NICK DUFRESNE. 1 1 p(x). To determine some formulas for the generating function of the Schröder numbers, r(x) = a(x) =

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1 AN INTRODUCTION TO SCHRÖDER AND UNKNOWN NUMBERS NICK DUFRESNE Abstract. I this article we will itroduce two types of lattice paths, Schröder paths ad Ukow paths. We will examie differet properties of each, ad attempt to relate the two sets i a few differet ways.. Itroductio to the Schröder ad Ukow umbers Defiitio. A Schröder path is a lattice path that starts o the x-axis, eds o the x-axis, ever goes below the axis ad is costructed by the steps, ),, ) ad 2, 0). We will let R deote the umber of Schröder paths of legth 2. Defiitio 2. A peak is a up-step,, ), followed immediately by a dow-step,, ). We will say that the height of a peak is k if the top of the peak is at height k i the path. Defiitio 3. A set of umbers, which we will call the Ukow umbers, cout the umber of Schröder paths which have o peaks at height oe, ad o horizotal edges at height zero. We will let U deote the th Ukow umber, which couts such Schröder paths of legth 2. These types of paths will be referred to as Ukow paths. Defiitio 4. We will call a o-empty path prime i some set of paths S if it caot be factored ito smaller paths p ad p 2 such that both p ad p 2 are i S. 2. Some geeratig fuctios It is kow see []) that if ax) is a geeratig fuctio that couts some set of paths S that ca all be uiquely factored ito primes, ad if px) is the geeratig fuctio that couts the prime paths i S the ax) = px). To determie some formulas for the geeratig fuctio of the Schröder umbers, rx) = R x, lets look at the prime Schröder paths. We will cosider the primes of the Schröder paths to be ay path which starts ad eds o the x-axis ad ever touches it ibetwee. It is clear that a path of this type ca ot be factored further. We will weight each step i the geeratig fuctio for Schröder paths by x. We kow that a sigle horizotal step has legth 2 ad is prime. So the horizotal step is couted by x. Also, it is clear that a up-step followed by a arbitrary Schröder path of legth 2 2 followed by a dow-step is prime path of legth 2. It should be easy to covice yourself that these are the oly prime paths. This iterpretatio leads us to the geeratig fuctio of the prime Schröder paths px) = x xsx). Ad fially this gives us that Date: 5//2003.

2 2 NICK DUFRESNE We ca rearrage this to get that rx) = x xrx). 2.) rx) = xrx) xrx) ) Solvig 2.2) for rx) usig the quadratic formula ad selectig the solutio that gives a power series we get that rx) = x 6x x ) 2x Usig the same approach as for the Schröder umbers we ca discover some formulas for the geeratig fuctio for the Ukow umbers ux) = U x. We kow that the prime Ukow paths caot have ay horizotal steps o the x-axis or ay peaks of height. It is ot too difficult to see that the oly type of prime Ukow path is of the form up-step followed by a o-empty Schröder path followed by a dow-step. That is, a prime path of legth 2 is a up-step followed by a Schröder path of legth 2 2 followed by a dow-step, which is couted by R x sice there is a uique prime for every Schröder path of legth 2 2. This gives us that the geeratig fuctio for the primes is px) = =2 R x, or equivaletly px) = xrx) ). Usig the secod versio we see that this gives us ux) = xrx) ). 2.4) If we substitute 2.3) i for rx) we fid that ux) = 3x 6x x ) 2x3 2x) 3. Paths with a eve umber of peaks I this sectio we will see that half of the Ukow paths of legth 2 have a eve umber of peaks. To prove this lets cosider the subset S of the paths couted by U where we cosider oly paths with a eve umber of peaks. Now we eed to show that S is couted by U 2. To do this lets cosider ay path p i S. Lets take p to be the path of p where all peaks i p are replaced by horizotal steps. We kow that p is still a path couted by U, because if ay of the ew horizotal steps were at height zero that would imply that p had a peak at height. Now, lets say that p has horizotal steps. We ca replace ay umber of horizotal steps i p with peaks a up-step followed by a dow-step) ad the result is a path i U. This tells us that there are k) paths with k peaks resultig from p. So if is eve the umber of paths with a odd umber of peaks resultig from p is ) 3) ) which is equal to 0) 2) ) which is the umber of paths resultig from p with a eve umber of peaks. Ad if is odd similarly the umber of odd peaks resultig from p is ) 3) ) = 0) ) 2 ) which is agai the umber of paths with a eve umber of peaks resultig from p. Sice every path couted by U ca be reduced ito a path with oly horizotal steps, ad every path of this type results i a equal umber of paths with ad odd umber of peaks ad eve umber of peaks, we have that S is couted by U 2.

