1 Introduction↩︎

The so-called Somos-4 recurrence is a three-term discrete quadratic equation of fourth order \[S_{n+4}S_{n} =\alpha S_{n+3}S_{n+1} + \beta S_{n+2}^2, \label{somoseq4}\tag{1}\] where \(\alpha\) and \(\beta\) are constant parameters, while the Somos-5 recurrence is defined as a discrete quadratic equation \[S^*_{n+5}S^*_{n} =\alpha S^*_{n+4}S^*_{n+1} + \beta S^*_{n+2}S^*_{n+1}.\label{somoseq5}\tag{2}\] They can both be interpreted as discrete integrable systems [1]–[3]. In fact, they could be regarded as reductions of the discrete Hirota equation (also referred to as the bilinear discrete KP, or the octahedron recurrence) [4]–[6], which is one of the most important discrete integrable systems. Along with their high-order analogues, these two recurrences have attracted considerable attention.

As for the Somos-4 recurrence, an early and remarkable observation is that, if the initial values are set to be \((S_0,S_1,S_2,S_3)= (1,1,1,1)\), then the bilateral Somos-4 sequence \(\{S_n\}_{n \in \mathbb{Z}}\) produced by the recurrence 1 with \(\alpha=\beta=1\) consists of integers[7]–[9] \[\ldots,314,59,23,7,3,2,1,1,1,1,2,3,7,23,59,314,\ldots . \label{somos4sequence}\tag{3}\] One efficient and impressive way to prove this fact is to fit the iteration 1 into a big picture called cluster algebras, which was drawn by Fomin and Zelevinsky. In their theory of cluster algebras [10], [11], the Somos-4 recurrence is interpreted as a path of a 4 regular tree \(\mathbb{T}_{4}\) driven by a quiver with 4 vertices. In fact, they proved that, for a large class of cluster algebras of geometric type, the cluster variables exhibit the Laurent property, that is, any cluster variable is a Laurent polynomial when written as a rational function in terms of any other cluster. As special cases, certain Somos recurrences admit the Laurent property, from which, the fact that the sequence 3 produced by the Somos-4 recurrence 1 are all integers immediately follows as a straightforward corollary.

In addition to the Laurent phenomenon closely related to the theory of cluster algebras, Somos recurrences also exhibit some other distinctive features. As mentioned above, certain Somos recurrences and their higher-order analogues can be thought of discrete integrable systems, serving as integrable symplectic mappings [12], as reductions of the discrete Hirota equation [4], [6] or Miwa equation (also known as the bilinear discrete BKP, or the cube recurrence) [13], [14]. Somos sequences naturally appear in some problems in number theory [15]–[19]. They may also arise in solvable models within the realms of statistical mechanics and quantum field theory, such as the hard hexagon model [20], dimer models and quiver gauge theory [21].

It is interesting that general solutions to certain Somos recurrences can be derived. It has been shown that certain Somos recurrences admit general explicit solutions in terms of sigma functions associated with hyperelliptic curves [1], [3], [13], [22], [23]. In particular, the general solutions to the Somos-4 and Somos-5 recurrences can be constructed in terms of Weierstrass sigma function [1], [3], and can also be expressed using closed forms in terms of Hankel determinants [2], [24]–[28] with the matrix entries from elliptic curves. For some other discussions on the related topics, see e.g. [29]–[32].

The Somos-4 recurrence enjoys a structure of Hankel determinants. Recall that a sequence of Hankel determinants \(\{H_{n}\}_{n \in \mathbb{N}}\) is defined according to \[H_{n} := \left\{ \begin{align} &\det(s_{i+j-2})_{i,j= 1,\ldots,n}, \quad & n \ge 1,\\ &1, \quad & n = 0, \end{align} \right.\] for the matrix entries \(\{s_n\}_{n \in \mathbb{N}}\), which, in some sense, can be regarded as a transformation from one sequence \(\{s_n\}_{n \in \mathbb{N}}\) to another sequence \(\{H_n\}_{n \in \mathbb{N}}\). It was originally conjectured by Somos [7], later proved by Xin [27] that, when the sequence \(\{s_n\}_{n \in \mathbb{N}}\) is chosen to be the coefficients of the Taylor expansion of a certain function \(G(z)\) \[G (z):= \frac{y(z)}{z}=\frac{\frac{1}{2}+\sqrt{\frac{1}{4}-z+z^3}}{z} - 1 = \sum_{n \ge 1} s_{n-1}z^n\] obtained from the function \(y=y(z)\) satisfying the restriction \(y-y^2=z-z^3\), then the corresponding sequence of Hankel determinants \(\{H_{n}\}_{n \in\mathbb{N}}\) exactly gives a half of the Somos-4 sequence \[1,1,2,3,7,23,59,314,\ldots .\] It was further conjectured by Barry that, if the sequence \(\{s_n\}_{n \in \mathbb{N}}\) satisfies a particular convolution recursion [33], then the corresponding sequence of Hankel determinants \(\{H_{n}\}_{n \in \mathbb{N}}\) also gives a half of the Somos sequence with specific \(\alpha\) and \(\beta\). This conjecture was proved by Hu and one of the authors in [24]. Later in [25], together with Xin, they derived the Hankel determinant formulae for the half of the Somos-4 recurrence with arbitrary \(\alpha\) and \(\beta\). All of these results are devoted to the half of the Somos-4 recurrence.

As for the bilateral Somos-4 recurrence 1 with \(n\in \mathbb{Z}\), at first glance, Hankel determinants may not be the ideal tool to describe its general solution, since it is generally not natural to extend the definition of Hankel determinants from natural numbers to all integers. To obtain an overall expression of the bilateral Somos-4 sequences, in [2], Hone made an attempt based on the connections with continued fraction expansions of functions on hyperelliptic curves, which are also related to the geometry of the Jacobians of the curves [34], [35]. He expressed the positive sequence and the negative sequence in terms of Hankel determinants independently, and then glue them together to get an overall expression using gauge transformations; see [2]. However, his process seems to break the overall structure, and the relationship between the positive part and the negative part still remain unclear. In fact, as mentioned in [2], a “mismatch" problem arises, that is,”in general one cannot just take \(S_n=H_n\) for nonnegative \(n\) and \(S_n=H_{-n-1}^\dagger\) for negative \(n\), because there will be a mismatch at the values of \(d_0\), \(d_1\), and \(v_0\) that are left unspecified."

This paper is mainly motivated by the above mismatch problem [2]. To tackle it, by digging up the interior symmetries hidden in the continued fraction recursions, we introduce a new concept called quadratic orthogonal pair for a pair of generating functions associated with hyperelliptic curves.

More precisely, we start from a concept called “proper" quadratic field extension over \(\mathbb{C}(X)\) and show that the quadratic field extensions induced by hyperelliptic curves are all proper so that the continued fraction theory can apply. Then, based on the Galois symmetry of a proper quadratic field extension, by introducing the new notion called quadratic orthogonal pairs for hyperelliptic functions, and combining it with the continued fraction theory for the Laurent series field \(\mathbb{C}((X^{-1}))\), we obtain the consistent expressions 28 and 29 together with Lemma 30 that are defined for all integers, which eventually solves the mismatch problem raised by Hone in [2]. It is noted that this notion also reveals the relation between the two generating functions from which one can get an overal expression of solutions to a particular nonlinear recurrence. A correspondence between iterations of continued fraction type for generating functions and guage transformations for Somos 4 sequences is also given in Theorem 12.

As further applications of our approach, we derive the general solution to the initial value problem for the bilateral Somos-4 recurrence using Hankel determinants defined on a quadratic orthogonal pair for elliptic curves. The explicit Hankel determinant expressions also allow us to verify the Laurent property of the bilateral Somos-4 recurrence. In addition, the initial value problem of the bilateral Somos-5 recurrence is also solved using Hankel determinants defined on two quadratic orthogonal pairs. We also show how our approach on quadratic orthogonal pairs is used in generalizing Somos’ original conjecture on half of Somos-4 sequence to the bilateral case. In particular, we find a “symmetric" curve closely related to the fundamental curve \(y-y^2=z-z^3\).

Furthermore, the integrability of the continued fraction recursions for hyperelliptic functions is also studied. In particular we show that every hyperelliptic function is naturally associated with a projective transformation with two fixed points and the continued fraction expansion can also be regarded as a projective transformation. To understand the Lax integrability, we find that the compatibility conditions of these two transformations are closely related to the symmetries of the continued fraction expansions.

The paper is organized as follows. In Section 2, we present some facts on continued fractions in \(\mathbb{C}((X^{-1}))\) with a particular focus on J-fractions and the corresponding Hankel determinant representations. In Section 3, we investigate the quadratic field extensions induced by hyperelliptic curves with a concentration on genus 1. We demonstrate that every hyperelliptic function is naturally associated with a Lax pair and the compatibility condition is equivalent to a certain restriction on the continued fraction expansions of the hyperelliptic functions. The geometric meaning of the Lax matrices is also clarified. By further introducing pairs of hyperelliptic functions called quadratic orthogonal pairs, we solve the mismatch problem posed by Hone in [2]. Section 4 is devoted to further applications of our approach to initial value problems of the bilateral Somos-4 and Somos-5 recurrences, based on which Laurent properties are also discussed. Additional comments on quadratic orthogonal pairs for hyperelliptic functions with higher genus are collected in Appendix 6.

2 On J-fractions and Hankel determinants↩︎

In this section, we introduce some facts on continued fractions with a particular focus on J-fractions. The related materials can be found in e.g. [36], [37].

