Lecture notes for Enumerative Combinatorics

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1 Lecture otes for Eumerative Combiatorics Aa de Mier Uiversity of Oxford Michaelmas Term 2004 Cotets Subsets, multisets, ad balls-i-bis 3. Words ad permutatios Subsets ad biomial umbers Multisets ad iteger compositios Balls-ad-bis ad multiomial umbers Mappigs The Priciple of Iclusio ad Exclusio 0 3 Iteger ad set partitios; Stirlig umbers 4 3. Iteger partitios Set partitios Decompositio of permutatios ito disjoit cycles Geeratig fuctios ad recurreces Formal power series Liear recurreces A o-liear recurrece: Catala umbers The geeratig fuctio for iteger partitios Please ay commets or suggestios to ademier@gmail.com; thaks to the oes that have already doe it. Last correctio: 5 Jue 2006.

2 5 The symbolic method for ulabelled structures Costructios Compositios revisited Rooted plae trees The symbolic method for labelled structures 4 6. Costructios Labelled graphs Set partitios revisited Permutatio decompositios revisited

3 A good way to covey what a area of Mathematics is about is by givig a list of problems. This is particularly true i Combiatorics. I this course, we attempt to solve problems like: I how may ways... ca we pick 6 umbers from to 5 so that o two are cosecutive? ca we climb a ladder if we move up either oe or two steps at a time? ca 7 balls be placed i 4 boxes if o box is to be left empty? ca we give chage of a poud? As suggested by the title of the course, we will be maily cocered with coutig problems, although o our way we will ecouter algebraic ad structural questios. The first thig we lear about maths is to cout, but as we shall see coutig ca become quite tricky ad requires techiques. The course has two parts. The first oe itroduces the basic objects, ideas, ad priciples i eumeratio. The secod part is devoted to geeratig fuctios ad their powerful uses. The begiig of the course is sort of liear, sice we will describe a good variety of relatively simple objects; but those are ot to be forgotte, sice as we go our way ito the subject they will appear oce ad agai uder ew lights ad perspectives. Eumeratio is better uderstood by example; for this reaso, we ecourage visual ad combiatorial proofs, ad the examples treated should be used as ispiratio to solve further problems. Notatio. Uless otherwise stated, all sets cosidered will be fiite; a -set is a set with elemets; a k-subset of a set is a subset with k elemets. The set of the first itegers, that is {, 2, 3,..., }, will be deoted by []. Subsets, multisets, ad balls-i-bis This chapter deals with choice problems that might be familiar from probability courses; they are sometimes called combiatios ad variatios, with or without repetitio. I this course we do ot use this otatio ad we oly refer to coutig permutatios, subsets, multisets, etc.... Words ad permutatios Let us start by lookig at a very simple object. Defiitio.. A word over the alphabet X is a fiite sequece of elemets of X. Example. Some words over the alphabet [5] are 23, 23, 25, 435, 443,... Theorem.2. There are k words of legth k over a alphabet of symbols. 3

4 Proof. We have choices for the first letter, choices for the secod letter, ad so o, util we have choices for the last letter. Hece, there are = k words of legth k. Words allow symbols to be repeated. Permutatios are sequeces where all elemets are differet. Defiitio.3. A permutatio of a set X is a total liear orderig of the elemets of X; we represet it as x x 2 x. Example. The set [3] has 6 permutatios: 23, 32, 23, 23, 32, 32. Theorem.4. The umber of permutatios of a -elemet set X is!. Proof. We have choices for the first elemet x. Oce this is chose, we have choices for x 2 ; the, 2 choices for x 3, ad so o, util we have oly oe choice for the last elemet x. Therefore the total umber of permutatios of X is ( ) ( 2) 2 =!. Defiitio.5. A k-permutatio of a set X is a total liear orderig of a k-subset of X; we represet it as x x 2... x k. Example. The set [3] also has 6 2-permutatios: 2, 3, 2, 23, 3, 32, but it oly has 3 -permutatios:, 2, 3.! Theorem.6. A -set X has ( k)! = ( ) ( k + ) k-permutatios. Proof. We proceed as i the proof of Theorem.4. There are choices for x, choices for x 2, ad so o, util we have k + choices for x k. The set of all permutatios of X is deoted S X ; if X = [], we simplify to the usual otatio from group theory: S. Permutatios ca ot oly be viewed as arragemets of elemets, but also as bijective maps from [] oto X. If π is the permutatio x x 2 x, it defies also a map π : [] X as π(i) = x i. The group S has a particularly rich structure that has bee extesively studied; the algebraic poit of view does ot have a importat role for our purposes, but some basic results o represetig permutatios as products of cycles will be eeded later i the course..2 Subsets ad biomial umbers This sectio is cocered with the umber of ways i which we ca select a k-elemet subset of a -set (regardless of the order of the elemets, i oppositio to permutatios). We start by examiig some properties of this umber ad later we derive a formula for it. Defiitio.7. ( k) deotes the umber of subsets of size k of []; or, equivaletly, the umber of ways i which we ca select k diferet elemets from a -elemet set. 4

5 ( ( ( ( ( ( ( ) 0) ) 2) 3) 4) 5) Table : Pascal s Triagle. The umber ( k) is read choose k ad is called a biomial umber or a biomial coefficiet. We start with some basic properties of biomial umbers. The first three follow directly from the defiitio. () ( ( 0) = ) = (2) ( ) ( = ) = (3) ( ) ( k = ) k The followig is the key recurrece relatio for biomial umbers. (4) ( ) ( k = ) ( k + k ) for > k Proof. We prove this equality by coutig subsets of [] accordig to whether or ot they iclude a fixed elemet, say. Let A be a k-subset of []; i total there are ( k) ways of choosig A. If A does ot iclude the elemet, the A ca be chose i ( ) k ways; o the other had, if A does cotai, the the remaiig k elemets of A ca be chose from [ ] i ( k ) ways. The proof above is a example of what is called a combiatorial proof, i costrast to algebraic proofs. It is ofte the case that a result ca be proved i a variety of ways, some of them usig algebraic tools, some others based o bijectios or o structural properties. Of course all proofs are correct ad valid, although i geeral combiatorial proofs ted to be more beautiful ad elightig (ad sometimes quite difficult to fid!). Recurrece (4) allows us to compute biomial umbers recursively; table is usually called Pascal s triagle. I further sectios we will ecouter other combiatorial umbers that satisfy similar relatios that are proved usig aalogous ideas. The followig is Newto s famous Biomial Theorem (from which biomial umbers take their ame). 5

