλ ) − 2πiθ dθ =: e dθ, 1 2 1 1 2πiθ 1 − 2 −c 2 −c n n λ +iλ 1 2 (4.3) for some c > 0, say. We claim that we can apply Theorem 2.1 with x → 0,A(x) → 1and B as above. Here, B(0) = 0 and expanding the geometric series gives 4π 2 2 B(θ) =− θ + o θ , θ → 0, λ + iλ 1 2 61 Page 22 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 4π with Re > 0. Finally, we claim that Re(B(θ)) ≤ 0 with equality if and only if λ +iλ 1 2 θ = 0. Indeed, ⎛ ⎞ 2 iψ λ + iλ |λ + iλ | e 1 2 1 2 ⎝ ⎠ Re(B(θ)) = Re − λ = Re − λ , 1 1 2πiθ λ 2πθ λ 2 1 − 1 1 + i − λ +iλ 1 2 λ λ 1 1 i π π where λ + iλ = |λ + iλ | e and ψ ∈ − , . Lemma 2.7 applies to show 1 2 1 2 2 2 2 2 |λ + iλ | ψ λ 1 2 2 1 Re(B(θ)) ≤ cos − λ = − λ = 0, 1 1 λ 2 λ 1 1 with equality if and only if λ 2πθ ψ Arg 1 + i − = = Arg (λ + iλ ) , 1 2 λ λ 2 1 1 thus if and only if θ = 0, as claimed. Now by Theorem 2.1, we conclude that (4.3)is asymptotic to λ + iλ 1 2 √ 3 2 πn and overall ⎛ ⎞ θ 2 h,k √ √ (λ + iλ ) 0 1 2 −hn a ⎝ ⎠ ζ ω ζ × exp −2λ n + + n(λ + iλ ) − 2πinθ dθ h,k 1 1 2 0 b k0 λ +iλ 1 2 −θ − 2πiθ h,k 1≤h<k n 0 0 (h,k)=1 λ + iλ 1 2 2iλ n −hn a ∼ e ζ ω ζ . h,k √ 3 k 0 b 2 πn 1≤h<k (h,k )=1 2λ n When the e is brought back to the right-hand side, this is the right-hand side of Theorem 3.2. For k = k , we follow Parry in Lemma 5.2 of [18] and write ⎛ ⎞ ⎛ ⎛ ⎞ ⎞ ak ak Li ζ Li ζ iψ 2 2 k b b e ⎝ ⎠ ⎝ ⎝ ⎠ ⎠ Re + nt = λ n Re √ + 1 , θ 1 2 2 2 2πθ n k t k λ 2 1 1 + i − λ λ 1 1 ak ak iψ where ψ satisﬁes Li (ζ ) =|Li (ζ )|e . Arguing as for the major arcs, the expression 2 2 b b Re(·) above is at most cos .Now let ⎛ ⎞ ⎛ ⎛ ⎞ ⎞ ak Li (ζ ) ⎜ ⎟ ⎝ ⎝ ⎠ ⎠ := inf 1 − Re > 0. ⎝ ⎠ k=k kλ Then ⎛ ⎞ ak ak Li ζ Li (ζ ) √ √ b ψ b 2 ⎝ ⎠ Re + nt ≤ λ n cos + 1 ≤ λ n(2 − ). θ 1 1 2 2 k t k λ 2 Thus, θ k h,k √ Li (z ) √ √ √ −hn ζ exp −2λ n + + nt + δO( n) dθ ≤ exp −λ n + δO n . 1 θ 1 k 2 k t −θ θ h h,k ∈F k=k 0 W. Bridges et al. Res Math Sci (2022) 9:61 Page 23 of 31 61 We can choose δ small enough so that the constant in the exponential is negative, and the minor arcs are exponentially smaller than the major arc(s). This completes the proof of cases (1), (2) and (3). Li (1) Forcase(4),wehave λ + iλ = λ = = . Exactly as in case (3) the minor 1 2 1 3 6 arcs are those with k = 3, and these are shown to be exponentially smaller than for k = 3. Thus, by Lemma 4.1 part (4), we have √ 1 1,3 √ −2λ n −n 6 1 e Q (ζ ) = ζ C t exp −2λ n + + nt + o(1) dθ n 3 1,3 1 θ 3 θ −θ 1,3 θ 2 2,3 √ λ −2n 1 + ζ C t exp −2λ n + + nt + o(1) dθ, 2,3 1 θ −θ θ 2,3 where 1 1 − 1 2 3 1 1 1 2 2 6 6 2 2 C := 1 − ζ (1 − ζ ) √ ,C := 1 − ζ (1 − ζ ) . 