An upper bound for the size of the largest antichain. in the poset of partitions of an integer. University of Georgia.

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1 An pper bond for the size of the largest antichain in the poset of partitions of an integer E. Rodney Caneld Department of Compter Science University of Georgia Athens, GA 3060, USA Konrad Engel Fachbereich Mathematik Universitat Rostock 805 Rostock, Germany Janary 998 Abstract Let Pin be the poset of partitions of an integer n, ordered by re- nement. Let b(pin) be the largest size of a level and d(pin) bethe largest size of an antichain of Pin. Weprove that d(pin) b(pin) e + o() as n! The denominator is determined asymptotically. In addition, we show that the incidence matrices in the lower half of Pin have fll rank, and we prove a tight pper bond for the ratio from above if Pin is replaced by any graded poset P.

2 Proposed rnning head Antichains of integer partitions Send proofs to Prof. Konrad Engel Universitat Rostock FB Mathematik 805 Rostock Germany

3 Introdction Let P be a graded poset, i.e. a partially ordered set which can be partitioned into levels N 0 N r(p ) sch that N 0 (resp. N r(p ) ) is the set of all minimal (resp. maximal) elements of P and p N i pl q imply q N i+.herep l q means that p<qand there is no element q 0 with p<q 0 <q.wesay that in this case q covers p. Note that the partition of P into levels is niqe if it exists. The nmber r(p )iscalledtherank of P. Let b(p ) be the largest size of a level of the graded poset P. An antichain in P is a set of pairwise incomparable elements of P.Let d(p ) be the largest size of an antichain in P.Obviosly, for each graded poset P, d(p ) b(p ) After Sperner [9], it was proven for many interesting classes of graded posets that the ineqality is in fact an eqality, cf.[5]. j j j j j j j j j j j j j j Figre Bt there exist graded posets where the ratio is arbitrarily large. E.g., for the class of graded posets which is illstrated in Figre for r(p )=5we have d(p ) b(p ) = jp j 8 + We will show that there is no graded poset with a larger ratio if jp j. 3

4 Theorem Let P be a graded poset. Then d(p ) b(p ) max jp j 8 + Some similar reslts have been obtained in [6]. Let n be the (graded) poset (lattice) of partitions of [n] =f ng, ordered by renement. From [] and [4] we know (all logarithms are natral) Theorem Let a = ( ; e log )=4 Then for sitable constants c c,and n> c n a (log n) ;a;=4 d( n) b( n ) c n a (log n) ;a;=4 Moreover, corresponding limit theorems (cf. [5, p. 36]) imply Theorem 3 We have b( n ) p log n j p n j p as n! n In this paper we will stdy a qotient of the partition lattice n, namely the poset Pi n of nordered partitions of an integer n Apartition of the integer n into k parts, k = n,isanintegral soltion to the system n = x + + x k x x k > 0 We obtain all partitions in Pi n which arecovered by this partition by taking one smmand x l ( l k) and partitioning x l into exactly two parts and nally ordering the two new parts together with the old npartitioned parts in a nonincreasing way. The Hasse diagram of the poset Pi 7 is illstrated in Figre. The main reslt of the paper is the following Theorem 4 We have d(pi n) b(pi n ) e + o() as n! We will give a proof of the following theorem, since it follows by the same method we se to prove Theorem 9 it was rst shown by Alck, Chowla, and Gpta []. 4

5 Figre Theorem 5 We have b(pi n ) e p 6 jpi n j p n as n! For a graded poset P, the incidence matrix M k, k =0 r(p ) ;, is an (jn k jjn k+ j) 0{{matrix whose rows and colmns are indexed by the elements of N k and N k+, respectively, and whose element inrow p N k and colmn q N k+ eqals i p l q. The following reslt is de to Kng [8] (see also [8] for frther backgrond) Theorem 6 Let P = n and k< n;.then rank(m k )=jn k j We will prove that the theorem remains tre for the poset of partitions of an integer Theorem 7 Let P = Pi n and k< n;.then rank(m k )=jn k j 5

