Báo cáo toán học: "The asymptotic number of set partitions with unequal block sizes" pot

The asymptotic number of set partitions with unequal block sizes A. Knopfmacher Department of Computational and Applied Mathematics University of Witwatersrand Johannesburg, South Africa e-mail: arnoldk@gauss.cam.wits.ac.za A. M. Odlyzko AT&T Labs - Research Florham Park, New Jersey 07932, USA e-mail: amo@research.att.com B. Pittel 1 Department of Mathematics Ohio State University Columbus, Ohio 43210, USA e-mail: bgp@math.ohio-state.edu L. B. Richmond Department of Combinatorics and Optimization University of Waterloo Waterloo, Ontario N2L 3G1, Canada e-mail: lbrichmo@watdragon.uwaterloo.ca D. Stark 2 Department of Mathematics and Statistics University of Melbourne Parkville, Victoria 3052, Australia e-mail: dstark@maths.mu.oz.au G. Szekeres Department of Mathematics University of New South Wales Kensington, New South Wales 2033, Australia e-mail: G.Szekeres@unsw.edu.au N. C. Wormald 3 Department of Mathematics and Statistics University of Melbourne Parkville, Victoria 3052, Australia e-mail: nick@maths.mu.oz.au Submitted: April 20, 1998; Accepted: October 21, 1998. 1 Research supported by the “Focus on Discrete Probability Program” at the Dimacs Center of Rutgers University in Spring of 1997, and under NSA Grant MDA 904-96-1-0053. 2 Current address: BRIMS, Hewlett-Packard Labs, Filton Road, Stoke Gifford, Bristol Bs12 6QZ, UK 3 Research supported by Australian Research Council the electronic journal of combinatorics 6 (1999), #R2 1 Abstract The asymptotic behavior of the number of set partitions of an n-element set into blocks of distinct sizes is determined. This behavior is more complicated than is typical for set par- tition problems. Although there is a simple generating function, the usual analytic methods for estimating coefficients fail in the direct approach, and elementary approaches combined with some analytic methods are used to obtain most of the results. Simultaneously, we obtain results on the shape of a random partition of an n-element set into blocks of distinct sizes. Mathematics Subject Classification (1991): 05A18, 05A16 1. Introduction The literature on enumerating set partitions is not as extensive as that on ordinary partitions, but it is large. We refer to [4, 11, 17] for references to recent papers. In this note we investigate b n , the number of partitions of an n-element set with blocks of unequal sizes. Carlitz [6] has shown that b n has the explicit generating function F (z)= ∞  n=0 b n n! z n = ∞  k=1  1+ z k k!  . (1.1) In addition, according to Wilf (p. 96 in [22]), F(z) is a special case of an enumerator in an exponential family for hands whose cards all have different weights. One interesting aspect of our work is that although F (z) is defined very simply and is entire, we do not obtain our estimates by the usual analytic methods which start by applying Cauchy’s formula to express a coefficient as a contour integral. There are other problems with similar generating functions for which the asymptotics are easy to derive. For example, evaluation of sums of multinomial coefficients (p. 126 in [7]) leads to the generating function G(z)= ∞  k=1  1 − z k k!  −1 . (1.2) The function G(z) has a first order pole at z = 1 with residue R = ∞  k=2  1 − 1 k!  −1 =2.529477 . the electronic journal of combinatorics 6 (1999), #R2 2 The next smallest singularities are the first order poles at ± √ 2. Therefore [z n ]G(z), the coefficient of z n in the Taylor series expansion of G(z), satisfies [z n ]G(z)=R+O(2 −n/2 ) . The analysis of [z n ]G(z) is simple because G(z) has a single dominant singularity at z = 1. If we consider H(z)= ∞  k=1  1+ z k k  , (1.3) then [z n ]H(z) is the probability that a permutation on n letters will have all cycle lengths distinct. The unit circle is a natural boundary of analyticity for H(z) and there are no singularities of H(z)in|z|<1, so the situation is more complicated than for G(z). Greene and Knuth [12] used a Tauberian theorem to show that [z n ]H(z) ∼ e −γ as n →∞, where γ =0.577 is Euler’s constant. A generating function similar to H(z) arises when we consider the analogous problem of determining the probability that a polynomial over a finite field with q elements has only distinct degree irreducible factors. See [15]. The function F(z) is entire and has nonnegative coefficients, so at first glance it might appear that it should be easy to obtain the asymptotics of its coefficients, easier even than for H(z). The usual method for doing this is the saddle point method. It works well in many situations where the generating function is smooth and grows rapidly. However, it cannot be applied to F (z). The basic saddle point conditions are not satisfied, as the high order logarithmic derivatives of F (z)forzreal are not sufficiently small. At a more fundamental level, the saddle point method fails here because it requires that on a circle centered at the origin, the integrand can be large only in a small neighborhood of the positive real axis. The function F (z)haskevenly spaced zeros on each circle of radius (k!) 1/k ∼ k/e as k →∞. On the other hand, on the circle of radius r =(k+1/2)/e, a short analysis shows that |F (z)| >F(r)/10, say. Therefore the integrand is not small outside a small region, and the saddle point method cannot possibly work. The generating function F (z) is also one of the relatively rare cases where the simple bound [z n ]F (z) ≤ min x>0 F (x)x −n , the electronic journal of combinatorics 6 (1999), #R2 3 gives a poor result. This bound holds for all generating functions with nonnegative coeffi- cients, and it is often too weak by only a fractional power of n (cf. [16]). For our function F (z), defined by (1.1), this bound is off only by a constant factor when n = k +m (m+1)/2 for k either very small or very close to m.Itispoorwhenm ≤ k ≤ (1 −)m and m →∞, on the other hand. For most values of n, we shall use elementary estimates and some well-known bounds for ordinary partitions (of a number) to express most values of b n in terms of the number of ordinary partitions of k with bounded or determined largest part. For some particularly recalcitrant values of n, this requires more sophisticated analytic techniques. Analysis of the partitions of k completes the task. We use (1.1) and define a n by a n =[z n ]F(z)= b n n! =  h 1 <h 2 <···<h r  h j =n 1 r  j=1 h j ! . (1.4) We will show in Proposition 2.2 that the largest term in this sum, when n = m(m+1)/2+k, 0 ≤k ≤m,is (m+1−k)! m+1  j=1 j! , which comes from r = m, {h 1 , ,h m }= {1,2, ,m+1}−{m+1−k}. This proposition is not actually used as part of the proof of our asymptotic estimates, but rather to motivate the following definition. We define f(m, k)tobea n divided by this largest term, that is, b n = n! (m +1−k)!f(m, k) m+1  j=1 j! . (1.5) Figure 1 presents a graph of log f(200,k)for0≤k≤200, which represents the values of a n for 20100 ≤ n ≤ 20300. It shows that the oscillations of f(m, k) are large even for small values of m. One disadvantage