Integrability and non-integrability of periodic non-autonomous Lyness recurrences

This paper studies non-autonomous Lyness-type recurrences of the form x n+2 = (a n + x n+1)/x n , where {a n } is a k-periodic sequence of positive numbers with primitive period k. We show that for the cases k ∈ {1, 2, 3, 6}, the behaviour of the sequence {x n } is simple (integrable), while for the remaining cases satisfying this behaviour can be much more complicated (chaotic). We also show that the cases where k is a multiple of 5 present some different features.


Introduction and main results
The dynamical study of the Lyness difference equation ( [2,5,16,28]) and its generalizations to higher order Lyness type equations ([4, 7, 8, 18]), or to difference equations with periodic coefficients ( [11,17,22,23,25]), has been the focus of an active research activity in the last two decades.In more recent dates, Lyness type equations have also been approached using different points of view: from algebraic geometry ([6, 15, 27]) to the theory of discrete integrable systems ( [10,19,21,25,26]).This paper deals with the problem of the integrability and non-integrability of nonautonomous planar Lyness difference equations of the form where {a n } is a cycle of k positive numbers, i.e. a n+k = a n for all n ∈ N, being k the primitive period and we consider positive initial conditions x 1 and x 2 .As we will see, the behavior of the sequences {x n } can be essentially different according to whether k ∈ {1, 2, 3, 6}, k is a multiple of 5 or it is not.
In this section we summarize our main results on (1) in terms of k.We also give an account of the tools that we have developed for this study that we believe might be interesting by themselves.We start by introducing the notations and definitions used in the paper.

Notations and definitions
Given a periodic sequence {a n } of primitive period k we will say that its rank is m if Card{a 1 , a 2 , . . ., a k } = m ∈ N.
The values a 1 , . . ., a k will be usually called parameters.In our context the recurrence (1) is called persistent if for any sequence {x n } there exist two real positive constants c and C, which depend on the initial conditions, such that for all n, 0 < c < x n < C < ∞.
For each k, the composition maps are where each F a i is defined by and a 1 , a 2 , . . ., a k are the k elements of the cycle.When there is no confusion, for the sake of shortness, we also will use the notation F [k] := F a k ,...,a 2 ,a 1 .Note that these maps are birational maps and are always well-defined in the open invariant set Q + = {(x, y) : x > 0, y > 0} ⊂ R 2 .Moreover (x 1 , x 2 ) and in general, There are two concepts coexisting in this context, the non-autonomous invariants and the first integrals, that we will use in this paper.Given a difference equation of the form (1), a non-autonomous invariant is a function V (x, y, n), such that for all initial conditions and all n ∈ N. On the other hand, when the difference equation has k-periodic coefficients a first integral is a function H, which is a first integral for the discrete dynamical system generated by F [k] , that is H(F [k] (x, y)) = H(x, y), for all points in an open set.In terms of the recurrence for all initial conditions (x n , x n+1 ).We will relate both concepts in Section 3.
Two analytic functions P, Q : U ⊂ C 2 → C are said to be coprime if the points of the set {(x, y) ∈ U : P (x, y) = Q(x, y) = 0} are isolated.A function H = P/Q, with P and Q coprime, will be called a meromorphic function.A meromorphic first integral of an analytic map F : U → C 2 is a meromorphic function H = P/Q such that P (F (x, y))Q(x, y) = P (x, y)Q(F (x, y)) for all (x, y) ∈ U.
Observe that from this definition H(F (x, y)) = H(x, y) for all points of U for which both terms of this last equality are well-defined.When P and Q are polynomials then it is said that H is a rational first integral.Similarly we can talk about meromorphic or rational invariants, and in this sense we will talk about rational or meromorphic integrability.
Finally, we will say that a planar map F has structurally stable numerical chaos (SSNC) when studying numerically several of its orbits, we observe that it presents all the features of a non-integrable perturbed twist map, that is: many invariant curves and, between them, couples of orbits of p-periodic points (for several values of p), half of them of elliptic type and the other half of hyperbolic saddle type.Moreover the separatrices of these hyperbolic saddles intersect transversally, for instance see [1,Chapter 6].