6 Theorem.8. (Biomial Theorem) For all itegers 0, (a + b) = k=0 ( ) a k b k. k Proof. Write (a + b) as (a + b)(a + b) (a + b). To expad this product we have to choose either a a or a b from each of the factors (a + b). Hece each term i the expasio is of the form a k b k for some k betwee 0 ad. This term will appear as may times as ways of pickig k a s from the product above; this is the same as selectig k of the factors (a + b), ad this ca be doe i ( k) ways. Hece the coefficiet of a k b k i the biomial is ( k). As a applicatio of the biomial theorem, we prove the followig summatio formulas for biomial coefficiets. (5) ( ) k=0 k = 2 (6) k=0 ( )k( k) = 0 Proof. By the biomial theorem, k=0 ( ) = k k=0 ( ) k k = ( + ) = 2, k ad ( ) ( ) k = k k=0 k=0 ( ) ( ) k k = ( + ) = 0. k This proof is a archetypical example of a algebraic proof ; the meaig of ( k) plays o role, oly its algebaric properties. Equality (5) is equivalet to sayig that the total umber of subsets of a -set is 2. If X is a set, we deote by P(X) the set of all subsets of X (icludig ad X). Hece, we have just proved that P(X) = 2 X. Equality (6) above ca be phrased as the umber of subsets of eve size of a -set equals the umber of subsets of odd size. Ca you fid combiatorial proofs for these equalities? Fially we come to the well-kow formula for biomial umbers. 0! =.) (By defiitio, Theorem.9. ( ) = k!, for k 0. k!( k)! Proof. A k-subset of a -set ca be see as a k-permutatio i which the order of the elemets does ot matter. Hece, to pick a k-subset, just pick a k-permutatio ad forget 6

7 about the order of the elemets. Sice there are!/( k)! k-permutatios ad a set of size k ca be ordered i k! ways, we have that the total umber of k-subsets of a -set is! ( k)! k! =! k!( k)!..3 Multisets ad iteger compositios Up to ow we have oly cosidered the case of selectig distict elemets of a set (with or without order). The ext step is to allow repetitios i the elemets we select. Towards this ed we have to itroduce the cocept of a multiset; a multiset is like a set, but we allow each elemet to be repeated a (fiite) umber of times. Defiitio.0. Let X be a set. A multiset M o X is a fuctio ν : X N such that ν(x) is fiite for all x X. The umber ν(x) is the umber of copies (or repetitios) of x, ad x X ν(x) is the size of M. Example. Let X = {a, b, c, d}. The multiset correspodig to the fuctio ν(a) = 2 ν(b) = 0 ν(c) = ν(d) = 3 ca be represeted as {a, a, c, d, d, d}. The size of M is 6. We ow cout how may multisets of size k does a -set have. I other words, i how may ways we ca pick k elemets from a -set if we ca repeat elemets. Let x,..., x be the elemets of X. To select a multiset of size k we have to select o-egative umbers a,..., a such that a + + a = k. Here, a i is the umber of copies of x i we pick. Theorem.. The umber of solutios to the equatio a + + a = k, a i 0, a i N, is ( ) ( k+ k. I other words, a -set has k+ ) k multisets of size k. Proof. The problem is the same as fidig the umber of ways of placig k udistiguishable balls i umbered boxes (a i represets the the umber of balls i box i). Put the k balls i a row. To represet the boxes we use vertical bars, meaig the separatio betwee two cosecutive boxes. Hece we eed bars. A distributio of the balls i the boxes is the a arragemet of bars ad balls, such as I this arragemet we have five boxes, of which the first has oe ball, the secod ad the last oes are empty, the third has two balls, ad the fourth has six balls. So, i total we 7

8 have k + positios that ca be either or, ad k of the positios are. Sice there is o further restrictio, the solutio is the umber of ways of selectig k elemets from a set of k +, ad therefore the theorem follows. Multisets are strogly related to iteger compositios. Defiitio.2. A compositio of a iteger is a expressio of as a ordered sum = k of strictly positive itegers. Example. Let us fid all compositios of the first itegers. 2 = + 3 = 2 + = + 2 = = 3 + = + 3 = = = = = Let c() be the umber of compositios of the iteger ad let c k () be the umber of compositios of i exactly k parts. The example above suggests that c() = 2. We shall prove this by fidig first a formula for c k (). Theorem.3. The umber of compositios of i k parts is c k () = ( k ). The total umber of compositios of is 2. Proof. Observe that c k () is the umber of solutios to the equatio + + k = with i. We proceed as i the proof of Theorem.: we have to distribute balls i k boxes, but ow o box ca be left empty. I terms of the balls ad bars diagram, we have balls ad k separatios with the extra coditio that o two bars ca be cosecutive. Hece, betwee ay two cosecutive we ca place at most oe. Hece, from the spaces betwee two we have to select k to put a. This ca be doe i ( k ) ways. The formula for c() is foud by summig all the c k () ad applyig the biomial theorem. c() = c () + c 2 () + + c () = k= ( ) = k ( ) = 2 j j=0 Our proof for the formula c() = 2 is agai algebraic, but there are also combiatorial proofs. Try to fid oe (there are several, but oe follows icely usig balls ad bars diagrams as above). 8

9 .4 Balls-ad-bis ad multiomial umbers Suppose we have m umbered balls that we wat to place i r umbered bis. I how may ways ca this be doe if we do ot impose ay extra coditio? We have r choices for where to put ball ; for ball 2 we have agai r choices; ad agai r choices for each of the balls 3 to m. Hece, i total we have r r r = r m possibilities. Now suppose that we are give the umber of balls that must go ito each of the bis, that is, we have umbers m,..., m r so that bi i has to cotai m i balls. (Implicit i the defiitio is that m + m m r = m.) Let us cout the umber of ways of placig the balls by coutig the umber of possibilities for each bi. I bi we have to put m balls, so there are ( ) m m choices. Bi 2 must cotai m 2 balls, but of course we caot choose amog the oes that we have already put i bi. Hece, there are ( m m ) ( m 2 choices. Similarly, we have m m m 2 ) m 3 choices for bi 3, ad so o, util the last bi. Thus the umber of ways of placig m balls i r bis with m i balls i bi i is ( )( )( ) ( ) m m m m m m 2 m m m r = m m 2 m 3 m! (m m )! (m m m 2 )! m!(m m )! m 2!(m m m 2 )! m 3!(m m m 2 m 3 )! (m m m r )! = m r!0! m! m!m 2! m r!. This umber is deoted by ( ) m m,m 2,...,m r ad it is called a multiomial umber. Notice that whe r = 2 we recover the usual formula for biomial umbers ( ) m = m! ( ) ( ) ( ) m m m m, m 2 m!m 2! = = =. m m m m 2 Ideed, to place m balls i two bis such that the first cotais m balls ad the other m 2 = m m balls, it is eough to choose which balls go ito the first bi, or which oes go ito the secod bi. Aalogous to the biomial theorem, we have the multiomial theorem. Theorem.4. (x + x x r ) m = m,m 2,...,mr P mi =m,m i 0 ( m m r m, m 2,..., m r ) x m xm 2 2 x m r r Proof. The proof follows the same idea as the proof of the biomial theorem ad it is left as a exercise. Observe that settig x i = for all i i the biomial theorem we recover the fact that the total umber of ways of placig m balls i r bis, regardless of the umber of balls i each bi, is r m. 9

