HANKEL DETERMINANT SOLUTIONS TO SEVERAL DISCRETE INTEGRABLE SYSTEMS AND THE LAURENT PROPERTY XIANG-KE CHANG∗†, XING-BIAO HU∗, GUOCE XIN‡ Abstract. ManydiscreteintegrablesystemsexhibittheLaurentphenomenon. Inthispaper,we investigate threeintegrable systems: the Somos-4recurrence, the Somos-5recurrence andasystem relatedtoso-calledA1Q-system,whosegeneralsolutionsarederivedintermsofHankeldeterminant. 5 As a result, we directly confirm that they satisfy the Laurent property. Additionally, it is shown 1 that the Somos-5 recurrence can be viewed as a specified B¨acklund transformation of the Somos-4 0 recurrence. TherelatedtopicsaboutSomospolynomialsarealsostudied. 2 n Key words. Hankeldeterminantsolution,Laurentproperty,discreteintegrablesystems a J AMS subject classifications. 11B83,37J35,15B05, 11Y65 5 2 1. Introduction. Laurent phenomenon is a crucial property behind integrality shared by a class of combinatorial models while integrability is a key feature for a ] h class of what we call integrable systems. An interesting observation is that many p discrete integrable systems exhibit the Laurent phenomenon, and many mappings - with the Laurent property are proved to be integrable. In the recent decade, a lot of h t relatedworkweredoneviathestudyofaclassofcommutativealgebras,calledcluster a algebras [19, 21], for its success in proving the Laurent phenomenon. For instance, m see[11,12,17,20,22,23,24,25,26,31,34]. Morerecently,ageneralizationofcluster [ algebras called Laurent phenomenon algebras [39] was also introduced in order to 2 study the Laurent phenomenon [1, 36]. v One question is whether we can prove the Laurent property (or a stronger prop- 9 erty) from other views. In this paper, we shall give one possible choice—the view of 7 explicit determinant solution. That is, the Laurent property of some discrete inte- 8 grablesystemscanbeprovedbytheirdeterminantsolutions. Ofcourse,howtoobtain 2 0 the desireddeterminantformulaeis anotherchallengingproblem. This approachsuc- . ceeds for the following three integrable systems: 1 0 The first two systems are the Somos-4 recurrence and the Somos-5 recurrence 5 defined by 1 : S S =αS S +βS2 , (1.1) v n n 4 n 1 n 3 n 2 − − − − i X and r a S S =α˜S S +β˜S S , (1.2) n n 5 n 1 n 4 n 2 n 3 − − − − − with nonzero parameters α,β,α˜,β˜ and arbitrary initial values, respectively. Here we assume S never vanish. By using their theory of cluster algebras [19, 21], Fomin i and Zelevinsky [20, 32, 35] provedthat the Somos-4and Somos-5recurrences exhibit the Laurent property, that is, S are Laurent polynomials in its initial values with n coefficients in Z[α,β] and Z[α˜,β˜], respectively. Hone constructed explicit solutions to the Somos-4 and Somos-5 recurrences in terms of the Weierstrass sigma function Emails: [email protected]; [email protected];[email protected] ∗LSEC,InstituteofComputationalMathematicsandScientificEngineeringComputing,AMSS, ChineseAcademyofSciences, P.O.Box2719,Beijing100190, PRChina. †UniversityofChineseAcademyofSciences,Beijing,PRChina. ‡Departmentofmathematics, CapitalNormalUniversity,Beijing100048, PRChina. 