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&esthenthFibonaumberandwefixf1=f2=1.Wecallsuulathatdefineseachmemberofasequespredecessorsareorarecerelation.
Howdoesthissequewasfirstintrodu1202byLeonardoofPisa，betterknownasFibonatheformofhiscelebratedRabbitProbleAfemalerabbitisborwomourityaergivesbirthtoadaughtereath.ThenumberoffemalerabbitswehaveatthebegihistheheFibonaumbers，forthereis1rabbitatthebeginnimonth，aatthestartofthethirdmonthshegivesbirthtoadaughtersowethes.hshehasan3ahafterthatwehave5buhmotheradaughterareehebegihthereafter，thenumberhtersequalsthenumberoffemaleswehadtwomonthsago，asonlytheyareoldenoughtobreed.Itfollowsthatthenumberoffemaleswehaveatthestartofeachsubsequehetotalofthepreviousmonth（Fibonacciísrabbitsareimmortal）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，anditarisesquiteofitseometrigsthatlookaworldawayfromFibonacci’srabbits.Forexample，τistheratioofthediagonalularpentagontoitsside（seeFigure4）.Eaeetsaapointthatdivideseatotwopartsthatarethemselvesiioτ:1.Pairssidesaingdiagonalsformthefoursidesofarhombus（a‘square’parallelogram）ABCDasshonalscross，theyformasmalleriagon.
&agonandtheGle
Anotherwaythisvaluationofτisthroughitsso-tiion，whichtiesτdirectlytotheFibonaumbers，andlorethisideainChapter7.
Inthelongrun，theFibonacebehaveslikeageressioheGoldenRatio.ItisthispretherwithitssimpleruleofformationthatcausestheFibooarisesopersistently.
StirlingandBellnumbers
&hebis，theStirlieingproblemsawovariables，nairlingnumberS（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.
5.Stirling&#le
&heisfyareeaningthateaberelatedtoearlierohearray.Ihthebis，eagnumberbeobtaihetwoaboveit，butitisnotsimplythesuWhatismore，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.
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（n，lyfromthedefinitionasfollows.Anarbitrarypartitiointoafirstsetaisdescribedbyabinarystrihhepresenceofa1indicatespresesetanda0intheseasimilaredthatthenumberofsubsetsofais2herefore2nsuchorderedpairsofsets.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）.OryregardingpartitionsisthattheitionsofnintompartsisequaltotheitionsofninwhichthelargestpartisOnewayofseeingthatthisistrueisthroughtheFerrar’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，thehailstonenumbersareintriguingastheyarealsodenedrecursivelybuthavemoreofaavourofthealiquotsequewemetihefollowiiongoesbyseveralzAlgorithm，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，ahasbeenveriedforallnupbeyondamillionmillion.Thingsaredifferentifyouddlewiththerules:forinstance，repla+1by3sinacycle:
7→20→10→5→14→7→….
Thesequenbersthatarisefromthesecalsbehavelikehailstotheyriseaicallyperiodbuteventually，itseems，alwayshitthegrou1000integers，morethaonemaximumheightof9232beforegto1.Thisenonintoapowerof2，fortheyareexaumbersthatcauseyouthtdowntogrouengas.
Allsortsfeaturesbedisgraphsandplotsbasedoonesequeofotherchaotisthatariseinmathsandphysig‘hailstonenumbers'intoyourfavouritesearewillprovideyouwithawealthofinformatioriguiimesspeculative，butgenerallyinclusive.
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