10 .5 Mappigs May of the results i this sectio ca be phrased i terms of maps. Let F(, m) be the set of all mappigs from [] to [m]. The followig coutig results are simple applicatios of the priciples of this sectio. F(, m) = m For m, F(, m) cotais m!/(m )! ijective maps. There are ( k) maps i F(, 2) such that the preimage of is a set of size k. The umber of maps f F(, m) such that f() < f(2) < < f() is ( m ). 2 The Priciple of Iclusio ad Exclusio The Priciple of Iclusio ad Exclusio (PIE) is a very useful tool to cout sets that ca be expressed as a uio or a itersectio of simpler sets. The idea is to geeralize the well-kow formula to compute the cardiality of a uio: A B = A + B A B. For the case of three sets it is also easy to fid a formula by ispectio: A B C A B C = A + B + C A B A C B C + A B C. As we see, to compute the size of the uio we sum the sizes of the compoets, substract the itersectios of pairs, ad sum back the size of the triple itersectio. The Priciple of Iclusio ad Exclusio is a geeralizatio of this idea. Theorem 2.. (PIE) Let A, A 2,..., A be subsets of a set X. The A A 2 A = A i A i A j + A i A j A k i= i<j i<j<k + + ( ) A A. 0

11 Proof. There are several proofs of the PIE. We choose oe that has a more combiatorial flavour (as a excersise, prove it by iductio). We check that the formula couts each elemet i A A 2 A just oce. Let x be i A A 2 A ; by relabellig the sets if ecessary, we ca assume that x belogs to A, A 2,..., A p but does ot belog to A p+,..., A. The i the RHS of the above formula, x cotributes with This equals p ( ) p + 2 p ( ) p ( ) i = i i= ( ) ( ) p p + + ( ) p. 3 p p ( ) p ( ) i + = 0 + =, i i=0 sice we kow that the sum of siged biomials is zero. Before lookig at the applicatios, let us start with some remarks. The PIE is stated i terms of uios, but ca also be used to cout itersectios. Ideed, A A 2 A = (A c A c 2 A c ) c = X A c A c 2 A c, where B c stads for the complemet of the set B i X. Now, usig PIE, A A = X A c i + i= i<j A c i A c j + + ( ) A c A c. Oe of the mai tricks i usig PIE is to choose the sets A i suitably so that computig the itersectios A i A ik is feasible. It is ofte the case that A i A ik does ot deped o the A ij but oly o k, the umber of sets i the itersectio. I this case the PIE has a simpler form. Let A k deote the size of the itersectio of ay k of the A i. The, ( ) ( ) A A 2 A = A A 2 + A ( ) A. 2 3 The rest of this sectio is devoted to examples of applicatio of the PIE. Example. Oe of the classic applicatios of the Priciple of Iclusio ad Exclusio is the deragemet problem. Suppose people leave their coats at the cloakroom of a theater. At the ed of the play, the attedat gives the coats back without lookig at the tickets. Which is the probability that obody gets their ow coat? Let us idetify the coats with the itegers from to. Each way of givig back the coats is a permutatio of []. For istace, the permutatio represets the case that each perso gets back their ow coat. Let π be a permutatio of []. If for all i we have

12 π(i) i, the obody gets their ow coat back. We call such permutatios deragemets. The the probability asked is umber of deragemets of [].! Hece our goal is to compute the umber of deragemets of []; we deote the set of deragemets by D ad its cardiality by d. Let S be the set of the! permutatios of elemets ad for each i with i let A i be the subset of all permutatios π such that π(i) = i. By the remarks above, we have that D = A c A c 2 A c = S (A A 2 A ). Hece, ad by PIE d =! A i + i= i<j d =! A A 2 A, A i A j i<j<k The oly thig left ow is to cout the sizes of the itersectios. A i A j A k + + ( ) A A. A i = {π S π(i) = i} = ( )!, sice the elemet i is fixed ad we ca permute the remaiig i ay way. Note that this is idepedet of the elemet i. A i A j = {π S π(i) = i, π(j) = j} = ( 2)!, sice two elemets are fixed ad the remaiig 2 ca be permuted arbitrarily. Agai, the result is the same for all pairs i, j. Ad i geeral, A i A i2 A ik = ( k)!, for i < i 2 < < i k. Therefore, d =! ( ) ( ) i ( i)! =! i i= ( ) i. i! Hece, the probability asked is ( ) i i=0 i!. Does this probability have a limit as teds to ifiite? Recall the series expasio for the expoetial fuctio e x = i 0 xi i! ; this series coverges for all real values of x. Hece, as, the probability that obody gets their ow coat gets closer to e 0.37; actually, the rate of covergece is really fast, sice the absolute error is bouded by /( + )!. i=0 2

13 Example. Euler s φ fuctio. Recall from elemetary umber theory the defiitio of the fuctio φ of Euler. Give a positive iteger, φ() is the umber of itegers smaller tha that are relatively prime to (icludig ). For istace, φ(2) = {} =, φ(3) = {, 2} = 2, φ(4) = {, 3} = 2, φ(5) = {, 2, 3, 4} = 4. Note that if is prime, the φ() =, sice all itegers smaller tha are relatively prime to. Our goal is to fid a formula for φ() for ay iteger. We assume that we have the decompositio of ito prime factors, = r i= where r is the umber of distict prime factors of, the p i are the distict prime factors, ad α i stads for their multiplicities. The itegers that are relatively prime with are those that do ot cotai ay of the p i as a factor. This suggests to defie B i = {m : m <, p i m}, that is, the set of itegers smaller tha that are divisible by p i. Hece, By PIE, φ() = r B i + i= i<j r p α i i, φ() = B c B c 2 B c. B i B j i<j<k r B i B j B k + +( ) r B B r. Agai, the problem reduces to computig the itersectios of B i s. It is ot difficult to show that B i B ik = {m : m <, p i p ik m} =. p i p ik Note that i this case the size of B i B ik ot oly depeds o k but also o the specific sets we itersect. Puttig this ito the formula give by PIE, we have φ() = ( ) r, p i p i p j p p r i r i<j r which ca be writte more compactly as ) ) ) φ() = ( ( p p2 ( pr. Example. The followig is left as a excersise. Show that the umber of surjective maps from [] oto [k] is k ( ) k (k ) + ( ) ( ) k k (k 2) + + ( ) k = 2 k k ( ) k ( ) k j j. j j= 3