1 2 TEXPRODUCTION in [30] and [33], respectively. He also indicated that the Somos-4 recurrence can be thoughtofasanintegrablesymplecticmapping[7,45]bymakingachangeofvariables andSomos-5be aintegrablePoissonmapping. Unfortunately,it is notobvioustosee the Laurent phenomenon from their Weierstrass sigma function formulae. Our result for the Somos-4 recurrence with initial values 1,1,x,y is as follows. Theorem 1.1. If we let S = 1,S = 1,S = x,S = y,S = αy + βx2 1 0 1 2 3 − by shifting the indices, then the Somos-4 recurrence (1.1) has the following explicit Hankel determinant representation: S =det(p ) , n i+j 0 i,j n 1 ≤ ≤ − where pm = βx2−y√2+αxαyx3−αypm−1 + β+αyx−xypm−2 +Pmk=−02pkpm−2−k(m ≥ 2) with initial values p =x,p = √α. 0 1 − The Laurentphenomenonclearlyholdsfor initialvalues 1,1,x,y by the theorem: i) the determinant formula asserts that S is a Laurent polynomial in √α,x,y and n polynomial in β; ii) the recursion (1.1) shows that S is rational in α and that S n n exists and not equal to 0 when α = 0, which can be checked by induction. To see that the Laurent phenomenon holds for arbitrary initial values, we observe that the recursion (1.1) holds for S if and only if it holds for S /c for constant c=0, if and n n 6 only if it holds for S tn for constant t = 0. This makes it sufficient to consider the n sequence Sn(S0)n , which starts w6ith 1,1. Note that the reduction to the case {S0 S1 }n≥0 1,1,x,y is a simple but important step, since the Somos-4 recurrence with arbitrary initial values does not seem to have the desired Hankel determinant representation. Similarly, it is sufficient to give the following result for the Somos-5 recurrence. Theorem 1.2. Ifwelet S =1,S =1,S =x,S =y,S =z,S =α˜z+β˜xy, 2 1 0 1 2 3 S =α˜2xz+α˜β˜x2y+β˜yz by −shifting th−e indices, then the Somos-5 recurrence (1.2) 4 has the following explicit Hankel determinant representation: S =xn+1det(p ) , (1.3) 2n i+j 0 i,j n 1 ≤ ≤ − S =yn+1det(q ) , (1.4) 2n+1 i+j 0 i,j n 1 ≤ ≤ − where p is recursively given by p = z ,p = β˜, m 0 x2 1 − α˜x2z+α˜x2y2+β˜x3y+zy2 z2 p = − − p m m 1 − xyz − yx4α˜2+x5α˜β˜+β˜zx3 z2y m−2 + x2yz − pm−2+ X pkpm−2−k(m≥2), k=0 and q is recursively given by q = α˜z+β˜xy,q = β˜, m 0 y2 1 − α˜x2z+α˜x2y2 z2 zy2+β˜x3y q = − − q m m 1 xyz − α˜y4 α˜z2+β˜xy3 β˜xzy m−2 + − y2z − qm−2+ X qkqm−2−k(m≥2). k=0 The third system is called the extended A Q-system: 1 S S =S2 +β, (1.5) n n 2 n 1 − − DeterminantSolutionsandtheLaurentProperty 3 with nonzero parameters β and arbitrary initial values. Here we assume S never i vanish. Thespecialcasewhenβ =1andalltheinitialvaluesare1sreducestotheA 1 Q-system [12, 14, 16]. From the view of integrable systems, (1.5) is C-integrable[8]. Note that the Laurent property of (1.5) was shown in [12, 20, 34]. By dividing (1.5) by S2, we may assume the initial value is S = 1,S = x (β 0 0 1 changes but do not affect the Laurent property). The determinant solution is as follows, from which the Laurent property is obviously satisfied. Theorem 1.3. If we let S = 1,S = x, then the extended A Q-system (1.5) 0 1 1 has the following explicit Hankel determinant representation: S =det(p ) , n i+j 0 i,j n 1 ≤ ≤ − where pm = β+1x−x2pm−1+Pmk=−01pkpm−1−k(m ≥ 2) with initial values p0 = x,p1 = 1. All three theorems can be proved by employing Sulanke and Xin’s method of quadratictransformationforHankeldeterminants[43,48]orclassicdeterminanttech- nique. Inparticular,thespecialcasex=y =α=β =1ofTheorem1.1wasobtained bythethirdnamedauthorin[48]. Itisalsoconfirmedbyusingtheclassicdeterminant techniquein[9],whichisusedtosolveaconjecturethatacertaindeterminantsatisfies a specified Somos-4 recurrence [3, 4]. In Section 3, we shall give a detailed proof for the Somos-4 case by using Sulanke and Xin’s method of quadratic transformation, and prove the result for the extended A Q-system by applying the classic determi- 1 nant technique. As for the Somos-5 recurrence, we can solve it via the B¨acklund transformation[29,40]oftheSomos-4recurrence,whichwillbeintroducedinSection 2. As a result, the Laurent property for the three models are confirmed from their determinant solutions. In Section 4, we prove the Somos-4 polynomials result in [41] in our own way and give more Somos polynomial sequences. 2. Preliminaries. 2.1. Sulanke and Xin’s quadratic transformation for Hankel determi- nants. There are many classical tools for evaluating Hankel determinants, such as orthogonalpolynomialsapproach(cf. e.g.[38,10,46]),Gessel-Viennot-Lindstro¨mthe- orem (cf. ep.g. [5, 27, 43]),the J -fractions (cf. e.g. [38, 47]) and the S-fractions (cf. e.g.[37]). Recently, Sulanke and Xin proposed a method of quadratic transformation for Hankel determinants [43] developed from the continued fraction method of Ges- sel and Xin [28]. In [48], Xin applied the special quadratic transformation to solve Somos’ conjecture about the Hankel determinant solution to the Somos-4 recurrence with initial values S = 1,i = 0,1,2,3 and α = 1,β = 1. We restate the iterative i transformation as follows. Lemma 2.1. Given the initial values a ,b ,c ,d ,e ,f , let the generating func- 0 0 0 0 0 0 tion Q (x) be the unique power series solution of 0 a +b x 0 0 Q (x)= . 0 1+c x+d x2+x2(e +f x)Q (x) 0 0 0 0 0 4 TEXPRODUCTION If a ,b ,c ,d ,e ,f are recursively defined by n+1 n+1 n+1 n+1 n+1 n+1 a3e +a2d a b c +b2 a = n n n n− n n n n, n+1 − a2 n a4f +c a3d a2c2b +2a c b2 a2b d b3 b = n n n n n− n n n n n n− n n n− n, n+1 − a3 n c =c , n+1 n a2d 2a b c +2b2 d = n n− n n n n, n+1 − a2 n e = 1, n+1 − b n f = , n+1 −a n where we suppose that a =0,n=0,1,2, , then n 6 ··· det(Hn(Q0))=an0a1n−1···an−1. Here we remark that H (R) denotes the Hankel determinant det(r ) for n i+j 0 i,j n 1 any power series R(x)=P∞n=0rnxn. ≤ ≤ − The basic toolin ourapproachis the followingsurprisingresult, whichsolvesthe above recursion system. Lemma 2.2 (Theorem 2 in [48]). Suppose c = c,e = 1, then the recursion n n − system in Lemma 2.1 satisfies a 2(f +c+f )2 0 0 1 a a +a a =2a a +a (2f +c)(f +c+f ) . (2.1) n+2 n+1 n+1 n 0 1 0 1 0 1 − a n+1 2.2. B¨acklund transformation. A B¨acklund transformation is a transforma- tion between a solution u of a given differential or difference equation, (u)=0, 1 L and another solution v of another differential or difference equation, (v)=0, 2 L where maybethesameas,ordifferentfrom, . Itisanimportanttoolforfinding 2 1 L L new solutions in soliton theory and integrable systems. See [29, 40] for more details. Here we shall give an example, which is concerned about B¨acklund transformation in bilinear form between two solutions of the Somos-4 recurrence. It can be used for solving the Somos-5 recurrence. Firstly, we rewrite