A Stirling Encounter with Harmonic Numbers

Transcription

1 VOL 75, NO, APRIL A Stirlig Ecouter with Haroic Nubers ARTHUR T BENJAMIN GREGORY O PRESTON Harvey Mudd College Clareot, CA bejai@hcedu gpresto@hcedu JENNIFER J QUINN Occidetal College 1600 Capus Road Los Ageles, CA jqui@oxyedu Haroic ubers are defied to be partial sus of the haroic series For 1, let H The first five haroic ubers are H 1 1, H 3/, H 3 11/6, H 4 5/1, H 5 137/60 For coveiece we defie H 0 0 Sice the haroic series diverges, H ca get arbitrarily large, although it does so quite slowly For istace, H 1,000, Haroic ubers eve appear i real life If you stack -ich log playig cards to overhag the edge of a table as far as possible, the axiu distace that cards ca hag off the edge of the table is H [5] For exaple, 4 cards ca be stacked to exted past the table by just over iches, sice H 4 5/1 Haroic ubers satisfy ay iterestig properties For oegative itegers ad, we list soe idetities below: 1 k 1 k ( k 1 H k H (1) ( k ) H k ) 1 k ( )( H 1 ) () ( ) (H H ) (3) Although all of these idetities ca be proved by algebraic ethods (see [5]), the presece of bioial coefficiets suggests that these idetities ca also be proved cobiatorially A cobiatorial proof is a coutig questio, which whe aswered two differet ways, yields both sides of the idetity Cobiatorial proofs ofte provide ituitive ad cocrete explaatios where algebraic proofs ay ot For exaple 1 k k! 1 is a stadard exercise i atheatical iductio But to a cobiatorialist this idetity couts perutatios i two differet ways The right side couts the uber of ways

2 96 MATHEMATICS MAGAZINE to arrage the ubers 1 through, excludig the atural arrageet 1 3 The left side couts the sae quatity by coditioig o the first uber that is ot i its atural positio: for 1 k 1, how ay arrageets have k as the first uber to differ fro its atural positio? Such a arrageet begis as 1 3 k 1 followed by oe of k ubers fro the set { k + 1, k +,,} The reaiig k ubers (ow icludig the uber k) ca be arraged k! ways Thus there are k k! ways for k to be the first isplaced uber Suig over all feasible values of k yields the left side of the idetity Although H is ever a iteger for > 1[5], it ca be expressed as a ratioal uber whose uerator ad deoiator have cobiatorial sigificace Specifically, for 0 we ca always write H p as a (typically oreduced) fractio where p is a oegative iteger Now p 0 H 0 0 For 1, H H 1 + 1/ leads to Hece for 1, p p 1 ( 1)! + 1 p 1 + ( 1)! (4) p p 1 + ( 1)! (5) The cobiatorial iterpretatio of these ubers is the topic of the ext sectio Stirlig ubers For itegers k 1, let [ k] deote the uber of perutatios of eleets with exactly k cycles Equivaletly [ k] couts the uber of ways for distict people to sit aroud k idetical circular tables, where o tables are allowed to be epty [ k] is called the (usiged) Stirlig uber of the first kid As a exaple, [ 3 ] 3sice oe perso ust sit aloe at a table ad the other two have oe way to sit at the other table We deote these perutatios by (1)(3), (13)(),ad(1)(3) We ca copute the ubers [ k] recursively Fro their defiitio, we see that for 1, [ ] ( 1)!, 1 sice the arrageet (a 1 a a 3 a ) is the sae as arrageets (a a 3 a a 1 ) ad (a 3 a 4 a 1 a ) ad so o Now for k, we will see that [ ] [ ] [ k k 1 k ] (6) O the left, we are directly coutig the uber of ways to seat + 1 people aroud k circular tables O the right we cout the sae thig while coditioig o what happes to perso + 1 If + 1 is to be aloe at a table, the the reaiig people ca be arraged aroud k 1tablesi [ k 1] ways If + 1 is ot to be aloe, the we first arrage 1 through aroud k tables (there are [ k] ways to do this); for each of these cofiguratios, we isert perso + 1 to the right of ay of the already-seated

