Hankel Determinants for Some Common Lattice Paths - garsia at york

HANKEL DETERMINANTS FOR LATTICE PATHSSince, by the i-t-config of Lemma 2.2, (H) i,j = |P i,j | counts the large Schröder paths from (0, 0) to(2i+2j −2, 0), immediately we have the the following corollary for the Hankel determinants of the weightednon-aerated Schröder numbers:Proposition 2.1. Let f n denote the weight of the path set of paths from (0, 0) to (2n, 0) which neverrun beneath the x-axis and where H = (2, 0) is weighted by t. Equivalently, let f n satisfyF(x) = ∑ n≥0f n x n = 1 + txF(x) + xF(x) 2 .Then the determinant of the n-th order Hankel matrix equals (1 + t) n(n−1)/2 .As a second corollary to Lemma 2.2, we haveLemma 2.3. For the lattice paths that use the steps U, H = (2, 0), and D, that never run below thex-axis, and that have the i-t-config with(x i , y i ) = (−2i + 1, 0) and (x ′ i, y ′ i) = (2i − 1, 0)for 1 ≤ i ≤ n, the sum of the signed weights for the nonintersecting n-tuples in ∪ σ∈Sn P σ is (1 + t) n(n+1)/2 .Proof of Lemma 2.3. We first translate all paths upwards one unit and then prepend a U-step andappend a D-step to every path. Next we add the point path at (0, 0). The sum of the signed weights of thenonintersecting n-tuples in the original configuration equals that of the nonintersecting n + 1-tuples in thisnew configuration, which in turn is given by Lemma 2.2.□Proof of Proposition 1.2. Suppose that n is even; the proof when n is odd is similar. Here theHankel matrix (|P i,j |) 1≤i,j≤n corresponds to the i-t-config with(x i , y i ) = (−i + 1, 0) and (x ′ i, y ′ i) = (i − 1, 0) for 1 ≤ i ≤ n.Since l = 2, no endpoint of a step on a path that originates from an oddly indexed initial point (i.e., a point(−i + 1, 0) for odd i) will intersect an endpoint of a step on a path that originates from an evenly indexedinitial point. Moreover, for any permutation σ corresponding to a nonintersecting n-tuple, σ(i) − i must beeven for each i, and hence sgn(σ) = 1. Thus the weight of the nonintersecting n-tuples is the product ofthe weight of those originating from oddly indexed initial points times the weight of those originating fromevenly indexed initial points.Hence, with m = n/2, let i-t-configA haveand let i-t-configB have(x i , y i ) = (−2i + 2, 0) and (x ′ i, y ′ i) = (2i − 2, 0) for 1 ≤ i ≤ m,(x i , y i ) = (−2i + 1, 0) and (x ′ i, y ′ i) = (2i − 1, 0) for 1 ≤ i ≤ m.Applying Lemmas 2.2 and 2.3 to these configurations yields the weight of nonintersecting n-tuples of theoriginal configuration as(1 + t) m(m−1)/2 (1 + t) m(m+1)/2 = (1 + t) m2 = (1 + t) n2 /4 .□Next we consider Hankel determinants for sequences of path weights that ignore the initial term. Forthe sequence f(1, t, l), f(2, t, l), . . ., we will let H 1 n denote the matrix where the entries satisfy (H1 n ) i,j =f(i + j − 1, t, l). See Figure 2.Proposition 2.2. For l = 1 (Motzkin case again), the sequence of determinants satisfies the recurrencesubject to det(H1 1) = t and det(H1 2 ) = (t − 1)(t + 1).det(H 1 n ) = t det(H1 n−1 ) − det(H1 n−2 )