the Somos-4 recurrence as a bilinear form (e2Dn αeDn β)S S =0, (2.2) n n − − · where the Hirota’s bilinear operators [29] are defined as follows: ∂ ∂ ∂ ∂ DzmDtka·b≡(∂z − ∂z )m(∂t − ∂t )ka(z,t)b(z′,t′)|z′=z,t′=t, ′ ′ ∂ ∂ eδDnan bn exp [δ( )]anbn′ n′=n =an+δbn δ. · ≡ ∂n − ∂n′ | − DeterminantSolutionsandtheLaurentProperty 5 Then, we have the following lemma. Lemma 2.3. The Somos-4 recurrence has the following bilinear B¨acklund trans- formation: (e Dn λeDn µ)f g =0, (2.3) − n n − − · (µe3/2Dn ηe 1/2Dn αe1/2Dn)f g =0, (2.4) − n n − − · where µ,λ,η are arbitrary constants. Proof. Let f be a solution of the Somos-4 recurrence. What we need to prove is n that g satisfying (2.3) and (2.4) is another solution of the Somos-4 recurrence, i.e., n P (e2Dn αeDn β)g g =0, n n ≡ − − · Then, using the bilinear operator identities (e2Dnf f )g2 f2(e2Dng g )=2sinh(D )(eDnf g ) (e Dnf g ), n· n n− n n· n n n· n · − n· n (eDnfn·fn)gn2 −fn2(eDngn·gn)=2sinh(1/2Dn)(e1/2Dnfn·gn)·(e−1/2Dnfn·gn), sinh(D )(eDnf g ) (f g )=sinh(1/2D )(e3/2Dnf g ) (e 1/2Dnf g ), n n n n n n n n − n n · · · · · · sinh(δD )f f =0, n n n · we have Pf2 =[(e2Dn αeDn β)f f ]g2 f2(e2Dn αeDn β)g g − n − − n· n n− n − − n· n =[(e2Dnf f )g2 f2(e2Dng g )] α[(eDnf f )g2 f2(eDng g )] n· n n− n n· n − n· n n− n n· n =2sinh(D )(eDnf g ) (e Dnf g ) n n n − n n · · · 2αsinh(1/2Dn)(e1/2Dnfn gn) (e−1/2Dnfn gn) − · · · =2µsinh(D )(eDnf g ) (f g ) n n n n n · · · 2µsinh(1/2Dn)(e3/2Dnfn gn) (e−1/2Dnfn gn) − · · · =0. Thus the proof is completed. 3. Determinant solution and Laurent property. 3.1. On the Somos-4 recursion system. TheHankeldeterminantformulain Theorem 1.1 is discovered by employing Sulanke and Xin’s quadratic transformation for Hankel determinants. The proof is given below. Proof of Theorem 1.1. Let P(t) = P∞n=0pntn be the generating function of the sequence p , we only need to prove S = det(H (P)). Additionally, it is easy { n}∞n=0 n n to see that P(t) satisfies a +b t 0 0 P(t)= (3.1) 1+c t+d t2+t2(e +f t)P(t) 0 0 0 0 6 TEXPRODUCTION where a =x, 0 βx2 y2+αx3 b = − , 0 − √αy βx2 y2+αx3 αy c = − − , 0 − √αxy β+αx xy d = − , 0 − y e = 1, 0 − f =0. 0 By Lemma 2.1, the recursion (1.1) we need to prove is transformed to a a a =α+β/a . (3.2) n n 1 n 2 n 1 − − − Lemma 2.1 also gives a1 = y/x2 and f1 = βx2−√yα2x+yαx3. Thus applying Lemma 2.2 gives βx2+y2+αx3+αy a =( )/a α/a2 a . (3.3) n+2 xy n+1− n+1− n By substituting (3.3) with n replaced by n 2 into (3.2), we need to show that − βx2+y2+αx3+αy T(n):=( )a a αa a2 a2 αa β =0 (3.4) xy n−1 n−2− n−2− n−1 n−2− n−1− holds for n 2. ≥ Weprovethisbyinduction. ItiseasytoconfirmT(2)=0. AssumeT(n 1)=0. − Replacing a by (3.3) with n replaced by n 3 in T(n), we have n 1 − − βx2+y2+αx3+αy T(n)=( )a a αa a2 a2 αa β =T(n 1)=0. xy n−2 n−3− n−3− n−2 n−3− n−2− − This completes the proof. 