The Hankel Transform of the Sum of Consecutive Generalized ...

The Hankel Transform of the Sum ofConsecutive Generalized Catalan NumbersPredrag Rajković, Marko D. Petković,University of Niš, Serbia and Montenegroe-mail: pedja.rajk@gmail.com, dexterofnisgmail.comPaul BarrySchool of Science, Waterfall Institute of Technology, Irelande-mail: pbarry@wit.ieAbstract. We discuss the properties of the Hankel transformation of a sequencewhose elements are the sums of consecutive generalized Catalan numbers.1. IntroductionLet A = {a 0 , a 1 , a 2 , ...} be a sequence of real numbers. The Hankel matrix generatedby A is the infinite matrix H = [h i,j ], where h i,j = a i+j−2 , i.e.,⎡H =⎢⎣a 0 a 1 a 2 a 3 ...a 1 a 2 a 3 a 4 ...a 2 a 3 a 4 a 5 ...a 3 a 4 a 5 a 6 ...a 4 a 5 a 6 a 7 .......⎤⎥⎦.. .The Hankel matrix H n of order n is the upper-left n × n submatrix of H andthe Hankel determinant of order n of A, denoted by h n , is the determinant of thecorresponding Hankel matrix.For a given sequence A = {a 0 , a 1 , a 2 , ...}, the Hankel transform of A is the correspondingsequence of Hankel determinants {h 0 , h 1 , h 2 , . . . } (see Layman [?]).The elements of the sequence in which we are interested are the sums of twoadjacent generalized Catalan numbers with parameter L:a n = a n (L) = c(n; L) + c(n + 1; L) (n = 0, 1, 2, . . .), (2)c(n; L) = T (2n, n; L) − T (2n, n − 1; L), (3)1(1)

2with∑n−k( )( )k n − kT (n, k; L) =L j . (4)j jj=0Herea 0 = L + 1. (5)Example 1.1. Let L = 1. Vandermonde’s convolution identity implies that( n=k)∑ ( )( )k n − k.j jjHence we haveT (2n, n; 1) =wherefrom we get Catalan numbers( ) 2nc(n) = −nanda n = c(n) + c(n + 1) =( )( )2n 2n, T (2n, n − 1; 1) = ,n n − 1( ) 2n= 1 ) 2nn − 1 n + 1(n(2n)!(5n + 4)n!(n + 2)!(n = 0, 1, 2, . . .).In the paper [?] the authors have proved that the Hankel transform of a n equalssequence of Fibonacci numbers with odd indices1h n = F 2n+1 = √{( √ 5 + 1)(3 + √ 5) n + ( √ 5 − 1)(3 − √ }5) n .5 2n+1Example 1.2. For L = 2 we get like a n (2) the next numbersand the Hankel transform h n :One of us, P. Barry conjectured thath n (2) = 2 n2 −n2 −23, 8, 28, 112, 484, . . . ,3, 20, 272, 7424, 405504, . . . .{(2 + √ 2) n+1 + (2 − √ 2) n+1 }.In general, he made the conjecture, which we will prove through this paper.Theorem 1.1. (The main result) For the generalized Pascal triangle associatedto the sequence n ↦→ L n , the Hankel transform of the sequencec(n; L) + c(n + 1; L)