Main results
This subsection collects the general outlines of all our results about the recurrence (1), in terms of k.Cases k ∈ {1, 2, 3, 6} and other concrete integrable cases.For k ∈ {1, 2, 3} it is already known that the recurrences (1) are persistent.Moreover, either each sequence {x n } is periodic, with period a multiple of k or it densely fills at most k disjoint intervals of R + , see [11].A key point for the proof is the existence of a rational first integral for F and others such that their adherence consists of k points, see Theorem 6.The existence for k = 5 of values of a j , j = 1, . . ., 5, for which the sequence {x n } has the first behavior has been already established in previous works, see [7,Example 5.43.1] or [14], but only for very concrete initial conditions and parameters a 1 , . . ., a 5 .
Moreover in this case we can prove that for most values of the parameters the map has no meromorphic first integral (see Theorem 20).Furthermore the phase portrait of the map F [k] does not always coincide with the ones found in all the rational integrable cases, see for instance the second picture in Figure 1.In this case, apart of the celebrated Lyness map F 1 which satisfies F 5 1 = F 1,1,1,1,1 = Id, there are values of the parameters a 1 , . . ., a k such that F m k [k] = Id (see Corollary 14) and other for which the number of fixed points of the maps is a 1-dimensional manifold, or 2, 1, or 0 points, see Lemma 12. Finally, when k ≥ 15, there are cases presenting SSNC, see Section 6.
Cases k ∈ {1, 2, 3, 5, 6, 10}.When k ∈ {4, 7, 11, 15}, for some values of the parameters, a 1 , . . ., a k , we have numerically found SSNC, see Section 6.In fact we prove in Lemma 23 that based on these examples we can obtain values a 1 , . . ., a k , all different, with a similar behavior for all the remaining values of k.So for all these values of k there are situations for which the sequence {x n } can have different behaviors to those given in the above situations.For instance there appear sequences which fill more than k intervals.Some concrete examples for k = 4 are shown in Section 6.
Observe that as an application of our results we can show an interesting and curious phenomena that can be understood as a kind of "chaos regularization": Consider a map G = F [2] , which is a rationally integrable map, and a map H = F [4] which has chaotic behavior, then both maps G • H and H • G are rationally integrable because are of type F [6] .Thus G regularizes H.
Finally note that the above results show that the only cases for which recurrence (1) can have a rational invariant for all values of the parameters a 1 , . . ., a k are k ∈ {1, 2, 3, 6}.