1,3 3 2,3 3 3 3 1 3 3 2π 6 3 2π Noting t = √ 1 − 6i 6nθ , 3 6n we have −2λ n e Q (ζ ) n 3 1 2 6 1,3 √ (π) λ −n 1 = ζ C 1 − 6i 6nθ exp −2λ n + + nt + o(1) dθ 1,3 1 θ 3 1 1 6 12 −θ θ 3 (6n) 1,3 1 1 θ 2 6 12 2,3 − √ 3 (6n) λ −2n 1 + ζ C 1 − 6i 6nθ exp −2λ n + + nt + o(1) dθ. 2,3 1 θ 3 1 6 −θ θ (π) 2,3 (4.4) Setting θ → θn and arguing as before, both integrals are asymptotic to λ 1 = . √ 3 5 3 3 4 4 4 4 2 πn 2 3 n Hence, the second term in (4.4) dominates and gives the claimed asymptotic formula. 5 Proof of Lemmas 4.1, 4.2 and 4.3 We prove Lemma 4.2 ﬁrst then make use of these ideas in the proof of Lemma 4.1.We ﬁnish the section by proving Lemma 4.3. 5.1 Proof of Lemma 4.2 We rewrite E as a sum of two functions: a function to which we can apply Euler– h,k Maclaurin summation, and another to which we apply the tools in Proposition 2.5 and Lemma 2.9. 61 Page 24 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 Lemma 5.1 For Re(t) > 0 and z = ζ ,wehave ⎛ ⎞ k j − + jmh −jh 2 k ma 2 a −jt ⎝ ⎠ E (z, t) = ζ ζ kt g t(bk + km) + Log 1 − ζ ζ e , h,k j,k b b k k 1≤m≤bk ≥0 j=1 1≤j≤k (5.1) where j j − w − w k k e 1 1 j e g (w):= − − − . j,k −w 2 w(1 − e ) w 2 k w Proof In E , we expand the logarithm using its Taylor series as h,k m(ka+bjh) a νh −νt −(bk+m)(νk+j)t −1 ζ ζ e ζ e a h −t h −t b k bk Log ζ ζ e ; ζ e = = b k k ∞ bk + m ν≥1 ν≥0 1≤j≤k ≥1 ≥0 1≤m≤bk −jt(bk+m) mjh ma = ζ ζ . b k 2 −t(bk +km) (bk + m)(1 − e ) 1≤j≤k ≥0 1≤m≤bk This corresponds to the left term in g . For the middle term in g , we compute (using j,k j,k gcd(h, k) = 1) 1 k 1 1 mjh ma ma ka ζ ζ kt = ζ = Li ζ . b k b b 2 2 2 2 2 2 t (bk + km) t (bk + mk) tk ≥0 ≥0 1≤j≤k 1≤m≤bk 1≤m≤bk m≡0(mod k) Finally, it is simple to show that the logarithm of the product in (5.1) cancels with the sum of the right term in g , simply by expanding the logarithm into its Taylor series. j,k We estimate the ﬁrst term in (5.1) using Euler–Maclaurin summation. First, we need a technical deﬁnition. Since gcd(h, k) = 1, there is at most one j in the sum in (5.1) for j h which ζ ζ = 1; i.e., such that ak + bj h ≡ 0 (mod bk). Deﬁne b k S :={(h, k) ∈ N : there exists j ∈ [1,k]with ak + j bh ≡ 0 (mod bk)}. 0 0 a,b Lemma 5.2 Let j be as above. For k ≤ n and −θ ≤ θ ≤ θ ,wehave h,k h,k jmh ma 2 ζ ζ kt g t (bk + km) θ j,k θ ≥0 1≤m≤bk 1≤j≤k 3 5 j 1 j j 1 k k 0 0 0 = log + − log − log(2π) 1 + O √ + O . (h,k)∈S a,b k 2 k k 2 n n Proof Note that the function g (w) is holomorphic at 0 and in any cone |Arg(w)|≤ − η. j,k 1 1 Also, θ , θ ≤ = O implies that t lies in such a ﬁxed cone (see also [1]onp. h,k h,k kN 2 m 75). Thus, we can apply Theorem 2.3 to g (z)with w → t bk , a → , and N → 0, j,d θ bk ∞ 2 2 m 1 1 j 1 m k g t bk + = g (w)dw − − − + O √ . j,k θ j,k 2 2 bk t bk 12 2k 2 bk n θ 0 ≥0 W. Bridges et al. Res Math Sci (2022) 9:61 Page 25 of 