Prim2006 Prezentacija

International conference PRIM 2006HANKEL TRANSFORM OF NARAYANAPOLYNOMIALS AND GENERALIZED CATALANNUMBERSMarko D. Petković,Predrag M. Rajković,Department of MathematicsUniversity of NišSerbia.

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 21 IntroductionFirst we will define Hankel transform of a integer sequence and show some examples.Definition 1 The Hankel transform of a given sequence A = {a 0 , a 1 , a 2 , ...} is thesequence of Hankel determinants {h 0 , h 1 , h 2 , . . . } where h n = |a i+j−2 | n i,j=1 , i.e∣ ∣∣∣∣∣∣∣∣∣∣∣ a 0 a 1 · · · a na 1 a 2 · · · a n+1A = {a n } n∈N0 → h = {h n } n∈N0 : h n =. . . . . ... (1.1).∣ a n a n+1 · · · a 2nM.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 3Example 1 The Hankel transform of a Catalan sequence given by c(n) = 1is the sequence of all 1’s. Thus each of the determinants has value 1:n+1( 2nn)|1| ,∣1 11 2∣ ∣∣∣∣∣∣∣ 1 1 2∣ , 1 2 52 5 14, . . . (1.2)∣Example 2 Sequence of central binomial coefficients defined by a n = ( 2n) n has Hankeltransform h n = 2 n , i.e.∣ ∣∣∣∣∣∣∣ 1 2 61 2|1| = 1,∣ 6 20 ∣ = 2, 2 6 20= 4, . . . (1.3)20 70 252 ∣M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 4Example 3 In paper [3], A. Cvetković, P. Rajković and M. Ivković have proven thatHankel transform of sequence A005087 in On-Line Encyclopedia of Integer Sequences[11] defined by:a n = c(n) + c(n + 1) = 1n + 1( 2nn)+ 1n + 2( ) 2n + 2n + 1equals to the sequence A001906, i.e. bisection of Fibonacci sequence F(2n + 1).(1.4)We generalized previous result, i.e. computed the Hankel transform of the generalizedsequence a n (L) = c(n; L)+c(n+1; L), where c(n; L) is a sequence of generalizedCatalan numbers.M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 52 Hankel transform of Narayana PolynomialsIn this section we will present the metod for computing Hankel transform of thesequence of Narayana polynomials based on Krattenthaler formula in [6]. Similarmethod will be used also for generalized sequence from [3].• First we will find the real measure whose moments are values of Narayana polynomials• Then we will construct the sequence of orthogonal polynomials with respect tofound measure• Finally, from the three-terms recurrence relation we will derive Hankel transformin the closed form.M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 62.1 Narayana numbers and polynomialsWe will consider the sequence of the Narayana and shifted Narayana numbersN(n, k) = 1 ( )( ) n n, Ñ(n, k) = N(n + 1, k).n k k + 1To this sequence we can join the Narayana trianglesN = [ N(n, k) ] n,k∈N , Ñ = [ Ñ(n, k) ] n,k∈N .and the Narayana polynomialsa(n; r) =n∑k=0Ñ(n, k)r k , a 1 (n; r) =n∑k=0N(n, k)r k .It is valida(n; r) = a 1 (n + 1; r)(n ∈ N).M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 7Definition 2 For a given function y = f(x), f(0) = 0, the series reversion is thesequence {s k } such thatx = f −1 (y) = s 0 + s 1 y + · · · + s n y n + · · · ,where x = f −1 (y) is the inverse function of y = f(x).In the paper [1], P. Barry showed the next facts.Lemma 1 The series reversion of the next functions are Narayana and shifted Narayananumbersy = f(x) =x1 + (r + 1)x + rx 2 ⇒ f −1 (y) =y = g(x) =x(1 − rx)1 − (r − 1)x⇒ g −1 (y) =+∞ ∑n=0+∞ ∑n=0a(n; r)y n ,a 1 (n; r)y n .M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 8From the previous Lemma, we can easily derive the generating functions of thesequences a(n; r) and a 1 (n; r).Corolary 1 The generating functions of the sequences a(n; r) and a 1 (n; r) are