Main tools
In this subsection we present several results that we have obtained which we believe that are interesting by themselves.Other technical results will be given in Section 2.
The first result is a necessary condition for the meromorphic integrability of planar maps near a fixed point.Our approach follows the guidelines of Poincaré when he studied the same problem for ordinary differential equations, see [24] and the references there in for the approach to ordinary differential equations.In Section 5 we will apply the result below to study the case k = 5.
Assume that F has a meromorphic first integral H in U.
When the map F : U ⊂ R 2 → R 2 is real valued and of class C 2 (U) the proof of Theorem 1 can be adapted following the same steps.Taking into account that in this case, when λ ∈ C is an eigenvalue of DF (0, 0), then λ is also, and we have to deal with the resonant condition λ p λq = 1, we obtain the following result: a C 2 (U) map such that F (0, 0) = (0, 0) ∈ U and DF (0, 0) is diagonalizable, with eigenvalues λ and µ.Assume that F has a meromorphic first integral H in U.
The above results will be applied to prove the meromorphic non-integrability of many cases when k = 5, see Theorem 20.On the other hand, next result will be the key point to prove the existence of rational integrable cases for all k = 5, see Theorem 19.
For the recurrence (1) we look for non-autonomous invariants of the form where with all the sequences of positive numbers.This method is introduced in [17] and the special form of V is inspired by this paper and the known invariant of the Lyness recurrences, see [2,22,23].We prove: Theorem 3. If the recurrence (1) has an invariant of the form (3) then a n+6 = a n and where (ii) The first integrals of the maps F [k] , for k ∈ {1, 2, 3, 6}, corresponding to the invariants given in Theorem 3 are: Remark 5. Our proof of Theorem 3 does not require the sequence of parameters {a n } to be periodic.
Observe that xy is the usual first integral (invariant) of the map F a associated to the classical Lyness recurrence, see for instance [2].It is already known, see [11,22,23], that in the two and three periodic cases the functions V b,a and V c,b,a are first integrals of the maps F b,a and F c,b,a , respectively.These first integrals play a crucial role for the understanding of the recurrence (1) when k = 2, 3, see again [11].To the best of our knowledge the existence of a first integral for the general non-autonomous 6-periodic case was not known.In Section 4 we use it to describe the dynamics in this case.
Also observe that, due to the form of the invariants, all the maps F [k] , for k ∈ {1, 2, 3, 6}, preserve a foliation of the plane given by biquadratic curves, which are elliptic except for a finite number of level sets.In fact these maps are particular cases of the celebrated QRT family of planar maps.Perhaps a further algebraic-geometric approach, like the one presented in [15], by studying the maps induced by each F [k] on the corresponding elliptic surface could give more information about the reason why the cases k ∈ {1, 2, 3, 6} are special.
It is also interesting to notice that other integrable QRT maps with periodic coefficients have been found recently [25], as well as another major family of maps: the Hirota-Kimura-Yahagi type ones [20].
As we have already commented, it is known that for very concrete values of a 1 , . . ., a 5 and suitable initial conditions, the behavior of {x n } is different to the ones appearing when k ∈ {1, 2, 3} and in particular ( 1) is non-persistent, see [7,Example 5.43.1] or [14].This behavior can also be seen considering ) it is clear that for a > 1 the orbits of the points of the form (1, y) with y = 0 are unbounded.
Our next result allows to establish, when k = 5, the non-persistence of the recurrence (1) for many values a 1 , . . ., a k .
In the above result, the open set of initial conditions for which the result holds is sometimes the whole first quadrant Q + .For instance, this is the case when k = 5 and a 1 = a, a 2 = ac, a 3 = c, a 4 = 1/a and a 5 = 1/(ac), when a > 1 and ac > 1, because is a linear map with a stable node at the origin.
The rest of the paper is organized as follows.In Section 2 we introduce some preliminary results, while in Section 3 we prove the main tools described in Section 1.3.Section 4 is devoted to the cases for which we find rational integrability, while in Section 5 we prove the non-integrability results when k is a multiple of five.Finally, in Section 6 we present some numerical evidence of chaos.

Preliminary results
This section contains some technical preliminary results and other known results that we will use in the proofs given in subsequent sections.
The next result will be useful for our numerical simulations.As we will see, some new variables allows to "observe" much better the numerical non-integrability studied in Section 6.Its proof is straightforward.
Lemma 7. The sequence (1) in the variables z n := log(x n ) becomes and the corresponding maps where and each G a is defined on the whole plane, R 2 .
Observe that the maps G [k] are area preserving.In fact, it is easy to see that the maps The next result proves that all the fixed points p of some map F a j ,a j−1 ,...,a 1 (p), for all j ≤ k, is well defined are resonant.As we will see in Proposition 21, the hypothesis on the maps F a j ,a j−1 ,...,a 1 (p) is unavoidable because for j = 5 there appear some cancelations that make that the maps F [5j] have as fixed point p = (0, 0) and this point can be of saddle type with arbitrary eigenvalues.In fact this property will be the key point for our proof of non-existence of meromorphic first integrals for most Proposition 10.Let p ∈ R 2 be a fixed point of a composition map F [k] and such that F a j ,a j−1 ,...,a 1 (p) is well defined for all j ≤ k.Then Proof.From Lemma 8 we know that G Proof. (i) The set of fixed points of F d,c,b,a is exactly the set of points satisfying but it is not easy to handle these two equations.On the other hand, the equivalent condition lead us to the system of the statement.Clearly both parabolas meet at a unique point in Q + .
(ii) Studying the condition we obtain the system of the statement.The x-coordinate of a fixed point has to satisfy the quadratic equation From this equation we easily obtain the result.Notice that simple cases having infinitely many fixed points appear for instance when b = d = 1 and e = a.
(iii) The two conditions given by directly lead to the system of the statement.The set of singular points of V f,e,d,c,b,a is formed by the points satisfying The two equations describing the above set exactly coincide again with the two equations given in the statement.So Finally, we will also use the following results: Lemma 13.The map F 1/a,c,ac,a is conjugate to the Lyness' map F 1/(ac 2 ) .