14 3 Iteger ad set partitios; Stirlig umbers 3. Iteger partitios We itroduced compositios as ordered sums, hece regardig 2 + ad + 2 as differet compositios of 3. But as partitios, we cosider them the same. Defiitio 3.. A partitio of a iteger is a expressio of as a sum of positive itegers = λ + λ λ k with λ λ 2 λ k. We usually deote this partitio by (λ, λ 2,..., λ k ). Example. Let us look at the partitios of the first itegers. 2 = + 3 = 2 + = = 3 + = = = = 4 + = = = = = We deote by p() the umber of partitios of the iteger ad by p k () the umber of partitios of with k parts. From the example above we have p() =, p(2) = 2, p(3) = 3, p(4) = 5, p(5) = 7. There are o explicit formulas kow for p(). But evertheless iteger partitios are oe of the icest 2 ad richest objects i combiatorics. We start with a recursio for the umbers p k (). Propositio 3.2. For k 2, p k () = p k ( ) + p k ( k). Proof. Let λ be a partitio of with k parts. We compute p k () by coutig partitios accordig to whether the last part λ k is or greater tha. There are as may partitios of with k parts ad λ k = as partitios of with k parts. If λ k >, the all parts are greater tha, hece we ca substract from each part ad get a partitio of k with k parts. This recurrece ad the trivial cases p () = ad p 0 (0) = allow us to compute the umbers p k () (Table 2), i the same spirit as Pascal s triagle. 4

15 p() p () p 2 () p 3 () p 4 () p 5 () p 6 () p 7 () p 8 () Table 2: Iteger partitios accordig to the umber of parts. A very useful way to represet partitios is by the meas of Ferrers diagrams. For a partitio (λ, λ 2,..., λ k ), its Ferrers diagram is costructed by placig, left justified ad from top to bottom 3, λ dots, λ 2 dots,..., λ k dots. This is better uderstood with a example. The Ferrers diagram for the partitio is As a excersise, iterpret Propositio 3.2 above i terms of Ferrers diagrams. Give a partitio λ, its cojugate µ is the partitio whose Ferrers diagram is obtaied by reflectig the Ferrers diagram of λ alog the diagoal y = x. That is, istead of readig the diagram by rows, we read it by colums. For istace, the cojugate of (5, 3,, ) is (4, 2, 2,, ). The followig descriptio of the cojugate of a partitio follows easily by lookig at the Ferrers diagram. The asymptotic behaviour is kow. If this souds iterestig, you may look at Thm. 5.7 of Va Lit ad Wilso, A course i combiatorics, Cambridge UP. You may wat to wait util we have studied geeratig fuctios though. 2 Well, this is a persoal opiio. 3 This is the Eglish otatio. I Frech otatio, Ferrers diagrams are draw with the largest part at the bottom.. 5

16 Propositio 3.3. If λ is a partitio of, the its cojugate µ is also a partitio of ad µ j = {i λ i j}. The followig result also follows by reasoig o Ferrers diagrams ad cojugate partitios. Propositio 3.4. The umber of partitios of with k parts equals the umber of partitios of whose largest part is k. A partitio is called self-cojugate if it equals its cojugate; or, equivaletly, if the Ferrers diagram is symmetric with respect to its diagoal. For istace, (4, 3, 3, ) is a self-cojugate partitio. Propositio 3.5. The umber of self-cojugate partitios of equals the umber of partitios of all whose parts are odd ad distict. Proof. We give a proof by picture. We defie a bijectio betwee self-cojugate partitios ad partitios whose parts are odd ad distict, see Figure. Figure : Bijectio betwee self-cojugate partitios ad partitios all whose parts are odd ad differet. This bijectio ca be defied formally i the followig way. Let (λ,..., λ t ) be a selfcojugate partitio. Cosider the partitio give by µ i = 2(λ i (i )) for i such that λ i > i. Check that all parts of µ are odd ad differet. Coversely, give a partitio (µ,..., µ s ) with all parts odd ad differet, defie λ j = (µ j + )/2 + (j ) for j s. For k s +, let λ j = {λ i : i s, µ i j} (as log as this is o-zero). The partitio λ is self-cojugate. 6

17 For the momet, we fiish our discussio about partitios with a quite surprisig result. We delay its proof util we develop some geeratig fuctio tools later i the course. But of course you are allowed (ad ecouraged) to thik about a combiatorial proof 4. Theorem 3.6. The umber of partitios of ito odd parts is the same as the umber of partitios of ito differet parts. 3.2 Set partitios Oe of the first questios we studied was choosig a k-subset of a -set. Observe that pickig a subset A of a set X is the same as partitioig X ito two disjoit subsets, A ad A c. So we ca ask the questio: i how may ways ca we partitio a -set X ito two disjoit o-empty subsets? (By a partitio of X ito two subsets we mea pickig A ad B such that A B = ad X = A B.) The aswer is quite simple. We have 2 2 choices for the first subset, say A ( ad X are ot valid choices), ad oce A is determied, we have that B = A c. Observe though that the order of A ad B is irrelevat, hece we have to divide by 2 ad the result is (2 2)/2 = 2. We ow geeralize these ideas to partitios of sets ito k blocks. Defiitio 3.7. A partitio of a set X is a decompositio of X of the form X = A A k with A i for all i ad A i A j = for all i j. The sets A i are called the blocks of the partitio. Note that partitios are defied regardless of the order of the blocks. Example. Let us list all possible partitios of the sets [], [2] ad [3]. [] = {} [2] = {, 2} = {} {2} [3] = {, 2, 3} = {, 2} {3} = {, 3} {2} = {} {2, 3} = {} {2} {3} The umber of partitos of a -set ito k blocks is deoted by { k} ad it is called a Stirlig umber of the secod kid 5. (A alterative otatio, though less commo owadays, is S(, k).) The total umber of partitios of a -set is deoted by B() ad it is called a Bell umber. By defiitio, B() = { k= k}. Example. The followig values of Stirlig ad Bell umbers are easily deduced or computed ( ). { } { } { } { } { } ( ) { } 0 = = 0 = = 2 = = B() = B(2) = 2 B(3) = 5 B(4) = 5 Like the umber of iteger partitios, Stirlig umbers of the secod kid satisfy a liear recurrece that allows a easy recursive computatio. 4 Or cheat by readig Sectio 3.3 of Stato ad White, Costructive Combiatorics, Spriger, The first kid will appear soo. 7