Sulanke and Xin(0,0) (3,0)Figure 2. Three of the 4-tuples of paths for l = 1 and for i-t-config with[(0, 0), (−1, 0), (−2, 0), (−3, 0)] and [(1, 0), (2, 0), (3, 0), (4, 0)]. The first and second 4-tuplesare both nonintersecting. The first has a signed weight of t 4 while the second has a signedweight of −t 2 . The third is intersecting only at the point (0, 1).Proof. Aigner [1] considered the case for t = 1. For arbitrary t, our proof considers how the particularpaths must look in the nonintersecting case. Observe that det(Hn 1 ) is the sum of the weights of thenonintersecting n-tuples for the i-t-config(n) taken as[(0, 0), (−1, 0), . . ., (n − 1, 0)] and [(1, 0), (2, 0), . . ., (n, 0)].Each of these nonintersecting n-tuples belongs to one of two types: (1) those containing the path from (0, 0)to (1, 0) with all other paths forced to begin with U, end with D, and have ordinate at least one elsewhere;(2) those containing the path UD from (0, 0) to (2, 0) and the path UD (−1, 0) to (1, 0) with all other pathsforced to begin with UU, end with DD, and have ordinate at least two elsewhere. The set of the first typehas a total weight t times the sum of the weights of the nonintersecting (n−1)-tuples on the i-t-config(n-1),which is t det(Hn−1). 1 Since each n-tuple of the second type has the defined crossing of the path from (0, 0)with that from (−1, 0), the set has total weight is the sign of the corresponding permutation times the sumof the weights of the nonintersecting (n − 2)-tuples on the i-t-config(n-2), which is − det(Hn−2).1 □For l = 2, we will indicate how Lemma 2.3 provesProposition 2.3. For n ≥ 0, l = 2, and arbitrary t, the sequence of determinants satisfies{ 0det(Hn 1 ) = (−1) n/2 (1 + t) n(n+2)/4if n is oddif n is evenProof. Here the Hankel matrix can correspond to i-t-config with(x i , y i ) = (−i + 1, 0) and (x ′ i , y′ i ) = (i, 0) for 1 ≤ i ≤ n.Since l = 2, if there is a path from (x i , y i ) to (x ′ j , y′ j ), then i − j is odd. It follows that, if n is odd, therecan be no n-tuples of paths for the configuration. If n is even and m = n/2, the sign of any permutationfor an nonintersecting n-tuples can be shown to be (−1) m . Thus the weight of the nonintersecting n-tuplesis (−1) m times the weight of those originating from oddly indexed initial points times the weight of thoseoriginating from evenly indexed initial points. The proof is completed by applying 2.3 to i-t-configA withand to i-t-configB with(x i , y i ) = (−2i + 2, 0) and (x ′ i , y′ i ) = (2i, 0) for 1 ≤ i ≤ m,(x i , y i ) = (−2i + 1, 0) and (x ′ i, y ′ i) = (2i − 1, 0) for 1 ≤ i ≤ m.3. Periodicity fourteen and continued fractionsHere we will repeated apply the “continued fractions method” recently developed by Gessel and Xin [6] todetermine the periodicity of the sequence of Hankel determinants for l = 3 and t = 1. This method, presentedmore formally in the next section, transforms both generating functions and corresponding determinants. Inthis section we will concentrate on the specific generating function F(x) satisfyingF(x) = 1 + x 3 F(x) + x 2 F(x) 2 .From this functional equation, or from the related recurrence for its coefficients, there appears to be no cluewhy the associated sequence of Hankel determinants should have a period of 14.□

