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Chapter 5 Numbers that count（第2页）

Howdoesthissequewasfirstintrodu1202byLeonardoofPisa，betterknownasFibonatheformofhiscelebratedRabbitProblem.Afemalerabbitisborwomourityaergivesbirthtoadaughtereath.ThenumberoffemalerabbitswehaveatthebegihistheheFibonaumbers，forthereis1rabbitatthebeginnimonth，aatthestartofthethirdmonthshegivesbirthtoadaughtersowethes.hshehasan3ahafterthatwehave5buhmotheradaughterareehebegihthereafter，thenumberhtersequalsthenumberoffemaleswehadtwomonthsago，asonlytheyareoldenoughtobreed.Itfollowsthatthenumberoffemaleswehaveatthestartofeachsubsequehetotalofthepreviousmonth（Fibonacci’srabbitsareimmortal）plusthehemohereforetheruleofformationoftheFibonaumbersexactlymatchesthebreedingpatternofhisrabbits.

&hefactthatrealrabbitsdohistrivedfashion，Fibonaumbersariseinnatureiyofways，ingplantgrowth.Thereasonsforthisarewelluoodbutarerelatedtomoresubtleattributesofthesequeheso-calledGoldenRatio，aweareabouttointroduce.

&typesressioietricprogressionsihefirstse.AlthoughtheFibonaeitherofthese，itdoeshorisinglinkwiththelattertype.IfweformthesequenceofdiffereheFibonace，becauseofthewaythesequenceisdefi0，1，1，2，3，5，8，13，···，thatiswerecovertheFibonaexceptthistimebeginningat0.Thishappenspreciselybecauseofthewaythesequened:thedifferewosecutiveFibonaumbersistheoelypregbothioseethisalgebraically，subtra-1frombothsidesoftheFibonaccireceabove.）NoristhesequericprogressioioofsecutiveFibonaumbersisnott.Allthesame，whetheratioofsuccessivetermsweseethatitdoesseemtosettledowntoalimitihisablebehaviouroftheratioesaboutquitequickly，asweseeaswedivideeaberbyitspredecessor:

Butwhatisthemysteriousnumber，1.6180...，whichweseeemerging?ThisnumberτisknownastheGoldenRatio，aeofitseometrigsthatlookaworldawayfromFibonacci’srabbits.Forexample，τistheratioofthediagonalularpentagontoitsside（seeFigure4）.Eaeetsaapointthatdivideseatotwopartsthatarethemselvesiioτ:1.Pairssidesaingdiagonalsformthefoursidesofarhombus（a‘square’parallelogram）ABCDasshonalscross，theyformasmalleriagon.

&agonandtheGle

Anotherwaythisvaluationhitsso-tiion，whichtiesτdirectlytotheFibonaumbers，andlorethisideainChapter7.

Inthelongrun，theFibonacebehaveslikeageressioheGoldenRatio.ItisthispretherwithitssimpleruleofformationthatcausestheFibooarisesopersistently.

StirlingandBellnumbers

&hebis，theStirlieingproblemsawovariables，irlingnumberS（n，r）isthenumberofartitiooforblooblockempty，andtheorderoftheblodwithintheblocks，isimmaterial）.（StrictlythesearecalledStirlihesed.Thoseofthefirstkind，whicharerelated，ethingquitedifferehenumberofermuteorcycles.）Foriwithmembersa，b，bepartitiohreeblojustoneway:{a}，

{b}，

{c}，

intotwoblothreeways{a，b}，

{c};{a}，

{b，c}，

and{a，c}，

{b}，

andintoasinglebloewayonly:{a，b，c};itfollowsthatS（3，1）=1，S（3，2）=3andS（3，3）=1.Siofnmembersbepartitionedioeither1blotonblocks，wealwayshaveS（n，1）=1=S（n，n）.IfthetriairlierthefashionofPascal’sTriangle，wearriveatthearrayofFigure5，andwenowexplairiaed.

&heisfyareeaningthateaberelatedtoearlierohearray.Ihthebis，eagnumberbeobtaihetwoaboveit，butitisnotsimplythesum.Whatismore，therowsymmetrywesawiiglethatgehebisisirling’sTriangle.Forexample，S（5，2）=15butS（5，4）=10.Theruleofreceissimpleenough，however.Theentry90，forexample，isequalto15+3×25.Thisisihegeuation:tofihebodyle，takethetwoimmediatelyaboveit，aotheseultipliedbythehepositionintherowyouareat.（Thistime，uigle，startyourrowtat1.）Inasimilarway，theentryS（5，4）=10=6+4×1.ItisooftheruleinitalicsthatdiffersfromthatoftheArithmetigle.Thatisenough，however，tomakethestudynumberssiderablymoredifficulttothatofthebis.Forinstance，wederivedasimpleexpliulaforeaialtihefactorials.Similarly，thereisaformulaforthenthFibonaumberintermsofpowersoftheGoldenRatio，butnothingofthekisfnumbers.