2.1 Continued fraction expansions in \(\mathbb{C}((X^{-1}))\)↩︎

Let \(\mathbb{K}\) be a field. For a sequence \(\{a_{n}\}_{n\in \mathbb{N}}\) in \(\mathbb{K}\), its partial continued fraction \(\langle a_0,\ldots,a_n \rangle\) is defined by \[\langle a_0,\ldots,a_n \rangle := a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 +\cfrac{1}{ \cdot\cdot\cdot +\frac{1}{a_n}} } } } .\] In general, the partial continued fraction can be written as a fraction, that is, \[\langle a_0,\ldots,a_n \rangle = \frac{p_n}{q_n},\] where \(p_n\) and \(q_{n}\) are polynomials in \(\{a_{n}\}_{n\in \mathbb{N}}\) recursively defined by \[\left\{\begin{matrix} p_n= a_n p_{n-1} +p_{n-2} \\ q_n= a_n q_{n-1} +q_{n-2} \end{matrix}\right.,\quad n \geqslant 0,\label{polyreprensentation}\tag{4}\] with initial values \[\left\{\begin{matrix} p_{-1}= 1 \\ q_{-1}= 0 \end{matrix}\right., \qquad \left\{\begin{matrix} p_{-2}= 0 \\ q_{-2}= 1 \end{matrix}\right. .\]

The above procedure can be generalized to function field cases when \(\mathbb{K}\) is taken to be a certain function field and \(\{a_n\}_{n \in \mathbb{N}}\) to be a series of functions. In particular, if we choose \(\{a_n\}_{n \in \mathbb{N}}\) to be a series of linear functions of one variable, then the corresponding continued fractions are said to be J-fractions[36].

For later use, we now choose \(\mathbb{K} = \mathbb{C}((X^{-1}))\) to be the Laurent series field in one variable over \(\mathbb{C}\), and examine the continued fraction expansion of a given function \(f = \sum_{n \in \mathbb{Z}} c_n X^n\) in \(\mathbb{C}((X^{-1}))\), where \(c_n\) are complex numbers satisfying the condition that there exists an integer \(N\) such that \(c_n = 0\) for all \(n \ge N\). Here we make the convention that, throughout the paper, whenever we mention an element in \(\mathbb{C}((X^{-1}))\), this condition is always assumed. Under this assumption, the following map is well defined \[\begin{align} \max : \mathbb{C}((X^{-1})) &\longrightarrow \mathbb{Z} \\ f = \sum_{n \in \mathbb{Z}} c_n X^n & \longmapsto \max\{n \in \mathbb{Z}| f_n \ne 0\} . \end{align} \label{max}\tag{5}\]

For elements in \(\mathbb{C}((X^{-1}))\), define an operation as follows \[\begin{align} \left \lfloor \cdot \right \rfloor : \mathbb{C}((X^{-1})) &\longrightarrow \mathbb{C}[X] \\ f = \sum_{n \in \mathbb{Z}} c_n X^n & \longmapsto \left \lfloor f \right \rfloor=\sum_{n \ge 0 }c_n X^n . \end{align}\] This operation maps an element to its polynomial part so that one has a natural decomposition of elements in \(\mathbb{C}((X^{-1}))\) \[f= \left \lfloor f \right \rfloor + (f-\left \lfloor f \right \rfloor) .\label{decompose}\tag{6}\] Using this decomposition, one can easily construct a continued fraction expression for each element in \(\mathbb{C}((X^{-1}))\). More precisely, given an arbitrary element \(f_0 \in \mathbb{C}((X^{-1}))\), decomposing \(f_0\) using 6 , we have \[f_0= \left \lfloor f_0 \right \rfloor + (f_0-\left \lfloor f_0 \right \rfloor) =: a_0 + b_0,\] where we use \(a_0\) to denote the polynomial part and \(b_0\) to denote the negative series part. If \(b_0\) is not zero, then it is invertible in \(\mathbb{C}((X^{-1}))\). Thus we have a new element in \(\mathbb{C}((X^{-1}))\) \[f_1 := b_0^{-1} .\] Note that the same operation can be applied to \(f_1\) \[\begin{align} &f_1= \left \lfloor f_1 \right \rfloor +(f_1-\left \lfloor f_1 \right \rfloor) =: a_1+ b_1,\\ &f_2 := b_1^{-1}, \end{align}\] if \(b_1\neq0\). Repeating this operation, we have, for all natural number \(n\) \[\begin{align} &f_n= \left \lfloor f_n \right \rfloor +(f_n-\left \lfloor f_n \right \rfloor) =: a_n+ b_n, \\ &f_{n+1} := b_n^{-1}, \end{align}\] once \(b_n\neq0\). Thus we have, for any \(0 \le m \le n\), \[f_m = \langle a_m,a_{m+1},\ldots,a_{n},f_{n+1} \rangle .\] If follows from 4 that \(f_m\) can be expressed in terms of a fraction of two polynomials in \(\{a_m,\ldots,a_{n},f_{n+1}\}\). In addition, it is clear that, by definition of \(\left \lfloor \cdot \right \rfloor\), we have \[\left \lfloor -f \right \rfloor = -\left \lfloor f \right \rfloor .\] This implies that, if \(f\) and \(\tilde{f}\) satisfying \(f+ \tilde{f}=0\) are expanded in continued fractions as \[\begin{align} f = \langle f_0,f_1,\ldots,f_n,\ldots \rangle, \qquad \tilde{f} =\langle \tilde{f}_0,\tilde{f}_1,\ldots,\tilde{f}_n,\ldots \rangle, \end{align}\] then \[f_n+ \tilde{f}_n=0,\quad \forall\, n \in \mathbb{N},\] which is quite different from the continued fraction theory for real numbers.

2.2 Hankel determinant representations for J-fractions↩︎

As a particular type of continued fractions, J-fractions are known to be useful in dealing with Hankel determinants[37], [38]. In general, the quantities in J-fractions can be represented in terms of Hankel determinants with the coefficients of the corresponding series expansion as elements.

Given an element \(f\) in \(\mathbb{C}((X^{-1}))\), if the corresponding continued fraction expansion of \(f\) is a J-fraction, then \(f\) is said to be J-expressible, that is, \(f\) can be expressed as a J-fraction as follows \[f = \langle a_0,a_1,a_2,\ldots \rangle, \label{fractions}\tag{7}\] where \(a_0 = aX+b\), \(a_n= \alpha_n X +\beta_n\), \(\alpha_n\) and \(\beta_n\) are all complex numbers, \(\alpha_n \ne 0\). Conversely, if \(f\) is J-expressible, then \(f\) must have the following form \[f = aX+b+ s_0 X^{-1} +s_1 X^{-2}+\cdot\cdot\cdot . \label{series}\tag{8}\] Combining 8 and 7 , we have \[\begin{align} aX+b + s_0 X^{-1} +s_1 X^{-2}+\cdot\cdot\cdot &= \langle a_0,a_1,a_2,\ldots \rangle \\ &= a_0 +\langle 0,a_1,a_2,\ldots \rangle \\ &= aX + b + \langle 0,a_1,a_2,\ldots \rangle, \\ \end{align}\] which gives \[\langle 0,\alpha_1X+\beta_1,\alpha_2X+\beta_2,\ldots \rangle = s_0 X^{-1} +s_1 X^{-2}+\cdot\cdot\cdot .\label{eq}\tag{9}\] This means that there must exist a close relation between the set \(\{\alpha_n,\beta_n\}_{n \in \mathbb{N}^*}\) and the set \(\{s_n\}_{n \in \mathbb{N}}\), where \(\mathbb{N}^*\) means the set of all positive integers. Actually, the relations between these two sets are given by the following theorem (see e.g. [36],[37] or [38]). The proof can be achieved in various ways; see e.g. [36] based on the underlying orthogonality or [2] based on identities for determinants.

3 Continued fractions and quadratic orthogonal pairs for hyperelliptic functions↩︎

In this section, we investigate continued fraction and Laurent series expansions for certain hyperelliptic functions. We also propose the concept of quadratic orthogonal pairs for hyperelliptic functions, for which intimate relationships among their continued fraction expansions are revealed.

3.1 Proper quadratic extensions over \(\mathbb{C}(X)\)↩︎

We first introduce some preliminary results on quadratic extensions. Since one can always complete the square in general, every quadratic extension over \(\mathbb{C}(X)\) can be reduced to the following case \[\mathcal{F} = \mathbb{C}(X)[Y]/(Y^2-w),\] where \(w \in \mathbb{C}(X)\) is a rational function in one variable over \(\mathbb{C}\).

On the other hand, due to the fact that every rational function can be expanded into a Laurent series at \(\infty\), we have the following embedding of fields \[\begin{align} \pi_0 : \mathbb{C}(X) &\longrightarrow & \mathbb{C}((X^{-1})) \\ f & \longmapsto & \sum_{n \in \mathbb{Z}} f_n X^n, \end{align}\] where \(f_n\) are complex numbers given by \[f_n = \frac{1}{2\pi i}\oint_{\infty} X^{-n-1} f dX .\] The following lemma provides a sufficient and necessary condition to judge whether a given quadratic field extension \(\mathcal{F}\) over \(\mathbb{C}(X)\) can be embedded into the Laurent series field \(\mathbb{C}((X^{-1}))\).

For later use, we call a quadratic field extension over \(\mathbb{C}(X)\) a proper quadratic extension if and only if it satisfies the condition 10 and \(w\) is not a square. It turns out that all quadratic extensions induced by hyperelliptic curves [2], [39]–[41] are proper.

