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Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg Asymptotic analysis of expectations of plane partition statistics
Asymptotic analysis of expectations of plane partition statistics
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Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg
DOI:
10.1007/s121880180191z
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April, 2018
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Abh. Math. Semin. Univ. Hambg. (2018) 88:255–272 https://doi.org/10.1007/s121880180191z Asymptotic analysis of expectations of plane partition statistics Ljuben Mutafchiev1,2 Received: 9 October 2017 / Published online: 6 February 2018 © The Author(s) 2018 Abstract Assuming that a plane partition of the positive integer n is chosen uniformly at random from the set of all such partitions, we propose a general asymptotic scheme for the computation of expectations of various plane partition statistics as n becomes large. The functions that arise in this study are of the form Q(x)F(x), where Q(x) = ∞ generating j − j is the generating function for the number of plane partitions. We show how j=1 (1 − x ) asymptotics of such expectations can be obtained directly from the asymptotic expansion of the function F(x) around x = 1. The representation of a plane partition as a solid diagram of volume n allows interpretations of these statistics in terms of its dimensions and shape. As an application of our main result, we obtain the asymptotic behavior of the expected values of the largest part, the number of columns, the number of rows (that is, the three dimensions of the solid diagram) and the trace (the number of cubes in the wall on the main diagonal of the solid diagram). Our results are similar to those of Grabner et al. (Comb Probab Comput 23:1057–1086, 2014) related to linear integer partition statistics. We base our study on the Hayman’s method for admissible power series. Keywords Plane partition statistic · Asymptotic behavior · Expectation Mathematics Subject Classification 05A17 · 05A16 · 11P82 Communicated by M. Schacht. B Ljuben Mutafchiev ljuben@aubg.edu 1 American University in Bulgaria, 2700 Blagoevgrad, Bulgaria 2 Institute of Mathematics and Informatics of the Bulgarian Academy of Sciences, Sofia, Bulgaria 123 256 L. Mutafchiev 1 Introduction A plane partition ω of the positive integer n is an array of nonnegative integers ω1,1 ω1,2 ω1,3 · · · ω2,1 ω2,2 ω2,3 · · · ··· ··· ··· (1; ) that satisfy h, j≥1 ωh, j = n, and the rows and columns in (1) are arranged in decreasing order: ωh, j ≥ ωh+1, j and ωh, j ≥ ωh, j+1 for all h, j ≥ 1. The nonzero entries ωh, j > 0 are called parts of ω. If there are λh parts in the hth row, so that for some l, λ1 ≥ λ2 ≥ ... ≥ λl > λl+1 = 0, then the (linear) partition λ = (λ1 , λ2 , ..., λl ) of the integer m = λ1 + λ2 + · · · + λl is called the shape of ω, denoted by λ. We also say that ω has l rows and m parts. Sometimes, for the sake of brevity, the zeroes in array (1) are deleted. For instance, the abbreviation 5 4 1 1 3 2 1 2 1 (2) represents a plane partition of n = 20 with l = 3 rows and m = 9 parts. Any plane partition ω has an associated solid diagram = (ω) of volume n. It is defined as a set of n integer lattice points x = (x1 , x2 , x3 ) ∈ N3 , such that if x ∈ and x j ≤ x j , j = 1, 2, 3, then x = (x1 , x2 , x3 ) ∈ too. (Here N denotes the set of all positive integers.) Indeed the entry ωh, j can be interpreted as the height of the column of unit cubes stacked along the vertical line x1 = h, x2 = j, and the solid diagram is the union of all such columns. Figure 1 represents the solid diagram of the plane partition in the example (2). Plane partitions are originally introduced by Young [29] as a natural generalization of integer partitions in the plane. The problem of enumerating plane partitions was studied first by MacMahon [14] (see also [15]), who showed that, for any parallelepiped B(l, s, t) = {(h, j, k) ∈ N3 : h ≤ l, j ≤ s, k ≤ t} and any x < 1, x  = ⊂B(l,s,t) (h, j,k)∈B(l,s,t) 1 − x h+ j+k−1 , 1 − x h+ j+k−2 (3) where is the solid diagram defined above and  denotes its volume (a more explicit version of (3) may be found in [23]). Let q(n) denote the total number of plane partitions of the positive integer n (or, the total number of solid diagrams of volume n). It turns out that (3) implies the