3 VOL 75, NO, APRIL people This gives us [ k] differet perutatios where + 1 is ot aloe Suig gives equatio (6) Notice that whe k, equatio (6) becoes [ ] [ ] ( 1)!, (7) which is the sae as recurrece (5) with p [ ] +1 Sicep1 1 [ ], it follows that for all 1, p [ ] +1 Cobiig with the defiitio of p i (4) gives THEOREM 1 For 1, H 1 [ ] + 1 Next we show how to cout Theore 1 directly without relyig o a recurrece First we set soe otatioal covetios Let T deote the set of arrageets of the ubers 1 through ito two disjoit, oepty cycles Thus T [ ] We always write our cycles with the sallest eleet first, ad list the cycles i icreasig order accordig to the first eleet For exaple, T 9 icludes the perutatio (18574)(396), but ot (195)(487)(36) or (13)(4567)(8)(9) By our covetio, the cycle cotaiig 1 is always writte first; cosequetly we call it the left cycle The reaiig cycle is called the right cycle All perutatios i T are of the for (a 1 a a j )(a j+1 a ),where1 j 1, a 1 1, ad a j+1 is the sallest eleet of the right cycle For a purely cobiatorial proof of Theore 1 that does ot rely o a recursio, we ask, for 1 k, how ay perutatios of T +1 have exactly k eleets i the right cycle? To create such a perutatio, first choose k eleets fro {,, + 1} ( ( k) ways), arrage these eleets i the right cycle ((k 1)! ways), the arrage the reaiig k eleets i the left cycle followig the uber 1 (( k)! ways) Hece there are ( k) (k 1)! ( k)! /k perutatios of T+1 with k eleets i the right cycle Sice T +1 has [ +1 [ + 1 ] perutatios, it follows that ] k H, as desired Aother way to prove Theore 1 is to show that for r + 1, there are r 1 perutatios i T +1 that have r as the iiu eleet of the right cycle Here, the perutatios beig couted have the for (1 )(r ) where eleets 1 through r 1 all appear i the left cycle, ad eleets r + 1 through + 1 ca go i either cycle To cout this, arrage eleets 1 through r 1 ito the left cycle, listig eleet 1 first; there are (r )! ways to do this Place eleet r ito the right cycle Now we isert eleets r + 1 through + 1, oe at a tie, each iediately to the right of a already placed eleet I this way, eleets 1 ad r reai first (ad sallest) i their cycles Specifically, the eleet r + 1 ca go to the right of ay of the eleets 1 through rnext,r + ca go to the right of ay of the eleets 1 through r + 1 Cotiuig i this way, the uber of ways to isert eleets r + 1 through + 1isr(r + 1)(r + ) /(r 1)! This process creates a perutatio i T +1 with r as the sallest eleet i the right cycle Thus, there are (r )! (r 1)! r 1

