A. Sárközy

J.-L. Nicolas

Journal de Théorie des Nombres de Bordeaux

JOURNAL DE THÉORIE DES NOMBRES DE BORDEAUXJ.-L. NICOLASA. SÁRKÖZYOn partitions without small partsJournal de Théorie des Nombres de Bordeaux, tome 12, no 1 (2000),p. 227-254<http://www.numdam.org/item?id=JTNB_2000__12_1_227_0>© Université Bordeaux 1, 2000, tous droits réservés.L’accès aux archives de la revue « Journal de Théorie des Nombresde Bordeaux » (http://jtnb.cedram.org/) implique l’accord avec les condi-tions générales d’utilisation (http://www.numdam.org/conditions). Toute uti-lisation commerciale ou impression systématique est constitutive d’uneinfraction pénale. Toute copie ou impression de ce fichier doit conte-nir la présente mention de copyright.Article numérisé dans le cadre du programmeNumérisation de documents anciens mathématiqueshttp://www.numdam.org/227-On partitions without small partspar J.-L. NICOLAS et A. SÁRKÖZYRÉSUMÉ. On désigne par r(n, m) le nombre de partitions del’entier n en parts supérieures ou égales à m. En appliquantla méthode du point selle à la série génératrice, nous donnonsune estimation asymptotique de r (n, m) valable pour n ~ ~, et1 ~m~c1n (logn)c2.ABSTRACT. Let r(n, m) denote the number of partitions of n intoparts, each of which is at least m. By applying the saddle pointmethod to the generating series, an asymptotic estimate is givenfor r(n, m), which holds for n ~ ~, and 1 ~ m ~ c1 n (log n)c2 1. INTRODUCTIONLet r(n, m) (resp. p(n, m)) denote the number of partitions of the positiveinteger n into parts, each of which is at least m (resp. at most m), that is,the number of partitions of n of the typeand let q(m, n) denote the number of partitions of n into distinct parts,each of which is at least m, i.e., the number of solutions ofThe function p(n, m) has been extensively studied by Szekeres in [6] and[7].In the last several years, Dixmier and Nicolas [1], [2] have studied thefunction r(n, m) while Erd6s, Nicolas and Szalay [4], [3] have estimated thefunction q(n, m); in all these papers m is relatively small in terms of n. In[5], Freiman and Pitman gave an asymptotic formula for q(n, m) in termsof a certain parameter a = a(n, m) in a quite wide range for m:Manuscrit reçu le 6 mai 1999.Partially supported by Hungarian National Foundation for Scientific Research, Grant No. T029759, by CNRS, Institut Girard Desargues, UMR 5028 and Balaton, 98009. This paper waswritten while the second author was visiting the Université Claude Bernard Lyon 1.228Theorem A (G.A. Freiman and J. Pitman, ~5~). As n - +oo we havewhere o- and B are given byandanduniformly with respect to m such thatHere Ko and the implied constants in the estimate of E are effective positiveconstants independent of m and n.They used a probabilistic approach in proving this theorem, however,their proof could be presented as well in terms of the saddle point method,without any reference to probability theory. Moreover, in certain intervalsfor m, they succeeded in making (1.1) more explicit by eliminating theparameter a, i.e., by giving an asymptotics for q(n, m) in terms of m andn only.In this paper our goal is to prove the r {n, m) analog of Theorem A and,in fact, we will prove a sharper result with a series expansion instead of afactor of type 1-~- E as in ( 1.1 ) . Before formulating our main theorem, firstwe have to introduce a parameter a (playing the same role as the parametera in Theorem A).We shall write e"’° = e(a) and N, R and C will denote the set of thepositive integers, real numbers and complex numbers, respectively. For alla E R and 0  R  1, writing z = R e(a) clearly we have229where r* (~, m) - for 0  1  n and 0  r*(t,m) :5 forn  .