31 61 When summing the O-term, we get 2 5 k k kt O √ = O , n n 1≤m≤bk 1≤j≤k where we used the fact that t lies in a cone. Summing ﬁrst over m gives 1 j 1 m jmh ma ζ ζ − − = O (k) . b k 12 2k 2 bk 1≤m≤bk Hence, jmh ma 2 ζ ζ kt g t (bk + km) θ θ j,k b k 1≤m≤bk ≥0 1≤j≤k ∞ 3 2 5 k k k = g (w)dw 1 + O √ + O √ + O j ,k (h,k)∈S 0 a,b n n 3 5 j 1 j j 1 k k 0 0 0 = log + − log − log(2π) 1 + O √ + O , (h,k)∈S a,b k 2 k k 2 n n 3 5 1 k k where the last step follows by Lemma 2.4.If ε< , then the terms O and O are 4 n √ √ smaller than n as needed since N = O(δ n) where δ is chosen small and independently of n. It remains to estimate the product term in (5.2). Lemma 5.3 Uniformly for k ≤ n and −θ ≤ θ ≤ θ ,wehave h,k h,k ⎛ ⎞ k j − + −jh 2 k a −jt ε ⎝ ⎠ Log 1 − ζ ζ e = O (n ) . b k j=1 Proof Recall the sums G deﬁned in Lemma 2.9. Using the Taylor expansion for the logarithm followed by Proposition 2.5, we write ⎛ ⎞ m(ak+bjh) k 1 j k − + 1 j ζ −jh 2 k a −jt bk −jmt θ θ ⎝ ⎠ Log 1 − ζ ζ e = − e 2 k m j=1 j=1 m≥1 1 j ak + bhj −jt −mjt θ θ = − (1 − e ) G e . 2 k bk m≥1 j=1 It is elementary to show that at most one of {ak +bjh} is divisible by bk. Suppose that 1≤j≤k ak+bhj this happens at j (if it never happens, then the argument is similar). Then G = 0 m bk − log(1−x) H ,the m-th harmonic number. Applying the formula H x = ,the m m m≥1 1−x above is a hj −j t −mj Re(t ) −jt −mjRe(t ) 0 θ 0 θ θ θ 1 − e H e + 1 − e max G + e m m m≥1 b k m≥1 1≤j≤k m≥1 j=j −j t −jt 0 θ θ 1 − e |1 − e | a hj −j t 0 θ log 1 − e + max G + . −j Re(t ) −jRe(t ) 0 θ θ m≥1 b k 1 − e 1 − e 1≤j≤k j=j 0 61 Page 26 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 The fact that t lies in a cone |Arg(t )|≤ − η with j ≤ k ≤ n < n gives θ θ −jt 1 − e |t | = O = O(1). −jRe(t ) Re(t ) 1 − e θ Thus, using Lemma 2.10 ⎛ ⎞ ⎛ ⎞ k j − + ⎜ ⎟ a hj 2 k −jh a −jt θ ⎜ ⎟ ⎝ ⎠ Log 1 − ζ ζ e = O(log(n)) + O max G + k ⎝ ⎠ m≥1 b k j=1 1≤j≤k j=j = O(log(n)) + O(k) = O(n ), as claimed. Lemma 4.2 now follows from Lemmas 5.1, 5.2 and 5.3 by recalling that k ≤ n where 0 <ε ≤ : j 1 j j 1 0 0 0 E (ζ ,t ) = log + − log − log(2π) 1 h,k (h,k)∈S b a,b k 2 k k 2 3 5 k k + O √ + O + O (n ) n n 1 1 3ε− ε 3ε− ε 2 2 log(n) + n + n = O n + O (n ) . 5.2 Proof of Lemma 4.1 To prove Lemma 4.1, we need an elementary fact about the sets S . a,b Lemma 5.4 Let 1 ≤ a < with gcd(a, b) = 1 and b ≥ 3.Then (1, 1) ∈ / S , (h, 2) ∈ / S , a,b a,b and (h, 3) ∈ S if and only if (a, b) = (1, 3). a,b Proof We prove the case (h, k) = (h, 2) and note that the remaining cases are analogous. We have that (h, 2) ∈ S if and only if 2b divides 2a +b or 2a +2b. Clearly, 2b (2a +2b), a,b and since 2a + b < 3b < 2 · (2b), 2b | (2a + b) ⇐⇒ 2a + b = 2b ⇐⇒ 2a = b ⇐⇒ (a, b) = (1, 2), which is a contradiction. Proof of Lemma 4.1 Cases (1), (2) and (3) are simple consequences of Lemmas 5.1 and 5.2 and 5.4. For case (4), we suppose ζ = ζ . Then one ﬁnds j (1, 3, 1, 3) = 2 and Lemmas 5.1 3 0 and 5.2 imply ⎛ ⎞ 3 j − + 2 1 2 1 j 2 3 −jt ⎝ ⎠ E (ζ ,t ) = Log 1 − ζ ζ e + log − log − log(2π) + o(1) 1,3 3 θ 3 3 6 3 2 j=1 (1 − ζ ) 1 1 1 = Log t + Log + log + log (3) − log(2π) + o(1), 3 6 2 (1 − ζ ) as claimed, whereas j (1, 3, 2, 3) = 1, and so 3 j − + 1 1 1 1 2j 2 3 a −jtθ E (ζ ,t ) = log 1 − ζ ζ e + log + log − log(2π) + o(1) 2,3 θ 3 b 3 3 6 3 2 j=1 1 1 1 1 1 1 1 6 6 = log t + log 1 − ζ (1 − ζ ) + log + log − log(2π) + o(1), θ 3 3 6 3 2 as claimed. W. Bridges et al. Res Math Sci (2022) 9:61 Page 27 of 31 61 5.3 Proof of Lemma 4.3 In preparation for the proof of Lemma 4.3, we rewrite E as in Lemma 5.1, this time h,k using only the ﬁrst two terms of g . The proof is analogous. j,k Lemma 5.5 For Re(t) > 0 and z = ζ ,wehave jmh ma 2 E (z, t) = ζ ζ kt % g t bk + km , (5.2) h,k j,k b k 1≤m≤bk ≥0 1≤j≤k where − w e 1 % g (w):= − . j,k −w 2 w(1 − e ) w We will need to estimate the sum in (5.2) separately for ≥ 1and = 0. For ≥ 1, we can ﬁrst compute the sum on j as mh − ζ e jmh ζ g (z) = − · 1 . j,k k|m mh − z(1 − ζ e ) 1≤j≤k k Thus, writing m = νk with 1 ≤ ν ≤ b when k | m,wehave jmh ma 2 ζ ζ kt % g t(bk + km) j,k b k ≥1 1≤m≤bk 1≤j≤k ν m νka ma = t ζ f bkt + + t ζ f bkt + =: S + S , 1 2 1 2 b b b bk ν=1 ≥1 1≤m≤bk ≥1 km say, where −z e 1 f (z):= − −z 2 z(1 − e ) z and mh −z ζ e f (z):= . mh −z z(1 − ζ e ) Lemma 5.6 For k ≥ n , −θ ≤ θ ≤ θ and t → t ,wehave |S |= O (log(n)). θ 1 h,k h,k Proof We use Theorem 2.2 (with N →∞) to write f bkt 1 + ν 1 θ ν f bkt + = + f bkt x + dx 1 θ 1 θ b 2 b ≥1 ν 1 + bkt f bkt x + {x}− dx θ θ b 2 f kt b + ν 1 ( ( )) 1 θ = + f (z) dz 2 bkt θ kt (b+ν) & ' z − 1 + f (z) − dz. bkt 2 kt (b+ν) θ θ 61 Page 28 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 Here, f (z) as z → 0, thus f (kt (b + ν)) = O . 1 θ k|t | t∞ Furthermore c f (z)dz = O(1) for |Arg(t)|≤ − η, uniformly for any η,c > 0. As t· 2 |t| noted before, t lies in such a cone, so 2b ∞ t |t | 1 1 1 | log(kt )| θ θ f (z) dz = O dz = O . bkt k|t | z k|t | θ kt (b+ν) θ kt (b+ν) θ θ θ Similarly, one has −z −z −2z e e e 2 f (z) =− − − + , −z 2 −z −z 2 3 z(1 − e ) z (1 − e ) z(1 − e ) z t∞ π 1 so c f (z)dz = O(1) for |Arg(t)|≤ − η, for any η,c > 0. And one has f (z) as t· 1 2 1 |t| z → 0, thus & ' 2b ∞ t z − |t | 1 1 1 f (z) − dz = O dz = O . bkt 2 z k|t | θ θ kt (b+ν) kt (b+ν) θ θ The above bounds are all clearly uniform in 1 ≤ ν ≤ b, thus overall | log kt | |S |= |t |O = O (| log(kt )|) = O(log n), 1 θ θ k|t | 1≤m≤bk km as claimed. ε −ε Lemma 5.7 For k ≥ n , −θ ≤ θ ≤ θ and t → t ,wehave |S |= O n . θ 2 h,k h,k Proof For Re(z) > 0, we see immediately that −Re(z) |f (z)|≤ =: f (Re(z)). 