3is given by−n)/2h n =L(n2 2 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 .(6)From now till the end, let us denote byNow, we can writeξ = √ L 2 + 4, t 1 = L + 2 + ξ, t 2 = L + 2 − ξ. (7)h n = Ln(n−1)/22 n+1 ξ· ((ξ + L)t n 1 + (ξ − L)t n 2) . (8)Or, introducingϕ n = t n 1 + t n 2, ψ n = t n 1 − t n 2 (n ∈ N 0 ), (9)the final statement can be expressed byh n = Ln(n−1)/22 n+1 ξ· (Lψ n + ξϕ n ) . (10)Lemma 1.1. The values ϕ n and ψ n satisfy the relationsandϕ j · ϕ k = ϕ j+k + (4L) j ϕ k−j , ψ j · ψ k = ϕ j+k − (4L) j ϕ k−j (0 ≤ j ≤ k) (11)ϕ j · ψ k = ψ j+k + (4L) j ψ k−j , ψ j · ϕ k = ψ j+k − (4L) j ψ k−j (0 ≤ j ≤ k). (12)Corollary 1.1. The function h n = h n (L) is the next polynomialh n (L) = 2 −n L n(n−1)/2⎧⎨[(n−1)/2]∑·⎩i=0( n2i+1)L(L + 2) n−2i−1 (L 2 + 4) i +[n/2]∑i=0⎫( n) ⎬2i (L + 2) n−2i (L 2 + 4) i ⎭ .

4Proof. By previous notation, we can write(L + ξ)(L + 2 + ξ) n − (L − ξ)(L + 2 − ξ) nn∑ (= (L + ξ) n)n∑k (L + 2) n−k ξ k − (L − ξ) (−1) k( )nk (L + 2) n−k ξ k== 2n∑k=0k=0(1 − (−1)k )( nk)L(L + 2) n−k ξ k +[(n−1)/2]∑i=0⎧⎨= 2ξ⎩∑[(n−1)/2]i=0n∑k=0( n2i+1)L(L + 2) n−2i−1 ξ 2i+1 + 2( n2i+1)L(L + 2) n−2i−1 ξ 2i +k=0( )( )1 + (−1)k nk (L + 2) n−k ξ k+1[n/2]∑i=0[n/2]∑wherefrom immediately follows the expression for h n .✸k=0i=02. The generating function( n2i)(L + 2) n−2i ξ 2i+1⎫( n) ⎬2i (L + 2) n−2i ξ 2i ⎭ ,The Jacobi polynomials are given byP n (a,b) (x) = 1 n∑( )( )n + a n + b(x − 1) n−k (x + 1) k (a, b > −1).2 n k n − kAlso, they can be written in the form( x − 1) nn∑( )( ) n + a n + b (P n (a,b)x+1(x) =2k n − kFrom the factL = x + 1x − 1we conclude thatandk=0⇔ x = L + 1L − 1T (2n, n; L) = (L − 1) n · P (0,0)nT (2n + 2, n; L) = (L − 1) n · P (2,0)nx−1) k.(x ≠ 1, L ≠ 1).The generating function G(x, t) for the Jacobi polynomials is∞∑G (a,b) (x, t) = P n (a,b) (x)t n 2 a+b=φ · (1 − t + φ) a · (1 + t + φ) , (13)bwheren=0(L+1L−1()L+1L−1).φ = φ(x, t) = √ 1 − 2xt + t 2 .

5Now,Also,∞∑T (2n, n; L) t n =n=0∞∑T (2n + 2, n; L) t n =n=0∞∑n=0∞∑n=0P (0,0)nP (2,0)n((L+1L−1L+1L−1∞∑T (2n, n − 1; L) t n = t ·n=0∞∑T (2n + 2, n + 1; L) t n = 1 t ·n=0) ((L− 1)t) n= G(0,0)(L+1L−1 , (L − 1)t ),) ((L− 1)t) n= G(2,0)(L+1L−1 , (L − 1)t ).() }{G (2,0) L+1, (L − 1)t − 1 ,L−1() }{G (0,0) L+1, (L − 1)t − 1 .L−1The generating function G(t; L) for the sequence {a n } n≥0 is given byG(t; L) =∞∑a n t nn=0(14)= t + 1 ( )( )G (0,0) L+1L−1t, (L − 1)t − (t + 1)G (2,0) L+1, (L − 1)t − 1 L−1t .The function()L+1ρ(t; L) = φ , (L − 1)t = √ 1 − 2(L + 1)t + (L − 1)L−1 2 t 2 (15)has domainorD ρ =(−∞, 1 − 2√ L + L) (√1 + 2 L + L)∪1 − 2L + L 2 1 − 2L + L , +∞ 2D ρ = (−∞, 1/4) (L = 1).(L ≠ 1),Theorem 2.1. The generating function G(t; L) for the sequence {a n } n≥0 isG(t; L) = t + 1 { }1ρ(t; L) t − 4− 1 (1 − (L − 1)t + ρ(t; L)) 2 t . (16)Example 2.1. For L = 1, we getG(t; 1) =and for L = 2, we find∞∑a n (1) t n = 1 tn=0( (1 −√ 1 − 4t)(1 + t)2t)− 1 . (17)G(t; 2) =∞∑n=0a n (2) t n = − 1 { }t + t + 1 1√t2 − 6t + 1 t − 4(1 − t + √ . (18)t 2 − 6t + 1) 2