following generating function identity: ∞ ∞ Q(x) = 1 + q(n)x n = (1 − x j )− j (4) n=1 j=1 (more details may be also found, e.g., in [24, Corollary 18.2] and [2, Corollary 11.3]). Subsequent research on the enumeration of plane partitions was focused on bijective interpretations and proofs of MacMahon’s formula (4) (see, e.g., [13,22]). The asymptotic form of the numbers q(n), as n → ∞, has been obtained by Wright [28] (see also [20] for a little correction). It is given by the following formula: q(n) ∼ 123 (ζ (3))7/36 211/36 (3π)1/2 n −25/36 exp (3(ζ (3))1/3 (n/2)2/3 + 2γ ), (5) Asymptotic analysis of expectations of plane partition statistics 257 Fig. 1 The solid diagram of plane partition (2) where ζ (z) = ∞ j −z j=1 is the Riemann zeta function and γ = 0 ∞ u log u 1 du = ζ (−1). 2π u e −1 2 (6) (The constant ζ (−1) = −0.1654... is closely related to the Glaisher–Kinkelin constant; see [6]). Remark 1 In fact, Wright [28] has obtained an asymptotic expansion for q(n) using the circle method. Next, we introduce the uniform probability measure P on the set of plane partitions of n, assuming that the probability 1/q(n) is assigned to each plane partition. In this way, each numerical characteristic of a plane partition of n becomes a random variable (a statistic in the sense of the random generation of plane partition of n). In the following, we will discuss several different instances of plane partition statistics. Our goal is to develop a general asymptotic scheme that allows us to derive an asymptotic formula for the nth coefficient [x n ]Q(x)F(x) of the product Q(x)F(x), where Q(x) is defined by (4) and the power series F(x) is suitably restricted on its behavior in a neighborhood of x = 1. We will show further that expectations of plane partition statistics we will consider lead to generating functions of this form. From one side, our study is motivated by the asymptotic results of Grabner et al. [9] on linear partition statistics. Their study is based on general asymptotic formulae for the nth coefficient of a similar product of generating functions with Q(x) replaced by the Euler partition generating function P(x) = ∞ (1 − x j )−1 . The second factor F(x) satisfies j=1 123 258 L. Mutafchiev similar general analytic conditions around x = 1. In addition, our interest to study plane partitions statistics was also attracted by several investigations during the last two decades on shape parameters of random solid diagrams of volume n as n → ∞. Below we present a brief account on this subject. Cerf and Kenyon [4] have determined the asymptotic shape of the random solid diagram, while Cohn et al. [5] have studied a similar problem whenever a solid diagram is chosen uniformly at random among all diagrams boxed in B(l, s, t), for large l, s and t, all of the same order of magnitude. Okounkov and Reshetikhin [21] rediscovered Cerf and Kenyon’s limiting shape result and studied asymptotic correlations in the bulk of the random solid diagram. Their analysis is based on a deterministic formula for the correlation functions of the Schur process. The joint limiting distribution of the height [the largest part in (1)], depth [the number of columns in (1)] and width [the number of rows in (1)] in a random solid diagram was obtained by Pittel [23]. The onedimensional marginal limiting distributions of this random vector were found in [17]. The trace of a plane partition is defined as the sum of the diagonal parts in (1). Its limiting distribution is determined in [12]. Bodini et al. [3] studied random generators of plane partitions according to the size of their solid diagrams. They obtained random samplers that are of complexity O(n log3 n) in an approximatesize sampling and of complexity O(n 4/3 ) in exactsize sampling. These random samplers allow to perform simulations in order to confirm the known results about the limiting shape of the plane partitions. In the proof of our main asymptotic result for the coefficient [x n ]Q(x)F(x) we use the saddle point method. In contrast to [9], we base our study on a theorem due to Hayman [11] for estimating coefficients of admissible power series (see also [7, Section VIII.5]). We show that Q(x) (see (4)) is a Hayman admissible function and impose conditions on F(x) which are given in terms of Hayman’s theorem. In the examples we present, we demonstrate two different but classical approaches for estimating