4 98 MATHEMATICS MAGAZINE such perutatios Sice T +1 has [ ] +1 perutatios, ad every perutatio i T+1 ust have soe sallest iteger r i the right cycle, where r + 1, we get [ ] r 1 1 k H r A alterate way to see that /(r 1) couts perutatios of the for (1 )(r ) is to list the ubers 1 through + 1 i ay order with the provisio that 1 be listed first There are ways to do this We the covert our list 1 a a 3 r a +1 to the perutatio (1 a a 3 )(r a +1 ) by isertig paretheses This perutatio satisfies our coditios if ad oly if the uber r is listed to the right of eleets, 3,,r 1 This has probability 1/(r 1) sice ay of the eleets, 3,,r have the sae chace of beig listed last aog the Hece the uber of perutatios that satisfy our coditios is /(r 1) Algebraic coectio The Stirlig ubers ca also be defied as coefficiets i the expasio of the risig factorial fuctio [3]: [ ] x(x + 1)(x + ) (x + 1) x (8) Usig this defiitio, Theore 1 ca be derived algebraically by coputig the x coefficiet of x(x + 1)(x + ) (x + ) To show that this algebraic defiitio of Stirlig ubers is equivalet to the cobiatorial defiitio, oe typically proves that both satisfy the sae iitial coditios ad recurrece relatio However, a ore direct correspodece exists [1], which we illustrate with a exaple By the algebraic defiitio, the Stirlig uber [ ] 10 3 is the coefficiet of x 3 i the expasio x(x + 1)(x + ) (x + 9) The cobiatorial defiitio says [ ] 10 3 couts the uber of ways that eleets 0, 1,,,9 ca sit aroud 3 idetical circular tables Why are these defiitios the sae? Each ter of the x 3 coefficiet is a product of seve ubers chose fro aog 1 through 9 Surely this ust be coutig soethig What is a ter like coutig? As illustrated i FIGURE 1, this couts the uber of ways eleets 0 through 9 ca seat theselves aroud 3 idetical tables where the sallest eleets of the tables are the issig ubers 0, 4, ad 7 To see this, we pre-seat ubers 0, 4, 7the seat the reaiig ubers oe at a tie i icreasig order The uber 1 has just oe optio sit ext to 0 The uber the has two optios sit to the right of 0 or sit to the right of 1 The uber 3 ow has three optios sit to the right of 0 or 1 or The uber 4 is already seated Now uber 5 has five optios sit to the right of 0 or 1 or or 3 or 4, ad so o A geeral cobiatorial proof of equatio (8) ca also be doe by the precedig (or should that be pre-seatig?) arguet With this uderstadig of the iteractios betwee haroic ad Stirlig ubers, we ow provide cobiatorial explaatios of other haroic idetities 1 Recoutig haroic idetities I this sectio, we covert idetities (1), (), ad (3) ito stateets about Stirlig ubers ad explai the cobiatorially We view each idetity as a story of a coutig proble waitig to be told Each side of the idetity recouts the story i a differet, but accurate way Both of our cobiatorial proofs of Theore 1 were

5 VOL 75, NO, APRIL Figure 1 How ay ways ca the ubers 1,, 3, 5, 6, 8, 9 seat theselves aroud these tables? obtaied by partitioig the set T+1 accordig to the size of the right cycle or the iiu eleet of the right cycle, respectively I what follows, we shall trasfor haroic equatios (1), () ad (3) ito three Stirlig uber idetities, each with o the left-had side The right-had sides will be cobiatorially explaied by partitioig T accordig to the locatio of eleet, the largest of the last t eleets, or the eighborhood of the eleets 1 through Our first idetity, after applyig Theore 1, ad re-idexig ( : 1) gives us I DENTITY 1 For ( )! k + 1 ( 1)! + k! To prove this cobiatorially, we ote that the left side of the idetity,, couts the uber of perutatios i T O the right, we kow fro our secod cobiatorial proof of Theore 1, that ( 1)! couts the uber of perutatios i T where the uber appears i the right cycle It reais to show that the suatio above couts the uber of perutatios i T where is i the left cycle Ay such perutatio has the for (1 a1 a a k b1 b b j 1 )(b j bk ), for soe 1 k ad 1 j k We assert that the uber of these perutatios with exactly k ters to the right of is give by the kth ter of the su