of f (m, k) as a measure of the behavior of b n is that it compares b n to the contribution of the largest term, which does not behave smoothly. Table 1 presents another measure of the irregularity in the behavior of the coefficients b n .Itshowsthe asymptotic form of b n+1 /b n for n near m(m +1)/2asm→∞. It is of interest to note that b n grows roughly like the square root of the total number of partitions of an n element set, denoted say by B n , in the sense that log b n ∼ 1 2 log B n as n →∞. the electronic journal of combinatorics 6 (1999), #R2 4 k log f(200,k) 0 50 100 150 200 01234567 Figure 1: log f(200,k) j −3 −2 −1012 3 ratio 1 8 m 2 1 4 m 2 1 2 m 2 1 2 mm 3 4 m 5 6 m Table 1: Asymptotic behavior of the ratio b N+j+1 /b N+j for N = m(m +1)/2asm→∞. Equation (1.4) shows that b n can be interpreted as a certain weighted sum over the ordi- nary partitions of n with distinct parts. The analogous combinatorial sum over unrestricted partitions of n has as its exponential generating function the function G(z) of Eq. (1.2). The same sum over compositions (ordered partitions) of n is B(n)=  n k=1 k!S(n, k), the number of ordered partitions of an n element set. Finally, by the multinomial theorem, the same sum over ordered n-tuples of nonnegative integers is n n . Forthesimilarweightedsum   h j =n 1 r  j=1 h j over partitions with distinct parts we have the generating function H(z)ofEq.(1.3)con- sidered by Greene and Knuth. The corresponding sums over unrestricted partitions and the electronic journal of combinatorics 6 (1999), #R2 5 compositions are treated in [14]. The shape of an unrestricted set partition was studied in [8] and [19]. It was proved, for instance, that in a typical set partition almost all elements of the set are in blocks of size close to log n ([8]). This situation contrasts sharply with the present topic of partitions with distinct block sizes. In this case, we show that a typical partition has blocks of sizes 1, 2, ,s, where s is approximately √ 2n, with a few missing. We obtain various precise results about the distribution of the missing sizes, from which the shape is determined completely. 2. Main results Let p(n)andQ(x) be defined by Q(x)= ∞  k=0 p(n)x n = ∞  l=1 1 1 − x l , so that p(n) denotes the number of partitions of (the number) n,andQis its ordinary generating function. Also, define p(n, k) to be the number of partitions of n with largest part at most k. We will have use of the rough bound p(n) ≤ exp(π  2n/3) (2.1) (valid for all n ≥ 1; see Apostol [2] for instance) and the more precise p(n) ∼ 1 4 √ 3n exp  π  2n/3  (2.2) as n →∞. Moreover, it is well known from Erd˝os and Lehner [10], that for almost all partitions of n the largest part is asymptotic to (π  2/3) −1 n 1/2 log n;thus p(n, [n 1/2 log n]) ∼ p(n) (2.3) as n →∞. Theorem 2.1. Let n = k + m(m +1)/2 and 0 ≤ k ≤ m. Then with f(m, k) defined by (1.5), we have as m →∞the asymptotic relations f(m, k) ∼              [m] k m k p(k) , if k = o(m 2/3 / log m) , 1 (s +1)!  