6 Proof of the general ratio bond Proof of Theorem. We proceed by indction on r(p ). The case r(p )= 0 is trivial, ths consider the step <r(p )! r(p ). Let briey b = b(p )and let A be a maximm antichain in P. Case. There is some k f0 r(p )g sch that ja \ N k j = jn k j Since P is graded, we have A = N k and ths d(p ) b(p ) = max jp j 8 + Case. There is some k f r(p ) ; g sch that ja \ N k j = jn k j;. Let A l = [ k; i=0 (A \ N i )anda = r(p ) [ i=k+ (A \ N i ) Let p be the (niqe) element ofn k n A. Since P is graded, all elements of A l and A are comparable with p, hence A l = or A =. Let w.l.o.g. A =. Let Clearly, P 0 is also graded and P 0 = d(p )=jaj d(p 0 ) d(p ) b(p 0 ) b(p ) k[ i=0 Conseqently, by the indction hypothesis N i d(p ) b(p ) d(p 0 ) jp b(p 0 ) max 0 j 8 + jp j max 8 + Case 3. Not Case and not Case. Then Obviosly, Hence d(p )=jaj jp j;(r(p ) ; ) ; =jp j;(r(p )+)+ jp jb(r(p )+) i.e., r(p )+ jp j b d(p ) jp j; jp j b +=jp jb ; b 6 +

7 and conseqently (since b; b attains its maximm at b =4) d(p ) b(p ) b ; b jp j + b max jp j Estimation of the size of the largest antichain in Pi n Let Pi n be the set of all nordered partitions of n into parts which are all greater than. Theorem 8 We have d(pi n ) jpi n j Proof. Let ' Pi n n Pi n! Pi n be the mapping that assigns to the partition p (having a smmand ) the partition p 0 that can be obtained from p by combining a smmand and the largest smmand of p. Clearly, for all p Pi n n Pi n p l '(p) The mapping ' is injective since p can be recovered from '(p) (partition the largest smmand s of '(p) into (s ; ) + ). Let l(p) be the rst natral nmber for which ' l(p) (p) Pi n. In addition, let for p Pi n, ' 0 (p)=p. If p and q are incomparable elements in Pi n, then ' l(p) (p) 6= ' l(q) (q) since otherwise (say for l(p) l(q)) by the injectivity of' ' l(p);l(q) (p) =q i.e., p q. Hence, for any antichain A in Pi n, jaj = jf' l(p) (p) p Agj jpi n j Theorem 9 We have jpi n j p 6 jpi n j p n as n! 7

8 Note that Theorem 4 follows from Theorems 5, 8, and 9. Ths it remains to prove Theorems 5 and 9. We will prove them almost simltaneosly. Let P (n k) (resp. p(n k)) be the nmber of partitions of n into k or fewer (resp. into exactly k) parts and let p(n) =P (n n) =jpi n j We need the following reslt of Szekeres [0, ] which was reproved in [3] with a new recrsion method in a more or less elementary way Theorem 0 Let >0 be given. Then, niformly for k n =6 P (n k) = f() n ep ng()+o(n ;=6+ ) Here, = k= p n, and the fnctions f() g() are f() = v p 8 ; e ;v ; e ;v ;= () g()= v ; log( ; e;v ) () where v(= v()) is determined implicitly by = v Z v 0 t dt (3) e t ; With standard calcls one may verify that the RHS of (3), and ths also is an increasing (continos) fnction of v, hence the inverse fnction exists. We know from [3] (sing (e t ; ) ; = P m= e ;mt and P m= m ; = =6) that, with C = p 6, Z 0 t e t ; dt = C (4) which implies that with also v tends to innity (and vice versa) and that Lemma We have for!(or v!) v lim = C (5)! v = C ; v + C e;v + O(v e ;v ) Proof. It is easy to verify that for t te ;t t e t ; te;t +te ;t 8