3.2 Hyperelliptic functions of genus 1↩︎

Following van der Poorten[39]–[41] and Hone[2], we consider hyperelliptic curves over \(\mathbb{C}\). For our purposes of solving the initial problems of Somos-4 and Somos-5 recurrences, we restrict our attention to the genus 1 case, i.e. the elliptic curve. The constructions in these sections do not actually depend on the explicit formula of the curve and can be easily generalized to an arbitrary proper quadratic field extension over \(\mathbb{C}(X)\).

Consider a curve over \(\mathbb{C}\) \[\mathcal{C}:Y^2 = (X^2+f)^2+4u(X-v) \label{curve}\tag{11}\] where \(f\),\(u\),\(v\) are complex numbers satisfying a certain non-degenerating condition so that the roots of the polynomial on the right hand side are all distinct. It is evident that the genus of the curve \(\mathcal{C}\) is 1 and it is nothing but an elliptic curve. For convenience, we also call \(\mathcal{C}\) a hyperelliptic curve of genus 1. As an algebraic variety, the coordinate ring \(\mathcal{R}\) of the hyperelliptic curve over \(\mathbb{C}\) of genus 1 is given by \[\mathcal{R} = \mathbb{C}[X,Y]/(Y^2 - (X^2+f)^2 - 4u(X-v))\] and the associated function field is computed by \[\mathcal{F} = Frac(\mathcal{R}) \cong \mathbb{C}(X)[Y]/(Y^2-(X^2+f)^2 - 4u(X-v)) = \{aY+b | a,b \in \mathbb{C}(X) \} .\label{function32field}\tag{12}\] Elements in \(\mathcal{F}\) are said to be hyperelliptic functions of genus 1.

3.2.1 Laurent series expansions↩︎

According to 12 , the function field \(\mathcal{F}\) can be viewed as a two dimensional vector space over \(\mathbb{C}(X)\), with \(\{1,Y\}\) to be its basis. The only information we know about \(Y\) is the algebraic relation given by 11 . Actually, there are exactly two such functions satisfying the same algebraic relation 11 . These two functions are symmetric under the action of the Galois group \(Gal(\mathcal{F}/\mathbb{C}(X))\). Without loss of generality, we simply pick \(Y\) to be one of these two functions formally defined by \[Y = \sqrt{(X^2+f)^2+4u(X-v)},\] while the other such function is given by \[Y^*:= -Y = -\sqrt{(X^2+f)^2+4u(X-v)} .\]

Recall that, for any \(f \in \mathbb{C}(X)\), we can expand it into a Laurent series when \(X \to \infty\), that is, we have the following ring homomorphism \[\begin{align} \pi_0 : \mathbb{C}(X) &\longrightarrow & \mathbb{C}((X^{-1})) \\ f & \longmapsto & \sum_{n \in \mathbb{Z}} f_n X^n, \end{align}\] where \(f_n\) are complex numbers given by \[f_n = \frac{1}{2\pi i}\oint_{\infty} X^{-n-1} f dX.\] Similarly, when \(X \to \infty\), since the right hand side of 11 satisfies the condition 10 , it follows from Lemma 1 that \(Y\) can also be expanded into a Laurent series \[Y = \sqrt{(X^2+f)^2+4u(X-v)} = \sum_{n \in \mathbb{Z}}y_n X^n =: (Y)_{\infty},\] where \(y_n\) are complex numbers given by \[y_n = \frac{1}{2\pi i}\oint_{\infty} X^{-n-1} Y dX.\] Consequently, we can naturally lift \(\pi_0\) to \(\mathcal{F}\) \[\begin{tikzcd} {\mathcal{F}} & {\mathbb{C}((X^{-1}))} \\ {\mathbb{C}(X)} \arrow["\pi", from=1-1, to=1-2] \arrow["i", hook, from=2-1, to=1-1] \arrow["{\pi_0}"', from=2-1, to=1-2] \end{tikzcd}\] where \(\pi(f) = \pi(aY+b) := \pi_0(a)(Y)_{\infty} +\pi_0(b)\), for any hyperelliptic function \(f = aY+b\). This means that the ring homomorphism \(\pi\) maps a hyperelliptic function to its Laurent series expansion at \(\infty\). In the sequel, we also use \((\cdot)_{\infty}\) to denote \(\pi\) for convenience.

3.2.2 Continued fraction expansions↩︎

In the previous subsection, we have shown that the hyperelliptic function field \(\mathcal{F}\) can be embedded into the Laurent series field \(\mathbb{C}((X^{-1}))\). Recall that, as indicated in Section 2.1, every Laurent series has a continued fraction expansion. Therefore, we conclude that every hyperelliptic function can be expanded into a continued fraction.

The continued fraction expansions of hyperelliptic functions preserve nice properties. To see this, we start from an arbitrary element \(Y_0= r_{0}Y+s_{0}\) in \(\mathcal{F}\), where \(r_{0}\) and \(s_{0}\) are elements in \(\mathbb{C}(X)\), \(r_{0}\) is nonzero. Under the action of the Galois group \(Gal(\mathcal{F}/\mathbb{C}(X))\), we have another hyperelliptic function \(Y_{0}^*\) defined by \[Y_{0}^* := r_{0}Y^*+s_{0} = r_{0}(-Y)+s_{0},\] where \(Y_{0}^*\) is said to be the symmetry of \(Y_0\).

As is seen in Section 2.1, the continued fraction expansion of \(Y_0\) is equivalent to the following recursion \[Y_n = a_n + \frac{1}{Y_{n+1}} \quad, n \in \mathbb{N},\label{recursion}\tag{13}\] where \(a_n := \left \lfloor Y_n \right \rfloor\) for all natural number \(n\). Since \(r_{0}\) is nonzero and \(Y\) is not rational, the above recursion does not terminate. In other words, \(Y_{0}\) has an infinite continued fraction expansion. Note that if \(Y_n= r_{n}Y+s_{n}\), then \(r_{n}\) is also nonzero, otherwise \(Y\) would be rational, which is a contradiction.

From a geometric viewpoint, the above recursion can be interpreted as a sequence of projective transformations in the projective space \(\mathbb{P}^1(\mathcal{F})\). To be precise, \(Y_n\) can be regarded as a point \((Y_n:1)\) in \(\mathbb{P}^1(\mathcal{F})\). Thus the above recursion becomes \[\begin{align} \begin{pmatrix} Y_{n+1}\\ \cdot \cdot\\ 1 \end{pmatrix} = \begin{pmatrix} (Y_{n}-a_n)^{-1}\\ \cdot \cdot\\ 1 \end{pmatrix}= \begin{pmatrix} 1\\ \cdot \cdot\\ Y_{n}-a_n \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & -a_n \end{pmatrix} \begin{pmatrix} Y_{n}\\ \cdot \cdot\\ 1 \end{pmatrix} \end{align},\] from which, we obtain \[\begin{pmatrix} Y_{n+1}\\ \cdot \cdot\\ 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & -a_n \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & -a_{n-1} \end{pmatrix} \cdots \begin{pmatrix} 0 & 1 \\ 1 & -a_0 \end{pmatrix}\begin{pmatrix} Y_{0}\\ \cdot \cdot\\ 1 \end{pmatrix}.\] If we introduce \[\begin{align} M_n := \begin{pmatrix} 0 & 1 \\ 1 & -a_n \end{pmatrix} \end{align},\] for \(n \in \mathbb{N}\), then recursion 13 is equivalent to the sequence of projective transformations \(\{M_n\}_{n \in \mathbb{N}}\). Note that \(\{M_n\}_{n \in \mathbb{N}}\) are uniquely determined by \(\{a_n\}_{n \in \mathbb{N}}\), thus uniquely determined by \(Y_0\).

In general, although \(M_n\) transforms \((Y_n:1)\) to \((Y_{n+1}:1)\), it does not certainly transform its symmetries \((Y_n^*:1)\) to \((Y_{n+1}^*:1)\). The following lemma provides a sufficient and necessary condition for that \(M_n\) transforms \((Y_n^*:1)\) to \((Y_{n+1}^*:1)\).