18 { { { { { { } 2} 3} 4} 5} 6} B() Table 3: Stirlig umbers of the secod kid ad Bell umbers. Propositio 3.8. For k, { } = k { } + k k { Proof. We cout partitios of [] accordig to whether {} is a block or ot. If {} is a block of the partitio, the remaiig elemets have to be partitioed ito k blocks, hece there are { k } choices. If {} is ot a block, the elemet is i oe of the k blocks together with some other elemets. Partitio first the set [ ] ito k blocks, ad the choose ay of the k blocks ad adjoi to it. We have { } k choices for the partitio, ad k choices for the block that cotais. k }. This recurrece ca be used to compute some values of Stirlig ad Bell umbers (Table 3). Bell umbers ca also be computed recursively, although i this case the recurrece ivolves all previous terms. Propositio 3.9. B( + ) = k=0 ( ) B(k) k Proof. We cout partitios of [ + ] accordig to the size of the block that cotais +. Say that the block cotaiig + has size j, for some j with j +. There are ( j ) choices for the other elemets of the block, ad oce this is chose, the remaiig + j elemets have to be partitioed, which ca be doe i B(+ j) ways. Therefore, + ( ) + ( ) B( + ) = B( + j) = B( + j) = j j + j= j= k=0 ( ) B(k). k 8

19 There is actually a explicit formula for Stirlig umbers of the secod kid based i a correspodace betwee set partitios ad surjective mappigs. Let f be a surjective map from [] to [k]. The sets f (), f (2),..., f (k) form a partitio of [] (sice f is surjective, oe of these sets is empty). Observe that this correspodace gives each partitio ito k blocks a total of k! times, sice switchig the preimages gives the same partitio but a differet map. Hece, k! { k} is the umber of surjective maps from [] to [k], which we already kow. Therefore, { } = k k! k ( ) k ( ) k j j. j j= 3.3 Decompositio of permutatios ito disjoit cycles Fially, we briefly discuss the decompositio of permutatios ito disjoit cycles ad Stirlig umbers of the first kid. Usually we write a permutatio π of [] as a ordered sequece a a 2... a of the umbers {, 2,..., }. A permutatio ca be viewed as a bijectio from [] oto itself, defied by π(i) = a i. For istace, 2354 deotes the permutatio 2, 2 3, 3 5, 4 4, 5. Aother way of writig permutatios is the disjoit cycle otatio 6. The permutatio above i disjoit cycle otatio is (235)(4). Each permutatio ca be writte uiquely as a product of disjoit cycles up to the order of the cycles ad the cyclic order of the elemets i each cycle (a cycle of legth k ca be writte i k ways). We associate with each permutatio π of S a -tuple (c, c 2,..., c ), where c i is the umber of cycles of legth i i the disjoit cycle decompositio of π. This -tuple is called the type of the permutatio. The permutatio of our example has type (, 0, 0,, 0). Propositio 3.0. The umber of permutatios of type (c, c 2,..., c ) is! c! c! c 2 c 2 c Proof. Let π = x x 2 x be ay permutatio of []. Parethesize the word π so that the first c cycles have legth, the ext c 2 have legth 2, ad so o. The result is the disjoig cycle otaio of a permutatio of cycle type (c, c 2,..., c ). But this procedure gives the same permutatio several times. Let us cout how may times a permutatio of cycle type (c, c 2,..., c ) will appear. First, each cycle of legth c ca be writte i c differet ways; so just takig ito accout this, each permutatio is repeated c 2 c2 c times. But the relative order of the cycles of the same legth is irrelevat, ie, the c cycles of legth ca be ordered i c! ways, the c 2 cycles of legth 2 i c 2! ways, etc... Hece, each permutatio is couted c!c 2! c! c 2 c2 c times, hece the formula. If istead of the type of a permutatio we are oly iterested i the umber of cycles i its decompositio, we obtai Stirlig umbers of the first kid. 6 See ay basic group theory book if you are ot familiar with this. 9

20 Defiitio 3.. The umber of permutatios of [] whose cycle decompositio cotais k cycles is deoted by [ k] ad its called a Stirlig umber of the first kid. (I old otatio, s(, k).) Clearly [ ] =, sice the oly permutatio i S that decomposes as cycles is the idetity (all cycles must have legth oe). The umber [ ] couts permutatios that decompose as a uique cycle, hece permutatios that are cycles; their cycle type is (0,..., 0, ), ad by the previous result there are ( )! of them. It is also easy to determie [ ] ; if we have cycles, the cycle type must be ( 2,, 0,..., 0), hece [ ] =! 2!( 2)! = ( 2). The sum over k of the Stirlig umbers [ k] should be the total umber of permutatios of a -elemet set, which we kow is!. Like Stirlig umbers of the secod kid, the umbers [ k] also satisfy a recurrece. Propositio 3.2. For k, [ ] [ ] = k k [ + ( ) k Proof. We cout permutatios accordig to whether is a fix poit or ot (ie, accordig to whether () is a cycle of the decompositio). If () is a cycle, the remaiig itegers give a permutatio with k cycles, so this gives the term [ k ]. If () is ot a cycle, the the elemet is i a cycle of legth at least 2. Take a permutatio π S with k cycles; the we ca isert the elemet after ay of the umbers, 2,..., i the disjoit cycle decompostio of π. This yields a permutatio i S i which is ot a fix poit. So there are [ ] k choices for the permutatio ad choices for the positio i which to isert. ]. We ca ow use this recurrece to geerate a table of Stirlig umbers of the first kid (Table 4). 4 Geeratig fuctios ad recurreces Up to this poit, our aswers to eumerative problems have cosisted oly of closed formulas, such as c() = 2. But as we have see i the case of partitios, these closed formulas are ot always easy to fid, if kow. We have also studied some problems by meas of recurreces, that do ot provide closed formulas but allow us to compute as may terms as we like. These ad the followig sectios deal with aother way of expressig the result of a coutig problem, amely geeratig fuctios. With them we shall be able to give more iformatio about old ad ew coutig sequeces, ad solve a variety of problems that were out of reach with the basic techiques of the previous chapters. 20