3.2. On the Somos-5 recursion system. In finding the determinant solution of the Somos-5 recursion, the key observation is that the even and odd terms of the Somos-5 sequence are both Somos-4 sequence. This observation has been made by Hone (Proposition 2.8 in [33]), through making the connection with a second order nonlinearmappingwithafirstintegral. Herewegiveanalternativeexplanationfrom the view of B¨acklund transformation, which seems simpler and more intuitive. Now we prove Theorem 1.2 by using Lemma 2.3 and Theorem 1.1. Proof of Theorem 1.2. If we let f = S and g = S , then we get two n 2n n 2n+1 coupled equations g f α˜f g β˜g f =0, (3.5) n+1 n 1 n+1 n 1 n n − − − − f g α˜g f β˜f g =0. (3.6) n+2 n 1 n+1 n n+1 n − − − Recall that, as is indicated in Lemma 2.3, the Somos-4 recurrence (1.1) has the following B¨acklund transformation DeterminantSolutionsandtheLaurentProperty 7 g f λf g µg f =0, (3.7) n+1 n 1 n+1 n 1 n n − − − − µf g ηg f αf g =0, (3.8) n+2 n 1 n+1 n n+1 n − − − whereµ,λ,ηarearbitraryconstants. Whenµ=β˜,λ=α˜,η =α˜β˜,α=β˜2,itiseasyto checkthat(3.7),(3.8)arethesameas(3.5),(3.6). Thismeansthatsubsequencesf = n S andg =S oftheSomos-5recurrence(1.2)maysatisfythesameSomos-4re- 2n n 2n+1 currence(1.1)withα=β˜2andβ = α˜(α˜3yxz+α˜2β˜y2x2+α˜β˜y2z+α˜β˜z2+2α˜2zyx+α˜2β˜x2z+β˜2α˜yx3), xyz where β is determined by the first five values of f . In fact, by use of the meaning n of B¨acklund transformation, this can be confirmed by induction. Thus, it is easy to constructthegeneralsolutiontotheSomos-5recurrence(1.2). Andonecancomplete the proof after some substitutions and calculations. It is naturalto ask if we can givea more directproofto the Somos-4recursionof S (or S ). The answer is positive. We sketch the idea for S in the end of this 2n 2n 1 2n − subsection. It is easy to perform the detailed proof by Maple. We can eliminate the odd S to obtain the recursion n S2n 6S2n α˜2S2n 4S2n 2 β˜ (cid:0)S2n 6S2n 2+α˜S2n 42(cid:1) − − − − = − − − , (3.9) β˜ (cid:0)S2n−6S2n−2+α˜S2n−42(cid:1) S2n−8S2n−2−α˜2S2n−6S2n−4 using which we can prove S S =αS S +βS2 2n 2n 8 2n 2 2n 6 2n 4 − − − − by induction. Eliminating S in the above equation by using (3.9), we are left to show the 2n nullity of a relation, sayingtarget(S ,S ,S ,S ). Eliminating S in 2n 2 2n 4 2n 6 2n 8 2n 2 − − − − − target(S ,S ,S ,S )byusing (3.9)withnreplacedbyn 1,weobtain 2n 2 2n 4 2n 6 2n 8 − − − − − S2 S target(S2n−2,S2n−4,S2n−6,S2n−8)= S222nn−46S22nn−180target(S2n−4,S2n−6,S2n−8,S2n−10). − − Then the proof is completed after checking target(S ,S ,S S )=0. 6 4 2, 0 3.3. On the extended A Q-system. LetusfirstproveTheorem1.3byclassic 1 determinant technique. This approach gives no hint how we discovered the Hankel determinant formula, but this proof seems to be shorter than using the Sulanke and Xin’s quadratic transformation method. Proof of Theorem 1.3. Let H(l) denote det(p ) and we shall use the n i+j+l 0 i,j n 1 ≤ ≤ − conventions H(l) =1, 0 (3.10) H(l) =0 for n<0. n Firstly, we assert that there hold H(1) =1, n 0; (3.11) n ≥ H(0) =xH(0) +βH(2) , n 1. (3.12) n n−1 n−2 ≥ Employing the Jacobi determinant identity [2, 6], we get H(0)H(2) =H(0) H(2) (H(1) )2, n 2. (3.13) n n−2 n−1 n−1− n−1 ≥ 8 TEXPRODUCTION By use of (3.11) and (3.12), replace H(1) , H(2) and H(2) in (3.13), then we can n 1 n 2 n 1 obtain − − − H(0) H(0) =(H(0))2+β, n 1. n+1 n−1 n ≥ Noting that S =H(0), thus it suffices to confirm the formulae (3.11) and (3.12). n n It is noted that p also satisfy the following recurrence: m m 1 β+1 − pm = pm 1+ X pkpm 1 k, m 2. (3.14) x − − − ≥ k=1 We will prove (3.11) and (3.12) by employing row and column operations for the determinants and using this recurrence. (1) Let’s consider H firstly. n Step 1: Subtract the i-th column multiplied by p from the n-th column for n 1 i i = 1,2, ,n 1 and subtract the (n 1)-th colum−n −multiplied by β+1 from the ··· − − x n-thcolumn. Next,applyingasimilarproceduretothe(n 1)-th,(n 2)-th, ,2-nd − − ··· columns, by using the recursion relation (3.14), we have p 0 ... 