6It is known (for example, see Krattenthaler [?]) that the Hankel determinant h nof order n of the sequence {a n } n≥0 equalswhere {β n } n≥1 is the sequence given by:h n = a n 0β n−11 β n−22 · · · β 2 n−2β n−1 , (19)∞∑G(x) = a n x n a 0=βn=01 x 21 + α 0 x −β1 + α 1 x − 2 x 21 + α 2 x − · · ·(20)The sequences {α n } n≥0 and {β n } n≥1 are the coefficients in the recurrence relationQ n+1 (x) = (x − α n )Q n (x) − β n Q n−1 (x), (21)where {Q n (x)} n≥0 is the monic polynomial sequence orthogonal with respect to thefunctional L determined byL[x n ] = a n (n = 0, 1, 2, . . .). (22)In the next section this functional will be constructed for the sum of consecutivegeneralized Catalan numbers.3. The weight function corresponding to the functionalWe would like to express L[f] in the form:∫L[f(x)] = f(x)dψ(x),Rwhere ψ(x) is a distribution, or, even more, to find the weight function w(x) suchthat w(x) = ψ ′ (x).Denote by F (z; L) the functionF (z; L) =∞∑a k z −k−1 .k=0From the generating function (??), we have:Example 3.1. From (??), we have:F (z; 1) = z −1 G ( z −1 ; 1) = 1 2F (z; L) = z −1 G ( z −1 ; L). (23){z − 1 − (z + 1)√1 − 4 z}.

7Example 3.2. From (??), we have:{ ()}F (z; 2) = −1 1 + z 2 − z + (z + 1)√1 − 6 2zz + 1 .z 2∫F (z; 2)dz = z + 1 z(z − 1)ρ(1/z, 2) + log(z)4− 1 2 log (1 + z ( ρ(1/z, 2) − 3 )) − 7 2 log( z − 3 + zρ(1/z, 2) ) .When we replace relation (8) in previous relation and after some simplificationswe obtain that2(z + 1)F (z; L) = −1 +L − 1 + z + √ L 2 + (z − 1) 2 − 2L(z + 1)2(z + 1)= −1 +L − 1 + z(1 + zρ( 1, L)) zDenote R(z; L) = zρ( 1, L) = √ Lz 2 + (z − 1) 2 − 2L(z + 1). From the theory ofdistribution functions (see Chihara [?]), we have Stieltjes inversion formulaψ(t) − ψ(0) = − 1 π limy→0 + ∫ t0IF (x + iy; L)dx. (24)using which we can calculate the distribution function ψ(x) and weight functionw(x) = ψ ′ (x). It can be shown that holds:∫F (z; L)dz = 1 []z 2 − 2Lz − (z − L + 1)R(z; L) − l 1 (z) + l 2 (z) (25)4where we denoted:[]l 1 (z) = 2(3L + 1) log z − (L + 1) + R(z; L)l 2 (z) =[ −(L − 1)R(z; L) − (L − 1) 2 + z(L + 1)]2(L − 1) logz 2 (L − 1) 3Denote with F(z; L) = ∫ F (z; L)dz. Now we will replace z = x + iy and find a limitlim y→0 F(x + iy; L). First rewrite function R(z; L) in the following form:R(z; L) = √ (z − L − 1) 2 − 4LNow replace z = x + iy and let y tends to 0 + . Then we have:where{ √R(x; L) = lim R(x + iy; L) = √ i 4L − (x − L − 1)2 , x ∈ (a, b);y→0 + (x − L − 1)2 − 4L, otherwise,a = ( √ L − 1) 2 , b = ( √ L + 1) 2 . (26)