21 [ [ [ [ [ [ ] ] 2] 3] 4] 5] Table 4: Stirlig umbers of the first kid. Let us start by formalizig our goals. Give a problem, the aswer we look for is a sequece of umbers a 0, a, a 2,..., such as,, 2, 3, 5, 7,, 5,... if we are coutig iteger partitios. To this sequece we associate a geeratig fuctio. Defiitio 4.. The (ordiary) 7 geeratig fuctio (OGF) of a sequece {a } 0 is the formal power series a 0 + a z + a 2 z 2 + a 3 z 3 + = 0 a z. By a formal power series we mea that the variable z does ot take ay value ad we regard it just as a mark (we will formalize this soo). We do ot care about covergece issues either. So i priciple it does ot seem that a geeratig fuctio is goig to be more useful tha a recurrece... Example. Fix some iteger m. The sequece { ( ) m k }k 0 has as geeratig fuctio ( ) ( ) ( ) ( ) ( ) m m m m m z k = + z + z z m = ( + z) m. k 0 2 m k 0 So i this case, the geeratig fuctio is ot a ifiite series but a polyomial. Example. Cosider the sequece,,,,... Although it does ot seem combiatorially atractive, this sequece ad its geeratig fuctio will be of extreme importace to us. The geeratig fuctio is F (z) = + z + z 2 + z 3 + Observe that F (z) zf (z) =, hece F (z) = z. This is the well-kow formula for the sum of a geometric series; as metioed before, i this cotext we do ot care about aalytic or covergece properties of F (z). By makig the substitutio z az, oe shows that the geeratig fuctio for the sequece, a, a 2, a 3,... is az. We will soo justify that the preceedig operatios o formal power series are well defied ad soud, but before doig so we examie other examples ad start explorig the spirit of geeratigfuctioology, as it is sometimes called 8. 7 The other class of geeratig fuctios that we will study i this course are expoetial geerartig fuctios. 8 Herbert S. Wilf, Geeratigfuctioology, Academic Press,

22 Let a(z) be the OGF of the sequece {a } 0, that is a(z) = 0 a z. We might wat to express i terms of a(z) some slight variatios of this series, such as 0 a z or 0 a z. After some thought, the followig table is derived (Table 5). series 0 a z a z closed form for OGF a(z) za(z) 0 a +z a(z) a 0 z 0 a +kz a(z) a 0 a z a k z k 0 b z 0 (a + b )z z k b(z) a(z) + b(z) Table 5: Operatios o OGF s. These rules ca ow be used to fid the OGF of a sequece give by a recurrece relatio. For istace, cosider the sequece defied by a 0 =, a =, a +2 = a + + a for 0 Multiply both sides of the recurrece by z : a +2 z = a + z + a z ad sum over all 0: 0 a +2 z = 0 a + z + 0 a z. By usig the rules i the previous table, we get Solvig this equatio for a(z) we deduce a(z) z z 2 = a(z) + a(z). z a(z) = z z 2. How ca this geeratig fuctio help us uderstad better the sequece give by the recurrece relatio? Ideally, we would like to fid a closed formula for the umbers a. I some situatios, like the preset oe, this is possible. First of all, we expad i partial fractios. z z 2 We first decompose the deomiator i liear factors: z z 2 = ( αz)( βz). 22

23 Equatig the coefficiets of the powers of z oe fids α = + 5 2, β = 5. 2 (α ad β are the iverses of the roots of z z 2.) Now we fid costats A ad B such that z z 2 = A αz + B βz A + B (αa + βb)z =. ( αz)( βz) By equatig the powers of z we obtai the system of equatios { A + B = ; whose solutio is A = α α β, B = β α β. Hece, a(z) = αa + βb = 0. z z 2 = 5 ( α αz β ) βz Now recall that az = 0 a z. Hece, a(z) = α + z β + z By defiitio of a geeratig fuctio, a is the coefficiet of z i a(z). Therefore, we have that ( a = + ) + ( 5 ) By fidig their geeratig fuctio first, we have bee able to fid a formula for the umbers a. These umbers a are called the Fiboacci umbers 9. Imagie ow that we are ot that iterested i a exact formula for a, but rather we wat to kow aproximately how fast Fiboacci umbers grow. Note that for large values of, the term ( 5 2 ) + ca be eglected, ad actually will ever be as large as 0.5. Hece, a 5 ( + ) Eve for small this approximatio is extremely good (it will always be at most 0.5 off the true value; actually, sice we kow that a is a iteger, we coclude that a is the iteger earest to the approximate value). By usig Maple 0 we obtai the followig umerical results. 9 Most combiatorics textbooks cotai the problem of rabbit breedig that origially lead to Fiboacci umbers. 0 Or aother computer algebra package. 23

24 a a I other situatios we will ot be as lucky ad a closed formula for the coefficiets will ot be available. I those cases, there is still much that ca be said, especially i terms of asymptotic aproximatios. Ufortuately, this issue goes further beyod the scope of this course. 4. Formal power series The purpose of this sectio is to develop the theory of formal power series to provide a valid framework i which to carry out our computatios with geeratig fuctios. Defiitio 4.2. A formal power series is a expressio of the form a 0 +a z +a 2 z 2 +a 3 z 3 +, where a 0, a, a 2, a 3,... are ratioal umbers 2. The set of all power series is deoted by Q[[z]]. Agai, the variable z has to be iterpreted as a mark. Observe that ot all terms eed to be differet from zero. For istace 4, +z, ad z 3 are formal power series. If f(z) is a formal power series, we deote by [z ]f(z) the coefficiet of z i f(z). For example, [z]( + z) =, [z 0 ](4 + z 4 ) = 4, ad [z]( + z 2 ) = 0. If a formal power series is called a(z), we use the same letter for the coefficiets: a 0, a, a 2,... We ca defie operatios o formal power series as oe would expect. Their sum is defied as a(z) + b(z) = 0(a + b )z. The product is a bit trickier but also atural: a(z)b(z) = 0 0 i a i b i z. Observe that ( z)( + z + z 2 + ) =. Hece it makes sese to write z = + z + z2 + Defiitio 4.3. We say that b(z) is the (multiplicative) iverse of a(z) if a(z)b(z) =. Propositio 4.4. The formal power series a(z) has a multiplicative iverse if ad oly if a 0 0. The iterested reader will fid coutless examples i Flajolet ad Sedgewick s forecomig book Aalytic combiatorics, especially i the parts devoted to complex asymptotics. The prelimiary versio of the book is available o-lie at 2 Actually, ay other field as the real or complex umbers would do. 24