givenby:√A(x, r) = −1 + (r + 1)x + (1 − (r + 1)x) 2 − 4rx 2A 1 (x, r) =A(x, r) − 1x2rx 2(2.5)M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 10It can be shown that an integral of F(z; r) is equal to:∫F(z; L) = F(z; L)dz = 1 []z 2 +(1+r−z)ρ(z; r)−2z(r+1) +l 1 (z; r), (2.9)4rWhere we denoted:ρ(z; r) = √ (z − r − 1) 2 − 4rl 1 (z; r) = ln (−(r + 1) + z + ρ(z; r))(2.10)We can notice that in the complex plane, function ρ(z; r) has two branch pointsz = ( √ r − 1) 2 and z = ( √ r + 1) 2 , and l 1 (z) has one more, z = r + 1.M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 11Now by choosing appropriate regular branches of ρ(z; r) and l 1 (z; r) we can find thelimits:⎧⎨√4r − (x − r − 1)2, x ∈ ( ( √ r − 1) 2 , ( √ r + 1) 2)lim Iρ(x + iy; r) =,y→0 + ⎩ 0 , otherwise.andlim Il 1 (x + iy; r) =y→0 +⎧⎪⎨⎪⎩√4r−(x−r−1) 2π + arctan, x ∈ ( ( √ r − 1) 2 , r + 1 )√x−(r+1)4r−(x−r−1) 2arctan, x ∈ ( r + 1, ( √ r + 1) 2)x−(r+1)0 , otherwise.Now to complete the proof we need to substitute these values into the formula forw(x).M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 122.3 Orthogonal polynomials w.r.t. the weight w(x)In the paper [6],C. Krattenthaler proved that Hankel transform h n of a sequencea n = ∫ R xn w(x)dx is given with the following relation h n = a n ∏ n−10 i=1 βn−i i.Coefficients β i are from the three-terms recurrence relation between monic orthogonalpolynomials with respect to the weight ω(x).Q n+1 (x) = (x − α n )Q n (x) − β n Q n−1 (x), (2.11)Lemma 2 Coefficients α n and β n , n = 0, 1, . . . in the three-term recurrence relation(2.11) with respect to weight function:⎧ √⎨ 4r−(x−r−1) 2, x ∈ ( ( √ r − 1) 2 , ( √ r + 1) 2) ;w(x) =2πr⎩ 0, otherwise.are given with:β 0 = 1 β n = r, n ≥ 1 α n = r + 1, n ≥ 0M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 13Proof. Second kind Chebyshev polynomials are orthogonal w.r.t the weight w (1) (x) =√ 1 − x2 .Q (1)n (x) = S n(x) = sin( (n + 1) arccos x )2 n · √1 − x 2Corresponding coefficients are:β (1)0 = π 2 , β(1) n= 1 4 , n ≥ 1 α(1) n = 0, n ≥ 0Let we introduce new weight function:w (2) (x) = √ 4r − (x − r − 1) 2 = w( 12 √ r x − r + 1 )2 √ r= w(ax + b)Using the transformation formulas from [4] we obtain new coefficients:β (2)0 = √ rπ, β (2)n= β(1) n= r, n ≥ 1 α(2)a2 n= α(1) n− b = r + 1, n ≥ 0a 2M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 14Finally by dividing weight function w (2) (x) with constant1π √ rwe haveβ (3)n= β(2) n= r, n ≥ 1, β(3) 0 = 1, α (3)n= α(2) n = r + 1which completes the proof.M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 152.4 The Hankel transform of {a(n; r)} n∈N0Now we are ready to prove the main theorem of this section:Theorem 2 The Hankel transform of the sequence a(n; r) ish(n; r) = r 2) (n .Proof. Using Krattenthaler formula we have:n−1∏h(n; r) = a(0; r) ni=1β n−iin−1∏= 1 ni=1r n−i = r (n 2) .M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 163 Hankel Transform of sum of consecutive generalizedCatalan numbersIn this section we will consider the generalized Catalan numbers and we will find theHankel transform of a sequence a n (L), the generalization of the sequence A005087.• First we will define the generalized binomial coefficients and generalized Catalannumbers and consider its basic properties.• Then we will derive the generating function of a n (L).