3 AN INTRODUCTION TO SCHRÖDER AND UNKNOWN NUMBERS 3 Sice the Schröder paths have the same structure as the Ukow paths the same argumet applies for them. So we have that the umber of Schröder paths of legth 2 with a eve umber of peaks is R Paths with a iitial ru of up-steps of eve legth Defiitio 5. A ru is a series of cosecutive steps all of the same type. Similar to the previous sectio, we ca show that the umber of Ukow paths with a iitial ru of eve legth is couted by 2 U. To see this lets cosider the followig fuctio f from Ukow paths with a eve umber of iitial up-steps U eve to Ukow paths with a odd umber of iitial up-steps U odd. Take u to be some path i U eve with a iitial ru of legth 2k. We kow that the step followig the iitial ru must be either a dow-step or a horizotal step otherwise the step would be icluded i the ititial ru). If the step followig the iitial ru is a horizotal step, we replace it with a up-step followed by a dow-step a peak) to get u. Clearly this leaves u with a iitial ru of legth 2k so it is i U odd. If the step after the iitial ru is a dow-step we take u to be u where we replace the previous up-step ad the dow-step with a horizotal step. Note that this will ot place a horizotal step o the x-axis sice the iitial ru must have legth at least 2. This will leave u with a iitial ru of legth 2k so it will be i U odd. This process is uique so it is oe-to-oe. Also we ca see that applyig f to fu) gives back u, so for ay u i U odd we have fu ) is i U eve by a similar argumet as before) ad ffu )) = u. So f is oto. This implies that we have foud a bijectio betwee Ukow paths with a iitial ru of eve legth ad Ukow paths with a iitial ru of odd legth. So half of U couts U eve. Agai, because the Schröder paths have a similar structure to the Ukow paths this argumet holds for them also. So 2 R also couts Schröder paths with a iitial ru of eve legth. 5. A Recurrece Relatio for The Schröder Numbers From 2.3) we kow that rx) = R x = x 6x x 2. 2x For ease of calculatio let ax) = 2xrx). The a x) = ax)) x = 2 3 x) 6x x 2 6x x 2 Multiplyig a x) by it deomiator we see we are left with somethig similar to ax). We fid that 6x x 2 )a x) 3 x)ax) = 2 2x. 5.)

4 4 NICK DUFRESNE From this differetial equatio we ca derive a simple recurrece relatio for the Schröder umbers. To do this we should ote that the derivative of the geeratig fuctio rx) is r x) = R x. Let us defie R to be zero for less tha zero. Usig this fact ad that ax) = 2R x ad a x) = 2 )R x we ca ow write equatio 5.) to be 2x 2 = 6x x 2 ) 2 )R x 3 x) 2R x = = 2 )R x 2 )R x 2 )R x 2 2 )R x 6R x 2R x 2 )R 2 x 6R x 2R x 2 2R 2 x Now lookig at the coefficiets of x where > ) i this equatio we see that o the left side we get zero, ad o the right side we get [x ] 2 )R x [x ] 2R x [x ] 2 )R 2 x [x ] 6R x [x ] 2R 2 x = 2 2)R 2R 2 2)R 2 6R 2R 2 Now equatig both sides, dividig by 2 ad simplifyig we fid that From 2.5) we have )R = 6 3)R 2)R 2 5.2) 6. A Recurrece Relatio for The Ukow Numbers ux) = U x = 3x 6x x 2 6x 4x 2.

5 AN INTRODUCTION TO SCHRÖDER AND UNKNOWN NUMBERS 5 Agai, to make the calculatio a little easier we will take bx) = 6x 4x 2 )ux). Note that this makes ad bx) = 6U x 4U 2 x 6.) = =2 b x) = 6 )U x From the closed form of the geeratig fuctio ux) we see that 4 )U x 6.2) = b x) = 3 8x 3x 2 3 x) 6x x 2 This looks very similar to what we foud previously for the Schröder umbers. We fid that bx) satisfies differetial equatio 6x x 2 )b x) 3 x)bx) = 6 0x. We will defie U to be zero for less tha zero. Next substitutig i equatios 6.) ad 6.2) we fid that 6 0x = 6 )U x 4 )U x 36U x 24U 2 x 6 )U 2 x Now equatig the coefficiets of x i this equatio we fid that 4 )U 3 x. 3 3)U = 6 )U 9U 2 2 4)U ) 7. Combiig the Schroder ad Ukow Numbers We would like to be able to express the Schröder umbers i relatio to the Ukow umbers. Usig 2.3) ad 2.5) we see that 3 2x)ux) = rx) 2 Usig the same method i the previous two sectios to express relatios betwee coefficiets of x i both sides of the equatio we fid that R = 3U 2U 7.)