Proof of the main tools
This section is devoted to proving Theorems 1, 3 and 6.
Proof of Theorem 1. Write , where P ñ and Q m are homogeneous polynomials with degrees ñ ≥ 0 and m ≥ 0, respectively, and O(k) denotes terms of order at least k.Firstly we prove that it is not restrictive to assume that ñ ≥ m and that if ñ = m then P ñ(x, y)/Q m(x, y) is not constant.Notice that if H is a first integral then 1/H is also.Hence we can assume that ñ ≥ m.If ñ = m and ), as we wanted to see.
It is also clear that in a neighborhood of the origin we can assume that F (x, y) = (λx + O(2), µy + O(2)).
(i) We start studying the case λµ = 0.By imposing that H is a first integral of F in U we have that By taking the lower order terms of the above equality we obtain Define where n ≥ m are suitable non-negative integers.By using the homogeneity of P n and Q m , equation ( 8) becomes µ n y n P n (λx/(µy), 1)y m Q m (x/y, 1) = y n P n (x/y, 1)µ m y m Q m (λx/(µy), 1), (9) where notice that we have canceled the common factor of P ñ and Q m.By introducing the polynomials in one variable p n (w) = P n (w, 1), q m (w) = Q m (w, 1), with respective maximum degrees n and m, and ρ = λ/µ, w = x/y, equation ( 9) becomes µ n−m p n (ρw)q m (w) = p n (w)q m (ρw), (10) where we know that p n and q m are not identically zero and have no common root.
Notice that equality (10) implies that if w = w * is a root of p n then ρw * is also, and hence ρ l w * for any l ∈ N, is a root of p n .Since p n has at most n roots, if w * = 0 we have that ρ k = 1 for some k ≤ n, proving the theorem, because λ k µ −k = 1.A similar reasoning can be done for q m .Hence it only remains to study the cases for some complex numbers a and b, ab = 0. Remember that we know that both polynomials have no common roots.So, at least one of the two numbers n or m has to be zero.In any case, the equation ( 10) becomes giving µ n−m (λ/µ) n− m = λ n− mµ n+ m−m−n = 1, as we wanted to prove.
(ii) When λ = 0 and µ = 0 equation ( 8), after dropping the common factor of P ñ and Notice that P n (x, 0) = ax n , Q m (x, 0) = bx m , and (a, b) = (0, 0), because, otherwise P n and Q m would have y as a common factor.Hence and so ab = 0. Therefore P n (x, y) = aλ n−m x n−m Q m (x, y)/b.Since P n and Q m have no common factor, we get that m = 0 and so Q m = b.Hence this last equality becomes P n (x, y) = aλ n x n .Then ax n = P (x, 0) = aλ n x n , giving λ n = 1, as we wanted to prove.
To illustrate the above result in the next remark we present some examples of maps having (or not having) meromorphic first integrals.
Remark 15. (i) The linear maps F (x, y) = (λx, µy), with λ and µ satisfying the resonant condition λ p µ q = 1 are the simplest maps with meromorphic first integrals H(x, y) = x p y q .
(iii) For maps with identically zero linear part we can have existence or not of meromorphic first integrals.For instance the map F (x, y) = (x 2 , xy) has the first integral H(x, y) = x/y.On the other hand, by using the same tools that in our proof of Theorem 1, we can prove that the map F (x, y) = (x 2 , y 2 ) has no meromorphic first integral.
Proof of Theorem 3. The condition that a function V (x, y, n) of the form ( 3) is a nonautonomous invariant of the recurrence (1) becomes for all (x, y) ∈ Q + and all n ∈ N. Imposing that the each one of the coefficients of the 31 monomials x i y j vanishes identically and playing a little bit with these conditions we obtain that (4) holds and moreover that From the first equation we get that Plugging this equation in the second one we obtain that Clearly the above equation holds if either {F n } n is a 6-periodic sequence or In the first situation let us prove that if {F n } n is a p-periodic sequence, p ∈ {1, 2, 3, 6} then a n also has to be p-periodic.Assume for instance that p = 3, then using (11) we have as we wanted to see.The other cases follow similarly.