5.Stirling’sTriangle

Thereceruleisnothardtoexplain.Wearguesimilarlytothatforthereforthebis，andbydoihereedabovethatisidentiexceptflemultiplier.Ioformapartitioofsizenintoryblocks，rotwodistinctways.Wemaytakethefirsthesetandpartitionitintor-1yblo-1，r-1）ways，andthefihesetwillthehbloatively，artitiohesetintoryblocks，whibedoneinS（n-1，r）ways，andthendewhichoftherblockstoplaalmemberoftheset，givingamultiplierofrtothatnumber.Hehat

S（n，r）=S（n-1，r-1）+rS（n-1，r）forn=2，3，···

Usingthisreula，wemaycalculateeaeoftheStirlingTriaheo.Forexample，puttingn=7aain:

S（7，5）=S（6，4）+5S（6，5）=65+5×15=65+75=140.

uteS（n，2）andS（lyfromthedefinitionasfollows.Anarbitrarypartitiointoafirstsetaisdescribedbyabinarystrihhepresenceofa1indicatespreseanda0intheseasimilaredthatthenumberofsubsetsofais2herefore2nsuchorderedpairsofsets.Sihereishebloapartitiohisofindtheitiois，givingthenumber2n-1.Fiosubtrathisioexcludethecasewhereosisempty;hen，2）=2n-1-1.Youcheckthatthisrepresentstheseddiagonallineofnumbers1，3，7，15，31，63，···runningfromthethttothebottomleftinFigure5.

ThesumofanyrowoftheArithmetiglegivesthedihenumberofsubsetsofasetofagivensize.Similarly，summihr’sTriahenumberofwaysasetofoblodthisiscalledthenthBellnumber.

Partitionnumbers

Ifohehesettobepartitioidsootbedistinguishedfromohenumberoflittingthewholeupintoblocksisamuchsmallerinteger，knowitioicularpartitionasasumofpositiveihardtoorder:forexample，1+1+1+1+1isoionof5andtherearesixothers，forresent5as1+1+1+2，1+2+2，1+1+3，2+3，1+4，orsimplyas5.Thereforethe5thpartitiohatparestothe5thBellnumber，whitheStirlingTriaobe1+15+25+10+1=52）.Thereisformulaforthenthpartitiohereisaplexone，whichisitselfbasedoifulapproximatioheIndiangeniusSrinivasaRamanujan（1887–1920）.

Oryregardingpartitionsisthattheitionsofnintompartsisequaltotheitionsofninwhichthelargestpartism.OnewayofseeingthatthisistrueisthroughtheFerrar’sgraph（diagram）ofthepartition，whiorethaioionasadingarrayofdotsinwhichtherowsarelistedbydegsize.

IntheexampleshowninFigure6reseitionedas5+4+4+2+1+1.hensarealsolistedindelefttht.Ifwereflectthearrayalongthediagfromtoplefttht，werecoverasedFerrar’sgraphasshown，whibeihepartition17=6+4+3+3+1.Asimilarrefleofthesedgraphreturnsyoutothefirstathetwpartitiooohissymmetryallowsustoseethatthenumbersofpartitionsoftwtypesareequal:thedualofapartitioninwhichm，say，isthelargesthetoprowhasmdots）isapartitionhichdstoapartitionintomnumbers.Forexample，theitionsof17ihereforeequalstheitionsof17inwhich6isthelargestoccurs.

6.Dualpartitionsof17=5+4+4+2+1+1=6+4+3+3+1

Hailstonenumbers

Althoughnotagtool，thehailstonenumbersareintriguingastheyarealsodefinedrecursivelybuthavemoreofaflavourofthealiquotsequewemetihefollowiiongoesbyseveralzAlgorithm，theSyra，orsometimesjustthe3issimplytheobservationthat，beginningwithahefollowingprocessalwaysseemstoendwiththenumber1.Ifniseveby2，whileifnisodd，replaceitby3n+1.Forexample，beginningwithherulesthroughthefollowingsequence:

7→22→11→34→17→52→26→13→40→20→10→5→16→8→4→2→1

Aureistrueforn=7，ahasbeenverifiedforallnupbeyondamillionmillion.Thingsaredifferentifyoufiddlewiththerules:forinstance，repla+1by3sinacycle:

7→20→10→5→14→7→···.

Thesequenbersthatarisefromthesecalsbehavelikehailstotheyriseaicallyperiodbuteventually，itseems，alwayshitthegrou1000integers，morethaonemaximumheightof9232beforegto1.Thisenonintoapowerof2，fortheyareexaumbersthatcauseyouthtdowntogrouengas.

Allsortsfeaturesbedisgraphsandplotsbasedoonesequeofotherchaotisthatariseinmathsandphysig‘hailstooyourfavouritesearewillprovideyouwithawealthofinformatioriguiimesspeculative，butgenerallyinclusive.