COMBINATORIAL AND MODULAR SOLUTIONS OF SOME SEQUENCES WITH LINKS TO A CERTAIN CONFORMAL

If fn is a free parameter, we give a combinatorial closed form solution of the recursion (n+ 1)un+1 − fnun − nun−1 = 0, n ≥ 1, and a related generating function. This is used to give a solution to the Apéry type sequence rnn 3 + rn−1 { αn3 − 3α 2 n2 + { α 2 + 2θ } n− θ } + rn−2(n−1) = 0, n ≥ 2, for certain parameters α, θ. We show from another viewpoint two independent solutions of the last recursion related to certain modular forms associated with a problem of conformal mapping: Let f(τ) be a conformal map of a zero-angle hyperbolic quadrangle to an open half plane with values 0, ρ, 1, ∞ (0 < ρ < 1) at the cusps and define t = t(τ) := 1 ρ f(τ) f(τ)−ρ f(τ)−1 . Then the function E(τ) = 1 2πi f ′(τ) f(τ) 1 1− f(τ) ρ is a solution, as a generating function in the variable t, of the above recurrence. In other words, E(τ) = r0 +r1t+r2t2 + . . . , where r0 = 1, r1 = −θ, α = 2− 4 ρ .

(n + 1) 2 u n+1 − fnun − n 2 u n−1 = 0, n ≥ 1, and a related generating function.This is used to give a solution to the Apéry type sequence for certain parameters α, θ.
We show from another viewpoint two independent solutions of the last recursion related to certain modular forms associated with a problem of conformal mapping: Let f (τ ) be a conformal map of a zero-angle hyperbolic quadrangle to an open half plane with values 0, ρ, 1, ∞ (0 < ρ < 1) at the cusps and define t = t(τ ) := 1 ρ f (τ ) Then the function is a solution, as a generating function in the variable t, of the above recurrence.

Introduction
Let P (n) be the third degree polynomial in n defined by with α, θ complex or real numbers.
One should notice that This paper is devoted to the study of sequences (r) = (r 0 , r 1 , r 2 , . ..) defined by We will be interested in the solutions (a), (b) of the above recurrence starting with a 0 = 1, a 1 = −θ and b 0 = 0, b 1 = 1.Of course any solution (r) is a linear combination of (a) and (b).; see [15].Apéry used these sequences to prove the irrationality of ζ(3).
Example 2. One can find many sequences (a 0 , a 1 , a 2 , . . . ) solutions of the above recurrence having the notorious property of being integers for a long string before becoming rational numbers.A few examples are given in Table 1; in all cases α, θ are real and negative.Cases 1 and 11 (expanded) in that table are respectively This paper is, in some sense, an attempt to find the solutions (a), (b) of the recurrence (2) in closed form.
Our main results are Theorems 1-4, which we briefly discuss.We exhibit the solutions of (2) from two different viewpoints.Our first point of view is combinatorial and is developed in sections 2, 3 and 5. Firstly, in Theorem 1, which we believe is interesting in its own right, we solve the easier recursion (n + 1) 2 u n+1 − f n u n − n 2 u n−1 = 0, where f n is a free parameter.This result can be seen as a variant of a certain recursion given in an interesting paper of A. Schmidt [13] and should be compared to it.To solve the recursion we need to introduce certain combinatorial numbers linked to the Stirling numbers of first and second kind.In section 3, namely Theorem 2, we show how a particular case of Theorem 1 can be used to solve in closed form the recursion (2) and is, in some sense, a combinatorial solution of it.This solves also a particular case of Heun's equation.In Theorem 4 of section 5 we present a generating function related to the combinatorial numbers appearing in Theorem 1.
Our second point of view is a modular one: F. Beukers showed the connection of Apéry's sequences, that is those of Example 1, with modular forms.Section 4 is inspired by his remarkable paper [3] and this section can be read almost independently from sections 2 and 3.Here we begin with a problem of a certain conformal mapping: describe the function f (τ ) mapping a hyperbolic quadrangle, having angles all equal to zero at all four cusps, to a half plane.As shown in Theorem 3, we construct the solutions of the recursion (2) as a generating function of certain modular forms attached to f (τ ) with certain parameters α, θ depending on this last function.