In the second situation we have that Using this equation and equality (11) we have that which is a well-known 6-periodic recurrence, as we wanted to see.Finally, using (11) six times we get that F n+6 = F n .
Before proving Corollary 4, we explain here how non-autonomous invariants and first integrals are related.Consider a recurrence with k-periodic coefficients and having a non- is a non-autonomous periodic invariant of the recurrence.These relations are used in the next corollary for constructing the first integrals of F [k] , k = 1, 2, 3, 6 using the invariant found in the above theorem.Notice also that there is another way to relate both concepts.
Indeed, it is possible to replace the non-autonomous k-periodic recurrence by an autonomous map on an enlarged phase space of dimension R 2 × R k = R k+2 , simply considering the map (x n , x n+1 , a 1 , a 2 , ..., a k−1 , a k ) → (x n+1 , x n+2 , a 2 , a 3 , ..., a k , a 1 ).In this case, any nonautonomous invariant is just an ordinary first integral for the above map.This approach is not used in this paper.
Proof of Corollary 4. (i) This result is proved along the proof of Theorem 3, above.
(ii) We only give the details for k = 6.The other cases follow similarly.We introduce the following notations for the non-autonomous 6-periodic recurrences: Hence, with the notations of Theorem 3 we get that Hence V f,e,d,c,b,a (x, y) = eΦ 1 (x, y)/(xy) is a first integral of F [6] , as we wanted to prove.
We first prove Theorem 6 for k = 5.
Proof.A computation shows that F [5] (x, y) = (P 1 (x, y), P 2 (x, y)) where First observe that, contrary to what it happens for F [k] , k < 4, F [5] can be extended to a neighborhood of Q + .Note also that (0, 0) is a fixed point and So, under our hypotheses, the origin is a hyperbolic fixed point of F [5] .Arguing similarly with the shifted maps F a 1 ,a 5 ,a 4 ,a 3 ,a 2 , F a 2 ,a 1 ,a 5 ,a 4 ,a 3 , F a 3 ,a 2 ,a 1 ,a 5 ,a 4 , and F a 4 ,a 3 ,a 2 ,a 1 ,a 5 , we obtain that the origin is also a hyperbolic fixed point for these maps, with Jacobian matrices respectively.
Condition (12) implies that at least there exist a parameter with a value less than one, and other grater that one.Since there are no parameters with value equal to one and the sequence is cyclic we can choose two contiguous parameters and such that a i < 1 and a i+1 > 1.This implies that the origin is an attractive fixed point for some of the five shifted maps.For example, suppose that a 2 < 1 and a 3 > 1, then the origin is an stable node for Taking an initial condition (x 0 , y 0 ), with positive coordinates and in the basin of attraction of the origin for the corresponding shifted map we obtain that lim inf n→∞ x n = 0 for the solution of equation ( 1) with initial condition x 1 = x 0 and x 2 = y 0 .
Recall F a j (x, y) = (y, (a j + y)/x), a j = 0. Thus the fact that lim sup n→∞ x n = +∞ follows because if some {(x ns , y ns )} ns tends to (0, 0) then the second component of {F a j (x ns , y ns )} ns tends to +∞.
Proof of Theorem 6. Set F [k] for k = 5m.We can write can be extended to a neighborhood of Q + .Furthermore, observe that Similarly the Jacobian matrices of the shifted maps have the form Arguing as in Proposition 16, the relation ( 5) implies that at least there exists a couple (φ i , φ i+1 ) such that one of the values is greater than one and the other less than one.So the origin is an attractive fixed point for some of the m shifted maps.Now the proof follows again as in Proposition 16.