v≥0  d 0 <···<d v d 0 +···+d v =s+1 v  j=0 d j ! , if k = m −s, s fixed . the electronic journal of combinatorics 6 (1999), #R2 6 Theorem 2.2. Let n = k + m(m +1)/2 and 0 ≤ k ≤ m. Then with ω(m) denoting an arbitrary function such that ω(m) →∞as m →∞, we have as m →∞ f(m, k) ∼ k  t=0 [m +1−k+t] t m t p(t, k −t) for √ m log m<k<m−ω(m). This formula is actually valid for the small k range as well, and when k is fairly large it can be expressed in an interesting way. Proposition 2.1. Let n = k + m(m +1)/2, 0 ≤k ≤m. Then with f (m, k) defined by (1.5) and ω(m) denoting an arbitrary function →∞as m →∞,wehaveasm→∞ f(m, k) ∼ k  t=0 [m +1−k+t] t m t p(t, k −t) if 0 ≤ k<m−ω(m),and f(m, k) ∼ Q(1 −k/m) provided Cm 3/4 log m<k<m−ω(m)for some C>0. Remark. Note that the formula given for f(m, k) in Theorem 2.1 or Proposition 2.1 is asymptotic to p(k)fork=o( √ m). It is curious that, as shown in Proposition 2.1, the asymptotics change from this Taylor series coefficient of a generating function to a value of that generating function as k varies. To obtain the asymptotics of f(m, k) it is still necessary to evaluate the summation involving p(t, k −t) in Theorem 2.2 for a wide range of k.For0<µ<1/3, define S 1 = S 1 (µ)= 1 (24µk) 1/4  1 − µ 1 − 3µ  1/2 exp  F (µk)+2c  µk  , (2.4) where c = π/ √ 6andF(t) is defined for 0 ≤ t ≤ k by F (t)=(m−k+t)log  1+ t m−k  +tlog  1 − k m  − t. Also, define S 0 = e F (k) p(k) ∼ 1 4 √ 3k exp  2c √ k +(m−k)log  m m−k  −k  , where we used (2.2). Our final asymptotic formula for f (m, k) in terms of simple functions isthefollowing. Notethatwepermitβ→0andβ→∞. the electronic journal of combinatorics 6 (1999), #R2 7 Theorem 2.3. Take m and k as in Theorem 2.1. Set β = cm/k 3/2 and determine ξ by β = 8 27 + 2 27c 4 log m m 1 3 + ξ m 1 3 . For 0 <β≤2/3 √ 3,letµbe the (unique) solution of µ 1/2 (1 − µ)=β, 0 <µ≤1/3. Then f(m, k) ∼                        S 0 , if ξ →∞, S 0 +S 1 , if ξ = O(1), S 1 , if ξ →−∞ and k = o(m), 1 (s +1)!  v≥0  d 0 <···<d v d 0 +···+d v =s+1 v  j=0 d j ! , if k = m −s, s fixed . Moreover, for ξ = O(1), f(m, k) ∼ √ 2 √ 3c 1 6 m 7 96 exp  15 32 c 4 3 m 1 3  ×  1 9 √ 2c exp  513 64 c 4 3 ξ − 243 128 c 2  + 2 1 4 3 1 4 exp  81 64 c 4 3 ξ − 217 384 c 2   . Remark on “continuity” of estimates. This theorem was proved by continuously adding unsuspecting collaborators until the proverbial camel’s back could hold no longer. Conceptually, what is a camel, if not a horse designed by a committee? In our case, it was not even a camel, just a function f (m, k). Its graph in Figure 1 has some suspicious humps, due, no doubt, to the tail behavior! So it would be comforting to check that the last theorem provides a piece-wise smooth asymptotic description for f(m, k) dependent on how large k is, compared to m. The first three formulas for f(m, k) indicate three distinct modes of asymptotic behavior, subcritical (ξ →∞), near-critical (ξ = O(1)), and supercritical (ξ →−∞). The ξ- parametrization and the last formula for the near-critical case ξ = O(1) provides a smooth interpolation (a magnified bridge) between the three modes. Indeed, it will be seen in the proof that, for ξ = O(1), the first of the two terms in the long parenthetical factor comes from S 0 ,the the electronic journal of combinatorics 6 (1999), #R2 8 second - from S 1 , and the first (second, resp.) term is dominant if ξ →∞(if ξ →−∞, resp.). Thus, for k = o(m), the various asymptotics gracefully merge into each other at the borders of their respective spheres of influence, and f (m, k) has no humps, except legal ones! (Actually we shall see that the estimate S 0 + S 1 is valid for ε 1 <β< 8 27 + ε 2 ,for ε 1 >0andε 2 sufficiently small.) Next consider large k, in particular the expression in Theorem 2.1 for k = m − O(1). If we let f(x)=  k≥1 k!x k we see that the terms with v ≥ 1 are bounded by [x s+1 ]f(x) v+1 . However [x s+1 ]  v≥0 f(x) v+1 ∼ (s + 1)! by Bender[3, Theorem 3]. Thus lim s→∞ f(m, m −s) = 1, and so the expression for k = m −O(1) merges with the expression for large k in Proposition 2.1. Further remarks. It