~. Substituting a = 0 and dividing by we obtainn -say. Clearly,and U(R) is continuous in (0,1). Thus U(R) has a minimum in (0,1).Moreover, U(R) is twice differentiable in (0,1), thus its minimum is attainedat a point Ro satisfying the equationor, in equivalent form,Setting(1.4) can be rewritten in the formClearly we haveand is decreasing in (1, +oo). Thus (1.4) determines R ( 1) uniquely.Define a by--230(with the R defined by (1.4)) so that cr &#x3E; 0, and (1.5) can be rewritten inthe formFor hEN, l E N writeso that by (1.6),We shall also writeClearly, we havesay. By estimating the integral J we will prove:Theorem 1. There exist real numbers d(x, y) with x E N, y E {1,2,...,x}such that writingfor any fixed k E N, k &#x3E; 3 as n -~ +oo u~e havewith231whereand, for a certain absolute constant H &#x3E; 2,uniformly with respect to m such thatIn particular, 2ve haveand for k = 6 (1.12) holds withwhere L2, ..., L6 are defined above and where232In Theorem 1 and throughout the rest of the paper, the constants cl, C2, ...and the constants implied by the notations O(...), «, ~ as well are effec-tively computable, and they may depend only on the parameter l~ in Theo-rem 1 but are independent of any other parameters, in particular, of m andn. (We write f « g if f - O(g), and f ~ g means that both f « g andg « f hold.)Note that an upper bound will be given for the numbers A(h, l) with1  .~  h and thus also for Lh in Lemma 2, and the order of magnitude ofB = L12 will be determined in Lemma 3.2The rest of the paper will be devoted to the proof of Theorem 1, and inPart II of this paper the result will be made more explicit by eliminatingthe parameter car in a wide range for m and giving a sharp estimate for thenumbers L~ in Theorem 1.In [2], it has been proved that there exists a function 9 such that, forA &#x3E; 0, the following asymptotic estimation holds :Moreover, the function X e g(A) - a log A was proved to be analytic in aneighbourhood of 0, so that it can be writtenand, for j &#x3E; 2, the values of the coefficients aj are rational fractions of- r~’The following relation was also given in [2] :where, in the bracket, the coefficient a2 shows up. In Part II, it will beproved thatwhere = In the above relation, whenever m n 1/2-é, 9 can be replaced by the first terms of its power series expansion,233for instance :In Part II, the error term in the asymptotic estimation (1.19) of r(n,m)will also be precised.2. ESTIMATE OF QWe shall proveLemma 1. Let k &#x3E; 1 be an integer. There is a positive number H = H(k)such that H &#x3E; 2 andandProof of Lemmas 1. For large n we havesince otherwise by (1.15) for large m we should havewhich contradicts (1.6).On the other hand, we havesince otherwise we hadagain in contradiction with (1.6).We split the sum on the left hand side of (1.6) into two parts:234say. Clearly we haveandMoreover, writing M = we havesay. The function xe-x is decreasing for x &#x3E; 1 and thus the terms in thesum E’ 2 decrease as j increases. Thus we haveSincethus it follows from M &#x3E; 1/c, (2.3), (2.4) and (2.9) thatandThe function = xe-ax is decreasing for x &#x3E; 1/Q and by (1.15) and(2.3) we haveand thusBy (2.11) and (2.12) we haveIt follows from (2.8), (2.10) and (2.13) thatNow we have to distinguish two cases.235Case 1. Assume first thatThen by (2.5), (2.7), (2.14) and (2.15) we havewhenceIt follows from (2.15) and (2.17) that. I-Case 2. Assume now thatThen by (2.5), (2.6), (2.14) and (2.19) we haveWriting(2.20) can be rewritten in the formBy (2.19) here we have ma &#x3E; 1, and f (x) &#x3E; 1 for x &#x3E; 1. Thus (2.22)implies -whenceMoreover, for z - +oo (2.21) impliesand thus it follows from y - +oo, f (x) ~ y (note that this implies x -7+cxJ) that236Thus it follows from (2.22) thatNow we are ready to prove (2.1) and (2.2). By (2.23), there is a positivenumber K such that m  Knl/2 implies 1/Q &#x3E; m and then by (2.17), (1’satisfies (2.1) (for m  Kn 1/2). On the other hand, if H is large enoughand  m, then 1/Q  m by (2.18) and thus (2.2) holds by (2.24)(for m &#x3E; Hnl/2). Finally, if m  Hn1/2, then for m, (2.1)holds by (2.17) while for  m it follows from (2.22) and this completesthe proof of Lemma 1. D3. ESTIMATE OF THE SUMS Lemma 2. For all K E N‘ there is a positive nurraber C2 = c2(K) such thatfor h = l, 2, ..., K and t = 1, 2, ..., h (with H defined in Lemma 1).Proof of Lemma 2. We split the sum in the definition of A(h, t) into twoparts:say. Clearly,while for m  Hen1/2 by (2.1) in Lemma 1 we