Rational integrability and associated dynamics
As we have already explained in Subsection 1.2 the cases k = 1, 2, 3 are very similar and totally understood.From Corollary 4 we can prove in the next proposition a similar result when k = 6.Before stating the result we introduce the following notation: c,b,a := {(a, b, c, d, e, f ) ∈ (R + ) 6 : system (13) has a unique solution in Proof.We follow the same steps as in the proof of [11,Thm. 1].To prove the persistence of (1), it suffices to show that each level curve {(x, y) : By Lemma 12.(iii), under our hypotheses, the set of fixed points of F [6] and the set of singular points of V [6] coincide and consists of a single point.Following again the same guidelines of the proof of [11,Thm. 1], which in turn is based on [3, Prop.2.1], we prove that all the level curves of V [6] in Q + , apart from the fixed point, are diffeomorphic to circles.
Hence by using Lemma 8 and Theorem 9.(i) the proposition follows.
Remark 18.We believe that P f,e,d,c,b,a is the whole of (R + ) 6 but we have not been able to prove this equality.In any case it is easy to find sufficient conditions to ensure that some (ii) For any k < 15, k = 5, there exist sequences {a n } of prime period k with the ranks as in Table 1, such that F [k] is rationally integrable and the corresponding recurrence (1) is persistent.
(iii) Moreover it is possible to take in all the above cases parameters a 1 , a Proof.We start by introducing some notation.Given a i and c i positive, we consider the sets S i := Assume that a i and c i are such that Card(S i ) = 5 and consider where we have used expression (6).Notice that 1 / ∈ S i .Given any natural number m ≥ 1, we also consider m sets S 1 , S 2 , . . ., S m , and define Φ When m = 0 we consider Φ [0] (x, y) = (x, y).Finally, choosing the values of the parameters such that m i=1 a i = 1 and m i=1 c i = 1, we obtain that for all m ≥ 3, Φ [m] (x, y) = (x, y).Moreover, for these values of m, the parameters a i and c i can be chosen such that Card(∪ m i=0 S i ) = 5m.Observe also that when m = 1 it is not possible to choose a 1 = 1 and c 1 = 1.When m = 2 it is again possible but with Card(∪ 2 i=0 S i ) = 5.Now we can start the proof of the theorem.First we study the cases k ≤ 4 and k ≥ 15.
Proof.For simplicity, we prove result for the case k = 5, being the proof in the general case similar.Note that the above condition reads as {b, c, d, e} ⊂ {a r , r ∈ Q}.
If F [5] = F e,d,c,b,a has a meromorphic first integral, the same holds for all the other maps for some n i , m i ∈ Z, i = 1, . . ., 5. From these equalities we get that {b, c, d, e} ⊂ {a r , r ∈ Q}.
So the result follows.
A simple corollary of the above result is: In fact we prove: Lemma 23.If there exist maps of the form F [4] ,F [7] and F [11] exhibiting SSNC then, for any k ≥ 7, k ∈ {10, 15} there exist maps of the form F [k] with rank({a n }) = k having also SSNC.will also have SSNC.Note that, since 20 ≡ 0, 11 ≡ 1, 7 ≡ 2, 8 ≡ 3 and 4 ≡ 4 (mod 5), the above maps cover all the values of k given in the statement.Since one of the features of these maps is the existence of transversal homoclinic points, which is a structurally stable property, we can perturb each of the corresponding a j by a j + ε j , with all the ε j sufficiently small, to obtain k-periodic sequences of parameters having SSNC and rank({a n }) = k, as we wanted to prove.
We have (only numerically) shown the existence of SSNC for k ∈ {4, 7, 11, 15}, but note that the above lemma allows to reduce all the other cases to these four ones.
Although the complicated behavior of the maps, as in Figure 2, leads one to believe that even there may be no upper bound for the number of intervals given by the adherence of a sequence (formed by the sequence itself and its accumulation points), we only present here some simple examples.Concretely, for k = 4 we give a map F [k] and two sets of initial conditions such that the adherence of the sequences {x n } generated by (1) consists of more than k intervals.
We have that for a = 2, b = 4, c = 7 and d = 0.001, • the sequence starting at 13.35, 7.27 is formed by 20 intervals; • the sequence starting at 14.8, 8.25 is formed by 7 intervals.
For instance, this last assertion can be seen by making the phase portraits of the orbit of G 0.01,7,4,2 starting at (14.8, 8.25), which is formed by 5 islands, together with their images through G 2 , G 4,2 and G 7,4,2 and their projections in the x-axis, see Figure 3.The property of the existence of sequences {x n } generated by ( 1) such that their adherence consists of more that k intervals should be true for all the values of k given in Lemma 23.We also want to comment that, for these values of k, the initial conditions lying on the stable manifolds of the q-periodic saddle points of G [k] , also have a curious behavior: the adherence of {x n } is the sequence itself together with q more points, corresponding to the saddle points.In general q also is greater than k.On the other hand, the most complicated orbits, that is the ones between two big invariant curves, whose adherence seems to fill a region of positive measure give rise to a single interval when we consider their projections given by the sequence {x n }.