HANKEL DETERMINANTS FOR LATTICE PATHSFor an arbitrary generating function D(x, y) = ∑ ∞i,j=0 d i,jx i y j , let [D(x, y)] n denote the n by n determinantdet((d i,j ) 0≤i,j≤n−1 ). For any A(x) = ∑ n≥0 a nx n , define the Hankel matrix for A of order n, n ≥ 1,by H n (A) = (a i+j−2 ) 1≤i,j≤n . It is straight forward to show that the Hankel determinant det(H n (A)) can beexpressed as][ xA(x) − yA(y)det(H n (A)) =x − yWe will use an easily-proven “product rule” of [6] for transforming the generating functions: If u(x) isa formal power series with u(0) = 1, then.n[u(x)D(x, y)] n = [D(x, y)] n = [u(y)D(x, y)] n .We will make five transformations showing, for n ≥ 8,⎛ ⎛ ⎡det(H n (F)) = det⎝diag⎝[1], [1],⎣ 0 0 10 1 01 0 −2⎤⎞⎞⎦,[1], [1], H n−7 (F) ⎠⎠ ,where the right side is the determinant of a block-diagonal matrix consisting of six blocks along the diagonal,four of which are 1 by 1 identity matrices, and having entry 0 elsewhere. It then follows that det(H n (F)) =− det(H n−7 (F)). This implies that the period for det(H n (F)) is 14, and Proposition 1.3 will be proved.We start with F 0 (x) = F(x), and define F i (x) from F i−1 (x) according to a transformation where eachHankel determinant for F i (x) are derived from one for F i−1 (x) with the aid of the product rule, which is notalways mentioned. In the following, F i (x) will always satisfy a quadratic functional equationwhich is equivalent to the continued fractiona(x)F i (x) 2 + b(x)F i (x) + c(x) = 0,F i (x) =−c(x)b(x) + a(x)F i (x) .In particular, for l = 3,1F 0 (x) =1 − x 3 − x 2 F 0 (x) .Transformation 1: Using this continued fraction of F 0 , substitution, and simplification we obtain[ ] [xF0 (x) − yF 0 (y) −xy 2 F 0 (y) + yx 2 F 0 (x) + (x − y) ( yx 2 + xy 2 + 1 ) ]det(H n (F)) ==x − y(1 − x 3 − x 2 F 0 (x))(1 − y 3 − y 2 .F 0 (y))(x − y)nMultiplying by (1 − x 3 − x 2 F 0 (x))(1 − y 3 − y 2 F 0 (y)), which will not affect the value of the determinant bythe product rule, we can write the determinant as[1 + xy xF ]1(x) − yF 1 (y)x − ynwhere(3.1) F 1 (x) = F 0 (x) + x.The associated matrix is block-diagonal with two blocks: the matrix [1] and the Hankel matrix for F 1 (x).Certainly,det(H n (F 0 )) = det(H n−1 (F 1 )).From (3.1) and the functional equation for F 0 (x), we obtain the functional equationF 1 (x) =1 + x1 + x 3 − x 2 F 1 (x) .Transformation 2: Using this continued fraction for F 1 , substituting in xF 1(x) − yF 1 (y), and multiplyingby (1 + x 3 − x 2 F 1 (x))(1 + y 3 − y 2 F 1 (y)) yieldsx − y[ ] [ xF1 (x) − yF 1 (y) −xy 2 (x + 1)F 1 (y) + yx 2 ](y + 1)F 1 (x) − (y + 1)(x + 1)(xy − 1)(x − y)=x − yx − ynn.n

Sulanke and XinUpon multiplying by (1 + x) −1 (1 + y) −1 , the determinant is equal to[1 + xy xF ]2(x) − yF 2 (y)x − ywhere(3.2) F 2 (x) = F 1 (x)/(1 + x) − 1.The associated matrix being block diagonal shows,ndet(H n−1 (F 1 )) = det(H n−2 (F 2 )).From (3.2) and the functional equation for F 1 (x), we obtainF 2 (x) =x 21 − 2x 2 − x 3 − (x 3 + x 2 )F 2 (x) .Transformation 3: Substituting for F 2 with the above fraction, simplifying, [ and multiplying]by (1 +x)(1 −x−x 2 −x 2 F 2 (x))(1+y)(1 −y −y 2 −y 2 xF2 (x) − yF 2 (y)F 2 (y)) shows that the determinantequalsx − yn[y 2 x 3 (y + 1)F 2 (y) − x 2 y 3 (x + 1)F 2 (x) − (x − y) ( 2 y 2 x 2 − x 2 − xy − y 2) ]x − ywhich can be rewritten as[x 2 + xy + y 2 − 2x 2 y 2 + x 3 y 3 xF ]3(x) − yF 3 (y),x − ynwhere F 3 (x) is indeed a power series satisfying(3.3) F 3 (x) = (x + 1)F 2 (x)/x 2 .This time the corresponding matrix is a block-diagonal matrix with the blockthe Hankel matrix for F 3 (x). Hencedet(H n−2 (F 2 )) = − det(H n−5 (F 3 )).From (3.3) and the functional equation for F 2 (x), we obtainF 3 (x) =1 + x1 − 2x 2 − x 3 − x 4 F 3 (x) .⎡n⎣ 0 0 10 1 01 0 −2⎤⎦ followed byTransformation 4: Substituting for F 3 with the fraction, [ simplifying, ] and multiplying by (1 − 2x 2 −x 3 − x 4 F 3 (x))((1 − 2y 2 − y 3 − y 4 xF3 (x) − yF 3 (y)F 3 (y)) the determinantequalsx − yn[ −xy 4 (x + 1)F 3 (y) + yx 4 ](y + 1)F 3 (x) + (y + 1)(x + 1)(xy + 1)(x − y)x − yBy multiplying the generating function by (1 + x) −1 (1 + y) −1 , this determinant becomes[1 + xy xF ]4(x) − yF 4 (y)x − ywhere(3.4) F 4 (x) = 1 + x 2 F 3 (x)/(1 + x).Therefore,,ndet(H n−5 (F 3 )) = det(H n−6 (F 4 )).From (3.4) and the functional equation for F 3 (x), we obtainF 4 (x) =11 + x 3 − (x 3 + x 2 )F 4 (x) ..n

HANKEL DETERMINANTS FOR LATTICE PATHSTransformation 5: Substituting for [ F 4 with the above ] fraction, simplifying, and multiplying by (1 −x 2 F 4 (x))(1 − y 2 xF4 (x) − yF 4 (y)F 4 (y)) the determinantequalsx − yn[−xy 2 (y + 1)F 4 (y) + x 2 y (x + 1)F 4 (x) − (x − y) ( yx 2 + xy 2 − 1 ) ] [= 1 + xy xF ]5(x) − yF 5 (y),x − yx − ynwhere F 5 (x) = (1 + x)F 4 (x) − x. Hence, det(H n−6 (F 4 )) = det(H n−7 (F 5 )).Finally, it is routinely checked that F 5 (x) = F 0 (x).4. The quadratic transformation for Hankel determinantsOne can use the method introduced in the previous section to evaluate the Hankel determinants forgenerating functions satisfying a certain type of quadratic functional equation. The generating functionsF(x) in this section are the unique solution of a quadratic functional equation satisfying(4.1) F(x) =x du(x) + x k v(x)F(x) ,where u(x) and v(x) are rational power series with nonzero constants, d is a nonnegative integer, and k is apositive integer. Note that if k = 0, F(x) is not unique. Our task now is to derive a transformation T sothat det(H n (F)) = a det(H n−d−1 (T (F))) for some value a and nonnegative integer d. In addition to Hankelmatrices for the power series A = ∑ n≥0 a ix i , we will consider shifted Hankel matrices: H k n(A) denotes thematrix (a i+j+k−2 ) 1≤i,j≤n . Shifted matrices have appeared in Proposition 2.2 and 2.3.The first proposition is elementary:Proposition 4.1. If F satisfies (4.1), then G = u(0)F satisfiesdet(H n (G)) = u(0) n det(H n (F)), and G(x) =nx du(0) −1 u(x) + x k u(0) −2 v(x)G(x) .Proposition 4.2. Suppose F satisfies (4.1) with u(0) = 1. We separate u(x) uniquely as u(x) =u L (x) + x d+2 u H (x), where u L (x) is a polynomial of degree at most d + 1 and u H (x) is a power series.(i) If k = 1, then there is a unique G such thatMoreover,G(x) =and a shifted matrices appears with−v(x) − xu L (x)u H (x)u L (x) − x d+2 u H (x) − x d+1 G(x) ,G(x) = −xu H (x) − x −d v(x)F(x)det(H 1 n−d−1 (G(x))) = (−1)d(d+1)/2 det(H n (F(x))).