• Finally we will find the Hankel transform similarly as in the case of Narayanapolynomials.M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 173.1 Definitions and basic propertiesDefinition 3 For a given sequence {b n } n∈N0 define the generalized binomial coefficientwith:T(n, k, {b m }) = ∑ ( )( ) k n − kb j .j jjAlso define the sequence of generalized Catalan numbers with:c(n; {b m }) = T(2n, n; {b m }) − T(2n, n − 1; {b m })It can be directly verified that holds T(n, k, {b m }) = T(n, n − k, {b m }).Example 4 For the sequence b m = 1, we have that T(n, k; {1}) = ( nk). This comesfrom the Vandermode convolution identity:( n=k)∑ ( )( ) k n − k.j jjIn that case also holds c(n) = c(n; {1}).M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 18Now consider the sequence b m = L m where L is positive real number. To simplifynotation, we will denote:T(n, k; {L m }) = T(n, k; L), c(n; {L m }) = c(n; L)Definition 4 Denote with a n (L) generalization of the sequence A005087 defined by:a n (L) = c(n + 1; L) + c(n; L)Our goal is to find the Hankel transform h n (L) of this sequence.M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 193.2 The generating functionProposition 1 The generalized binomial coefficient T(2n+a, n+a; L) can be rewrittenusing Jacobi polynomial Pn(a,b) (x) by:( ) L + 1T(2n + a, n; L) = (L − 1) n P (a,0)nL − 1The generating function of Jacobi polynomials is given by:G (a,b) (x, t) =∞∑n=0P (a,b)n(x)t n =2 a+bφ · (1 − t + φ) a · (1 + t + φ) b , (3.12)where φ = φ(x, t) = √ 1 − 2xt + t 2 .Now we can derive the generating functions of T(2n + a, n; L) and also a n (L):∞∑n=0T(2n + a, n; L) t n = G (a,0)( L+1L−1 , (L − 1)t ), (3.13)M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 20whereG(t; L) =+∞ ∑n=0= t + 1t= t + 1ρ(t; L)a n (L)t n)L−1 , (L − 1)t − (t + 1)G (2,0)( )L+1L−1 , (L − 1)t − 1 t{ 1t − 4G (0,0)( L+1(1 − (L − 1)t + ρ(t; L)) 2 }− 1 t(3.14)( )ρ(t; L) = φ L+1L−1 , (L − 1)t = √ 1 − 2(L + 1)t + (L − 1) 2 t 2 (3.15)M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 213.3 The Hankel transform of a n (L)Theorem 3 Numbers a n (L) are the moments of the following weight function:√L (ω(x; L) = 1 + 1 π x) √ ( x − L − 1 ) 21 −2 √ (3.16)LNow we need to describe the orthogonal polynomials {Q n (x)} corresponding to thisweight function.Example 5 For L = 4, we can find the first membersQ 0 (x) = 1, ‖Q 0 ‖ 2 = 5,Q 1 (x) = x − 24 5 , ‖Q 1‖ 2 = 1045 ,Q 2 (x) = x 2 − 12713 x + 25613 , ‖Q 2‖ 2 = 108813 ,Q 3 (x) = x 3 − 54117 x2 + 109617 x − 134417 , ‖Q 3‖ 2 = 569617 ,M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 22We can again start from the second kind Chebyshev polynomials orthogonal w.r.t.the weight p (1/2,1/2) (x) = √ 1 − x 2 . They satisfy the three-term recurrence relation[2]:S n+1 (x) = (x − α ∗ n ) S n(x) − β ∗ n S n−1(x) (n = 0, 1, . . .), (3.17)with initial valuesS −1 (x) = 0, S 0 (x) = 1,whereα ∗ n = 0 (n ≥ 0) and β∗ 0 = π 2 , β∗ n = 1 4(n ≥ 1).M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 23Let we introduce new weight function ŵ(x) = (x − c) p (1/2,1/2) (x). Correspondingcoefficients ˆα n and ˆβ n can be evaluated as follows [4]:λ n = S n (c),ˆα n = c − λ n+1− β ∗ λ nn+1,λ n λ n+1(3.18)ˆβ n = β ∗ nλ n−1 λ n+1λ 2 n(n ∈ N 0 ).If we choose c = − L+22 √ , it can be shown that holds:Lwhere:λ n =ψ n =(−1) n√2 · 4 n L n 2 L2 + 4 ψ n+1 (n = −1, 0, 1, . . .).(L + 2 + √ ) n (L 2 + 4 − L + 2 − √ nL 2 + 4).M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 24The next transformation will be ˜w(x) = ŵ(ax + b), where a = 12 √ Lexchanging we obtain:( )√x − L − 1˜w(x) = ŵ2 √ = 1L 2 √ L (x + 1) 1 −Coefficients of three-term relation are now:˜α n = ˆα n − baand b = −L+12 √ L . After( x − L − 1) 2.