9 Taking the integral from v to innity yields Z (v +)e ;v t v e t ; dt = C ; (v +)e ;v + e;v (v +) and hence Conseqently, v v = C ; (v +)e ;v + O(ve ;v ) v = C ; v + = C e;v + O(ve ;v ) = C ; v + C e;v + O(v e ;v ) Lemma We have for!(or v!) g()=c ; C e;v + O(v e ;v ) Proof. We have ; log( ; e ;v )=e ;v + O(e ;v ) and conseqently by () and Lemma g() =C ; v + C e;v + O(v e ;v )+e ;v + O(e ;v ) Moreover, by Lemma Hence and nally v = C + O(ve ;v ) v C e;v = e ;v + O(v e ;v ) g()=c ; C e;v + O(v e ;v ) 9

10 Lemma 3 Let 0 << and I =[( 4C C ; )p n log n ( + C )p n log n] Then, niformly for k I as n! Here = k= p n. P (n k) p(n)e ;p n C e;c p(n k) p(n)e ;C; p n C e;c p(n ; k k) p(n)e ;C;p n C e;c Proof. Obviosly (sbtract from each part a one) p(n k) =P (n ; k k) (6) p(n ; k k) =P (n ; k k) (7) All the following estimates are niform for k I and taken for n!.let i f0 g. Let i = k= p n ; ik. Since i!we have Moreover, by Theorem 0 f( i ) P (n ; ik k) C p 8 C p 8n ep n;ikg(i ) (8) We have p p n ; ik = n ; ik = = p n ; i + o() (9) n i = ; ik ;= = + O(log n= p n) (0) n Let < < <. Then, for large n, 4C C ; log n< i < C + log n Let v i = v( i ). From (5) it follows ; C log n<v i < + C log n Conseqently, e ;v i < n =;C 0

11 From Lemma we obtain (noting (0)) v i = C i + O e ;v i = e ;C +O log n n =;C! log n n =;C = C + O!! log n n =;C! Obviosly, e ;C = O n =;C and ths e ;v i = e ;C + O log n n ;C(+ )! = e ;C + o(= p n) Lemma yields and from (9) we derive g( i )=C ; C e;c + o(= p n) p n ; ikg(i )= pn ; i C ; C e;c + o() Note that by the Hardy{Ramanjan formla [7] (pt in Theorem 0 = p n) p(n) C p 8n e p nc () Now we obtain from (8) { () P (n ; ik k) p(n)e ;ic; p n C e;c +o() and the assertion follows from (6) and (7). In the following let only i f g. Note that U i = C log n ; log ic C is the niqe point at which the fnction h i () =;ic ; p n C e;c

12 achieves its maximm. For = U i + t we have e h i() = (ic)i e;ict;ie;ct () i= n Let 0 << 4C and let U i = U i ; log n U i = U i + log n. Frther let p p k i = bu i nc ki = bu i nc k p i = bu i nc and i = k i = p n v i = v( i ) Lemma 4 We have for i f g P (n k i )=o(p(n)= p n) Proof. Since we have i = e ;p n C e;c i The assertion follows from Lemma 3. C ; log n + O() = e ;nc e O() =C = o(= p n) Lemma 5 We have for i f g p(n ; k i )=o(p(n)= p n) Proof. Let 0 < <. Then, for large n, pn n ; k n ; C + log n q n ; k p log n = n ; C + p = p n ; n From () we derive 4C + log n + o() ; ; p(n ; k i ).p(n)e C ; 4C + log n = p(n) p n ;C = o(p(n)= p n) n Proof of Theorem 5. By Lemma 3 and () (note t = o()) p(n k ) C e p n p(n) Becase h (U ) is the maximm of h () and again in view of Lemma 3 we have for k [k + k ; ] p(n k).p(n k )