power series around their main singularity. Our paper is organized as follows. In Sect. 2 we include Hayman’s admissibility conditions and his main asymptotic result. We briefly describe a relationship between Hayman’s theorem and Meinardus’ approach [16] for obtaining the asymptotic behavior of the Taylor coefficients of infinite products of the form f b (x) = ∞ (1 − x j )−b j (7) j=1 under certain general conditions on the sequence {b j } j≥1 of nonnegative numbers. In Sect. 2, we also state our main result (Theorem 1) for the asymptotic of the nth coefficient [x n ]Q(x)F(x) under certain relatively mild conditions on F(x). The proof of Theorem 1 is given in Sect. 3. Section 4 contains some examples of plane partition statistics that lead to generating functions of the form Q(x)F(x). We apply Theorem 1 to obtain the asymptotic behavior of expectations of the underlying statistics. For the sake of completeness, in the Appendix we show how Wright’s formula (5) follows from Hayman’s theorem. Remark 2 An extended abstract of this work was presented at EUROCOMB 2017 (European Conference on Combinatorics, Graph Theory and Applications, Vienna, Austria, August– September, 2017) and published in [19]. 123 Asymptotic analysis of expectations of plane partition statistics 259 2 Some remarks on Hayman admissible functions and Meinardus theorem on weighted partitions; statement of the main result Our first goal in this section is to give a brief introduction to the analytic combinatorics background that we will use in the proof of our main result. Clearly, x n [Q(x)F(x)] with Q(x) given by (4) can be represented by a Cauchy integral whose integrand includes the product Q(x)F(x) (the conditions that F(x) should satisfy will be specified later). Its asymptotic behavior heavily depends on the analytic properties of Q(x) whose infinite product representation (4) shows that the unit circle is a natural boundary and its main singularity is at x = 1. The main tools for the asymptotic analysis of q(n) = x n [Q(x)] are either based on the circle method (see [28]) or on the saddlepoint method (see [3]). Both yield Wright’s asymptotic formula (5). An asymptotic formula in a more general framework [see (7)] was obtained by Meinardus [16] (see also [10] for some extensions). A proof of formula (5) that combines Meinardus approach with Hayman’s theorem for admissible power series is started in Sect. 3 and completed in the Appendix. Here we briefly describe Menardus’ approach, which is essentially based on analytic properties of the Dirichlet generating series D(z) = ∞ b j j −z , z = u + iv, (8) j=1 where the sequence of nonnegative numbers {b j } j≥1 is the same as in the infinite product (7). We will avoid the precise statement of Meinardus’ assumptions on {b j } j≥1 as well as some extra notations and concepts. The first assumption (M1 ) specifies the domain H = {z = u + iv : u ≥ −C0 , 0 < C0 < 1} in the complex plane, in which D(z) has an analytic continuation. The second one (M2 ) is related to the asymptotic behavior of D(z) whenever v → ∞. A function of the complex variable z which is bounded by O( (z)C1 ), 0 < C1 < ∞, in a certain domain in the complex plane is called function of finite order. Meinardus’ second condition (M2 ) requires that D(z) is of finite order in the whole domain H. Finally, Meinardus’ third condition (M3 ) implies a bound on the ordinary generating function of the sequence {b j } j≥1 . It can be stated in a way, simpler than Meinardus’ original expression, by the inequality: ∞ b j e− jα sin2 (π ju) ≥ C2 α − 0 , 0 < j=1 α 1 < u < , 2π 2 for sufficiently small α and some constants C2 , 0 > 0 (C2 = C2 ( 0 )) (see [10, p. 310]). The infinite product representation (4) for Q(x) implies that b j = j, j ≥ 1, and therefore, D(z) = ζ (z −1). It is known that this sequence satisfies the Meinardus’ scheme of conditions (see, e.g., [20] and [10, p. 312]). Now, we proceed to Hayman admissibility method [11], [7, Section VIII.5]. To present the idea and show how it can be applied to the proof of our main result, we need to introduce some auxiliary notations. n Consider a function G(x) = ∞ n=0 gn x that is analytic for x < ρ, 0 < ρ ≤ ∞. For 0 < r < ρ, we set a(r ) = r b(r ) = G (r ) , G(r ) r G (r ) G (r ) + r2 − r2 G(r ) G(r ) (9) G (r ) G(r ) 2 . (10) 123 260 L. Mutafchiev In the statement of Hayman’s result we use the terminology given in [7, Section VIII.5]. We assume that G(x) > 0 for x ∈ (R0 , ρ) ⊂ (0, ρ) and satisfies the following three conditions: Capture condition. lim r →ρ a(r ) = ∞ and limr →ρ b(r ) = ∞. Locality condition. For some function δ = δ(r ) defined over (R0 , ρ) and satisfying 0 < δ < π, one has 2 G(r eiθ ) ∼ G(r )eiθa(r )−θ b(r )/2 (11) as r → ρ, uniformly for θ  ≤ δ(r ). Decay condition. G(r ) G(r eiθ ) = o √ b(r ) (12) as r → ρ uniformly for δ(r ) < θ  ≤ π. Hayman Theorem. Let G(x) be a Hayman admissible function and r = rn be the unique solution of the equation a(r ) = n. (13) Then the Taylor coefficients gn of G(x) satisfy, as n → ∞, gn ∼ G(rn ) √ rnn 2πb(rn ) (14) with b(rn ) given by (10). In the next section we will show that Q(x), given by (4), is admissible in the sense of Hayman and apply Hayman’s Theorem setting G(x) := Q(x) (and ρ := 1). Furthermore, as in [9], we will introduce two rather mild conditions on the second factor F(x). Since we will employ Hayman’s method, the first one is given in terms of the solution rn of Eq. (13). The second one requires that F(x) does not grow too fast as x → 1. Condition A. Let r = rn be the solution of (13). We assume that lim n→∞ F(rn eiθ ) =1 F(rn ) (15) uniformly for θ  ≤ δ(rn ), where δ(r ) is the function defined by Hayman’s locality condition. Condition B. There exist two constants C > 0 and η ∈ (0, 2/3), such that, as x → 1, η F(x) = O(eC/(1−x) ). Before the statement of the main result we wish to make a few clarifying comments on the role of the function δ(r ). Remark 3 Revisiting Hayman’s locality condition, we conclude that the function δ(r ) defines an arc on the circle x = r that is very close to the main singularity of the underlying function. The MacMahon generating function (4) has infinitely many singularities in any neighborhood of the point x = 1 and thus in the case at hand we have δ(r ) → 0 as r = x → 1. Note that from (11) and (12) it follows that the asymptotic behavior of Q(r eiθ ) significantly changes when θ leaves the interval (−δ(r ), δ(r )). So, an important detail in the analysis of the asymptotic behavior of the product Q(x)F(x) around the point x = 1 is to determine the function δ(r ) explicitly. We will find an expression for δ(r ) in the next section. Condition A for the second factor F(x) is rather technical. It avoids pathological examples for F(x) (e.g., zeros or oscillations of F(r eiθ ) which provide the ratio in (15) with points of accumulation that are = 1 as r → 1 whenever θ ∈ (−δ(r ), δ(r ))). 123 Asymptotic analysis of expectations of plane partition statistics Theorem 1 Let {dn }n≥1 be a sequence with the following expansion: 2ζ (3) 1/3 1 dn = − + O(n −1−β ), n → ∞, n 36n 261 (16) where β > 0 is a certain fixed constant. Furthermore, suppose that the function F(x) satisfies conditions (A) and (B) and Q(x) is the infinite product given by (4). Then, there is a constant c > 0 such that, as n → ∞, 1 2/9 2 [x n ]Q(x)F(x) = F(e−dn )(1 + o(1)) + O(e−cn / log n ), q(n) where q(n) is the nth coefficient in the Taylor expansion of Q(x). The proof that we will present in the next section stems from the Cauchy coefficient formula in which the contour of integration is the circle x = e−dn +iθ , −π < θ ≤ π and dn is defined by (16). We have endn π n Q(e−dn +iθ )F(e−dn +iθ )e−iθ n dθ. (17) [x ]Q(x)F(x) = 2π −π The proof of Theorem 1 is divided into two parts: (i) Proof of Hayman admissibility for Q(x). (ii) Obtaining an asymptotic estimate for the Cauchy integral (17). 3 Proof of Theorem 1 Part (i). We will essentially use some more general observations established in [10,18]. First, we set in (9) and (10) G(x) := Q(x) and r = rn := e−dn . The next lemma is a particular case of a more general result due to Granovsky et al. [10, Lemma 2]. Lemma 1 For large enough n, the unique solution of the equation a(e−dn ) = n satisfies (16). Moreover, as n → ∞, b(e−dn ) ∼ 3n 4/3 . (2ζ (3))1/3 (18) Proof The Dirichlet generating series (8) of the sequence { j} j≥1 is ζ (z − 1), which is a meromorphic function in the complex plane with a single pole at z = 2 with residue 1 (see, e.g., [27, Section 13.13]). Granovsky et al. [10] showed that it satisfies Meinardus’ conditions (M1 ) and (M2 ). Hence, by formula (43) in [10], we have ζ (−1) dn = ((3)ζ (3))1/3 n −1/3 + + O(n −1−β ) 3n 1 2ζ (3) 1/3 + O(n −1−β ), β > 0, − = n 36n which is just (16) (here we have also used that ζ (−1) = −1/12 [27, Section 13.14]). Formula (18) follows from (16) and a more general result from [18, formula (2.6)]. 