0 (cid:12) 1 (cid:12) (cid:12) p p p ... p p (cid:12) (cid:12) 2 1 1 1 n 1 (cid:12) (cid:12) 2 2 − (cid:12) (cid:12) (cid:12) H(1) =(cid:12)(cid:12)(cid:12) p3 Xi=1pip3−i ... Xi=1pipn+1−i (cid:12)(cid:12)(cid:12) n (cid:12)(cid:12)(cid:12) ... ... ... ... (cid:12)(cid:12)(cid:12) (cid:12) n 1 n 1 (cid:12) (cid:12) − − (cid:12) (cid:12)(cid:12) pn Xpipn i ... Xpip2n 2 i (cid:12)(cid:12) (cid:12) i=1 − i=1 − − (cid:12) (cid:12) (cid:12) p p ... p p (cid:12) 1 1 1 n 1 (cid:12) (cid:12) 2 2 − (cid:12) (cid:12) (cid:12) (cid:12)(cid:12) Xpip3−i ... Xpipn+1−i (cid:12)(cid:12) (cid:12) i=1 i=1 (cid:12) =(cid:12)(cid:12)(cid:12) ... ... ... (cid:12)(cid:12)(cid:12). (cid:12) n 1 n 1 (cid:12) (cid:12) − − (cid:12) (cid:12)(cid:12) Xpipn i ... Xpip2n 2 i (cid:12)(cid:12) (cid:12) i=1 − i=1 − − (cid:12) (cid:12) (cid:12) Step 2: Perform row operations for the above determinant. For fixed k = 2,3, ,n 1, we subtract the i-th row multiplied by p /p for i=1, ,k 1 k+1 i 1 ··· − − ··· − from the k th row. Then it follows that (3.11) holds. − Now let’s turn to the proof of (3.12). Obviously, the result holds for n = 1. In the following we consider the case of n>1. Step 1: Subtract the i-th column multiplied by p from the n-th column for n 1 i i = 2,3, ,n 1 and subtract the (n 1)-th colum−n −multiplied by β+1 from the ··· − − x n-thcolumn. Next,applyingasimilarproceduretothe(n 1)-th,(n 2)-th, ,2-nd − − ··· DeterminantSolutionsandtheLaurentProperty 9 columns, by using the recursion relation (3.14), we have p β 0 ... 0 (cid:12) 0 − (cid:12) (cid:12) p p p p p ... p p (cid:12) (cid:12) 1 1 0 1 1 1 n 2 (cid:12) (cid:12) 2 2 2 − (cid:12) (cid:12) (cid:12) H(0) =(cid:12)(cid:12)(cid:12) p2 Xi=1pip2−i Xi=1pip3−i ... Xi=1pipn−i (cid:12)(cid:12)(cid:12) n (cid:12)(cid:12)(cid:12) ... ... ... ... ... (cid:12)(cid:12)(cid:12) (cid:12) n 1 n 1 n 1 (cid:12) (cid:12) − − − (cid:12) (cid:12)(cid:12) pn 1 Xpipn 1 i Xpipn i ... Xpip2n 3 i (cid:12)(cid:12) (cid:12) − i=1 − − i=1 − i=1 − − (cid:12) (cid:12) (cid:12) p p p p ... p p (cid:12) 1 0 1 1 1 n 2 (cid:12) (cid:12) 2 2 2 − (cid:12) (cid:12) (cid:12) (cid:12)(cid:12) Xpip2−i Xpip3−i ... Xpipn−i (cid:12)(cid:12) (cid:12) i=1 i=1 i=1 (cid:12) =x(cid:12)(cid:12)(cid:12) ... ... ... ... (cid:12)(cid:12)(cid:12) (cid:12) n 1 n 1 n 1 (cid:12) (cid:12) − − − (cid:12) (cid:12)(cid:12) Xpipn 1 i Xpipn i ... Xpip2n 3 i (cid:12)(cid:12) (cid:12) i=1 − − i=1 − i=1 − − (cid:12) (cid:12) (cid:12) p p p ... p p (cid:12) 1 1 1 1 n 2 (cid:12) (cid:12) 2 2 − (cid:12) (cid:12) (cid:12) (cid:12)(cid:12) p2 Xpip3−i ... Xpipn−i (cid:12)(cid:12) (cid:12) i=1 i=1 (cid:12) +β(cid:12)(cid:12)(cid:12) ... ... ... ... (cid:12)(cid:12)(cid:12) (cid:12) n 1 n 1 (cid:12) (cid:12) − − (cid:12) (cid:12)(cid:12) pn 1 Xpipn i ... Xpip2n 3 i (cid:12)(cid:12) (cid:12) − i=1 − i=1 − − (cid:12) (cid:12) (cid:12) Step 2: Perform row operations for the above two determinants. For fixed k = 2,3, ,n 1, we subtract the i-th row