haveMoreover, ifs we have237and thusFor m &#x3E; Hn 1/2 by (2.2) for all j in I:2 we haveso that (3.5) holds also in this case. By (3.5), writing mi = max(m, Hn 1/2)we haveAn easy consideration shows that uniformly forx  j we have - -- - -and thusso thatSince we have(which can be shown easily by induction on h) thus it follows from (3.6)thatBy Lemma 1, it follows from the definition of mi that 1. Moreover,by Lemma 1 we have a = 0(1). Thus it follows from (3.7) thatwhence, again by Lemma 1 and the definition of m1, we have1. /,,and238(3.1) follows from (3.2), (3.3), (3.4), (3.8) and (3.9) which completes theproof of Lemma 2. 04. ESTIMATE OF BLemma 3. We haveProof of Lemma 3. By Lemma 1 and (1.15) we have m + 1/ u  n and thus,again by Lemma 1,which completes the proof of the lemma.5. ESTIMATE OF THE INTEGRAND FAR FROM 0We split the integral J defined in (1.9) into two parts. Setwhere c* is a positive number, large enough in terms of k, which will bedetermined later. Letandso that239and writesay. We will show that the integral Ji gives the main contribution while J2gives only a small error term. Thus we shall need an asymptotic formulafor J1 and an upper bound for J2. First we will estimate the integrand inJ2.Lemma 4. For 1/2, rrc E N‘, 2 zve haveProof of Lemmas 4. This is Lemma 8 in [5].Lemma 5. Let 17 be defined by (5.1~. Uniformly forwe haveProof of Lemma 5. For all 0  r  1 and # E IEB, writinghavewhence240since (1 - r)2  1. It follows thatsince by (1.15) and Lemma 1 we haveFor m  j  m + [1/u], the following inequality holds:and this impliesFor 0  x  2 a simple calculation shows thatand, since , it follows from (5.6) thatBy Lemma 1, Lemma 4, (5.1) and (5.3), it follows thatNow, in the first case (m  by (2.1), am = 0(l), and from (2.1) and(5.1), r~2o~-3 &#x3E; c3c*2 log n so that, by choosing c* large enough (in terms ofk), (5.8) yields (5.4).241In the second case (m &#x3E; H ý1ï), by (2.2)and from (2.2) and (5.1)since m  n, and again, by choosing c* large enough, (5.8) yields (5.4),which completes the proof of the lemma. D6. ESTIMATE OF J2Lemma 6. We haveProof of Lemma 6. By Lemma 5 we haveMoreover, by Lemma 2 and (1.8) we have(6.1) follows from (6.2) and (6.3).7. ESTIMATE OF THE INTEGRAND NEAR 0For m  j  n, write242so thatand the integrand in J can be rewritten asnLemma 7. Uniformly for(7.3)we have(7.4)and(7.5)Proof of Lemma 7. For all a satisfying (7.3) we haveso thatwhich proves (7.4).Since ex - 1 &#x3E; x for x &#x3E; 0 thus it follows from (7.4) thatwhich proves the first inequality in (7.5). Finally, q la = o(l) follows from(1.15), (5.1) and Lemma 1, and this completes the proof of Lemma 7. 0243Lemma 8. Uniformly for a satisfying (7.3) and rrt  j  n we haveProof of Lemma 8. For all j we haveand thusFor j &#x3E; 1 /10q we haveThe first inequality in (7.6) follows from (7.7), (7.8) and (7.9).For i  1/cr clearly we havesince q la = 0(1) as we have seen in the proof of Lemma 7. Moreover, forj &#x3E; we haveFor 0  ~ the function fk(x) = e-I is maximal = k + 1, and thevalue of the maximum is fk(k + 1) = (k + 1)k+l e-(k+1). Thus it followsfrom (7.11) that(7.10) and (7.12) complete the proof of (7.6).244Lemma 9. If k e N,andthen we havewithProof of Lenrama 9. By (7.14) we havewhereandBy (7.14) clearly we have245Moreover, by (7.13), (7.14) and (7.15) we have10 i-1 1(7.16) and (7.17) follow from (7.18), (7.19) and (7.20), and this completesthe proof of the lemma. DLemma 10. Uniformly for lal  r~ the integrand in Ji iswhere(7.22)andProof of Lemma 10. By (7.5) in Lemma 7 and by Lemma 8, (7.13) and(7.14) in Lemma 9 hold with uj and uj in place of u and u~, respectively.Thus by Lemma 9 we havewhere, by (7.4) in Lemma 7, Lemma 8 and (7.19) we have246It follows from (7.24) thatwhere by (7.25) the error term in the exponent isMoreover, the innermost term in the main term in the exponent in (7.26)can be rewritten asWriting herethe last expression becomes a polynomial, in the variables a and x, of theformh, then clearlyi.e., d(h, t, k) is independent of 1~ for 1~ &#x3E; h. Thus we may write247In particular, computing these numbers