123 262 L. Mutafchiev Lemma 1 shows that a(e−dn ) → ∞ and b(e−dn ) → ∞ as n → ∞, that is Hayman’s “capture” condition is satisfied with r = rn = e−dn . Our next argument deals with the function δ(r ) that was defined by Hayman’s locality condition. For the sake of convenience, we set δn := δ(e−dn ), where dn satisfies (16). Moreover, by log (.) we will denote the main branch of the logarithmic function satisfying log u < 0 for any u ∈ (0, 1). Further, to show that Hayman’s ”decay” condition is satisfied by Q(x), we also set 5/3 2ζ (3) 5/9 dn 1 δn = (1 + O(n −2/3 )). (19) = log n log n n The next lemma is a particular case of Lemma 2.4 in [18]. Lemma 2 For sufficiently large n, we have −2/3 Q(e−dn +iθ ) ≤ Q(e−dn )e−C3 dn uniformly for δn ≤ θ  ≤ π, where C3 > 0 is an absolute constant and dn satisfies (16). Proof First, we notice that Lemma 2.4 from [18] establishes a uniform bound for the ratio  f b (e−dn +iθ )/ f (e−dn ) in the general context of the infinite products (7) under the Meinardus’ scheme of assumptions on the sequence {b j } j≥1 . It turns out that this ratio decreases − 1 exponentially fast like e−Mdn for certain constants M, 1 > 0 whose values remain unspecified. Our goal here is to show that the plane partition generating function Q(x) is bounded in the same way with 1 = 2/3. To prove this, we follow an argument from [10] and [18]. Thus, setting θ = 2πu and taking logarithms of the infinite product (4) twice, for dn /2π ≤ u = θ /2π < 1/2, we get (log Q(e−dn +iθ )) − log Q(e−dn ) ⎞ ⎛ ∞ − jdn +2πiu j 1 − e ⎠ j log = ⎝− 1 − e− jdn j=1 =− =− ∞ 1 2 j=1 ∞ 1 2 j=1 1 − 2e− jdn cos (2πu j) + e−2 jdn j log (1 − e− jdn )2 4e− jdn sin2 (πu j) j log 1 + (1 − e− jdn )2 ∞ ≤− log 5 1 Sn , j log (1 + 4e− jdn sin2 (πu j)) ≤ − 2 2 (20) j=1 where Sn = ∞ j=1 je− jdn sin2 (π ju), u = θ , 2π and the last inequality follows from the fact that log (1 + y) ≥ log4 5 y if 0 ≤ y ≤ 4. To obtain a lower bound for Sn whenever dn < θ  ≤ π we will use a more general fact 123 Asymptotic analysis of expectations of plane partition statistics 263 established by Granovsky et al. [10, Lemma 1]. It is related to numerical sequences {b j } j≥1 , that satisfy the inequality b j ≥ α j w−1 uniformly for all j ≥ j0 ≥ 1, where α, w > 0 are some constants. The sequence { j} j≥1 obviously satisfies this inequality with α = w = j0 = 1. Then the condition stated in case (ii) [10, p. 312, Lemma 1]) is satisfied (i.e., we have w ≥ 1) and thus it follows that there exists a constant K > 0 such that Sn ≥ K dn−1 . Combining this inequality with (20), we finally conclude that there exists a constant C3 > 0 such that Q(e−dn +iθ ) −1 = exp ((log Q(e−dn +iθ ))) − log Q(e−dn ) ≤ e−C3 dn −d n Q(e ) (21) uniformly for dn < θ  ≤ π. Now, it remains to consider the interval δn ≤ θ  ≤ dn (i.e., δn /2π ≤ u ≤ dn /2π). For any real number y, we define y to be the distance from y to the nearest integer. To obtain a lower bound for Sn in this region, we will also use the inequality: sin2 (π y) ≥ 4y2 (22) (see, e.g., [8, p. 272]). Using the definition of y, we observe that u j = u j if u j < 1/2. This implies that, for 1 ≤ j ≤ π/dn , u j can be replaced by u j. Recalling that u ≥ δn /2π and applying (22) and (19), we conclude that there is a constant C3 > 0 such that, for large enough n, Sn = ∞ je− jdn sin2 (πu j) ≥ 4 j=1 ≥ 4u 2 je− jdn u j2 j=1 j 3 e− jdn 1≤ j≤π/dn ∼ ∞ 10/3 dn π 2 dn4 log2 n 0 π δ2 ≥ 2n 4 π dn dn ( jdn )3 e− jdn 1≤ j≤dn /π −2/3 y 3 e−y dy ≥ C3 dn / log2 n. Hence by (20), uniformly for δn ≤ θ  ≤ dn , Q(e−dn +iθ ) −2/3 2 ≤ e−C3 dn (log 5)/2 log n . −d Q(e n ) (23) Combining (21) and (23), we obtain the required uniform estimate for δn ≤ θ  < π with certain constant C3 > 0. −dn +iθ ) = This lemma, in combination with (18) and (16), implies that  Q(e −d −d n n o(Q(e )/ b(e )) uniformly for δn ≤ θ ≤ π, which is just Hayman’s ”decay” condition. Finally, since ζ (z − 1) satisfies Meinardus’ conditions (M1 ) and (M2 ), Lemma 2.3 from [18], implies Hayman’s ”locality” condition for Q(x). Lemma 3 We have, e−iθ n Q(e−dn +iθ ) 2 −dn = e−θ b(e )/2 (1 + O(1/ log3 n)) −d n Q(e ) as n → ∞ uniformly for θ  ≤ δn , where δn , dn and b(e−dn ) are determined by (19), (16) and (10), respectively. 