multiplied by p /p for i=1, ,k 1 k+1 i 1 ··· − − ··· − from the k th row. Then it follows that (3.12) holds. − Therefore, we complete the proof. Remark: If we let f = γn/2S , then f satisfy f f = f2 +βγn 1 with n n n n n 2 n 1 − f0 =1 and f1 =x√γ. And we can see that fn gives the −Fibonacc−i sequence when { } γ = 1, β = 1 and x = √ 1. It means that every Fibonacci number can be − − − expressed as a Hankel determinant. Moreover, it is noted that the extended A Q-system satisfy a three term re- 1 currence [34]. In the following we use the relations of Hankel determinants to show it. Theorem 3.1. The extended A Q-system with S = 1,S = x satisfy the 1 0 1 following three term recurrence x2+1+β S = S S . (3.15) n n 1 n 2 x − − − Proof. We shall firstly prove the formula H(0) =xH(2) H(2) , (3.16) n n−1− n−2 10 TEXPRODUCTION where the notations are the same as that in the previous proof. It is easy to see that H(0) =xH(2) +H˜ , where n n 1 n − 0 p p p (cid:12) 1 2 n 1 (cid:12) (cid:12) p p p ··· p− (cid:12) (cid:12) 1 2 3 n (cid:12) H˜ =(cid:12)(cid:12) p2 p3 p4 ··· pn+1 (cid:12)(cid:12). n (cid:12)(cid:12)(cid:12) ... ... ... ·.·.·. ... (cid:12)(cid:12)(cid:12) (cid:12) (cid:12) (cid:12) pn 1 pn pn+1 p2n 1 (cid:12) (cid:12) − ··· − (cid:12) Based on this observation, it suffices to prove that H˜ = H(2) . Step 1: Subtract the i-th column multiplied by pn −frno−m2the n-th column for n 1 i i = 1,2, ,n 1 and subtract the (n 1)-th colum−n −multiplied by β+1 from the ··· − − x n-thcolumn. Next, applyingasimilarproceduretothe (n 1)-th,(n 2)-th, ,3-rd − − ··· columns, by using the recursion relation (3.14), we have 0 p 0 ... 0 (cid:12) 1 (cid:12) (cid:12) p p 0 ... 0 (cid:12) (cid:12) 1 2 (cid:12) (cid:12) p p p p ... p p (cid:12) (cid:12) 2 3 1 2 1 n 1 (cid:12) (cid:12) 2 2 − (cid:12) (cid:12) (cid:12) H˜n =(cid:12)(cid:12)(cid:12) p3 p4 Xi=1pip4−i ... Xi=1pipn+1−i (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) ... ... ... ... ... (cid:12)(cid:12)(cid:12) (cid:12) n 2 n 2 (cid:12) (cid:12) − − (cid:12) (cid:12)(cid:12) pn 1 pn Xpipn i ... Xpip2n 3 i (cid:12)(cid:12) (cid:12) − i=1 − i=1 − − (cid:12) (cid:12) (cid:12) p p ... p p (cid:12) 1 2 1 n 1 (cid:12) (cid:12) 2 2 − (cid:12) (cid:12) (cid:12) (cid:12)(cid:12) Xpip4−i ... Xpipn+1−i (cid:12)(cid:12) (cid:12) i=1 i=1 (cid:12) =−(cid:12)(cid:12)(cid:12) ... ... ... (cid:12)(cid:12)(cid:12). (cid:12) n 2 n 2 (cid:12) (cid:12) − − (cid:12) (cid:12)(cid:12) Xpipn i ... Xpip2n 3 i (cid:12)(cid:12) (cid:12) i=1 − i=1 − − (cid:12) (cid:12) (cid:12) Step 2: Perform row operations for the above determinant. For fixed k = 2,3, ,n 2, we subtract the i-th row multiplied by p /p for i=1, ,k 1 k+1 i 1 from··th·e k−th row. Then it follows that H˜ = H(2) . − ··· − Next, −noting that S = H(0), the thnree t−ermn−r2ecurrence can be obtained by n n combining (3.12) and (3.16). Remark: When β =x2 1, S satisfy the three term recurrence n − { } S =2xS S . n n 1 n 2 − − − with initial data S =1 and S =x. This just meets the Chebyshev polynomials. 0 1 It is also noted that every sequence produced by the extended A Q-system is a 1 Somos-4 sequence [44]. More precisely, if S satisfy (1.5), then S also satisfy n n { } { } (x2+1+β)2 (x2+1+β)2 S S = S S +(1 )S2. n+2 n−2 x2 n+1 n−1 − x2 n