d(h, i) for h  6 we obtainUsing this notation, by (7.28) and (7.29) for ~a~  q the main term inthe exponent in (7.26) becomeswhere Lh and, in particular, L2, L3, L4, L5 and L6 are defined as in thetheorem, and by (1.7) and (7.30), the first term is248It follows from (7.2), (7.26), (7.27), (7.31) and (7.32) that for 71 theintegrand in J isBy (5.1) and Lemma 2, for m  in the error term we havewhile for m &#x3E; by k &#x3E; 3 we haveBy (1.15), both upper bounds in (7.34) and (7.35) are o(l) and thus in(7.33) we may writewith249Next in (7.33) we writewhereThen for we haveBy using (5.1) and Lemma 2, it follows in the same way as in (7.34) and(7.35) thatandBy (1.15), it follows from (7.41) and (7.42) that(uniformly for Thus the second factor in (7.38) can be written aswith250Moreover, the main term in (7.44) isConsider here the terms with t  v-2+2. Let max* denote the maximumtaken over the integers v, t, ~1, ..., ht with 1  v  1~2, 1  t  k, t  v-2+2 , 13  h 1, ... , ht  k, h 1 + ... + ht = v . Then uniformly for lal 1  q, by ( 5 .1 )and Lemma 2 each of the terms with t  v-2+2 isso that for m  Hnll2,while for m &#x3E; ,251Thus in (7.46) we may restrict ourselves to terms with at theexpense of a small error term:withNote that the summation over t is empty unlesswhenceso that the summation can be replaced by *Defining Z" by (7.22), it follows from (7.33), (7.36), (7.37), (7.38), (7.39),(7.43), (7.44), (7.45), (7.49), (7.50) and (7.51) that uniformly for ~a~  qthe integrand in J1 iswhereand this completes the proof of Lemma 10.2528. ESTIMATE OF JlLemma 11. We havewhereProof of Lemma 11. By Lemma 10 we havewith Eo satisfying (7.23). It remains to estimate the integralsIf v (&#x3E; 0) is odd then the integrand is odd and thusIf v = 2t is even, then substituting x = (2L2 )1/2 7ra we obtainwithIt is easy to see by induction (using integration by parts) that we have253Moreover, we havesay. By (1.16), (5.1) and Lemma 3 we haveIf now we fix c* to be large enough in terms of k (and so that (5.9) shouldalso hold) then it follows thatBy (8.3), (8.4), (8.5), (8.6), (8.7) and (8.8) (and since Zv = 0(l) for1  v  k2 by the proof of Lemma 10) we have254whence, by (1.16), (7.23) and (8.8), (8.1) and (8.2) follow. ~ 09. COMPLETION OF THE PROOF OF THEOREM 1The result follows from (1.9), (5.2), Lemma 6 and Lemma 11. In particu-lar, the formulas for L2, ..., L6 are obtained from (1.10) by using the numbersd(x, y) computed in the proof of Lemma 10, while (1.17) and (1.18) are thek = 6 special cases of (1.12) and (1.14), respectively.REFERENCES[1] J. Dixmier, J.-L. Nicolas, Partitions without small parts. Number theory, Vol. I (Budapest,1987), 9-33, Colloq. Math. Soc. János Bolyai 51, North-Holland, Amsterdam, 1990.[2] J. Dixmier, J.-L. Nicolas, Partitions sans petits sommants. A tribute to Paul Erdõs, 121-152,Cambridge Univ. Press, Cambridge, 1990.[3] P. Erdõs, J.-L. Nicolas, M. Szalay, Partitions into parts which are unequal and large. Numbertheory (Ulm, 1987), 19-30, Lecture Notes in Math., 1380, Springer, New York-Berlin, 1989.[4] P. Erdõs, M. Szalay, On the statistical theory of partitions. Topics in classical number theory,Vol. I, II (Budapest, 1981), 397-450, Colloq. Math. Soc. János Bolyai 34, North-Holland,Amsterdam-New York, 1984.[5] G. Freiman, J. Pitman, Partitions into distinct large parts. J. Australian Math. Soc. Ser. A57 (1994), 386-416.[6] G. Szekeres, An asymptotic formula in the theory of partitions. Quart. J. Math. Oxford 2(1951), 85-108.[7] G. Szekeres, Some asymptotic formulae in the theory of partitions II. Quart. J. Math. Oxford4 (1953), 96-111.J.-L. NICOLASInstitut Girard Desargues, UMR 5028Mathematiques, Bltiment 101Universit6 Claude Bernard (Lyon 1)F-69622 Villeurbanne Cedex, FRANCEjlnicolaQin2p3.frA. SARK6ZYDepartment of Algebra and Number TheoryE6tv6s Lorând UniversityKecskem6ti utca 10-12H-1088 Budapest, HungaryE-mail : sarkozyQcs . elte . hu

On partitions without small parts

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On partitions without small parts

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