(ii) If k ≥ 2, then there is a unique G such thatMoreover,andG(x) =−x k−2 v(x) − u L (x)u H (x)u L (x) − x d+2 u H (x) − x d+2 G(x) ,G(x) = −u H (x) − x k−d−2 v(x)F(x)det(H n−d−1 (G(x))) = (−1) d(d+1)/2 det(H n (F(x))).Proof. We prove only part (ii) as part (i) is similar. The generating function for H n (F) is given byxF(x) − yF(y)= 1 (x d+1x − y x − y u(x) + x k v(x)F(x) − y d+1 )u(y) + y k v(y)F(y)= −yd+1 u(x) − y d+1 x k v(x)F(x) + x d+1 u(y) + x d+1 y k v(y)F(y)(u(x) + x k v(x)F(x)) (u(y) + y k v(y)F(y))(x − y)

Sulanke and XinWe can multiply by (u(x)+x k v(x)F(x)) and by (u(y)+y k v(y)F(y)) without changing the above determinantby the product rule. Next we observe that x d divides F(x), and write u(x) = u L (x) + x d+2 u H (x) as in theproposition. The resulting generating function can be written as−y d+1 u L (x) + x d+1 u L (y)x − y+ (xy) d+1 −x(u H(x) − x k−d−2 v(x)F(x)) + y(u H (y) + y k−d−2 v(y)F(y)).x − yWe now set G(x) = −u H (x) − x k−d−2 v(x)F(x), which can be straightforwardly[ shown to agree ] withxF(x) − yF(y)the defining functional equations. Suppose that u L (x) = 1 + a 1 x + · · · a d+1 x d+1 , thenisx − ynequal to the determinant of the block-diagonal matrix⎛⎛⎞ ⎞0 · · · 0 1diag....⎜⎜⎟⎝⎝0 1 . a d−1⎠ , H(G(x)) ⎟⎠ .1 a 1 · · · a dThe determinant of the first block is easily seen to be (−1) d(d+1)/2 .□Given these propositions and that H 1 (A) = H(x −1 (A(x) − A(0))) for any series A, we can now defineour transformation T (F): For F satisfying (4.1),• if u(0) ≠ 1, then T (F) = G, as given in Proposition 4.1.• if u(0) = 1 and k = 1, then T (F) = x −1 (G(x) − G(0)), with G given in Proposition 4.2(i).• if u(0) = 1 and k ≥ 2, then T (F) = G, as given in Proposition 4.2(ii).Moreover, the relation between det(H n (F)) and det(H n (T (F))) is given in Propositions 4.1 and 4.2.Example 1: Other proofs of Propositions 1.1 and 2.2. For Motzkin paths with arbitrary t, thegenerating function F(x) satisfies1F(x) =1 − tx − x 2 F(x) .Applying Proposition 4.2 so F 1 = T (F) givesdet(H n−1 (F 1 )) = det(H n (F)) where F 1 (x) =11 − tx − x 2 F 1 (x) .Hence, F(x) = F 1 (x), and consequently det(H n (F(x))) = 1 for all n.Whereas the Gessel-Viennot-Lindström method leads to a proof in the shifted case for arbitrary t, as inProposition 2.2, we have been able to use the continued fractions technique only for t = 1 and t = 2.For t = 1 we will show that (det(Hn 1(F))) n≥1 = (1, 0, −1, −1, 0, 1, 1, . . .), continuing with period 6. LetG 1 (x) = (F(x) − 1)/x, so that det(Hn(F)) 1 = det(H n (G 1 )). Let G 2 = T (G 1 ) and G 3 = T (G 2 ), both underProposition 4.2(ii). Since1 + xG 1 (x) =1 − x − 2x 2 − x 3 G 1 (x)with d = 0, k = 3, u(x) = u L (x) = 1 − 2x, u H = 0, and v(x) = −(1 + x) −1 , we find thatxG 2 (x) =1 − x − 2x 2 − x 2 (1 + x)G 2 (x)with d = 1, k = 2, u(x) = u L (x) = 1 − x − 2x 2 , u H = 0, and v(x) = −(1 + x). Applying Proposition 4.2(ii)showsG 3 (x) = −x −1 (−(1 + x))G 2 (x)= −x −1 (−(1 + x))(−x(−(1 + x) −1 ))G 1 (x)= G 1 (x)and det(H n−3 (G 3 )) = − det(H n−1 (G 2 )) = − det(H n (G 1 )), which yields the periodicity of the sequence ofdeterminants.