A final remark
After the first version of this paper was finished, one of the authors (A.Cima) and S. Zafar have obtained a proof of the non rational integrability of generic maps F [4] by computing its dynamical degree, [13].

Figure 1
shows some typical behaviors of the orbits of F [k] .In fact we consider some maps G [k] , which are conjugate to F [k] , because the pictures are much more clear.See Lemma 7 for the definition of G [k] .

Figure 1 :
Figure 1: Different possible behaviors of the orbits of G [k] , according to k.Other behaviors are possible for k being a multiple of 5.

Lemma 8 .
G [k] satisfy the following properties: For every choice of positive numbers a 1 , . . ., a k , (i) The map F [k] preserves the measure m(B) = B 1 xy dx dy or, in other words, it preserves the symplectic form ω := 1 xy dx ∧ dy.In consequence, it holds that µ(F [k] (x, y)) = det(DF [k] (x, y)) µ(x, y), where µ(x, y) = xy.(ii) The map G [k] preserves the Lebesgue measure n(B) = B dx dy.That is, it preserves the canonical symplectic form dx ∧ dy.Hence, it holds that det(DG [k] (x, y)) ≡ 1.A related issue in connection with the above lemma is the fact ([6, Thm.1]) that the group of symplectic birational transformations of the plane (which is the group of birational transformations of C 2 which preserve the differential form ω) is generated by compositions of the Lyness map F 1 (the 5-periodic case of F [1] , with a = 1); an scaling; and a map of the form (x, y) → (x a y b , x c y d ) where the matrix a b c d ∈ SL(2, Z).This was conjectured by Usnich in [27] and recently proved by Blanc in the above mentioned reference.The next result allows to know the dynamics of each F [k] when the map has a smooth first integral, its applicability to characterize the dynamics of any integrable F [k] is guaranteed by Lemma 8. Theorem 9 ([9]).Let U ⊂ R 2 be an open set and let F : U → U be a diffeomorphism such that it has a smooth regular first integral V : U → R and there exists a smooth function µ : U → R + such that for any (x, y) ∈ U, µ(F (x, y)) = det(DF (x, y)) µ(x, y).Then the following holds: (i) If a level set Γ h := {(x, y) ∈ U : V (x, y) = h} is a simple closed curve invariant under F , then the map F restricted to Γ h is conjugate to a rotation.(ii) If Γ h is diffeomorphic to an open interval curve and invariant under F , then the map F restricted to Γ h is conjugate to a translation.