123 264 L. Mutafchiev So, all conditions of Hayman’s theorem hold, and we can apply it with gn := q(n), G(x) := Q(x), rn := e−dn and ρ = 2 to find that endn Q(e−dn ) , n → ∞. q(n) ∼ 2πb(e−dn ) (24) In the Appendix we will show that (24) implies the corrected version of Wright’s formula (5). Remark 4 The fact that MacMahon’s generating function Q(x) given by (4) is admissible in the sense of Hayman is a particular case of a more general result established in [18] and related to the infinite products f b (x) of the form (7). It turns out that the Meinardus scheme of assumptions on {b j } j≥1 implies that f b (x) is admissible in the sense of Hayman. Part (ii) We break up the range of integration in (17) as follows: [x n ]Q(x)F(x) = J1,n + J2,n , where J1,n endn = 2π J2,n = endn 2π δn −δn Q(e−dn +iθ )F(e−dn +iθ )e−iθ n dθ, δn ≤θ <π Q(e−dn +iθ )F(e−dn +iθ )e−iθ n dθ (25) (26) (27) and δn is defined by (19). The estimate for J1,n follows from Hayman’s ”locality” condition established in Lemma 3 and condition (A) for F(x). We have endn Q(e−dn )F(e−dn ) δn Q(e−dn +iθ ) −iθ n F(e−dn +iθ ) J1,n = e dθ 2π Q(e−dn ) F(e−dn ) −δn 1 endn Q(e−dn )F(e−dn ) δn −θ 2 b(e−dn )/2 e (1 + o(1))dθ = 1+O 2π log3 n −δn endn Q(e−dn )F(e−dn ) δn −θ 2 b(e−dn )/2 ∼ e dθ. (28) 2π −δn Substituting θ = y/ b(e−dn ), we observe that δn √b(e−dn ) δn 1 2 −d 2 n e−θ b(e )/2 dθ ∼ e−y /2 dy √ −d −d n n −δn b(e ) −δn b(e ) ∞ 2π 2 , n → ∞, e−y /2 dy = ∼ b(e−dn ) −∞ since by (18) and (19) √ n 1/9 δn b(e−dn ) ∼ 3(2ζ (3))7/18 → ∞, n → ∞. log n Hence, by (24), the asymptotic estimate (28) is simplified to J1,n = q(n)F(e−dn ) + o(q(n)F(e−dn )). 123 (29) Asymptotic analysis of expectations of plane partition statistics 265 To estimate J2,n , we first apply Lemma 2 with dn replaced by its expression (16). Thus we observe that there is a constant C4 > 0 such that the inequality Q(e−dn +iθ ) ≤ Q(e−dn )e−C4 n 2/9 / log2 n (30) holds uniformly for δn ≤ θ < π. We will combine this estimate with condition (B) on the function F(x). It implies that, for certain constants c0 , c1 > 0, we have −η F(e−dn ) = O(ec0 dn ) = O(ec1 n η/3 ). (31) Now, combining (27), (30), (18), (24) and (31), we obtain endn Q(e−dn +iθ )F(e−dn +iθ )dθ J2,n  ≤ 2π δn ≤θ <π endn η/3 2/9 Q(e−dn )O(ec1 n )(π − δn )e−C4 n / log n π endn Q(e−dn ) 2/3 C4 n 2/9 n exp − =O + c1 n η/3 log2 n 2πb(e−dn ) ≤ = O(q(n)e−cn 2/9 / log2 n ), (32) for some c > 0. Substituting the estimates obtained in (29) and (32) into (25), we complete the proof. 4 Examples 4.1 The trace of a plane partition The trace Tn of a plane partition ω, given by array (1), is defined as the sum of its diagonal parts: Tn = ω j, j . j≥1 The asymptotic behavior of Tn , as n → ∞, can be studied using the following generating function identity established by Stanley [25] (see also [2, Chapter 11, Problem 5]): 1+ ∞ n=1 q(n)x n n P(Tn = m)u m = 1 + m=1 ∞ q(n)ϕn (u)x n = n=1 ∞ (1 − ux j )− j , (33) j=1 where ϕn (u) denotes the probability generating function of Tn : ϕn (u) = E(u Tn ) ( u ≤ 1). Since ϕn (1) = E(Tn ), a differentiation of (33) with respect to u yields ∞ q(n)E(Tn )x n = Q(x)F1 (x), (34) n=1 where F1 (x) = ∞ j=1 jx j . 1−xj (35) 123 266 L. Mutafchiev For  θ ≤ δ(rn ), by Taylor’s formula we have F1 (rn eiθ ) = F1 (rn ) + O(θ F1 (rn )) = F1 (rn ) + O(δ(rn )F1 (rn )), where rn is the solution of (13). So, the function F1 (x) satisfies condition (A) if F1 (rn ) δ(rn ) → 0, n → ∞. F1 (rn ) Differentiating (35), we get F1 (x) = ∞ j 2 x j−1 . (1 − x j )2 (36) (37) j=1 Setting in (35) and (37) x = rn and interpreting the sums as Riemann sums with step size − log rn = − log (1 − (1 − rn )) = 1 − rn + O((1 − rn )2 ), it is easy to show that ∞ u2 F1 (rn ) = O (1 − rn )−3 du = O((1 − rn )−3 ). (eu − 1)2 0 Hence, with rn = e−dn , we have 1 − rn = dn + O(dn2 ) and F1 (x) x=e−dn = O(dn−3 ). (38) (e−dn ) we need a more precise estimate. In the same way, using the Riemann sum For F1 approximation, we obtain F1 (e−dn ) = ∞ ∞ jdn e− jdn je− jdn = dn−2 dn − jd 1−e n 1 − e− jdn j=1 j=1 ∞ u du = dn−2 ζ (2), n → ∞, ∼ dn−2 eu − 1 0 (39) where in the last equality we have used formula 27.1.3 from [1]. Now, the convergence in (36) follows from (38), (39) and (19) and thus F1 (x) satisfies condition (A). Condition (B) is also obviously satisfied, since an argument similar to that in (39) implies that F1 (x) = η O((1 − x)−2 ) = O(eC/(1−x) ), as x → 1, for any C > 0 and η ∈ (0, 2/3). Combining (39) with (16) and applying the result of Theorem 1 to (34), we obtain the following asymptotic equivalence for E(Tn ). Proposition 1 If n → ∞, then E(Tn ) ∼ κ1 n 2/3 , where κ1 = (2ζ (3))−2/3 π 2 /6 = 0.9166.... Remark 5 One can compare this asymptotic result with the limit theorem for Tn obtained in [12], where it is shown that Tn , appropriately normalized, converges weakly to the standard Gaussian distribution. 