2 2 −Re(z) Re(z)(1 − e ) Thus, since f is decreasing, we have |t | ˜ ˜ ˜ |S |≤|t |bk f bkRe(t ) ≤ f (x)dx + bkRe(t )f (bkRe(t )) , ( ) 2 θ 2 θ 2 θ 2 θ Re(t ) θ bkRe(t ) ≥1 by integral comparison. We can bound the right term as 1 1 bkRe(t )f (bkRe(t )) = ≤ . θ 2 θ bkRe(t ) e − 1 bkRe(t ) Furthermore, as η → 0 ,wehave ∞ 1 −x 1 e 1 1 f (x)dx = O(1) + dx ≤ O(1) + dx = O . −x 2 x(1 − e ) x η η η η Hence, overall, 1 n −ε |S |= O = O = O n , kRe(t ) k as claimed. W. Bridges et al. Res Math Sci (2022) 9:61 Page 29 of 31 61 It remains to estimate the double sum (5.2) for the term = 0; i.e., φ j (tkm) jmh ma k ζ ζ , 1≤m≤bk 1≤j≤k where −aw e 1 φ (w):= − . −w 1 − e w Note that φ is holomorphic at 0 and in the cone |Arg(w)|≤ − η, for any η> 0. We apply Lemma 5.8 to φ to bound diﬀerences as follows. Lemma 5.8 Let x be a complex number with positive imaginary part and |x|≤ 1.Then there is a constant c > 0 independent from a and x, such that for all m ≤ we have |x| φ (xm) − φ (x(m + 1)) ≤ c|x|. a a Proof The function φ (z) is holomorphic in B (0). The functions φ (z) are uniformly a 3 bounded on B (0) ⊂ B (0). Indeed, we have uniformly in a −az−z −az e ae 1 max |φ (z)|= max |φ (z)|≤ max + max + max 1. a a −z 2 −z 2 5 5 5 5 5 (1 − e ) 1 − e z |z|≤ |z|= |z|= |z|= |z|= 2 2 2 2 2 On the other hand, by Lemma 2.6 applied to f = φ , U = B (0), and B (0) ⊂ U,weﬁnd, a 3 since |xm|≤ 1and |x(m + 1)|≤|mx|+|x|≤ 2 |φ (xm + x) − φ (mx)|≤ max |φ (z)||xm + x − xm|= c|x|, a a |z|≤ where c does not depend on 0 < a ≤ 1. We also require the following lemma for large values of m, whose proof is a straightfor- ward calculation using that the denominators of the ﬁrst term in φ (w) are bounded away from 0. Lemma 5.9 Let x be a complex number with positive imaginary part. Then all m > |x| we have −mRe(x) −amRe(x) φ (xm) − φ (x(m + 1)) +|x|e + a|x|e . a a |x|m(m + 1) The following lemma, when combined with Lemmas 5.5–5.7, completes the proof of Lemma 4.3, and thus that of Theorem 3.2. Lemma 5.10 For n ≤ k ≤ Nand −θ ≤ θ ≤ θ ,wehave h,k h,k φ (t km) jmh ma k ζ ζ = O(k). b k 1≤m≤bk 1≤j≤k 61 Page 30 of 31 W. Bridges et al. Res Math Sci (2022) 9:61 |x| Proof Let x := kt and a := .Notethatwehave0 < a ≤ 1, Re(x) > 0, and 1 k Re(x) uniformly in k. We use Abel partial summation and split the sum into two parts: ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ = + . ⎝ ⎠ ( ) ( ) 1≤m≤bk j=1 1 1 0<m≤min bk, min bk, <m≤bk |x| |x| 1≤j≤k In the case |x| > 1, the ﬁrst sum is empty, so we can assume |x|≤ 1. We ﬁrst ﬁnd with