Since \(Y_0 \ne Y_0^*\), \(Y_0\) defines a unique projective transformation \(\mathcal{L}_0\) as \[\begin{align} \mathcal{L}_0 : \mathbb{P}^1(\mathcal{F}) &\longrightarrow \mathbb{P}^1(\mathcal{F}) \\ (Y_0:1)& \longmapsto (Y_0:1) \\ (Y_0^*:1)& \longmapsto (Y_0^*:1)\,. \end{align}\] The projective transformation \(\mathcal{L}_0\) can be represented by a linear transformation \(L_0\) with eigenvectors \((Y_0,1)\), \((Y_0^*,1)\) and the respective eigenvalues \(Y\), \(Y^*\). In other words, if we define \[L_0 = \begin{pmatrix} A_0&B_0 \\ C_0&D_0 \end{pmatrix},\] then \(L_0\) satisfies \[\begin{pmatrix} A_0&B_0 \\ C_0&D_0 \end{pmatrix} \begin{pmatrix} Y_0&Y_0^* \\ 1&1 \end{pmatrix} = \begin{pmatrix} Y_0&Y_0^* \\ 1&1 \end{pmatrix} \begin{pmatrix} Y&0 \\ 0&Y^* \end{pmatrix}.\] This uniquely determines \(L_0\). In fact, we have \[\begin{align} L_0 = \begin{pmatrix} A_0&B_0 \\ C_0&D_0 \end{pmatrix} &= \begin{pmatrix} Y_0&Y_0^* \\ 1&1 \end{pmatrix} \begin{pmatrix} Y&0 \\ 0&Y^* \end{pmatrix} \begin{pmatrix} Y_0&Y_0^* \\ 1&1 \end{pmatrix}^{-1} \\ &=\frac{1}{Y_0-Y_0^*} \begin{pmatrix} Y_0&Y_0^* \\ 1&1 \end{pmatrix} \begin{pmatrix} Y&0 \\ 0&Y^* \end{pmatrix} \begin{pmatrix} 1&-Y_0^* \\ -1&Y_0 \end{pmatrix} \\ &= \begin{pmatrix} \frac{YY_0-Y^*Y_0^*}{Y_0-Y_0^*}&-\frac{Y_0Y_{0}^*(Y-Y^*)}{Y_0-Y_0^*} \\ \frac{Y-Y^*}{Y_0-Y_0^*}&\frac{Y^*Y_0-YY_0^*}{Y_0-Y_0^*} \end{pmatrix} \\ &= \begin{pmatrix} \frac{s_0}{r_0}& \frac{r_0^2Y^2-s_0^2}{r_0}\\ \frac{1}{r_0}& -\frac{s_0}{r_0} \end{pmatrix}. \end{align}\] The above procedure can still hold when \(Y_0\) is replaced by \(Y_n=r_nY+s_{n}\), which gives us a sequence of matrices \(\{L_n\}_{n \in \mathbb{N}}\) defined by \[L_n = \begin{pmatrix} \frac{s_n}{r_n}& \frac{r_n^2Y^2-s_n^2}{r_n}\\ \frac{1}{r_n}& -\frac{s_n}{r_n} \end{pmatrix}\] satisfying \[L_n \begin{pmatrix} Y_n&Y_n^* \\ 1&1 \end{pmatrix} = \begin{pmatrix} Y_n&Y_n^* \\ 1&1 \end{pmatrix} \begin{pmatrix} Y&0 \\ 0&Y^* \end{pmatrix}.\] It turns out that the sequence \(\{L_n\}_{n\in \mathbb{N}}\) is related to the sequence \(\{M_n\}_{n \in \mathbb{N}}\), which is illustrated by the following theorem.

3.3 Quadratic orthogonal pairs in \(\mathcal{F}\)↩︎

Suppose that we have two hyperelliptic functions \(Y_0 = r_0Y+s_0\) and \(\tilde{Y}_0 = \tilde{r}_0Y+\tilde{s}_0\), where \(r_0\), \(s_0\), \(\tilde{r}_0\) and \(\tilde{s}_0\) are all rational functions in \(\mathbb{C}(X)\), \(r_0\) and \(\tilde{r}_0\) are nonzero. Following the previous section, one can construct \(\{Y_n=r_nY+s_n|r_n,s_n\in \mathbb{C}(X),r_n \ne 0 \}_{n\in \mathbb{N}}\), \(\{a_n\}_{n \in \mathbb{N}}\), \(\{L_n\}_{n\in \mathbb{N}}\) and \(\{M_n\}_{n \in \mathbb{N}}\) together with \(\{\tilde{Y}_n=\tilde{r}_nY+\tilde{s}_n|\tilde{r}_n,\tilde{s}_n\in \mathbb{C}(X),\tilde{r}_n \ne 0 \}_{n\in \mathbb{N}}\), \(\{\tilde{a}_n\}_{n \in \mathbb{N}}\), \(\{\tilde{L}_n\}_{n\in \mathbb{N}}\) and \(\{\tilde{M}_n\}_{n \in \mathbb{N}}\) satisfying \[\begin{align} &a_n = \left \lfloor Y_n \right \rfloor,&&Y_n = a_n +\frac{1}{Y_{n+1}},\\ &M_n = \begin{pmatrix} 0 & 1 \\ 1 & -a_n \end{pmatrix}, &&L_n = \begin{pmatrix} \frac{s_n}{r_n}& \frac{r_n^2Y^2-s_n^2}{r_n}\\ \frac{1}{r_n}& -\frac{s_n}{r_n}, \end{pmatrix}, \end{align}\] and \[\begin{align} &\tilde{a}_{n} = \left \lfloor \tilde{Y}_{n} \right \rfloor,&& \tilde{Y}_{n} = \tilde{a}_{n} +\frac{1}{\tilde{Y}_{n+1}},\\ &\tilde{M}_{n} = \begin{pmatrix} 0 & 1 \\ 1 & -\tilde{a}_{n} \end{pmatrix}, &&\tilde{L}_{n} = \begin{pmatrix} \frac{\tilde{s}_{n}}{\tilde{r}_{n}}& \frac{\tilde{r}_{n}^2Y^2-\tilde{s}_{n}^2}{\tilde{r}_{n}}\\ \frac{1}{\tilde{r}_{n}}& -\frac{\tilde{s}_{n}}{\tilde{r}_{n}} \end{pmatrix}, \end{align}\] for all natural number \(n\). Suppose that \(\tilde{Y}_{0}\) is related to \(Y_0\) according to \[\tilde{Y}_{0} Y_0^* = - 1,\label{symmetric}\tag{22}\] then it is natural to ask what are the relations between \(\{\{Y_n\}_{n\in \mathbb{N}},\{r_n\}_{n\in \mathbb{N}},\{s_n\}_{n\in \mathbb{N}},\{a_n\}_{n \in \mathbb{N}},\) \(\{L_n\}_{n\in \mathbb{N}},\{M_n\}_{n \in \mathbb{N}}\}\) and \(\{\{\tilde{Y}_n \}_{n\in \mathbb{N}},\{\tilde{r}_n \}_{n\in \mathbb{N}},\{\tilde{s}_n \}_{n\in \mathbb{N}}, \{\tilde{a}_n\}_{n \in \mathbb{N}}, \{\tilde{L}_n\}_{n\in \mathbb{N}}, \{\tilde{M}_n\}_{n \in \mathbb{N}}\}\). Here we partially answer this question for a special case that is examined by Hone in [2]. Note that when \(\tilde{Y}_{0}\) and \(Y_0^*\) are interpreted as the slopes of two lines in \(\mathbb{P}^1(\mathcal{F})\), the relation 22 is an analogue of the orthogonal condition for these two lines. For this reason, we call a pair of functions \((Y_0,\tilde{Y}_0)\) a quadratic orthogonal pair if the relation 22 is satisfied.

Now we introduce the choice of \(Y_0\) made by Hone in [2]. Consider the hyperelliptic curve of genus 1 given by \[\mathcal{C}:Y^2 = (X^2+f)^2+4u(X-v)\] as in Section 3.2. Then one can uniquely define a map \(\Phi\) as follows \[\begin{align} \Phi : \mathbb{C}^2 &\longrightarrow \mathbb{C}^2& \\ (x,y) & \longmapsto \left(-x-y^2-f,-\frac{u}{x+y^2+f}-y\right).& \end{align}\] It turns out in [2] that the map \(\Phi\) is birational, symplectic and integrable with a conserved quantity defined by \[H= x^2+fx+ xy^2-uy .\label{H}\tag{23}\]

Since \(\Phi\) is birational, its inverse \(\Phi^{-1}\) is well defined. Therefore if one starts with a point \((d_0,v_0)\) in \(\mathbb{C}^2\), one can generate a sequence of points \(\{(d_n,v_n)\}_{n \in \mathbb{Z}}\) by setting \[(d_n,v_n) := \Phi^n((d_0,v_0)),\quad n \in \mathbb{Z},\label{map}\tag{24}\] provided that \(\Phi^n((d_0,v_0))\) are well defined for all integers \(n\). Note that the well-definedness of \(\Phi^n((d_0,v_0))\) is equivalent to the non-zeroness of \(d_n\). Under this assumption, if we choose \(u_0\) to be another nonzero complex number, we can define a sequence \(\{u_n\}_{n \in \mathbb{Z}}\) via \(\{d_n\}_{n \in \mathbb{Z}}\) by setting \[u_nu_{n-1}+4d_n = 0, \quad n\in \mathbb{Z} .\label{u}\tag{25}\] We conclude that from an initial point \((d_0,v_0,u_0)\) satisfying \(d_0u_0\neq0\), we can generate a sequence of points \(\{(d_n,v_n,u_n)\}_{n \in \mathbb{Z}}\) satisfying 24 and 25 if \(\Phi^n((d_0,v_0))\) are well defined for all integers \(n\). Using this sequence, one can choose \(\{Y_n=r_n Y+s_n|r_n,s_n\in \mathbb{C}(X),r_n \ne 0 \}_{n\in \mathbb{Z}}\) and \(\{a_n\}_{n \in \mathbb{Z}}\) with \[\begin{align} &r_n := \frac{1}{u_n (X-v_n)},\qquad s_n :=\frac{X^2+f+2d_n}{u_n (X-v_n)},\qquad a_n :=\frac{2}{u_n}(X+v_n). \end{align} \label{choice}\tag{26}\]

When given specific conserved quantity \(H\), the above choice will satisfy the condition in Theorem 4 together with \(a_n = \left \lfloor Y_n \right \rfloor\), which also shows the existence of a Lax pair. This can be summarized in the following lemma.