4.2 The largest part, the number of rows and the number of columns of a plane partition Let X n , Yn and Z n denote the size of the largest part, the number of rows and number of columns in a random plane partition of n, respectively. Using the solid diagram interpretation (ω) of a plane partition ω, one can interpret X n , Yn and Z n as the height, width and depth 123 Asymptotic analysis of expectations of plane partition statistics 267 of (ω). Any permutation σ of the coordinate axes in N3 , different from the identical one, transforms (ω) into a diagram that uniquely determines another plane partition σ ◦ ω. The permutation σ also permutes the three statistics (X n , Yn , Z n ). So, if one of these statistics is restricted by an inequality, the same restriction occurs on the statistic permuted by σ . The one to one correspondence between ω and σ ◦ ω implies that X n , Yn and Z n are identically distributed for every fixed n with respect to the probability measure P. (More details may be found in [26, p. 371].) Hence, in the context of the expected value E with respect to the probability measure P, we will use the common notation E(Wn ) for Wn = X n , Yn , Z n . The starting point in the asymptotic analysis for E(Wn ) is the following generating function identity: 1+ ∞ P(X n ≤ m, Yn ≤ l)q(n)x n = n=1 l m (1 − x j+k−1 )−1 , l, m = 1, 2, .... k=1 j=1 It follows from a stronger result due to MacMahon [15, Section 495]. For more details and other proofs of this result, we also refer the reader to [24, Chapter V]. If we keep either of the parameters l and m fixed, setting the other one equal to infinity, we obtain 1+ ∞ P(Wn ≤ m)q(n)x n = n=1 m (1 − x k )−m k=1 = Q(x) ∞ ∞ (1 − x j )− j j=m+1 (1 − x j ) j−m , Wn = X n , Yn , Z n . j=m+1 This implies the identity ∞ E(Wn )q(n)x n = Q(x)F2 (x), (40) n=1 where F2 (x) = ∞ ⎛ ⎝1 − m=0 since E(Wn ) = form: n−1 m=0 P(Wn ∞ ⎞ j j−m ⎠ (1 − x ) , j=m+1 > m). For the sake of convenience, we represent F2 (x) in the ∞ F2 (x) = (1 − e Hm (x) ), (41) ( j − m) log (1 − x j ). (42) m=0 where Hm (x) = j>m Our first goal is to find the asymptotic of F2 (e−dn ). Then, we will briefly sketch the verification of conditions (A) and (B). So, in (41) and (42) we set x = e−dn and break up the sum representing F2 (e−dn ) into three parts: F2 (e−dn ) = 1 + 2 + 3 , (43) 123 268 L. Mutafchiev where 1 = (1 − e Hm (e −dn ) ), (44) 0≤m≤N1 2 = (1 − e Hm (e −dn ) ), (45) N1 <m≤N2 3 = (1 − e Hm (e −dn ) ). (46) m>N2 The choice of the the numbers N1 and N2 will be specified later. We will need first asymptotic expansions for dn−1 , dn−2 and log dn−2 . Using (16), it is not difficult to show that 1/3 n 1 dn−1 = + + O(n −1/3−β ), (47) 2ζ (3) 36(2ζ (3))2/3 n 1/3 2/3 n 1 + (48) dn−2 = + O(n −β ) 2ζ (3) 36ζ (3) and 2 2 (49) log n − log (2ζ (3)) + O(n −2/3 ). 3 3 We start with the analysis of 1 . In particular, we wish to find a lower bound for it. Hence, by (44), we need to find an upper bound for Hm (e−dn ). Using the fact that log (1 − e−kdn ) < −e−kdn and (42), we obtain log dn−2 = Hm (e−dn ) = ∞ k log (1 − e−(m+k)dn ) k=1 < −e −(m+1)dn ∞ u k u=e−dn = − k=1 e−(m+1)dn (1 − e−dn )2 e−mdn = − 2 + e−mdn + O(dn e−mdn ), m ≥ 1. dn (50) Furthermore, since log (1 − e jdn ) < 0 for all j ≥ 1, the sequence {Hm (e−dn )}m≥1 is monotonically decreasing in m. Hence, for all m ≤ N1 , 1 − e Hm (e −dn ) ≥ 1 − e HN1 (e −dn ) . Setting m = N1 in (50), we conclude that if N1 = N1 (n) → ∞ is such that N1 dn → ∞ as n → ∞, then the sum 1 in (44) satisfies the inequalities N1 ≥ 1 ≥ N1 − N1 e HN1 (e −dn ) −2 e−N1 dn = N1 − O(N1 e−dn ). (51) The last Oterm tends to 0 as n → ∞ if we set N1 = dn−1 (log dn−2 − log log (N1 log N1 )). dn−1 log dn−2 (52) This, of course, implies that N1 ∼ → ∞ [see (47) and (49)]. A more precise lower bound in (51) can be found using (52) and (16). We have 1 1 1 −2 −N1 dn . = O = N1 e− log (N1 log N1 ) = =O N1 e−dn e log N1 log n log dn−1 123 Asymptotic analysis of expectations of plane partition statistics 269 Hence (51) implies that N1 ≥ 1 ≥ N1 − O(1/ log n), or equivalently, 1 = N1 + O(1/ log n). (53) Once the asymptotic order of 1 was determined by (53), we need to find an asymptotic expression for N1 as a function of n. First, we analyze the log logterm in (52). We have log log (N1 log N1 ) = log (log N1 + log log N1 ) log log N1 = log log N1 + log 1 + log N1 log log N1 . = log log N1 + O log N1 (54) Next, we will apply (47) and (48). First, (47) implies the following estimate for log N1 : log log N1 log N1 = log dn−1 + log log dn−2 − log log N1 + O log N1 1/3 n = log + O(n −1/3 ) + O(log log dn−2 ) 2ζ (3) = 1 log n + O(log log n). 