Proposition 2.5 that & * +' 1 a hj 1 m(ak+hjb) ( ) ζ φ (xm) = G + φ x min bk, a a bk min bk, m b k |x| ( ) |x| m≤min bk, |x| a hj + G + (φ (mx) − φ ((m + 1)x)) . m a a b k ( ) m≤min bk, −1 |x| It follows that with Lemma 5.8 & * +' k k 1 a hj 1 mhj ma ( ) ζ ζ · φ (mx) ≤ G + φ x min bk, a 1 a min bk, ( ) m |x| b k |x| j=1 1 j=1 m≤min bk, |x| a hj + G + (φ (mx) − φ ((m + 1)x)) m a a b k ( ) j=1 1 0<m≤min bk, |x| k k a hj a hj ( ) G + + max G + |x|= O(k), 1 ( ) m min bk, b k b k |x| 1 m=1,...,min bk, j=1 j=1 |x| 0<m≤ |x| where we used Lemma 2.10 and G (1) = H = O(log(k)) in the last step. Similarly, we bk bk ﬁnd with Lemma 5.9 (without loss of generality we assume < bk) |x| mhj ma ζ ζ · φ (mx) b k j=1 <m≤bk |x| k k a hj a hj G + φ xbk + G + φ (mx) − φ ((m + 1)x) ( ) ( ) bk a m a a b k b k j=1 j=1 <m≤bk |x| a hj 1 −mRe(x) −amRe(x) O(k) + max G + +|x|e + a|x|e . b k |x|m(m + 1) m= ,...,bk |x| 1 j=1 <m≤bk |x| Note that we uniformly have φ (xbk) 1(as 1 |xbk| and x is part of a ﬁxed cone |Arg(x)|≤ − η)aswellas 1 1 ≤ 1 |x|m(m + 1) |x|m(m + 1) 1 1 <m≤bk <m<∞ |x| |x| and |x| −mRe(x) |x|e ≤ 1, −Re(x) 1 − e <m≤bk |x| W. Bridges et al. Res Math Sci (2022) 9:61 Page 31 of 31 61 |x| as 1, |x| 1. Similarly, Re(x) −amRe(x) a|x|e 1. <m≤bk |x| As a result, using Lemma 2.10 (again up to at most one summand in O(log(k))), k k 1 a hj mhj ma ζ ζ · φ (mx) O(k) + max G + = O(k), a m b k m b k m= ,...,bk 1 |x| j=1 j=1 <m≤bk |x| as claimed. Acknowledgements We would like to thank Kathrin Bringmann, Joshua Males, Caner Nazaroglu, Ken Ono and Wadim Zudilin for useful discussions and for making comments on an earlier version of this paper. We are grateful to the referees for their detailed comments that improved the exposition. Funding Open Access funding enabled and organized by Projekt DEAL. Declarations Conﬂict of interest There is no conﬂict of interest for the current study. Received: 5 May 2022 Accepted: 26 August 2022 Published online: 24 September 2022 References 1. Andrews, G.: The Theory of Partitions. Cambridge University Press, Cambridge (1984) 2. Beckwith, O., Mertens, M.: On the number of parts in integer partitions lying in given residue classes. Ann. Comb. 21, 558 (2017) 3. Beckwith, O., Mertens, M.: The number of parts in certain residue classes of integer partitions. Res. Number Theory A11, 558 (2015) 4. Boyer, R., Goh, W.: Partition Polynomials: Asymptotics and Zeros. (2007). 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Research in the Mathematical Sciences – Springer Journals
Published: Dec 1, 2022
Keywords: Circle method; Partitions; Asymptotics; Sign-changes; Secondary terms; 11P82; 11P83
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