is also worth checking the extent of overlap of the various formulae we have for f (m, k). If k = δm 2 3 then S 0 ∼ p(k) [m] k m k ∼ p(k)exp  − 1 2 δ 2 m 1 3 − 1 6 δ 3 +O(m − 1 3 )  and so by Theorem 2.3 the range of the first expression in Theorem 2.1 can be extended to o(m 2 3 ) and even ξ →∞, which is the limit of its range of validity. We next check the range of validity of the second formula in Proposition 2.1. If k = xm 3 4 then β = cx − 3 2 m − 1 8 = o(1) so f(k,m) ∼ S 1 . Now since √ µ(1 − µ)=βwe have µ = β 2 +2β 4 +7β 6 +O(β 8 )so µk = c 2 x −2 m 1 2 +2c 4 x −5 m 1 4 +7c 6 x −8 +O  m − 1 4  , (m−k+µk)log  1+ µk m − k  = µk + (µk) 2 2(m −k) + O(m − 1 2 ), µk log  1 − k m  = − µk 2 m − µk 3 2m 2 + O(m − 1 4 ). Thus F (µk)=− c 2 x m 1 4 + c 4 2x 4 +O(m − 1 4 ). Furthermore 2c(µk) 1 2 =2c 2 x −1 m 1 4 +2c 4 x −4 +O(m − 1 4 ) so exp(F (µk)+2c  µk)=exp  c 2 x m 1 4 + 5c 4 2x 4 +O(m − 1 8 )  . the electronic journal of combinatorics 6 (1999), #R2 9 As (24µk) 1 4 ∼ √ 2πm 1 8 x −1 2 we now have S 1 ∼ x 1/2 m −1/8 √ 2π exp  c 2 m 1/4 x + 5c 4 2x 4  . However Q(1 − k/m)=Q  e k m − k 2 2m 2 +O  k 3 m 3   and it is well known (see for instance Andrews [1]) that Q(e s )=e π 2 /6s  s/2π(1 + O(s)) (2.5) and hence Q(1 −k/m) ∼ f (k, m)ifk=xm 3/4 for fixed x. Note next that if k = o(m) but km −3/4 →∞then β → 0soµ∼β 2 and µk ∼ c 2 m 2 /k 2 = o(m 1/2 ). Thus (m − k + t)log  1+ t m−k  =t+O  t 2 m−k  =t+o(1), so the sum over those t for which t = µk + O  (µk) 7/8  in  t e F (t) p(t) is asymptotic to  t p(t)(1 − k/m) t (summed over these t). We shall see however that the sum over the other t in  t e F (t) p(t) is negligible. If |t − µk|≥(µk) 7/8 then p(t)(1 − k/m) t ≤ p(µk)(1 −µk/m) µk exp(−δ(µk) 1/4 ) for some δ>0 so the sum over these t of p(t)(1 −k/m) t is asymptotic to Q(1 − k/m). Thus the range of validity of f (k, m) ∼ Q(1 − k/m), the second formula in Proposition 2.1, is m −k →∞and km −3/4 →∞. As a final remark, it can be easily verified that, as functions of k,bothf(m, k)andS 0 attain their respective maxima at k ∼ c 2/3 m 2/3 . From now on, whenever we refer to n, m and k, we will always assume that n = m(m +1)/2+k,0≤k≤m.Let Π(n, r)=    (h 1 ,h 2 , ,h r ):1≤h 1 <h 2 <···<h r , r  j=1 h j = n    (2.6) be the set of partitions of the integer n into r distinct parts, and let Π(n)=  r≥1 Π(n, r) . Note that Π(n, r)=∅for r ≥ m +1. Forπ∈Π(n, r), π =(h 1 ,h 2 , ,h r ), let P (π)= 1 r  j=1 h j ! . [...]... partition of an n -set we mean the multiset of the cardinalities of its blocks As noted just above, for the partitions into distinct parts, the shape (which is a set) is characterised by the partition {λi } We can conclude various results about the shape of a random partition from our main results on asymptotics Define Ωn to be the probability space whose elements are partitions λ = {λ0 , , λv } of k, with. .. one (with the understanding that if d0 = 1, we merely fill this hole in rather than moving it to 0) The ratio of the contributions to (1.4) of the old and new partitions is d /d0 > ω 1/6 (n) The number of ways to reverse this operation is at most 2 and so we conclude the partitions with r ≤ m − 1 are negligible if m − k > ω(n) This establishes the first part of the lemma The proof of the second part of. .. Then the number of possible partitions qd0 , qd0 +1 , , qj of d0 − l into j − d0 + 1 nonnegative parts is at most p(d0 − l) Thus, the contribution of the original partitions to (1.4) is bounded by √ p(d0 )d0 2πd0 (1 + o(1)) ed0 times the contribution of all partitions with r + 1 parts