6 6 NICK DUFRESNE 8. Combiatorial proof of Equatio 7.) To see that R = 3U 2U we will fid it easier to prove that R U = 2U 2U. Let us take S to be the set of paths couted by R U. We kow that the Ukow umbers cout Schröder paths with o horizotal steps at height 0 ad o peaks at height. This would imply that S couts that umber of Schroder paths of legth 2 with at least oe horizotal step at height 0 or peak at height. Cosider the last occurrece of either i the path. Let S be the paths that have the last occurrece as a horizotal step o the x-axis. Let S be the paths that have the last occurrece as a peak at height. Lets let T be the set of paths couted by 2U 2U. Ad let T be oe half of the paths couted by U U, ad T be the other half couted by U U makig each path distict). Now cosider the followig fuctio f. We take a path s i S. If s is i S the we remove the last horizotal step at height 0 i s ad isert a up-step at the start of the path ad a dow-step where we remove the horizotal step to get fs). Figure. Example s S to s T This leaves o more horizotal steps at height 0 or peaks at height sice they will all be moved up oe level. So fs) will be a path couted by U sice it is of legth 2 still uless the path before the horizotal step was empty, i which case we are left with a peak of height at the start. We ca chop the peak off ad the remaiig path is couted by U sice it has o peaks at height or horizotal steps at height 0 ad has legth 2 2. So fs) is i T. Now if s is i S we remove the last peak ad place the up-step of the peak at the begiig of s ad the dow-step remais where it is to get fs). If s is o-empty before the peak the this will be couted by U ad if ot it will be couted by U. Agai s is i T. We see that ay path fs) i T is uiquely determied by s sice it is s uiquely rearraged. So this tells us that f is oe-to-oe. We already have see that U couts Ukow paths of legth 2 2 with a peak attatched to the frot. So for q i T, we just take the first up-step ad dow-step that touch the x-axis ad we remove them. Where the dow-step was, we put either a horizotal step which correspods to a path i S if q is i T or we put a peak of height which correspods to a path i S if q is i T. So we kow every path i T comes from some path i S. So f is oto. This bijectio tells us that R U = 2U 2U. 9. A fial equatio relatig the Ukow ad Schröder umbers Usig what we foud rx) to be i 2.3) ad ux) to be i 2.5) we ca verify that 2 We ca rewrite this to be ux) x 2 2x 2x 2 xrx)) =. 9.) 2 ux) x 2 = 2x 2x 2 xrx). 9.2)

7 AN INTRODUCTION TO SCHRÖDER AND UNKNOWN NUMBERS 7 Usig 9.2) ad defiig R to be zero for less tha zero we fid that 2 U 2x = 2x 2x 2 R x. 9.3) We ca ow fid a iterpretatio of the paths that U 2 2x couts based o the primes give o the right side of equatio 9.3). We see that the prime paths are very similar to the prime paths of the Schröder umbers except we have ad additioal x ad a additioal 2x 2. Let u deote a up-step ad d deote a dow-step ad h deote a horizotal step. We ca take the prime Schröder paths ad add i a additioal prime of legth 2, du, ad two additioal primes of legth 4, dduu ad dhu. Figure 2. Additioal Primes Now we see that ay path s costructed of these primes will ever go below 2 below the x-axis. By addig o two up-steps at the begiig of such a path ad two dow-steps at the ed to get s we have a path of legth 2 4 that will be a Ukow path of legth 2 4. We see this is true sice if there is a horizotal step o the x-axis i s the there is a horizotal step at 2 below the x-axis i s, which is impossible from the primes we have chose. Also a peak at height i s is impossible from the primes. Figure 3. Example s to s The path s has the property that ay horizotal step at height must be preceeded by a dow step ad followed by a up-step caused by the prime of the form dhu. Also, if s touches the axis at ay poit ot the startig or edig poit) the the poit must be preceeded by two dow steps ad followed by two up steps, which is caused by the prime dduu. What 9.3) tells us is that Ukow paths with these properties of legth 2 4 are exactly half of the paths couted by U 2. Refereces [] Nick Dufrese, Geeratig Fuctios for Lattice Paths i a Free Mooid, March, 2003