8In the case when x /∈(( √ L − 1) 2 , ( √ L + 1) 2 ), value R(x; L) is real. Therefore wecan calculate imaginary part of F(x; L) = lim y→0 + F(x + iy; L):IF(x; L) = I[l 2 (x) − l 1 (x)] = 0.Otherwise, if x ∈(( √ L − 1) 2 , ( √ )L + 1) 2 we have that:l 1 (x)⎧⎨Il 1 (x) =⎩= 2(3L + 1) log[x − (L + 1) ± i √ ]4L − (x − L − 1) 2√4L−(x−L−1) 22(3L + 1) arctan, x ≥ L + 1;x−(L+1)( √ )2(3L + 1) π + arctan, x < L + 14L−(x−L−1) 2x−(L+1)[ −(L−1) 2 +2x(L+1)−i(L−1)√4L−(x−L−1) 2l 2 (x) = 2(L − 1) logIl 1 (x)⎧⎪⎨ 2(L − 1)(2π + arctan √x(L+1)−(L−1)2=)⎪⎩ 2(L − 1)(π + arctan √x(L+1)−(L−1)24L−(x−L−1) 2x 2 (L−1) 3 ]4L−(x−L−1) 2 ), x ≥ (L−1)2L+1 ;, x < (L−1)2After substituting all considered cases in (7) we finally obtain valueIF(x; L) = limy→0 + IF(x + iy; L) = Il 2(x) − Il 1 (x) − (x − L + 1) √ 4L − (x − L − 1) 2From the relation (16) we conclude thatand finally we obtain ω(x; L):ω(x; L) = ψ ′ (x; L) = − 1 πω(x; L) = 12π (1 + 1 x )√ 4L − (x − L − 1) 2 =√LπL+1dIF(x; L) (27)dx(1 + 1 x) √ 1 −Previous formula holds for x ∈ (a, b), and otherwise we have ω(x; L) = 0.( x − L − 1) 22 √ (28)L4. Determining the three-term recurrence relationThe crucial moment in our proof of the conjecture is to determine the sequence ofpolynomials {Q n (x)} orthogonal with respect to the weight w(x; L) given by (??)on the interval (a, b) and to find the sequences {α n (x)} {β n (x)} in the three-termrecurrence relation.

9Example 4.1. For L = 4 we getwherefromQ 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 ,α 0 = 24 5 , β 0 = 5, α 1 = 32365 , β 1 = 10425 , α 2 = 1104221 , β 2 = 680169 .Hence( 104) 2h 1 = a 0 = 5, h 2 = a 2 0β 1 = 104, h 3 = a 3 0β1β 2 2 = 5 3 68025 169 = 8704.At the beginning, we will notice that in the definition of the weight appears thesquare root member.That’s why, let us consider the monic orthogonal polynomials {S n (x)} with respectto the p (1/2,1/2) (x) = √ 1 − x 2 on the interval (−1, 1). These polynomials aremonic Chebyshev polynomials of the second kind:S n (x) = sin( (n + 1) arccos x )2 n · √1− x 2They satisfy the three-term recurrence relation (Chihara [?]):with initial valueswhereS n+1 (x) = (x − α ∗ n) S n (x) − β ∗ nS n−1 (x) (n = 0, 1, . . .), (29)S −1 (x) = 0, S 0 (x) = 1,αn ∗ = 0 (n ≥ 0) and β0 ∗ = π 2 , β∗ n = 1 (n ≥ 1).4If we use the weight function ŵ(x) = (x − c) p (1/2,1/2) (x), then the correspondingcoefficients ˆα n and ˆβ n can be evaluated as follows (see, for example, Gautschi [?])λ n = S n (c),λ nˆα n = c − λ n+1− βn+1∗ ,λ n λ n+1ˆβ n = βn ∗ λ n−1 λ n+1(n ∈ Nλ 2 0 ).nFrom the relation (21), we conclude that the sequence {λ n } n∈N satisfies the followingrecurrence relation:(30)4λ n+1 − 4cλ n + λ n−1 = 0 (λ −1 = 0; λ 0 = 1). (31)