On the other hand, based on the sequence \(\{(d_n,v_n,u_n)\}_{n \in \mathbb{Z}}\) obtained by 24 and 25 , one can construct another pair of sequences \(\{\tilde{Y}_n=\tilde{r}_nY+\tilde{s}_n|\tilde{r}_n,\tilde{s}_n\in \mathbb{C}(X),\tilde{r}_n \ne 0 \}_{n\in \mathbb{N}}\) and \(\{\tilde{a}_n\}_{n \in \mathbb{N}}\) by setting \[\begin{align} &\tilde{r}_{n} := \frac{1}{u_{n-1} (X-v_{n-1})}, \quad \tilde{s}_{n} :=\frac{X^2+f+2d_n}{u_{n-1} (X-v_{n-1})}, \quad \tilde{a}_{n} :=\frac{2}{u_{n-1}}(X+v_{n-1}). \end{align} \label{choice2}\tag{27}\] The properties of \(\{\tilde{Y}_n=\tilde{r}_nY+\tilde{s}_n|\tilde{r}_n,\tilde{s}_n\in \mathbb{C}(X),\tilde{r}_n \ne 0 \}_{n\in \mathbb{N}}\) and \(\{\tilde{a}_n\}_{n \in \mathbb{N}}\) are summarized in the following lemma.

Moreover, we observe the intimate relationship between two pairs of sequences \(\{{a}_n\}_{n \in \mathbb{N}}\), \(\{{Y}_n\}_{n \in \mathbb{N}}\) and \(\{\tilde{a}_n\}_{n \in \mathbb{N}}\), \(\{\tilde{Y}_n\}_{n \in \mathbb{N}}\), which are given in the following theorem.

3.4 Hankel determinant expressions↩︎

Now we clarify why the notion of a quadratic orthogonal pair is useful in generalizing the Hankel determinant solution of the Somos-4 recurrence to the bilateral case.

Let \(n\) be an arbitrary integer. It follows from Lemma 3, 4 and Theorem 7 that \[\begin{align} (Y_{n})^{-1}&=(\langle a_n,a_{n+1},\ldots \rangle )^{-1} \\ &=\langle 0,a_n,a_{n+1},\ldots \rangle \end{align}\] and \[\begin{align} -(\tilde{Y}_{n})^{-1} &= -(\langle \tilde{a}_{n},\tilde{a}_{n-1},\ldots \rangle )^{-1} \\ &=(\langle -\tilde{a}_{n},-\tilde{a}_{n-1},\ldots \rangle )^{-1} \\ &=\langle 0,-\tilde{a}_{n},-\tilde{a}_{n-1},\ldots \rangle \\ &=\langle 0,-a_{n-1},-a_{n-2},\ldots \rangle . \end{align}\] Suppose that \((Y_{n})^{-1}\) and \(-(\tilde{Y}_{n})^{-1}\) are expanded into Laurent series as follows \[\begin{align} (Y_{n})^{-1} &= s_0^{(n)} X^{-1} + s_1^{(n)} X^{-2} + \cdots,\\ -(\tilde{Y}_{n})^{-1} & = \tilde{s}_0^{(n)}X^{-1} + \tilde{s}_1^{(n)}X^{-2} + \cdots, \end{align}\] then we have the following equalities \[\begin{align} \langle 0,a_n,a_{n+1},\ldots \rangle &= s_0^{(n)} X^{-1} + s_1^{(n)} X^{-2} + \cdots, \\ \langle 0,-a_{n-1},-a_{n-2},\ldots \rangle & = \tilde{s}_0^{(n)}X^{-1} + \tilde{s}_1^{(n)}X^{-2} + \cdots. \end{align}\] Noticing that \(a_k = \frac{2}{u_k}(X+v_k)\), it follows from Theorem 1 that \[\begin{align} - (-1)^{-1} \left(\frac{2}{u_n}\right)^{-1} &= \Delta_{1}^{(n)},\\ - \left(\frac{2}{u_{n+k-1}}\right)^{-1} \left(\frac{2}{u_{n+k}}\right)^{-1} &= \frac{\Delta_{k-1}^{(n)}\Delta_{k+1}^{(n)}}{(\Delta_{k}^{(n)})^2}, \quad \forall\, k \ge 1, \end{align}\] and \[\begin{align} - (-1)^{-1} \left(-\frac{2}{u_{n-1}}\right)^{-1} &= \tilde{\Delta}_{1}^{(n)},\\ - \left(-\frac{2}{u_{n-k}}\right)^{-1} \left(-\frac{2}{u_{n-k-1}}\right)^{-1} &= \frac{\tilde{\Delta}_{k-1}^{(n)}\tilde{\Delta}_{k+1}^{(n)}}{(\tilde{\Delta}_{k}^{(n)})^2}, \quad \forall\, k \ge 1, \end{align}\] where \[\begin{align} \Delta_{k}^{(n)} := \det(s^{(n)}_{i+j-2})_{1\le i,j \le k},\quad\qquad \tilde{\Delta}_{k}^{(n)} := \det(\tilde{s}^{(n)}_{i+j-2})_{1\le i,j \le k}, \quad \forall\, k \ge 1, \end{align}\] and we use the convention \(\Delta_0^{(n)} = \tilde{\Delta}_{0}^{(n)} = 1\). By virtue of 25 , the above equations become \[\begin{align} \frac{u_n}{2} = \Delta_{1}^{(n)}, \qquad d_{n+k} = \frac{\Delta_{k-1}^{(n)}\Delta_{k+1}^{(n)}}{(\Delta_{k}^{(n)})^2}, \quad \forall\, k \ge 1, \end{align}\] and \[\begin{align} -\frac{u_{n-1}}{2} = \tilde{\Delta}_{1}^{(n)} \qquad d_{n-k} = \frac{\tilde{\Delta}_{k-1}^{(n)}\tilde{\Delta}_{k+1}^{(n)}}{(\tilde{\Delta}_{k}^{(n)})^2}, \quad \forall\, k \ge 1. \end{align}\] If we introduce \(\{\tau_{n+k}^{(n)}\}_{k \in \mathbb{Z}}\) by setting \[\tau_{n+k}^{(n)} := \left\{ \begin{align} &\Delta_{k}^{(n)}, \quad && k \ge 1, \\ &1, \quad &&k= 0,\\ &\tilde{\Delta}_{-k}^{(n)}, \quad && k \le -1, \end{align} \right. \label{tauseq}\tag{28}\] then \(\{d_m\}_{m \in \mathbb{Z}}\) has the following consistent expressions \[d_{m} = \frac{\tau^{(n)}_{m-1}\tau^{(n)}_{m+1}}{(\tau^{(n)}_{m})^{2}},\quad \forall\, m \in \mathbb{Z}. \label{d}\tag{29}\] This solves the mismatch problem mentioned in [2] by Hone, as a result of which, a nicer expression is obtained for the initial problem to the bilateral Somos-4 recurrence as stated in the following section. It is noted that, for any integer \(n\), the sequence \(\{\tau_{m}^{(n)}\}_{m \in \mathbb{Z}}\) defined above by 28 is uniquely determined by \(Y_{n}\). For convenience, we call \(\{\tau_{m}^{(n)}\}_{m \in \mathbb{Z}}\) the associated tau sequence with \(Y_n\).

4 Applications to bilateral Somos-4 and Somos-5↩︎

The results in the above section can be directly applied to the initial value problems for the bilateral Somos-4 and Somos-5.

4.1 Connection formulae↩︎

We first present some connection formulae which will be useful in solving the initial value problems for the bilateral Somos-4 and Somos-5.

The above lemma immediately leads to the following corollary.

This corollary suggests that the sequence of quadratic orthogonal pairs \(\{(Y_n,\tilde{Y}_n)\}_{n \in \mathbb{Z}}\) is identically associated with a sequence of Somos-4 sequences \(\{(\ldots,\tau_{-1}^{(n)},\tau_{0}^{(n)},\tau_{1}^{(n)},\ldots)\}_{n \in \mathbb{Z}}\) with the same coefficients \(\alpha\) and \(\beta\). A further result on this family of tau sequences is given by the following theorem, which shows that all these tau sequences are essentially identical under the gauge transformations.

Based on the above theorem, we readily obtain the following corollary.

4.2 Initial value problem for Somos-4↩︎

The general initial value problem for the bilateral Somos-4 recurrence can be formulated as \[\begin{cases} & \tau^*_{n+2}\tau^*_{n-2} =\alpha \tau^*_{n+1}\tau^*_{n-1} + \beta (\tau^*_{n})^2, \, n\in \mathbb{Z},\\ & \tau^*_{-1} =A, \quad \tau^*_{0} =B, \quad \tau^*_{1} =C, \quad \tau^*_{2} =D, \end{cases} \label{initial40}\tag{34}\] where \(\alpha\), \(\beta\) and \(\{A,B,C,D\}\) are indeterminants. Since the above recurrence is invariant under gauge transformations \(\tau^*_{n} \mapsto ab^n\tau^*_{n}\) for all complex numbers \(a\) and \(b\), the problem 34 is equivalent to the problem

\[\begin{cases} & \tau_{n+2}\tau_{n-2} =\alpha \tau_{n+1}\tau_{n-1} + \beta (\tau_{n})^2, \, n\in \mathbb{Z},\\ & \tau_{-1} =x, \quad \tau_{0} =1, \quad \tau_{1} =y, \quad \tau_{2} =z . \end{cases} \label{initial4}\tag{35}\] We are going to solve the problem 35 .

Based on the argument in Subsection 3.3 and 4.1, in order to solve 35 , it suffices to make a choice of the parameters \(\{d_0,v_0,u_0,u,v,f\}\) such that the associated tau sequence of the hyperelliptic function \(Y_0\) is a solution to problem 35 . Recall that \(Y_0\) is defined by \[\begin{align} Y_0 =r_0Y+s_0= \frac{\sqrt{(X^2+f)^2+4u(X-v)}+X^2+f+2d_0}{u_0(X-v_0)}.\\ \end{align} \label{Y00}\tag{36}\] The following theorem shows that, for each triple of initial values \((x,y,z)\) leading to a unique solution to the problem 35 , there exist exactly two choices of the parameters \(\{d_0,v_0,u_0,u,v,f\}\) such that the solution of problem 35 is given by the associated tau sequence of the hyperelliptic function \(Y_0\).