Corollary 11 . 2
[k]  is a symplectic map that preserves the canonical form, hence det(DG [k] ) = 1.On the other hand, since F [k] and G [k] are conjugated maps (Lemma 7), the Jacobian matrices of these maps at their corresponding fixed points have the same eigenvalues and the result follows.As a consequence of the above Proposition we have:Let p ∈ R 2 be a m-periodic point of a composition map F [k] such thatF a j ,a j−1 ,...,a 1 (p) is well defined for all j ≤ km.Then det (DF m [k] (p)) = 1.The following lemma studies the number of fixed points of F [k] for k = 4, 5, 6. Lemma 12. (i) There is a unique fixed point of F d,c,b,a in Q + and it satisfies    x = y 2 + (a − c)y − d, y = x 2 + (d − b)x − a. (ii) There are either 0,1,2 or a continuum of fixed points of F e,d,c,b,a in Q + and they satisfy    (b − 1)x + (1 − d)y + (a − e) = 0, cxy − (e + x)(a + y) + bx + y + a = 0. (iii) Let F f,e,d,c,b,a ⊂ Q + be the set of fixed points of map F f,e,d,c,b,a and let S f,e,d,c,b,a ⊂ Q + be the set of singular points of its first integral V f,e,d,c,b,a given in Corollary 4. Then F f,e,d,c,b,a = S f,e,d,c,b,a and both sets coincide with the set of points of Q + satisfying = (f + x)(a + bx) e + dx , x 2 = (a + y)(f + ey) b + cy .Moreover card(F f,e,d,c,b,a ) ≥ 1.
F f,e,d,c,b,a = S f,e,d,c,b,a .That card(F f,e,d,c,b,a ) ≥ 1 can be seen studying the behavior of the functions, (f +x)(a+bx) e+dx and (a+y)(f +ey) b+cy , near 0 and +∞.

From the relations a n+1 F
n+2 − a n F n−3 = 0, we obtain that =

Proposition 17 .
For k = 6 the recurrence (1) is persistent.Moreover if (f, e, d, c, b, a) ∈ P f,e,d,c,b,a , any sequence {x n } generated by (1) is either periodic, with period a multiple of 6, or it densely fills at most 6 disjoint intervals of R + .
(a, b, c, d, e, f ) belongs to P f,e,d,c,b,a .For instance, since ∂ ∂x(f + x)(a + bx) e + dx = bdx 2 + 2bex + ae + bef − adf (e + dx) 2 ,∂ ∂y(a + y)(f + ey) b + cy = cey 2 + 2bey + bf + abe − acf (b + cy) 2 ,when both numerators have no positive real roots the point (a, b, c, d, e, f ) is in the set, because the functions that we have derived are both increasing and so the curves defined by system (13) cut in a single point.Next result collects our integrability results for any k = 5.Theorem 19.(i) For any k ≥ 15, there exist sequences {a n } of prime period k and rank k such that F [k] = F a k ,...,a 2 ,a 1 is rationally integrable and the corresponding recurrence (1) is persistent.

F
a,e,d,c,b , F b,a,e,d,c , F c,b,a,e,d , and F d,c,b,a,e .These five maps have the origin (0, 0) as a fixed point and are analytic in its neighborhood (see the proof of Proposition 16).Moreover the corresponding couples of eigenvalues of their linear parts at zero are 1/a, e; 1/b, a; 1/c, b; 1/d, c and 1/e, d, respectively.Hence, applying Theorem 1, we obtain the following necessary conditions for the existence of a meromorphic first integral

Table 1 .
2 , . . .a k such that each sequence {x n } is either periodic, with period a multiple of k, or it densely fills at most k disjoint intervals of R + .Possible ranks for integrable F [k] .