A second order recursion
The aim of this section is to prove Theorem 1, which solves, in a combinatorial way, a second order recursion.It is inspired by Asmus Schmidt's paper [13] and it could be seen as a generalization of Example 2 in [14].We need first some definitions.
We write s(i, k) for the Stirling numbers of first kind, which may be defined by the binomial Definition.We will write for short, if 0 By definition we put Observe that d n,0 = 1 for all n ≥ 0. One has We write for short Observe that this is a finite sum because Recall the well known fact that Stirling matrices are inverse to each other.This yields that the last equation can be inverted to give where S(i, j) are the Stirling numbers of second kind.Recall that these numbers may be defined by Thus the last equation is where M is the square matrix with k + 1 rows defined by Definition.We define the real numbers δ k by the equation Our objective is to prove the following result.
Theorem 1.Let (x 0 , x 1 , . . ., x j , . ..) be any sequence of complex numbers.Let and consider sequences (u) = (u 0 , u 1 , . ..) satisfying the recursion formula Then the recursion has two independent solutions (p), (q) as follows: The element p n is represented as where The element q n is represented as , where e 0 = 0, Our proof will follow from some lemmas.Firstly we have the following result of A. Schmidt as given in [14,Example 2,p. 366].
We will need the following lemma.
Proof.The identity of the last lemma can be written as Taking out the coefficient of x k in this recurrence one gets the desired identity.

Lemma 3. The following identity holds:
Proof.This is basically Lemma 1 of [13] which uses the Pfaff-Saalschütz identity.See page 196 of that paper.

Lemma 4.
Let α i,j,k be the real numbers defined at the beginning of this section.Then α i,j,k = 0 if k < i or k < j and α i,j,k = α j,i,k .Also Thus β i,j,k = β j,i,k and this implies α i,j,k = α j,i,k .Also Next we prove the stated identity.Firstly observe that if k ≤ n one may write ! .Now multiply the identity of Lemma 3 by s(u,i) u! s( ,j) ! and add from u = i up to n and = j up to n.The left-hand side gives using the definition of d n,k , while the right-hand side is equal to due to the definition of β i,j,k and because k ≤ n.That is, we have proved that By definition of α i,j,k one has that which proves the lemma.
Finally we prove Theorem 1.
Proof of Theorem 1. Set Our aim is to prove that rn = 0 for all n ≥ 1.
Writing the definition of p n without any explicit c k one has that rn is equal to We collect the terms with c k alone.Remembering that d n,n+1 = d n−1,n+1 = d n−1,n = 0, this can be rearranged to give that rn is equal to where we have used Lemma 2 and the fact that d n,−1 = 0. Putting the definition of c k+1 in the first sum one gets that rn is equal to By Lemma 4, the inner double sum in the first term could be summed up to n (in both summands i, j) instead of k because α i,j,k = 0 if i, j > k.Changing the order of summation and using the identity of Lemma 4 yields If one puts (q) then one obtains, with exactly the same proof, the additional term We record the first values of α i,j,k ; recall that α i,j,k = α j,i,k .One has

Connection with Apéry type sequences and Heun's equation
Our aim is to prove the following theorem which solves, in a certain closed form, the recursion (2).Theorem 2. Let θ be a complex number, α real and α < −2.Set and assume that (x 0 , x 1 , x 2 , . ..) is a complex sequence such that and B 1 := 1−θ 2 .Then there exists (u) = (u 0 , u 1 , u 2 , . ..) which is a linear combination of (p), (q), the solutions given in Theorem 1, such that if one writes ii) The coefficients of satisfy the recursion (2).Also, Note: The above equation ( 4) is a particular case of Heun's equation and is connected to the problem of mapping the half plane onto a hyperbolic quadrangle.
We first prove some lemmas.
Lemma 5. Let B 0 , K 0 be complex numbers and let (x 0 , x 1 , x 2 , . ..) be a complex sequence such that

) be a solution of the recursion
Then (u) is a linear combination of (p) and (q) of Theorem 1.
Proof.This lemma is immediate observing that the hypothesis gives Lemma 6.Let B 0 , K 0 be complex numbers.Set The sequence of complex numbers (u) = (u 0 , u 1 , u 2 , . ..) is a solution of the recursion Proof.The sequence of complex numbers (u) = (u 0 , u 1 , u 2 , . ..) satisfies the conditions of Lemma 6 if and only if x gives the required result after some algebraic manipulation.
Definition.We define the linear operators L, L 1 as follows: ( For the next two lemmas we will write for short . Then Proof.If all the functions involved are smooth enough, one has the following general formula.Set W (t) := g(x)V (x) where t := f (x).Then W (t) satisfies (here ˙denotes the derivative with respect to t) P COMBINATORIAL AND MODULAR SOLUTIONS OF SOME SEQUENCES . . .399 if and only if V (x) satisfies (here is the derivative with respect to x) Hint: Just put the derivatives of W (t = f (x)) := g(x)V (x) with respect to x into one equation to get the other.We note that this is a general formula valid for smooth functions P i .Now take as P i the polynomials defined by the linear operator x− 1and H(t) = 0.A tedious routine check gives the result.