13 For k k Lemma 4 implies, for large n, p(n k) P (n k )=o(p(n)= p n) <p(n k ) For k k wehave by Lemma 5, for large n, p(n k)=p (n ; k k) p(n ; k )=o(p(n)= p n) <p(n k ) Proof of Theorem 9. Obviosly (sbtract from each part of a member of Pi n a one) jpi n j = We divide the sm into 3 parts By Lemma 3 and () kx ; k=k + X = k X k= bx n c k= + p(n ; k k) (3) kx ; k=k + p(n ; k k) 4C k; X n p(n) k=k + + b X n c k=k e ;C(k=p n;u );e ;C(k=p n;u ) The sm on the RHS can be considered as an integral approximation with step size n ;=. Since k!;and k!this sm mltiplied by p n converges for n!to Z ; Conseqently, e ;Ct;e;Ct dt = 4C k ; X k=k + Moreover, by Lemma 4 k X k= Finally, by Lemma 5 b X nc e ;e;ct ;Ct + e ;e;ct ; = 4C p(n ; k k) C p n p(n) (4) p(n ; k k) P (n k )=o(p(n)= p n) (5) p p(n ; k k) p(n ; k )=o(p(n)= n) (6) k=k With (3){(6) the assertion is proved. 3

14 4 The proof of the incidence matrix reslt We represent the elements of Pi n as n-tples of natral nmbers a = (a a n ) where P n i= ia i = n (a i conts the nmber of smmands i). We have a l b i there are i j [n] sch that b i+j = a i+j + as well as b i = a i ; b j = a j ; ifi 6= j and b i = a i ; ifi = j. The kth level of Pi n is given by N k = fa Pi n a + + a n = n ; kg k=0 n; Proof of Theorem 7. First note that for a N k with k< n; necessarily a Indeed n = a +a + + na n a +(a + + a n ) (a + + a n ) ; a n (n ; k) ; a a n ; k > Now order the elements of N k lexicographically Let for a b N k a b if a i >b i for the smallest index i for which a i 6= b i. Dene N k! N k+ k< n;,by (a) =(a ; a + a n ) In contrast to the proof of Theorem 8 we do not combine here one smmand and the largest smmand, bt two smmands. Obviosly, a l (a) for every a, and is injective. Moreover, if a b then (a) (b). Let S = f (a) a N k g and consider the minor A of M k which is determined by allrows of M k and those colmns of M k which are indexed by elements of S. Here we sppose that the rows and colmns are ordered w.r.t.. From above weknow that A is sqare and that the diagonal elements of A are eqal to. It is enogh to show that A is lower trianglar. Assme that there are elements a b N k with a b and a l (b). It is easy to see that (a) is the greatest element w.r.t. which covers a (for all other sch elements the rst coordinate is greater since at most one is combined with another smmand). Conseqently, (b) (a) (b) a contradiction. References [] F. C. Alck, S. Chowla, and H. Gpta, On the maximm vale of the nmber of partitions of n into k parts, J. Indian Math. Soc. (N.S.) 6 (94) 05{. 4

15 [] E.R. Caneld, The size of the largest antichain in the partition lattice, J. Combin. Theory, Ser. A, to appear. [3] E. R. Caneld, From recrsions to asymptotics on Szekeres' formla for the nmber of partitions, Electron. J. Combin., 4 () (997) R6. [4] E.R. Caneld and L.H. Harper, Large antichains in the partition lattice, Random Strctres Algorithms 6 (995) 89{04. [5] K. Engel, Sperner Theory, Cambridge University Press, Cambridge, New York, 997. [6] K. Engel and N.N. Kzjrin, Abot the ratio of the size of a maximm antichain to the size of a maximm level in nite partially ordered sets, Combinatorica 5 (985) 30{309. [7] G.G. Hardy and S. Ramanjan, Asymptotic formlae in combinatorial analysis, J. London Math. Society 7 (98) 75{5. [8] J.P.S. Kng, The Radon transforms of the combinatorial geometry. II. Partition lattices, Adv. Math. 0 (993) 4{3. [9] E. Sperner, Ein Satz ber Untermengen einer endlichen Menge, Math. Z. 7 (98) 544{548. [0] G. Szekeres, An asymptotic formla in the theory of partitions, Qart. J. Math. Oxford, Ser. (), (95) 85{08. [] G. Szekeres, Some asymptotic formlae in the theory of partitions (II), Qart. J. Math. Oxford, Ser.() 4 (953) 96{. 5