25 Proof. Suppose first that a(z) has a iverse. The there exists a formal power series b(z) such that (a 0 + a z + a 2 z 2 + )(b 0 + b z + b 2 z 2 + ) =. By the defiitio of product of power series, we deduce that a 0 b 0 =, hece that a 0 0. Coversely, suppose ow that a 0 0 ad cosider the equatio (a 0 + a z + a 2 z 2 + )(b 0 + b z + b 2 z 2 + ) =. We show that this equatio ca be solved for the b s, ad hece that a(z) has a iverse. Expad the product ad equate coefficiets of z i both sides: a 0 b 0 = b 0 = a 0, sice a 0 0. Also, a 0 b + a b 0 = 0 b = a a 2, 0 a 0 b 2 + a b + a 2 b 0 = 0 b 2 = a 2a 0 a 2 a 3, 0 ad similary we ca iductively fid b 3, b 4,... Hece a(z) has a multiplicative iverse. At this poit we have edowed the set Q[[z]] with a sum ad a product; it is easy to check that these two opertios are associative ad commutative, ad that the sum is distributive with respect to the product. Hece Q[[z]] is a commutative rig with uity. The followig is the biomial theorem for egative iteger expoets. Propositio 4.5. As formal power series, for iteger k 0 ( z) k = ( ) + k z 0 Proof. ( z) k = ( + z + z2 + z 3 + ) k The coefficiet of z above equals the umber of ways of pickig oe power of z from each of the k factors, such that the sum of the expoets is. I other words, the coefficiet of z is the umber of solutios to the equatio e + e e k =, with e i 0. But we kow from Theorem. that this umber is ( ) +k. Aother operatio o power series is compositio. Give two power series a(z) ad b(z), their compositio is the series a(b(z)) = 0 a (b(z)). But is this operatio meaigful? Each term a (b(z)) is well defied, but takig a ifiite sum could lead to some problems. Suppose that b 0 0. The each term b(z) cotributes with a o-zero costat term b 0. Hece the costat term of a(b(z)) is b 0, ad this sum may easily be diverget. To 25

26 avoid this problem, we oly cosider compositio of series with costat term equal to zero, or such that a(z) is a polyomial. I this case, we have that [z ]a(b(z)) = [z ](a k b(z) k ). k=0 Example. Take a(z) = /( z) ad b(z) = z 2. The a(b(z)) = /( z 2 ) = + z 2 + z 4 + z 6 + The formal derivative of the power series a(z) = 0 a z is the power series a (z) = 0 a z. This formal derivate satisfies the usual rules with respect to sums, products, quotiets, compositio, etc... We use the familiar termiology from aalysis to deote some power series. For istace, we deote 0! z by exp(z) ad z by log( z ). I aalysis we have the equality exp(log( z )) = z. Is this equality true at the level of formal power series? Here is where we use some aalysis. Observe that the series 0 z coverges for z < ad the expoetial series coverges for all z. Hece the compositio exp(log( z )) coverges for z <. But if it coverges, it should coverge to the true value, that is, z. So we deduce that the series exp(log( z )) ad z are the same i a eighbourhood of the origi. But if they are the same i a o matter how small eighbourhood of the origi, they have the same coefficiets i their Taylor expasios, ad hece they also agree as formal power series. This same idea ca be used to prove the geeralized biomial theorem for ratioal expoets. Propositio 4.6. For all a Q, where ( ) a = a(a )(a 2) (a +)!. ( + z) a = 0 ( ) a z, So the workig rule to remember with power series is that, although the variable plays just a mark role ad takes o value, thigs behave as our ituitio from aalysis suggests. 4.2 Liear recurreces We have already see several recurreces i this course. I this part we show how geeratig fuctios ca help solvig recurreces. Mostly we will do this by example. But let us first defie recurreces formally. Defiitio 4.7. A recurrece for a sequece {a } 0 is a relatio of the form a +k = φ(a, a +,..., a +k ) valid for all 0. The values a 0, a,..., a k are the iitial coditios of the recurrece, ad k is the order or the recurrece. 26

27 For istace, the recurrece for the Fiboacci umbers has order 2. Example. Fid the umber b of biary words that do ot have two cosecutive zeros. Let us start by computig the first values of b. It is easy to check that b 0 = (the empty word), b = 2 (the words 0 ad ), ad b 2 = 3 (0, 0, ad ). I order to fid a recurrece for b, we have to somehow decompose a word with o two cosecutive zeros ito smaller words. Suppose that w w 2 w is a biary word with o two cosecutive zeros. We distiguish two cases. O the oe had, if w =, the w w 2 w is also a biary word with o two cosecutive zeros. O the other had, if w = 0, the we must have w =, ad hece w w 2 w 2 is a biary word with o two cosecutive zeros. Therefore, b = b + b 2 for 2 or equivaletly, b +2 = b + + b for 0. Is this is the same recurrece as for Fiboacci umbers. Observe that a recurrece also icludes the iitial coditios, ad i this case they are differet (b 0 =, b = 2). We solve the recurrece usig the same method as for Fiboacci umbers. We multiply both sides by z ad sum over all 0. b z 0 b +2 z = 0 b + z + 0 Let b(z) be the correspodig geeratig fuctio. Applyig the rules we deduce b(z) 2z z 2 = b(z) + b(z), z ad hece b(z) = + z z z 2. Observe that the deomiator is the same as we had for Fiboacci umbers, whereas the umerator is differet. As we shall see, it is always the case i dealig with liear recurreces that the iitial coditios determie the umerator ad the recurrece determies the deomiator. By usig the techique of partial fractios, we get to a explicit formula for b, amely b = + α α β α β α β β, where α = ad β = 5 2. Observe that the asymptotic behaviour is the same as for Fiboacci umbers. This is aother geeral fact, that the recurrece determies the growth rate, regardless of the iitial coditios. A recurrece is liear with costat coefficiets if it is of the form a +k + c a +k + c 2 a +k c k a = 0 for some ratioal umbers c, c 2,..., c k. Liear recurreces with costat coefficiets have very special geeratig fuctios. The followig is a basic theorem whose proof is left as a exercise (just mimic our procedure for solvig recurreces). 27