(3.19)2 √ L, ˜βn = ˆβ na 2 (n ≥ 0). (3.20)Multiplying the weight function ˜w(x) with the constant 2L we are only changing ˜βπ0 . Finally,we have that coefficients corresponding to the:˘w(x) = 2L √ √L( x − L − 1) 2π ˜w(x) = π (x + 1) 1 −2 √ (3.21)Lare given with:˘β 0 = L(L + 2),M.D. Petković, P.M. Rajković˘ β n = ˜β n = L ψ nψ n+2ψ 2 n+1(n ∈ N),α˘n = ˜α n = −1 + 1 2 · ψn+2+ 2L · ψn+1(n ∈ N 0 ).ψ n+1 ψ n+2(3.22)PRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 25w(x) ˘xFinal transformation will be ω(x; L) = . If we know all about the MOPS orthogonal withrespect to ˘w(x) what can we say about the sequence {Q n (x)} orthogonal w.r.t. a weightw d (x) = ˘w(x)x − d(d /∈ support( ˘w)) ?In the book [5], W. Gautshi has proved that, by the auxiliary sequence:∫r −1 = − w d (x) dx, r n = d − ᾰ n − ˘β n(n = 0, 1, . . .),r n−1it can be determined:Rα d,0 = ᾰ 0 + r 0 , α d,k = ᾰ k + r k − r k−1 ,β d,0 = −r −1 , β d,k = ˘β k−1r k−1r k−2We need the case d = 0. Next Lemma can be proved by induction:(k ∈ N).Lemma 3 The parameters r n have the explicit formr n = − ψ n+1· Lψ n+2 + ξϕ n+2(n ∈ N 0 ).ψ n+2 Lψ n+1 + ξϕ n+1(ϕ n = L + 2 + √ ) n (L 2 + 4 + L + 2 − √ n √L 2 + 4), ξ = L2 + 4(3.23)M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 26Now we have the coefficients β n = β 0,n . By exchanging and using Krattenthaler formula wefinally obtain:n−2r n−2∏h n (L) = β 0 β 1 β 2 · · · β n−2 β n−1 · h n−1 (L) = β 0r −1k=0˘β k · h n−1 (L)= Ln−12·Lψ n + ξϕ n· h n−1 (L) = Ln(n−1)/2Lψ n−1 + ξϕ n−1 2 n+1 ξ· (Lψ n + ξϕ n )= L(n2 −n)/22 n+1√ L 2 + 4·{( √ L 2 + 4 + L)( √ L 2 + 4 + L + 2) n + ( √ L 2 + 4 − L)(L + 2 − √ L 2 + 4) n} .(3.24)M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 27References[1] P. Barry, On Ineger Sequences Based Constructions of Generalized Pascal Triangles,Preprint, Waterford Institute of Technology, 2005.[2] T. S. Chihara, An Introduction to Orthogonal Polynomials, Gordon and Breach,New York, 1978.[3] A. Cvetković, P. Rajković and M. Ivković, Catalan Numbers, the Hankel Transformand Fibonacci Numbers, Journal of Integer Sequences, 5, May 2002, Article02.1.3.[4] W. Gautschi, Orthogonal polynomials: applications and computations, in Acta Numerica,1996, Cambridge University Press, 1996, pp. 45–119.[5] W. Gautschi, Orthogonal Polynomials: Computation and Approximation, ClarendonPress - Oxford, 2003.[6] C. Krattenthaler, Advanced Determinant Calculus,at http://www.mat.univie.ac.at/People/kratt/artikel/detsurv.html.M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 28[7] J. W. Layman, The Hankel Transform and Some of its Properties,Journal of Integer Sequences, Article 01.1.5, Volume 4, 2001.[8] P. Peart and W. J. Woan, Generating functions via Hankel and Stieltjes matrices,Journal of Integer Sequences, Article 00.2.1, Issue 2, Volume 3, 2000.[9] P.M. Rajković, M.D. Petković, P. Barry, The Hankel Transform of the Sum ofConsecutive Generalized Catalan Numbers, Journal of Integral Transforms, to appear.Available electronically at Mathematics ArXiv, 2006.[10] W. J. Woan, Hankel Matrices and Lattice Paths,Journal of Integer Sequences, Article 01.1.2, Volume 4, 2001.[11] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences. Publishedelectronically athttp://www.research.att.com/∼njas/sequences/[12] R. A. Sulanke, Counting Lattice Paths by Narayana Polynomials The ElectronicJournal of Combinatorics, Vol. 7(1), 2000.M.D. Petković, P.M. RajkovićPRIM2006

Hankel transform of Narayana polynomials and Generalized Catalan Numbers 29Thanks for your attention!M.D. Petković, P.M. RajkovićPRIM2006