The above theorem implies that the solution to the bilateral Somos-4 recurrence can be expressed by Hankel determinants. To be precise, suppose that we have \[\begin{align} &G:= (Y_0)^{-1} = \frac{\sqrt{(X^2+f)^2+4u(X-v)} -X^2-f-2d_0}{u_{-1}(X-v_{-1})}= s_0X^{-1}+s_{1}X^{-2}+\cdots,\\ &G^{\dagger}:= (-\tilde{Y}_0)^{-1} = \frac{-\sqrt{(X^2+f)^2+4u(X-v)} +X^2+f+2d_0}{u_{0}(X-v_{0})}=s^{\dagger}_0X^{-1}+s^{\dagger}_{1}X^{-2}+\cdots, \end{align}\] with \(u_{-1}\) and \(v_{-1}\) given by \[\begin{align} u_{-1} = -2x,\qquad v_{-1} = \frac{u}{d_0} -v_0 = \frac{u(-\beta y^2 +x^2z^2+\alpha z - \alpha x y^3)}{2\alpha xyz}. \end{align} \label{c-1}\tag{44}\] Furthermore, it is not hard to see that \(G\) and \(G^{\dagger}\) satisfy the following algebraic relations \[\begin{align} u_{-1}(X-v_{-1}) G^2 + 2 (X^2+f+2d_0) G &= u_0(X-v_0), \\ u_0(X-v_0)(G^{\dagger})^2 - 2 (X^2+f+2d_0) G^{\dagger} &= u_{-1}(X-v_{-1}), \end{align}\] from which, one can uniquely determine \(\{s_n\}_{n \in \mathbb{N}}\) and \(\{s^{\dagger}_n\}_{n \in \mathbb{N}}\) by recursion. In fact, we obtain \[\begin{align} s_0 &= \frac{u_0}{2},\\ s_1 &= -\frac{1}{2}u_0v_0,\\ s_2 &=-(f+2d_0)s_0- \frac{1}{2}u_{-1}\left(\sum_{i+j= 0}s_is_j\right), \\ s_k &= -(f+2d_0)s_{k-2} -\frac{1}{2}u_{-1}\left(\sum_{i+j= k-2}s_is_j\right)+\frac{1}{2}u_{-1}v_{-1}\left(\sum_{i+j= k-3}s_is_j\right),\quad k\ge 3, \end{align} \label{rec95s1}\tag{45}\] and \[\begin{align} s^{\dagger}_0 &= -\frac{u_{-1}}{2}, \\ s^{\dagger}_1 &= \frac{1}{2}u_{-1}v_{-1},\\ s^{\dagger}_2 &=-(f+2d_0)s^{\dagger}_0+ \frac{1}{2}u_{0}\left(\sum_{i+j= 0}s^{\dagger}_is^{\dagger}_j\right),\\ s^{\dagger}_k &= -(f+2d_0)s^{\dagger}_{k-2} +\frac{1}{2}u_{0}\left(\sum_{i+j= k-2}s^{\dagger}_is^{\dagger}_j\right)-\frac{1}{2}u_{0}v_{0}\left(\sum_{i+j= k-3}s^{\dagger}_is^{\dagger}_j\right),\quad k\ge 3. \end{align} \label{rec95s2}\tag{46}\]

Eventually, we have the following theorem.

4.3 Initial value problem for Somos-5↩︎

Now we consider the general initial value problem for the bilateral Somos-5 recurrence \[\begin{cases} & \tau^*_{n+5}\tau^*_{n} =\alpha \tau^*_{n+4}\tau^*_{n+1} + \beta \tau^*_{n+3}\tau^*_{n+2},\qquad n \in \mathbb{Z},\\ & \tau^*_{0} =A, \quad \tau^*_{1} =B, \quad \tau^*_{2} =C, \quad \tau^*_{3} =D,\quad \tau^*_{4} =E, \end{cases} \label{initial50}\tag{49}\] where \(\alpha\), \(\beta\) and \(\{A,B,C,D,E\}\) are indeterminants. Since the above recursion equation is invariant under gauge transformations \(\tau^*_{n} \mapsto ab^n\tau^*_{n}\) for all complex numbers \(a\) and \(b\), the problem 49 can be equivalently transformed into \[\begin{cases} & \tau_{n+5}\tau_{n} =\alpha \tau_{n+4}\tau_{n+1} + \beta \tau_{n+3}\tau_{n+2},\\ & \tau_{0} = 1, \quad \tau_{1} = 1, \quad \tau_{2} =x, \quad \tau_{3} =y,\quad \tau_{4} =z. \end{cases} \label{initial5}\tag{50}\]

In order to solve this problem, we recall a connection between Somos-4 and Somos-5, which was originally shown in [3] and then interpreted as a Bäcklund transformation in [25]. We claim that this relationship applies to not only the half case, but also the bilateral case and it enables us to solve the above problem for Somos-5.

By employing the above lemma, the problem 50 can be reformulated as the solution of the following two problems: \[\begin{cases} & \Delta^{(1)}_{n+2}\Delta^{(1)}_{n-2} =\tilde{\alpha} \Delta^{(1)}_{n+1}\Delta^{(1)}_{n-1} + \tilde{\beta} (\Delta^{(1)}_{n})^2,\\ & \Delta^{(1)}_{-1} =\frac{\alpha \beta x^2+\alpha^2 xy +\beta z }{yz},\\ & \Delta^{(1)}_{0} =1,\\ & \Delta^{(1)}_{1} =x,\\ & \Delta^{(1)}_{2} =z, \end{cases} \label{delta1}\tag{51}\] and \[\begin{cases} & \Delta^{(2)}_{n+2}\Delta^{(2)}_{n-2} =\tilde{\alpha}\Delta^{(2)}_{n+1}\Delta^{(2)}_{n-1} + \tilde{\beta}(\Delta^{(2)}_{n})^2,\\ & \Delta^{(2)}_{-1} =\frac{\beta x +\alpha y}{z},\\ & \Delta^{(2)}_{0} =1,\\ &\Delta^{(2)}_{1} =y,\\ &\Delta^{(2)}_{2} =\alpha z+ \beta xy, \end{cases} \label{delta2}\tag{52}\] where \(\tilde{\alpha}\) and \(\tilde{\beta}\) are given by \[\begin{align} \tilde{\alpha} = \beta^2, \qquad \tilde{\beta} = 2\alpha\beta^2+\alpha^2\beta \left(\frac{y^2z+z^2+\beta x^3y+\alpha x^2 z+\alpha x^2y^2}{xyz}\right)+\alpha^4 . \end{align}\] These two problems are nothing but two specific cases of the problem 34 which has been solved in the previous subsection. As a result, one can derive the solutions to 51 and 52 , and consequently solve the problem 49 .