6 100 MATHEMATICS MAGAZINE To see this, select a 1, a,,a k fro the set {3,,} i ay of ( )! /k! ways Fro the uchose eleets, there are [ ] k+1 ways to create two oepty cycles of the for ( b 1 b j 1 )(b j b k ) where 1 j k Multiplyig the two couts gives the kth ter of the su as the uber of perutatios i T with exactly k ters to the right of, as was to be show We apply a differet cobiatorial strategy to prove the ore geeral equatio (), which, after applyig Theore 1 ad re-idexig ( : 1, : t 1, ad k : k ), gives us IDENTITY For 1 t 1 [ ] ( 1)! [ ] k 1 ( 1 t)! + t t (k 1 t)! kt+1 The cobiatorial proof of this idetity requires a ew iterpretatio of ( 1)! /t For 1 t 1, we defie the last t eleets of (1a a j )(a j+1 a ) to be the eleets a, a 1,a +1 t, eve if soe of the are i the left cycle For exaple, the last 5 eleets of (18574)(396) are6,9,3,4,ad7 We clai that for 1 t 1, the uber of perutatios i T where the largest of the last t eleets is aloe i the right cycle is ( 1)! /t Here, we are coutig perutatios of the for (1a a 1 )(a ),wherea is the largest of {a +1 t, a + t,,a 1, a } Aog all ( 1)! perutatios of this for, the largest of the last t eleets is equally likely to be aywhere aog the last t positios Hece ( 1)! /t of the have the largest of the last t eleets i the last positio Next we clai that for 1 t 1, the uber of perutatios i T where the largest of the last t eleets is ot aloe i the right cycle is the suatio i Idetity To see this, we cout the uber of such perutatios where the largest of the last t eleets is equal to k Sice the uber 1 is ot listed aog the last t eleets, we have t + 1 k To costruct such a perutatio, we begi by arragig ubers 1 through k 1 ito two cycles The isert the uber k to the right of ay of the last t eleets There are [ ] k 1 t ways to do this The right cycle cotais at least oe eleet less tha k, sok is ot aloe i the right cycle (ad could eve be i the left cycle) So that k reais the largest aog the last t eleets, we isert eleets k + 1 through, oe at a tie, to the right of ay but the last t eleets There are (k t)(k + 1 t) ( 1 t) ( 1 t)! /(k 1 t)! ways to do this Multiplyig the two couts give the kth ter of the su as the uber of perutatios where the largest of the last t eleets equals k, ad it is ot aloe i the right cycle; suig over all possible values of k, we cout all such perutatios Sice for ay perutatio i T, the largest of the last t eleets is either aloe i the last cycle, or it is t, ad this establishes Idetity Notice that whe t 1, Idetity siplifies to Idetity 1 Whe t 1, Idetity essetially siplifies to equatio (7) For our fial idetity, we covert equatio (3) to Stirlig ubers usig Theore 1 ad re-idexig ( : 1, : 1, ad k : t 1) This gives us IDENTITY 3 For 1 [ ] [ ] ( 1)! 1 ( 1)! + t ( t 1 1 ) ( 1)! ( )! ( t) To prove this idetity cobiatorially, we coditio o whether ubers 1 through all appear i the left cycle First we clai that for 1, the first ter o

7 VOL 75, NO, APRIL the right i the idetity couts the uber of perutatios i T that do ot have eleets 1,, all i the left cycle: For these perutatios, the eleets 1 through ca be arraged ito two cycles i ways Isert the reaiig eleets + 1 through, oe at a tie, to the right of ay existig eleet, fidig that there are ( + 1) ( 1) ( 1)! /( 1)! ways to isert these eleets Multiplyig the two couts gives the first ter of the right-had side To coplete the proof, we ust show that the suatio o the right couts the uber of perutatios i T where eleets 1 through are all i the left cycle To see this, we clai that for t 1, the suad couts the perutatios described above with exactly t eleets i the left cycle ad t eleets i the right cycle To create such a perutatio, we first place the uber 1 at the frot of the left cycle Now choose 1 of the reaiig t 1 t 1 spots i the left cycle to be ways to select these 1 spots assiged the eleets {,, } There are 1 ad ( 1)! ways to arrage eleets,, 1 i those spots For exaple, to guaratee that eleets 1,, 3, 4 appear i the left cycle of F IGURE, we select three of the five ope spots i which to arrage, 3, 4 The isertio of 5, 6, 7, 8, 9 reais Now there are ( )! ways to arrage eleets + 1 through i the reaiig spots, but oly oe out of t of the will put the sallest eleet of the right cycle at the frot of the right cycle Hece, eleets + 1 through ca be arraged i ( )!/( t) legal ways Multiplyig gives the uber of ways to satisfy our coditios for a give t, ad the total is give by the desired suatio Figure I T9, a perutatio with 1,, 3, 4 i a left cycle cotaiig exactly six eleets is created by first selectig three of the five ope spots, ad the arragig, 3, 4 i the Subsequetly, 5, 6, 7, 8, 9 will be arraged i the reaiig spots We have already oted that haroic ubers arise i real life A further occurrece arises i calculatig the average uber of cycles i a perutatio of eleets Specifically, T HEOREM O average, a perutatio of eleets has H cycles There are perutatios of eleets, of which k have k cycles Cosequetly, Theore says k k H, or equivaletly, by Theore 1,