10The characteristic equationhas the solutions4z 2 − 4 c z + 1 = 0z 1,2 = 1 2and the integral solution of (23) isλ n = E 1 z n 1 + E 2 z n 2(c ± √ )c 2 − 1 .(n ∈ N).We evaluate values E 1 and E 2 from the initial conditions (λ −1 = 0; λ 0 = 1).i.e,In other to solve our problem, we will choose c = − L+22 √ L . HenceFinally, we obtain:z k = −t k4 √ L (k = 1, 2), where t 1,2 = L + 2 ± √ L 2 + 4.λ n =(−1) n2 · 4 n L n 2√L2 + 4λ n =After replacing in (22), we obtain:( )t n+11 − t n+12(λ = −1, 0, 1, . . .),(−1)n2 · 4 n L n 2 ξ ψ n+1 (λ = −1, 0, 1, . . .).ˆα n = − L + 22 √ L + 14 √ L · ψn+2ψ n+1+ √ L · ψn+1ψ n+2, (32)ˆβ n = ψ nψ n+2. (33)4ψn+12If a new weight function ˜w(x) is introduced bythen we have˜w(x) = ŵ(ax + b)˜α n = ˆα n − b, ˜βn = ˆβ n(n ≥ 0).aa 2Now, by using x ↦→ x−L−12 √ , i.e., a = 1L 2 √ L+1and b = −L 2 √ , we have the weight functionLThusand˜w(x) = ŵ( x − L − 12 √ L ) = 1 ( x − L − 12 2 √ L+ L + 2 ) √2 √ 1 −L( x − L − 1) 2.2 √ L˜α n = −1 + 1 2 · ψn+2ψ n+1+ 2L · ψn+1ψ n+2(n ∈ N 0 ), (34)

11Example 4.2. For L = 4 we getwherefrom˜β 0 = (L + 2) π 2 , ˜βn = L ψ nψ n+2ψ 2 n+1P 0 (x) = 1, ‖P 0 ‖ 2 = 3π,P 1 (x) = x − 173 , ‖P 1‖ 2 = 32π3 ,P 2 (x) = x 2 − 434 x + 1014 , ‖P 2‖ 2 = 42π,(n ∈ N). (35)P 3 (x) = x 3 − 33121 x2 + 157921 x − 218921 , ‖P 3‖ 2 = 3520π21 ,˜α 0 = 17 3 , ˜β0 = 3π, ˜α 1 = 6112 , ˜β1 = 32 9 , ˜α 2 = 42184 , ˜β2 = 6316 .Introducing the weight˘w(x) = 2L π ˜w(x)will not change the monic polynomials and their recurrence relations, only it willmultiply the norms by the factor 2L/π, i.e.˘P k (x) ≡ P k (x), ‖ ˘P k ‖ 2˘w =∫ ba˘P k (x) ˘w(x) dx = 2L π ‖P k‖ 2 (k ∈ N 0 ),˘β 0 = L(L + 2), ˘βk = ˜β k (k ∈ N), ᾰ k = ˜α k (k ∈ N 0 ).Here is˘β 0 ˘β1 · · · ˘β n−1 = Ln2 · ψn+1ψ n. (36)In the book [?], W. Gautschi has treated the next problem: If we know all aboutthe MOPS orthogonal with respect to ˘w(x) what can we say about the sequence{Q n (x)} orthogonal with respect to a weightw d (x) = ˘w(x) (d /∈ support( ˜w)) ?x − dW. 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−1Rit can be determinedα 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−2(k ∈ N).