First, upon comparing the problems 51 and 52 with the problem 34 together with the equivalent formulation 35 , we can set \[\begin{align} x^{(1)} &= \frac{\alpha \beta x^2+\alpha^2 xy +\beta z }{yz},\quad& y^{(1)} &=x,\quad &z^{(1)} &=z,\\ x^{(2)} &=\frac{\beta x+\alpha y}{z}, \quad&y^{(2)} &= y,\quad& z^{(2)} &=\alpha z + \beta xy, \end{align}\] and introduce the parameters \[\begin{cases} & d_0^{(r)} = x^{(r)}y^{(r)},\\ & v_0^{(r)} = \frac{\tilde{\beta} (y^{(r)})^2-(x^{(r)})^2(z^{(r)})^2+\tilde{\alpha} z^{(r)} + \tilde{\alpha} x^{(r)} (y^{(r)})^3}{2\sqrt{\tilde{\alpha}}x^{(r)}y^{(r)}z^{(r)}},\\ & u_0^{(r)}=2y^{(r)},\\ & u^{(r)} = \sqrt{\tilde{\alpha}},\\ & v^{(r)} = \frac{\tilde{\beta} (y^{(r)})^2+(x^{(r)})^2(z^{(r)})^2+\tilde{\alpha} z^{(r)} + \tilde{\alpha} x^{(r)} (y^{(r)})^3}{2\sqrt{\tilde{\alpha}}x^{(r)}y^{(r)}z^{(r)}},\\ & f^{(r)} = \frac{\tilde{\beta}}{\tilde{\alpha}} - \frac{(\tilde{\beta} (y^{(r)})^2+(x^{(r)})^2(z^{(r)})^2+\tilde{\alpha} z^{(r)} + \tilde{\alpha} x^{(r)} (y^{(r)})^3)^2}{4\tilde{\alpha} (x^{(r)})^2(y^{(r)})^2(z^{(r)})^2},\\ &u_{-1}^{(r)} = -2x^{(r)},\\ &v_{-1}^{(r)} = \frac{u^{(r)}}{d_0^{(r)}}-v_0^{(r)}, \end{cases}\] for \(r = 1,2\). Then we define two pairs of hyperelliptic functions with Laurent series expansions as follows \[\begin{align} &G_{r}:=\frac{\sqrt{(X^2+f^{(r)})^2+4u^{(r)}(X-v^{(r)})} -X^2-f^{(r)}-2d_0^{(r)}}{u_{-1}^{(r)}(X-v_{-1}^{(r)})}= s_{0,r}X^{-1}+s_{1,r}X^{-2}+\cdots,\\ &G_r^{\dagger}:=\frac{-\sqrt{(X^2+f^{(r)})^2+4u^{(r)}(X-v^{(r)})} +X^2+f^{(r)}+ 2d_0^{(r)}}{u_{0}^{(r)}(X-v_{0}^{(r)})}=s^{\dagger}_{0,r}X^{-1}+s^{\dagger}_{1,r}X^{-2}+\cdots. \end{align}\] Based on the intrinsic algebraic relations, we can show that \(\{s_{n,r},s_{n,r}^{\dagger}\}_{n \in \mathbb{N},r = 1,2}\) satisfy the following recursion relations \[\begin{align} s_{0,r} &= \frac{u_0^{(r)}}{2},\\ s_{1,r} &= -\frac{1}{2}u_0^{(r)}v_0^{(r)},\\ s_{2,r} &=-(f^{(r)}+2d_0^{(r)})s_{0,r}- \frac{1}{2}u_{-1}^{(r)}\left(\sum_{i+j= 0}s_{i,r}s_{j,r}\right), \\ s_{k,r} &= -(f^{(r)}+2d_0^{(r)})s_{k-2,r} -\frac{1}{2}u_{-1}^{(r)}\left(\sum_{i+j= k-2}s_{i,r}s_{j,r}\right)+\frac{1}{2}u_{-1}^{(r)}v_{-1}^{(r)}\left(\sum_{i+j= k-3}s_{i,r}s_{j,r}\right),\quad k\ge 3, \end{align} \label{rec-nodagger}\tag{53}\] \[\begin{align} s^{\dagger}_{0,r} &= -\frac{u_{-1}^{(r)}}{2},\\ s^{\dagger}_{1,r} &= \frac{1}{2}u_{-1}^{(r)}v_{-1}^{(r)},\\ s^{\dagger}_{2,r} &=-(f^{(r)}+2d_0^{(r)})s^{\dagger}_{0,r}+ \frac{1}{2}u_{0}^{(r)}\left(\sum_{i+j= 0}s^{\dagger}_{i,r}s^{\dagger}_{j,r}\right),\\ s^{\dagger}_{k,r} &= -(f^{(r)}+2d_0^{(r)})s^{\dagger}_{k-2,r} +\frac{1}{2}u_{0}^{(r)}\left(\sum_{i+j= k-2}s^{\dagger}_{i,r}s^{\dagger}_{j,r}\right)-\frac{1}{2}u_{0}^{(r)}v_{0}^{(r)}\left(\sum_{i+j= k-3}s^{\dagger}_{i,r}s^{\dagger}_{j,r}\right),\quad k\ge 3. \end{align} \label{rec-dagger}\tag{54}\] Finally, the solution to the problem 50 on the bilateral Somos-5 can be summarized in the following theorem.

4.4 A symmetric part for the Somos’ original conjecture for Somos-4↩︎

In the year 2000, Michael Somos examined the following curve \[\mathcal{C}_0 :y-y^2 = z- z^3. \label{somoscurve}\tag{56}\] He conjectured that if one expands \(\frac{y}{z}-1\) into a power series of \(z\) when \(z \to 0\), that is, \[\begin{align} \frac{y}{z}-1 &= \frac{\frac{1}{2}-\sqrt{z^3-z+\frac{1}{4}}}{z}-1 \\ &= 1z^1+1z^2+3z^3+8z^4+23z^5+68z^6+207z^7+644z^8+2040z^9+\cdots\\ &= s_0z^1+s_1z^2+s_2z^3 +\cdots, \end{align}\] where \(s_k = 2s_{k-1}+\sum_{i+j=k-2}s_is_j,k\ge 2\) with \(s_0 = s_1 = 1\), then the corresponding sequence of Hankel determinants \(\{\Delta_{n}\}_{n \in \mathbb{N}}\) defined by \[\Delta_{n} := \det(s_{i+j-2})_{1 \le i.j \le n}\] satisfies half of the Somos-4 recurrence \[\Delta_{n+4}\Delta_{n}=\Delta_{n+3}\Delta_{n+1}+(\Delta_{n+2})^2,\quad \forall\, n \in \mathbb{N}. \label{somoe}\tag{57}\] This conjecture can be proved using various techniques in [27] and [24].

It is noted that our scheme works for this particular curve. Indeed, if we take the birational transformation \[\begin{align} y &= \frac{1+(X+1)^{-2}Y}{2},\qquad z = (X+1)^{-1}, \end{align}\] the curve 56 becomes \[Y^{2} = (X^2 -3)^2 - 4(X+2),\] which is exactly in the form of the spectral curve 11 when the parameters are taken to be \(f= -3,u=-1,v= 2\). And the generating function \(\frac{y}{z} - 1= \frac{1}{2(X+1)} Y + \frac{X^2-1}{2(X+1)}\) corresponds to the choice of \(Y_0\) when the parameters take the values \(u_0 = 2, v_0 = -1, d_0 = 1\). Also see [2]. The action of the Galois group \(Gal(\mathcal{F}/\mathbb{C}(X))\) is independent of coordinates, therefore we are free to apply our results to the original coordinates \((y,z)\).

By using our methods based on quadratic orthogonal pairs, we can find another curve whose Taylor series expansion will give the negative part of the same Somos-4 recurrence. Indeed, let \(\frac{\tilde{y}}{z} - 1\) be the generating function for the corresponding negative sequence, by relation 22 , we have \[(1-\frac{\tilde{y}}{z})(\frac{y}{z}- 1)^* = -1.\] Substituting \(\frac{y}{z}- 1= \frac{\frac{1}{2}-\sqrt{z^3-z+\frac{1}{4}}}{z}-1\) into it and using the definition for the symbol \(*\), we immediately find the algebraic relation for \(\tilde{y}\) as: \[\tilde{\mathcal{C}}_0 :(1-z)\tilde{y}^2+(2z^2-1)\tilde{y} = z(z^2-1),\] which is the exact equation for the dual curve. If one expands \(\frac{\tilde{y}}{z}-1\) into a power series of \(z\) when \(z \to 0\), one has \[\begin{align} \frac{\tilde{y}}{z}-1 &= \frac{\sqrt{z^3-z+\frac{1}{4}}+z-\frac{1}{2}}{z(z-1)}\\ &= 1z^1+2z^2+5z^3+13z^4+36z^5+104z^6+311z^7+955z^8+2995z^9+\cdots\\ &= \tilde{s}_0z^1+\tilde{s}_1z^2+\tilde{s}_2z^3 +\cdots, \end{align}\] where \(\tilde{s}_{k} =3\tilde{s}_{k-1}-2\tilde{s}_{k-3} + \sum_{i+j = k-3}(\tilde{s}_{i}-\tilde{s}_{i+1}) (\tilde{s}_{j}-\tilde{s}_{j+1}),k\ge 3\) with \(\tilde{s}_0= 1, \tilde{s}_1 = \tilde{s}_1 = 2\). Define \[\tau_{n} = \left\{ \begin{align} &\det(s_{i+j-2})_{1 \le i,j\le n}, \quad && n \ge 1, \\ &1, \quad &&n= 0,\\ &\det(\tilde{s}_{i+j-2})_{1 \le i,j\le -n}, \quad && n \le -1, \\ \end{align} \right.\] then \(\{\tau_n\}_{n \in \mathbb{Z}}\) is a bilateral Somos-4 sequence satisfying \[\tau_{n+4}\tau_{n}=\tau_{n+3}\tau_{n+1}+(\tau_{n+2})^2,\quad \forall\, n \in \mathbb{Z},\] with \(\tau_{-2}=\tau_{-1}=\tau_0=\tau_1=1\).

5 Acknowledgement↩︎

We thank Professor A.N.W. Hone for his valuable communications. X.K. Chang’s research was supported by the National Natural Science Foundation of China (Grant Nos. 12288201, 12222119) and the Youth Innovation Promotion Association CAS.

6 Quadratic orthogonal pairs for general hyperelliptic curves↩︎

In this appendix let’s discuss quadratic orthogonal pairs for general hyperelliptic curves. Consider a hyperelliptic curve of genus \(g\) defined by \[\mathcal{C}:Y^2 = A(X)^2 + 4R(X),\] where \(g\) is an arbitrary positive integer and \(A(X)\),\(R(X)\) are polynomials defined by \[\begin{align} A(X) &= X^{g+1} +\sum_{j=0}^{g-1}k^{(j)}X^j,\qquad R(X) &= u\left(X^{g}+ \sum_{j=0}^{g-1}l^{(j)}X^j\right). \end{align}\]

In [2], Hone constructed two sequences of polynomials \(\{P_n\}_{n \in \mathbb{Z}}\) and \(\{Q_n\}_{n \in \mathbb{Z}}\) satisfying the following iteration \[\begin{align} P_{n+1} = a_{n}Q_{n} - P_{n},\qquad Q_{n+1} = Q_{n-1} +2a_{n}P_{n} - a_{n}^2Q_{n}, \qquad \forall\, n \in \mathbb{Z}, \end{align} \label{iteration}\tag{58}\] where \(\{P_n\}_{n \in \mathbb{Z}}\) and \(\{Q_n\}_{n \in \mathbb{Z}}\) are expressed by \[\begin{align} P_{n}(X) = A(X)+ \sum_{j=0}^{g-1}\pi_{n}^{(j)}X^j,\qquad Q_{n}(X) = u_{n}\left(X^g+\sum_{j=0}^{g-1}\rho_{n}^{(j)}X^j\right), \end{align}\] and \(\{a_n\}_{n \in \mathbb{Z}}\) is a sequence of linear functions uniquely determined by the coefficients of \(\{Q_{n}\}_{n \in \mathbb{Z}}\) via \[a_{n}(X) = \frac{2}{u_n}(X-\rho_{n}^{(g-1)}).\] By use of the iteration 58 , it is straightforward to compute \[\begin{align} P_{n+1}^2 + Q_{n+1}Q_{n} = (a_{n}Q_{n} - P_{n})^2+(Q_{n-1} +2a_{n}P_{n} - a_{n}^2Q_{n})Q_{n} = P_{n}^2+Q_{n}Q_{n-1}, \end{align}\] which shows that \(P_{n}^2+Q_{n}Q_{n-1}\) is a conserved quantity for the iteration 58 . Therefore if we choose initial values \(\{P_0,Q_0,Q_{-1}\}\) such that \[P_{0}^2+Q_{0}Q_{-1} = Y^2,\] then we will have \[P_{n}^2+Q_{n}Q_{n-1} = Y^2 \label{conserved}\tag{59}\] for any integer \(n\).