8 10 MATHEMATICS MAGAZINE IDENTITY 4 For 1, [ ] [ ] + 1 k k The left side couts the uber of perutatios of {1,,} with a arbitrary uber of cycles, where oe of the cycles is distiguished i soe way For exaple (184)(365)(79), (184)(365)(79), ad(184)(365)(79) are three differet arrageets with k 3 The right side couts the uber of perutatios of {0, 1,,} with exactly two cycles It reais to describe a oe-to-oe correspodece betwee these two sets of objects Ca you deduce the correspodece betwee the followig three exaples? ito (184)(365)(79) (079365)(184) (184)(365)(79) (079184)(365) (184)(365)(79) ( )(79) I geeral, we trasfor the perutatio with eleets (C k )(C k 1 ) (C j+1 )(C j )(C j 1 ) (C )(C 1 ) (0 C 1 C C j 1 C j+1 C k 1 C k )(C j ) The process is easily reversed Give (0 a 1 a j )(b 1 b j ) i T +1, the right cycle becoes the distiguished cycle (b 1 b j ) The distiguished cycle is the iserted aog the cycles C k 1,C, C 1, which are geerated oe at a tie as follows: C 1 (the rightost cycle) begis with a 1 followed by a ad so o util we ecouter a uber a i that is less tha a 1 Assuig such a a i exists (that is, a 1 1), begi cycle C with a i ad repeat the procedure, startig a ew cycle every tie we ecouter a ew sallest eleet The resultig cycles (after isertig the distiguished oe i its proper place) will be a perutatio of eleets writte i our stadard otatio Hece we have a oe-to-oe correspodece betwee the sets couted o both sides of Idetity 4 Notice that by distiguishig exactly of the cycles above, the procedure above ca be easily odified to prove the ore geeral k [ k ]( k ) [ Likewise by distiguishig a arbitrary uber of cycles, the sae kid of procedure results i [ ] k ( + 1)! k k0 ] Beyod haroic ubers We have oly scratched the surface of how cobiatorics ca offer ew isights about haroic ubers Other cobiatorial approaches to haroic idetities are preseted by Presto [6] We leave the reader with a challege: A hyperharoic uber

9 VOL 75, NO, APRIL H (k) is defied as follows: Let H (1) H ad for k > 1, defie H (k) i1 H (k 1) i Now cosider the followig geeralizatio of idetity (1) fro The Book of Nubers by Coway ad Guy [4]: ( ) + k 1 H (k) (H +k 1 H k 1 ) k 1 Such a idetity strogly suggests that there ust be a cobiatorial iterpretatio of hyperharoic ubers as well Ad ideed there is oe [] You ca cout o it! Ackowledget We thak Michael Raugh, David Gaebler, Robert Gaebler, ad the referees for helpful coets, ad Greg Levi for the illustratios We are grateful to Jaet Myhre ad the Reed Istitute for Decisio Scieces for supportig this research REFERENCES 1 Robert Beals, persoal correspodece, 1986 Arthur T Bejai, David J Gaebler, ad Robert P Gaebler, A Cobiatorial Approach to Hyperharoic Nubers, preprit 3 Louis Cotet, Advaced Cobiatorics, D Reidel Publishig Copay, Bosto, Joh H Coway ad Richard K Guy, The Book of Nubers, Copericus, Roald L Graha, Doald E Kuth, ad Ore Patashik, Cocrete Matheatics, Addiso Wesley, Greg Presto, A Cobiatorial Approach to Haroic Nubers, Seior Thesis, Harvey Mudd College, Clareot, CA 001 Giraffes o the Iteret RICHARD SAMUELSON