12In our case it is enough to take d = 0 to get the final weightHencer −1 = −(L + 1),w(x) = ˘w(x)x .(r n = − ᾰ n +˘β )nr n−1Lemma 4.1. The parameters r n have the explicit form(n = 0, 1, . . .). (37)r n = − ψ n+1ψ n+2· Lψ n+2 + ξϕ n+2Lψ n+1 + ξϕ n+1(n ∈ N 0 ). (38)Proof. We will use the mathematical induction. For n = 0, we really get theexpected valuer 0 = − L2 + 2L + 2(L + 1)(L + 2) .Suppose that it is true for k = n. Now, by the properties for ϕ n and ψ n , we have˜α n+1 · r n + ˜β n+1 = − ψ n+1ψ n+3· Lψ n+3 + ξϕ n+3Lψ n+1 + ξϕ n+1.Dividing with r n , we conclude that the formula is valid for r n+1 .Example 4.3. For L = 4 we getwherefromα 0 = 24r −1 = −5, r 0 = − 1315 , r 1 = − 5152 , r 2 = − 356357 ,5 , β 0 = 5, α 1 = 32365 , β 1 = 10425 , α 2 = 1104221 , β 2 = 680169 ,just the same as in the Example 4.1.Proof of the main result. Let us start from Krattenthaler formulah 1 = a 0 , h n = a n 0β1 n−1 β2 n−2 · · · βn−2β 2 n−1 (n = 2, 3, . . .). (39)Here is a 0 = β 0 = L + 1. This formula can be also written in the formh 1 = a 0 , h n = β 0 β 1 β 2 · · · β n−2 β n−1 · h n−1 . (40)From the theory of orthogonal polynomials, it is known thatwherefromHere,‖Q n−1 ‖ 2 = β 0 β 1 β 2 · · · β n−2 β n−1 (n = 2, 3, . . .), (41)h 1 = a 0 , h n = ‖Q n−1 ‖ 2 · h n−1 (n = 2, 3, . . .). (42)‖Q n−1 ‖ 2 = β 0r n−2r −1n−2∏k=0˘β k = Ln−12·Lψ n + ξϕ nLψ n−1 + ξϕ n−1. (43)

We will apply the mathematical induction again. The formula for h n is true forn = 1. Suppose that it is valid for k = n − 1. Thenh n = Ln−12wherefrom follows that the final statement·Lψ n + ξϕ n· L(n−1)(n−2)/2· (LψLψ n−1 + ξϕ n−1 2 n n−1 + ξϕ n−1 ) ,ξ13is true.h n = Ln(n−1)/22 n+1 ξ· (Lψ n + ξϕ n ) (n ∈ N)References[1] P. Barry, On Ineger Sequences Based Constructions of Generalized Pascal Triangles, Preprint,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 Transform andFibonacci Numbers, Journal of Integer Sequences, 5, May 2002, Article 02.1.3.[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, Clarendon Press -Oxford, 2003.[6] C. Krattenthaler, Advanced Determinant Calculus,at http://www.mat.univie.ac.at/People/kratt/artikel/detsurv.html.[7] 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.[8] W. J. Woan, Hankel Matrices and Lattice Paths, Journal of Integer Sequences, Article 01.1.2, Volume 4, 2001.[9] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences. Published electronically athttp://www.research.att.com/∼njas/sequences/.(Mentions sequences A005807, A001906, A001519.)Predrag Rajković, Marko D. Petković,University of Niš, Serbia and Montenegroe-mail: pedja.rajk@gmail.com, dexterofnisgmail.comPaul BarrySchool of Science, Waterfall Institute of Technology, Irelande-mail: pbarry@wit.ie