HANKEL DETERMINANTS FOR LATTICE PATHSFor t = 2 we will show that det(H 1 n(F)) = n + 1 for n ≥ 1. Define, G 1 to satisfy,G 1 (x) =2 + x1 − 2x − 2x 2 − x 3 G 1 (x) .One can easily see that G 1 (x) = (F(x) − 1)/x with G 1 (0) = u 1 (0) −1 = det(H 1 (G 1 )) = det(H 1 1(F)) = 2. Forn ≥ 2, define, G n to satisfy,G n (x) =(n − 1) 2 (n 2 + n + x)(n 2 − n)(n 2 − 2n 2 x − 2x 2 ) − n 2 (n 2 − n + x)x 2 G n (x) .By induction one can show that G n = T ◦ T (G n−1 ) (under Prop. 4.1 then under Prop. 4.2), and thatG n (0) = u n (0) −1 = (n − 1)(n + 1)/n 2 . Also by induction and Proposition 4.1, for n ≥ 2,[n−1∏( ) ] n+1−i (i − 1)(i + 1)det(H n (G 1 )) = 2 n i 2 det(H 1 (G n ))which simplifies to det(H n (G 1 )) = n + 1.i=2∏n ( (i − 1)(i + 1)= 2 ni=2i 2) n+1−iExample 2: Another proof of Proposition 1.2. For large Schröder paths arbitrary t, we haveApplying T givesApplying T again, we obtainF(x) =11 − tx 2 − x 2 F(x) .det(H n−1 (F 1 )) = det(H n (F)), where F 1 (x) =(1 + t) n det(H n−1 (F 2 )) = det(H n (F 1 )), where F 2 (x) =1 + t1 + tx 2 − x 2 F 1 (x) .11 − tx 2 − x 2 F 2 (x) .This implies F 2 = F, and hence the recurrence det(H n (F)) = (1+t) n−1 det(H n−2 (F)), with initial conditiondet(H 1 (F)) = 1, and det(H 2 (F)) = 1 + t.Example 3: Another proof of Proposition 2.1. Consider the continued fractionF(x) =11 − tx − xF(x) ,where F(x) is the generating function for the Catalan numbers for t = 0 and the large Schröder numbers fort = 1.Under Proposition 4.2(i) we have a unique G 1 such that G 1 (x) = F(x) and det(H 1 n−1(G 1 )) = det(H n (F)).Taking G 2 = (G 1 (x) − 1)/x = (F(x) − 1)/x, we haveG 2 (x) =(1 + t)1 − (2 + t)x − x 2 G 2 (x)where det(H n−1 (G 2 )) = det(Hn−1(F)) 1 and u(x) = (1 − (2 + t)x)/(1 + t).Under Proposition 4.1 we have a unique G 31G 3 (x) =1 − (2 + t)x − (1 + t)x 2 G 3 (x) ,with G 3 (x) = G 2 /(1 + t) and det(H n−1 (G 3 )) = (1 + t) −(n−1) det(H n−1 (G 2 )).Under Proposition 4.2(ii) we have a unique G 4 such that G 4 (x) = (1 + t)G 3 (x) and det(H n−2 (G 4 )) =det(H n−1 (G 3 )).We see that G 4 (x) = G 2 (x); thus det(H n−1 (G 2 )) = (1+t) n−1 det(H n−2 (G 2 )) with det(H 1 (G 2 )) = 1+t.Hence det(H n (F)) = det(H n−1 (G 2 )) = (1 + t) n(n−1)/2 .

Sulanke and XinExample 4: Another proof of Proposition 2.3. To compute det(H 1 n(F)), first we considerH 1 n (F) = H n(F 1 ), where F 1 =Applying T shows that det(H n (F 1 )) = −(1 + t) n det(H n−2 (F 1 )).Example 5: For l = 3, recall the functional equation1F 0 (x) =1 − tx 3 − x 2 F 0 (x) .(t + 1)x1 − (2 + t)x 2 − x 3 F 1.For arbitrary t, our transformation gives more and more complicated expressions. This is not surprisingsince the Hankel determinants do not factor nicely. However, for t = 1 and for k = 1, 2, 3, the transformationgives nice results similar to that of Proposition 1.3: indeed, sequences of det(H k n(F 0 )) also have period 14.For k = 4 there is an interesting result.Subexample 5i: The sequence for det(H 1 n (F 0)) starts with 0, −1, 0, 1, 1, 0, −1, 0, 1, 0, −1, −1, 0, 1. If wedefine F 1 so that F 0 (x) = 1 + xF 1 (x), thendet(H n (F 1 )) = det(H 1 n (F 0)), with F 1 =Then applying T repeatedly so T (F i ) = F i+1 , we obtaindet(H n−2 (F 2 )) = − det(H n (F 1 )), where F 