3 Hence log log N1 = − log 3 + log log n + O log log n log n . (55) Combining (47), (49) and (52)  (55), we finally establish that 1/3 n −1/3 + O(n ) 1 = 2ζ (3) 2 log log n n + O(n −2/3 ) + log 3 − log log n + O × log 3 2ζ (3) log n 1/3 n log log n 2 2 = log n − log log n − log (2ζ (3)) + log 3 + O . 2ζ (3) 3 3 log n (56) We will estimate 2 and 3 (see (45) and (46), respectively) setting N2 = dn−1 (log dn−2 + log log N1 ). Using the inequality 1 − e−u (57) ≤ u, u ≥ 0, we obtain in a similar way that −2 2 ≤ (N2 − N1 )O(dn−1 (log log N1 )dn−2 e− log dn −log log N1 ) log log N1 log log n = O dn−1 . = O n 1/3 log N1 log n (58) Finally, by (46), (47), (49), (50) and (57) dn−2 e−mdn = O(dn−2 e−N2 dn ) 3 ≤ m>N2 −1 log d −2 n = O(dn−2 e−dn ) = O(n 2/3 e−2n 1/3 (log n)/3 ). (59) 123 270 L. Mutafchiev Now, (43–46), (56), (58) and (59) imply that 1 presents the main contribution to the asymptotic of F2 (e−dn ) and we have 1/3 n −dn F2 (e ) = 2ζ (3) 2 log log n 2 log n − log log n − log (2ζ (3)) + log 3 + O , n → ∞. × 3 3 log n (60) This result implies that F2 (x) also satisfies condition (B). In fact, (60) shows that F2 (x)(≥ η F(x)) is of order O((1 − x)1/3  log (1 − x)) = O(eC/(1−x) ) for any C, η > 0. The verification of condition (A) is slightly longer. It is based on a convergence argument for F2 (rn ) similar to that in (36) for F1 (rn ). We omit the details and refer the reader to [17, formulas (3.9), (3.11)], which imply that the orders of growth of Hm (x) and Hm (x) are not larger than the third and second powers of (1 − x)−1 , respectively. An argument similar to that given in the proof of (60) yields the required convergence. Thus, using Theorem 1, (40) and (60), we obtain the following result. Proposition 2 If n → ∞, then E(Wn ) ∼ κ2 n 1/3 log n, where Wn = X n , Yn , Z n , and κ2 = 23 (2ζ (3))1/3 = 0.4976.... Remark 6 In [17] is shown that all three dimensions X n , Yn and Z n of the random solid diagram with volume n, appropriately normalized, converge weakly to the doubly exponential (extreme value) distribution as n → ∞. Remark 7 It is possible to obtain more precise asymptotic estimates (expansions) using the circle method. A kind of this method was applied by Wright [28] who obtained an asymptotic expansion for the numbers q(n) as n → ∞. His asymptotic expansion together with a suitable expansion for F(e−dn ) would certainly lead to better asymptotic estimates for the expectations of various plane partition statistics. Acknowledgements I am grateful the referee for the carefully reading the paper and for her/his helpful comments. Appendix In the Appendix we deduce Wright’s formula (5), using Hayman’s result (24). First, by (16) and (18) one has endn = exp ((2ζ (3))1/3 n 2/3 − 1/36 + O(n −β )), (6π)1/2 n 2/3 2πb(e−dn ) ∼ . (2ζ (3))1/6 (A.1) (A.2) An asymptotic expression for Q(e−dn ) can be obtained using a general lemma due to Meinardus [16] (see also [2, Lemma 6.1]). Since the Dirichlet generating series for the plane partitions is ζ (z − 1) (see also (8)), we get Q(e−dn ) = exp (ζ (3)dn−2 − ζ (−1) log dn + ζ (−1) + O(dnβ1 )) 1 log dn + 2γ + O(dnβ1 )) = exp (ζ (3)dn−2 + 12 123 Asymptotic analysis of expectations of plane partition statistics 271 where 0 < β1 < 1 and γ is given by (6) (more details on the values of ζ (−1) and ζ (−1) can be found in [27, Section 13.13] and [6, Section 2.15]). Using (48) and (49), after some algebraic manipulations, we obtain 2ζ (3) 1/36 Q(e−dn ) = exp ((ζ (3))1/3 (n/2)2/3 + 1/36 + O(n −β1 )). (A.3) n Combining (A.1–A.3), we find that 2ζ (3) 1/36 exp ((2ζ (3))1/3 − 1/36 + (ζ (3))1/3 (n/2)2/3 + 1/36 + 2γ ) q(n) ∼ n (3π)1/2 n 2/3 /(2ζ (3))1/6 (ζ (3))1/6+1/36 n −1/36−2/3 exp (3(ζ (3))1/3 (n/2)2/3 + 2γ ) 21/2−1/6−1/36 (3π)1/2 (ζ (3))7/36 −25/36 = 11/36 n exp (3(ζ (3))1/3 (n/2)2/3 + 2γ ). 2 (3π)1/2 = References 1. 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