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Chapter 5 Numbers that count（第2页）

&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。