2 =x(x + 1)1 − 2x 2 − x 3 − x 3 F 1and d = 1.x(x + 1)(1 − x − x 2 − x 3 F 2 )and d = 1;det(H n−2 (F 3 ))1 + x − x 2= − det(H n (F 2 )), where F 3 =1 − 2x 2 + x 3 − x 3 and d = 0;F 3det(H n−1 (F 4 ))x= det(H n (F 3 )), where F 3 =(1 + x − x 2 )(1 − x − x 2 and d = 1;F 3 )det(H n−2 (F 5 ))x(x + 1)= − det(H n (F 4 )), where F 5 =1 − 2x 2 − x 3 − x 3 .F 5The periodicity is established by noticing that F 5 = F 1 and det(H n−7 (F 5 )) = − det(H n (F 1 )).Subexample 5ii: The sequence for det(Hn(F 2 0 )) starts with 1, 1, 1, 1, 0, 0, −1, −1, −1, −1, −1, 0, 0, 1,. Ifwe define G 0 so that F 0 (x) = 1 + x 2 G 0 (x), then det(H n (G 0 )) = det(Hn 2(F 0)),1 + xG 0 =1 − 2x 2 − x 3 − x 4 .G 0One can establish the periodicity using Proposition 4.2. However, this generating function has appeared inTransformation 3 of section 3, where one can see that(4.2) det(H n (G 0 )) = − det(H n+5 (F 0 )).Subexample 5iii: The sequence for det(Hn(F 3 0 )) starts with 1, −1, −1, 0, 0, 0, −1, −1, 1, 1, 0, 0, 0, 1 andcontinues with period 14. The verification for this case uses Proposition 4.2(ii) occasionally interspersed withProposition 4.1. Here we will only sketch the verification. By defining F 1 so that F 0 (x) = 1 + x 2 + x 3 F 1 (x),one finds that1 + 2x + x 2 + x 3F 1 =1 − 2x 2 − x 3 − 2x 4 − x 5 .F 1For the first transformation, with F 2 = T F 1 , we findF 2 =1 − 2x + x 3−1 + 4x 2 + x 3 + 2x 4 − x 2 (1 + 2x + x 2 + x 3 )F 2,in which u(x) = (−1+4x 2 + x 3 + 2x 4 )/(1 − 2x+x 3 ). Now, since u(0) = −1, one needs to apply Proposition4.1 for the next transformation. One proceeds until a generating function equal to F 1 appears to establishthe periodicity. We remark that d = 0 for each transformation until the final one which uses Proposition4.2(ii) with d = 3 (This corresponds to a fourth order block).

HANKEL DETERMINANTS FOR LATTICE PATHSSubexample 5iv: The sequence for det(H 4 n(F 0 )) begins with2, 3, 4, 0, 0, −4, −5, −6, −7, −8, 0, 0, 8, 9, 10,11, 12, 0, 0, −12, −13, −14, −15, −16, 0, 0, 16, . . ..For n ≥ 8, an essence of periodicity can be gleaned from the recurrencedet(H 4 n (F 0)) = 4 det(H n−1 (F 0 )) − det(H 4 n−7 (F 0)),for which we sketch a proof, often omitting the functional equations.We will be applying the transformation T eight times, alternating its definition to be first underProposition 4.1 and then under Proposition 4.2(ii). Let F 1 satisfy F 0 = 1 + x 2 + x 3 + x 4 F 1 . Hence,det(H n (F 1 )) = det(Hn 4(F 0)), andF 1 =2 + 3x + 2x 2 + 2x 3 + x 41 − 2x 2 − x 3 − 2x 4 − 2x 5 − x 6 F 1.Here u(0) = 1 2 , where u(x) is for F 1. Thus, with F 2 = T F 1 , det(H n (F 2 )) = ( 1 2 )n det(H n (F 1 )). Nowd = 0, where d is for F 2 . With F 3 = T F 2 , det(H n−1 (F 3 )) = det(H n (F 2 )).Here u(0) = 4 3 , where u(x) is for F 3. Thus, with F 4 = T F 3 , det(H n−1 (F 4 )) = ( 4 3 )n−1 det(H n−1 (F 3 )).Now d = 0, where d is for F 4 . With F 5 = T F 4 , det(H n−2 (F 5 )) = det(H n−1 (F 4 )).Here u(0) = 9 8 , where u(x) is for F 5. Thus, with F 6 = T F 5 , det(H n−2 (F 6 )) = ( 9 8 )n−2 det(H n−2 (F 5 )).Now d = 0, where d is for F 6 . With F 7 = T F 6 , det(H n−3 (F 7 )) = det(H n−2 (F 6 )).Here u(0) = 4 3 , where u(x) is for F 7. Thus, with F 8 = T F 7 , det(H