The above two sequences of polynomials \(\{P_n\}_{n \in \mathbb{Z}}\) and \(\{Q_n\}_{n \in \mathbb{Z}}\) give rise to a choice of sequences of hyperelliptic functions. To be precise, let us choose one sequence \(\{Y_n=r_n Y+s_n|r_n,s_n\in \mathbb{C}(X),r_n \ne 0 \}_{n\in \mathbb{Z}}\) and \(\{a_n\}_{n \in \mathbb{Z}}\) by setting \[\begin{align} &r_n := \frac{1}{Q_{n}},\qquad s_n :=\frac{P_n}{Q_{n}},\qquad a_n := a_n. \end{align} \label{choiceg1}\tag{60}\] and the other sequence \(\{\tilde{Y}_n=\tilde{r}_nY+\tilde{s}_n|\tilde{r}_n,\tilde{s}_n\in \mathbb{C}(X),\tilde{r}_n \ne 0 \}_{n\in \mathbb{N}}\) and \(\{\tilde{a}_n\}_{n \in \mathbb{N}}\) by setting \[\begin{align} &\tilde{r}_{n} := \frac{1}{Q_{n-1}},\qquad \tilde{s}_{n} :=\frac{P_n}{Q_{n-1}},\qquad \tilde{a}_{n} :=a_{n-1}. \end{align} \label{choiceg2}\tag{61}\] As a result, we can obtain generalizations of Lemma 3, Lemma 4 and Theorem 7.

With Lemma 7, Lemma 8 and Theorem 23, one can go through Section 3.4 and Section 4.1 in the case of any genus \(g\) and obtain the generalized version of all results of Section 3.4 and Theorem 12 and Corollary 13 without much effort. However, Lemma 5 and Corollary 11 will be replaced by a much more complicated version that is still unknown in general. Even so, the notion of quadratic orthogonal pairs is still valid in resolving the mismatch problem. As an example to show how this works in higher genus, we reconsider the curve of genus 2 examined in [2] by Hone: \[Y^2 = (X^3-5X-1)^2-4(X^2+2X+3).\] The generating function for the positive half sequence he picked on this curve is denoted by \(G\) whose explicit formula and series expansion at \(X \to \infty\) and \(Y \sim X^3\) are given by \[\begin{align} G&=\frac{X+\frac{1}{2}+\frac{1}{4}(X^3-5X-1-Y)}{X^2+\frac{1}{2}X-\frac{3}{2}} \\ &= X^{-1} + 0X^{-2}+ 2X^{-3} + 0X^{-4}+7X^{-5} + 2X^{-6} + \cdots, \\ &= s_0 X^{-1} + s_1 X^{-2} + s_2 X^{-3} + \cdots, \end{align}\] and the moments \(\{s_{j}\}_{j \in \mathbb{N}}\) are recursively produced by \[s_j = s_{j-2} - s_{j-3} + 2 \sum_{i=0}^{j-2} s_i s_{j-2-i} + \sum_{i=0}^{j-3} s_i s_{j-3-i} - 3 \sum_{i=0}^{j-4} s_i s_{j-4-i}, \quad j \geq 3,\] with initial values \(s_0 = 1\), \(s_1 = 0\), \(s_2 = 2\). The corresponding sequence of Hankel determinants determined by \(G\) now reads: \[\label{somos8positive} 1,1,2,6,31,319,5810,147719,8526736,\ldots.\tag{62}\] It was shown in [2] with the use of computer algebra that this sequence is the positive half of a particular Somos-8 sequence: \[7\tau_{n+8}\tau_n + 137\tau_{n+7}\tau_{n+1} + 2504\tau_{n+6}\tau_{n+2} - 43424\tau_{n+5}\tau_{n+3} - 26959\tau_{n+4}^2 = 0.\] To obtain the negative half sequence, he picked another function denoted by \(G^{\dagger}\) whose explicit formula and series expansion at \(X \to \infty\) and \(Y \sim X^3\) is given by \[\begin{align} G^{\dagger} &= -\frac{1}{4}\left( \frac{-Y+X^3-\frac{5}{2}X+\frac{1}{2}}{X^2+\frac{1}{2}X -\frac{3}{2}} \right) \\ &= -\frac{5}{8}X^{-1} - \frac{1}{16}X^{-2} - \frac{45}{32}X^{-3} - \frac{25}{64}X^{-4} - \frac{757}{128}X^{-5} - \frac{801}{256}X^{-6} - \cdots, \\ &= s_0^{\dagger} X^{-1} + s_1^{\dagger} X^{-2} + s_2^{\dagger} X^{-3} + \cdots, \end{align}\] as in [2]. Here we use \(-Y\) to replace the \(Y\) there for consistency. The recursion for the moments \(\{s^{\dagger}_{j}\}_{j \in \mathbb{N}}\) is: \[s_j^\dagger = \frac{5}{2} s_{j-2}^\dagger - \frac{1}{2} s_{j-3}^\dagger - 2 \sum_{i=0}^{j-2} s_i^\dagger s_{j-2-i}^\dagger - \sum_{i=0}^{j-3} s_i^\dagger s_{j-3-i}^\dagger + 3 \sum_{i=0}^{j-4} s_i^\dagger s_{j-4-i}^\dagger, \quad j \geq 3,\] with initial values \(s_0^\dagger = -5/8, s_1^\dagger = -1/16, s_2^\dagger = 45/32\). The corresponding sequence of Hankel determinants determined by \(G^{\dagger}\) reads: \[\label{somos8negative} 1,-\frac{5}{2^3}, \frac{7}{2^3}, -\frac{303}{2^7}, \frac{4091}{2^9}, -\frac{63805}{2^{10}}, \frac{3496637}{2^{12}}, \ldots.\tag{63}\] Clearly, one cannot simply put the sequences 62 and 63 together to derive a bilateral Somos-8 sequence, therefore a gauge transformation of the form [2] is needed. However, based on the notion of quadratic orthogonal pairs that \((-\tilde{G})^{-1} (G^{-1})^*=-1\), it is rather easy to find the precise generating function \(\tilde{G}\) whose sequence of Hankel determinants gives exactly the corresponding negative sequence. More precisely, we have \[\begin{align} \tilde{G} = \frac{1}{G^*} &= \frac{X^2+\frac{1}{2}X-\frac{3}{2}}{X+\frac{1}{2}+\frac{1}{4}(X^3-5X-1+Y)} \\ &= 2 X^{-1}+ 1 X^{-2} + 3X^{-3} + 3X^{-4} + 11 X^{-5}+ 14X^{-6} + 54X^{-7} + 81X^{-8} + 300 X^{-9}+ \cdots,\\ &= \tilde{s}_{0} X^{-1} + \tilde{s}_{1} X^{-2} + \tilde{s}_{2} X^{-3} +\cdots, \end{align}\] where the moments \(\{\tilde{s}_{j}\}_{j \in \mathbb{N}}\) are recursively produced by \[\tilde{s}_j = \tilde{s}_{j-2} - \tilde{s}_{j-3} + \sum_{i=0}^{j-2} \tilde{s}_i \tilde{s}_{j-2-i} -\sum_{i=0}^{j-4} \tilde{s}_i \tilde{s}_{j-4-i}, \quad j \geq 3,\] with initial values \(\tilde{s}_0 = 2, \tilde{s}_1 = 1, \tilde{s}_2 = 3\), and the corresponding sequence of Hankel determinants determined by \(\tilde{G}\) reads: \[\label{somos8negativeour} 1,2,5,28,303,4091,127610,6993274,562196701,89628659609,26028173673848, \ldots.\tag{64}\] Combining the sequences 62 and 64 will exactly yield the doubly infinite sequence presented in [2], that is, \[\label{somos8bilateral} \begin{align} &\ldots,26028173673848,89628659609,562196701,6993274,127610,4091,\\&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,303,28,5,2,1, 1,2,6,31,319,5810,147719,8526736,\ldots. \end{align}\tag{65}\] It is also noted that, using our recursion for the moments \(\{s_{j}\}_{j \in \mathbb{N}}\) and \(\{\tilde{s}_{j}\}_{j \in \mathbb{N}}\), one can immediately see that the doubly infinite sequence 65 consists entirely of integers. This is not obvious when using the recursion for \(\{s^{\dagger}_{j}\}_{j \in \mathbb{N}}\) and applying the gauge transformation.

References↩︎