28 Theorem 4.8. Let {a } 0 be a sequece. The followig are equivalet. {a } 0 satisfies a liear recurrece with costat coefficiets a +k + c a +k + c 2 a +k c k a = 0. P (z) +c z+c 2 z 2 + +c k z k, where P (z) is a poly- Its geeratig fuctio a(z) is of the form omial i z of degree at most k. Geeratig fuctios that are of the form P (z) Q(z) for some polyomials P (z), Q(z) are called ratioal 3. They are the simplest kid of geeratig fuctios, ad they arise from liear recurreces. If we have a sequece whose GF is ratioal, we ca use the method of partial fractios to fid a explicit formula for the terms of the recurrece. Let us sketch the procedure briefly. We first decompose the deomiator as Q(z) = ( α z) d ( α 2 z) d2 ( α r z) d r. That is, the αi are the roots of Q(z) ad the d i are their multiplicities. By the theory of partial fractios, we kow that there exist complex umbers A i,j for i r ad j d i such that P (z) Q(z) = A, α z + A,2 ( α z) A,d ( α z) d + A 2, α 2 z + + A r,dr ( α r z) d r From this expressio we wat to extract the coefficiet of z. Recall that ( αz) d = ( ) + d α z. 0 Hece, [z ] P (z) Q(z) = ( ( ) ( )) d A, + A,2 + + A,d α + ( ( )) ( ( )) + d2 + A 2, + + A 2,d2 α2 dr + + A r, + + A r,dr α r So we have just proved oe directio of the followig theorem. follows easily by workig backwards. The other directio Theorem 4.9. If the sequece {a } 0 has a ratioal geeratig fuctio P (z) Q(z) with deg(p (z)) < deg(q(z)), the a = P ()α + P 2 ()α2 + + P r ()αr, where α,..., α r are the roots of Q(z) ad each P i () is a polyomial of degree d i, where d i is the multiplicity of the root αi. Coversely, every sequece with a geeral term of this form has a ratioal geeratig fuctio. 3 I geeral, we will assume that deg(p (z)) < deg(q(z)); if this is ot the case, divide P (z) by Q(z) ad observe that by modifyig a fiite umber of terms i our sequece we ca assume that the geeratig fuctio is P (z)/q(z) with deg(p (z)) < deg(q(z)). 28

29 Let us see how Theorems 4.8 ad 4.9 work. The advatage is that we do ot have to carry out all the steps. Example. Cosider the recurrece a +3 = 4a +2 5a + + 2a, with iitial coditios a 0 = 2, a = 2, a 2 = 3. Sice this is a liear recurrece with costat coefficiets, we kow that its geeratig fuctio is ratioal ad it is of the form P (z) 4z + 5z 2 2z 3, where P (z) is a polyomial of degree at most 2. Now we apply the theory of ratioal geeratig fuctios to fid a closed formula for a. First we eed the roots of the deomiator, which are ad 2, with multiplicities 2 ad, respectively. The the a are of the form P () + P 2 ()2, where P is a polyomial of degree i ad P 2 is a costat. Hece, a = (A+B)+C2, for some costats A, B, ad C. We fid these costats by usig the iitial coditios. A + C = 2 A + B + 2C = 2 A + 2B + 4C = 3 This system gives A = C = ad B =. Hece, a = A o-liear recurrece: Catala umbers Cosider the followig path coutig problem. We have two types of steps: U = (, ) ad D = (, ). I how may ways ca we go from (0, 0) to (2, 0) usig these steps ad without crossig the horizotal axis? These paths are usually called Dyck paths. See Figure 2. (0,0) (8,0) Figure 2: A Dyck path. Let us deote by C the umber of such paths. The first values of C ca be computed by had (Figure 3). Notice that C 0 = sice there is oe way of goig from (0, 0) to (0, 0). We ow look for a way of decomposig the paths so that we ca obtai a recurrece (see Figure 4). Cosider the first poit where the path returs to the horizotal axis; this 29

30 C = 0. C = C = 2 2 C = 5 3 Figure 3: Dyck paths of legth at most 6. poit is of the form (2i, 0) for some i with i. Now our path is split ito two smaller paths, oe of legth 2i ad oe of legth 2( i). Observe that the first path, before the first retur, ca be decomposed as UP D, where P is a path that does ot go below height. Hece the umber of paths of legth 2 whose first retur is at (2i, 0) is C i C i. Sice the first retur ca be i ay poit of the form (2i, 0) for i betwee ad, we have that which is valid for. C = C 0 C + C C C C 0 = C i C i, i=0 (0,0) (8,0) (2i, 0) C i C i Figure 4: Decompositio of a Dyck path. We chage idices so that the recurrece is valid for 0: C + = C i C i, ad ow we take geeratig fuctios: C(z) z i=0 30 = C(z) 2.

31 Solvig this quadratic equatio o C(z) we get ± 4z. 2z Which of the two ( solutios are we iterested i? Observe that the expasio of the term 4z is /2 ) 0 ( 4) z = +. Therefore, if we take the + sig above, the first term of C(z) would be /z, ad this is ot a power series. Hece we have to take the sig. So, C(z) = 4z. 2z Now, Now we wat to extract the cofficiet of z above to have a closed formula for C. [z ]C(z) = 2 [z+ ] 4z = ( ) /2 2 ( 4)+ + ( ) /2 = ( + )! = ( ) 4 (2 )! ( + )!( )!, ad fially C = [z ]C(z) = ( ) (2 )! ( 4)+ 2 4 ( + )!( )! = = ( ) 2. + The umbers C are called Catala umbers (although they first appeared i Euler s work). The first Catala umbers are,, 2, 5, 4, 42, 32, 429, 430,... We will see that Catala umbers are oe of the most ubiquitous i combiatorics sice they cout a very large variety of objects. The iterested reader will fid a challege i exercise 6.9 of Staley s Eumerative Combiatorics, Volume 2, where 66 differet objects eumerated by Catala umbers are give. Observe fially that the geeratig fuctio for Catala umbers is ot ratioal. Geeratig fuctios that satisfy a polyomial equatio are called algebraic. 4.4 The geeratig fuctio for iteger partitios Amog all quatities we have studied, the umber of partitios p() seems to be oe of the most out of reach. But its geeratig fuctio will prove to us more tractable. Our goal is to fid 0 p()z. Let us start by a simple case. Let p k () be the umber of partitios of all whose parts are k or less. We study first the GF for p (). Well, there is just oe partitio of all whose parts are, hece 0 p ()z = z. 3