n−3 (F 8 )) = ( 4 3 )n−3 det(H n−3 (F 7 )).Now d = 2, where d is for F 8 . With F 9 = T F 8 , det(H n−6 (F 9 )) = − det((H n−3 (F 8 )) =( 5 4 , 6 4 , 7 4 , 8 4 , 0, 0, −8 4 , −9 4 , −10 4 , . . .).Thus, (surprisingly)det(H n−6 (F 9 )) = −( 1 2 )n ( 4 3 )n−1 ( 9 8 )n−2 ( 4 3 )n−3 det(H 4 n (F 0))(4.3)= − 1 4 det(H4 n (F 0))Moreover,F 9 =20 + 16x − 8x 2 − 4x 3 + x 48(2 − 4x 2 − 2x 3 + x 4 ) − 16x 4 F 9= 5 4 + x + 2x2 + 3x 3 + 6x 4 + 10x 5 + · · ·It is easily verified that F 9 (x) and 1 4 +G 0(x), where G 0 appears in Subexample 5ii, satisfy the same functionalequation, and hence are equal. Therefore,[ ]xF9 (x) − yF 9 (y)det(H n−6 (F 9 )) =x − yn−6[ 1=4 + xG ]0(x) − yG 0 (y)x − yn−6= 1 4 det(H4 n−7(F 0 )) + det(H n−6 (G 0 ))where 1 4 det(H4 n−7(F 0 )) is 1 4 times the determinant of the 1,1-minor of H n−6(G 0 ), equivalently of H 2 n−6(F 0 ).Combining this with identity (4.3) and noting det(H n (G 0 )) = − det(H n+5 (F 0 )) from (4.2) proves the initialrecurrence of this subexample.Acknowledgments. We, particularly Xin, are most grateful to Ira Gessel. We also appreciate thesuggestions made by Christian Krattenthaler, Lou Shapiro, and referees.References[1] M. Aigner, Motzkin Numbers, Europ. J. Comb. 19 (1998), 663-675.[2] C. Banderier and S. Schwer, Why Delannoy numbers?, J. Stat. Plan. and Infer., 135 (2005), 40-54.[3] D. Bressoud, Proofs and Confirmations: The Story of the Alternating Sign Matrix Conjecture, Cambridge University Press,1999.

Sulanke and Xin[4] R. A. Brualdi and S. Kirkland, Aztec diamonds and digraphs, and Hankel determinants of Schröder numbers, J. Comb.Th., ser. B, 94 (2005), 334-351.[5] I. M. Gessel and G. X. Viennot, Binomial determinants, paths, and hook length formulae. Adv. in Math. 58 (1985), no. 3,300-321.[6] I. M. Gessel and G. Xin, The generating function of ternary trees and continued fractions, Electron. J. Combin., to appear,ArXiv:math.CO/0505217.[7] W. B. Jones and W. J. Thron, Continued Fractions: Analytic Theory and Applications, Encyclopedia of Mathematics andits Applications, vol. 11, Addison-Wesley Publishing Co., Reading, Mass., 1980.[8] C. Krattenthaler, Advanced determinant calculus: a complement, Linear Algebra Appl. 411 (2005), 68-166The Andrews Festschrift (Maratea, 1998). Sem. Lothar. Combin. 42 (1999), Art. B42q, 67 pp. (electronic).[9] E. Pergola, R. Pinzani, S. Rinaldi, and R. A. Sulanke, Lattice Path Moments by Cut and Paste. Adv. in Appl. Math. 30(2003), no. 1-2, 208-218.[10] Shapiro, L. W. A Catalan triangle. Discrete Math. 14 no. 1 (1976), 83–90.[11] R. A. Sulanke, Moments of generalized Motzkin paths, J. Int. Seq., 3 (2000), Article 00.1.1[12] R. A. Sulanke, Objects counted by the central Delannoy numbers. J. Int. Seq. 6 (2003), no. 1, Art. 03.1.5, 19 pp.[13] G. Viennot, Une théorie combinatoire des polynômes orthogonaux généraux, Lecture notes, Univ. Quebec, Montreal, Que.,1983.[14] H. S. Wall, Analytic Theory of Continued Fractions, Van Nostrand, New York, 1948.Department of Mathematics, Boise State University, Boise, ID, USA 83725E-mail address: sulanke@math.boisestate.eduDepartment of Mathematics, University of Kentucky, Lexington, KY, USA 40506E-mail address: gxin@ms.uky.edu