Since the sum over d0 of this expression converges (by (2.1), the lemma is proven for d0 → ∞ Suppose now, with m −... size can be ignored when counting the partitions It follows that the restriction can be dropped from Proposition 2.4 the electronic journal of combinatorics 6 (1999), #R2 26 Proof of Proposition 2.6 This follows immediately from the fact that in the proof of Theorem 2.1 for the case k = m − s, the di represent the holes in the partition 4 Proofs of lemmas Proof of Lemma 3.1 If r ≥ m + 1, then r i=1... probability proportional to the number of partitions of an n -set with hole partition λ Note that r can be regarded as a random variable on Ωn since it is determined in a natural way from λ by the fact that the r smallest integers other than those holes determined by λ sum to n We say that a random partition π is distributed asymptotically uniformly as a partition of a number (possibly with a bound on the largest... is fixed and k = m − s then in a random partition of n with distinct block sizes, hr = m + 1 almost surely Furthermore, the holes d0 , , dv are a random partition of s + 1 into distinct parts in which the probability is asymptotically proportional v j=0 dj ! to These propositions have immediate corollaries for random partitions of an n -set into blocks of distinct sizes For instance, the largest part... Lehner, The distribution of the number of summands in the partitions o of a positive integer, Duke Math J (1941), 335–345 [11] W M Y Goh and E Schmutz, Random set partitions, SIAM J Discrete Math 7 (1994), 419–436 [12] D H Greene and D E Knuth, Mathematics for the Analysis of Algorithms, 2nd ed., Birkh¨user Boston, 1982 a [13] A Guttman and M Hirschhorn, Comment on the number of spiral self-avoiding... , λv ) is distributed asymptotically uniformly as a random partition of t, where the distribution of t and the distribution of a random variable X with P(X = i) proportional to [m + 1 − k + i]i p(i) mi have total variation distance tending to 0 In fact, from the proof of Theorem 2.3, it is easy to deduce more about the distribution of X For the remaining, very large, values of k, we have the following... (m + 1) > n, so this case never arises We first deal with the statement about m − k → ∞ The first idea of the proof is to analyse a mapping from partitions with r parts to those with r + 1 parts Consider a partition with r ≤ m − 1 parts Such a partition must have at least two holes, since a partition with at most one hole has m parts by the definition of m Recall that d0 denotes the smallest hole So hd0... e t=k−k 5/8 ∼ eF (k) p(k) exp[G(k)] √ ∼ 4 3k (3.21) 22 the electronic journal of combinatorics 6 (1999), #R2 (We have used the formula p(a, b) = P (a+b, b), where P (a+b, b) is the number of partitions of a + b with the largest part equal to b exactly We also know that for almost all partitions √ of ν the largest part is of order O( ν log ν).) Comparing (3.20) and (3.21) we get k eF (t) p(t, k − t) . Council the electronic journal of combinatorics 6 (1999), #R2 1 Abstract The asymptotic behavior of the number of set partitions of an n-element set into blocks of distinct sizes is determined values of n, we shall use elementary estimates and some well-known bounds for ordinary partitions (of a number) to express most values of b n in terms of the number of ordinary partitions of k with. The asymptotic number of set partitions with unequal block sizes A. Knopfmacher Department of Computational and Applied Mathematics University of Witwatersrand Johannesburg,