Lemma 8. Let W (t) be a function such that
In particular, if L 1 W (t) = 0 then L W (t) 2 = 0.
Proof.We write for short P i = P i (t), W = W (t) and P 4 (t) := −t − θ.Then where the last equality follows checking that P 1 + P 2 = 6t 3 + 9 2 αt 2 + 3t and so on.Also, where we have used in the last equality the hypothesis L 1 W (t) = H(t), that is, P 1 W + P 2 W = −P 3 W/4 + H, and written H = H(t) for short.The last formula is equal to which, noticing that and the lemma follows.
Lemma 9. Assume P (n) is the polynomial defined by (1).Then a holomorphic function around zero if and only if the coefficients r i satisfy the recurrence for n ≥ 2 with inital conditions r 0 , r 1 .
Proof.After grouping the coefficients of t n in the operator L one obtains the above recursion.
Finally we give the proof of Theorem 2.
Proof of Theorem 2. Assuming that α < −2 and putting In the notation of Lemma 6 this gives A 1 = 2−α 4 B 1 = 1−θ 2 .By Lemmas 5 and 6 one has that satisfies part (i) of the theorem.Part (ii) of the theorem follows from Lemmas 7, 8 and 9.

Connection with modular forms and conformal mapping
In this section we start anew and we connect our sequences with a certain conformal mapping f (τ ) described below and certain modular forms E(τ ), E(τ )F 0 (τ ) related to f (τ ).We show in Theorem 3 that, choosing constants α, θ in (1) depending on f (τ ), the coefficients of these modular forms (viewed in an appropriate variable) are the sought sequences (a), (b) solutions of (2) described in the introduction.
By the Riemann mapping theorem there exists a conformal mapping f (τ ) of this region Q 0 onto the upper open half plane which can be extended to the boundary of the region.Moreover, by applying a bilineal map from the upper half plane into itself, one may normalize this mapping sending i∞ → 0, 0 → ρ, 2r 2 → 1, 1/2 → ∞ with 0 < ρ < 1.As in the construction of the modular invariant one may apply the Schwarz reflection principle an infinite number of times to the sides to get a function which is an extension of f (τ ) which we call in the same way.This function is the Hauptmodul of the discrete group generated by the bilinear transformations (not necessarily related to the modular group): Lemma 10.Under the above construction one has that the function f (τ ) : H → C is a holomorphic function, mapping Q 0 conformally onto the upper plane, where for i = 1, 2, 3 and (0 < ρ < 1) Also, f (τ ) takes real values on the lines iτ 2 and 1/2 + iτ 2 (0 ≤ τ 2 ) and the half circles C 2 , C 1 .Moreover, it has the mirror symmetry and f (τ ) = 0, ρ, 1 in the open upper half plane H. See figure 1.
The last two statements follow from the construction of f (τ ).Writing q = e 2πiτ one may write f (τ ) as a Taylor series in q with radius of convergence 1, because f (T 1 τ ) = f (τ + 1) = f (τ ).Such series will be of the form e 0 q + O(q 2 ) with e 0 > 0, because f (τ ) is univalent at τ = i∞ (that is at q = 0) and f (τ ) is real and increasing if τ moves from i∞ to i0 (on the line iτ 2 ) or if τ moves from 1/2 to 1/2 + i∞ (on the line 1/2 + iτ 2 ).All this gives that f (τ ) is real and increasing at q = 0 if q is real and then forces that all the coefficients of its Taylor series must be real.Moreover one can see that there exists e n ∈ R, e 0 > 0 such that (see [6]) In this section our aim is to show how f (τ ), e i ρ and the radius r 2 are related to the solutions (a), (b) of the recursion (2).
If we choose the circle C 0 as the circle centered at zero of radius √ r 2 , then C 0 is orthogonal to the circle C 1 (defined at the begining of this section), see figure 2.
We define the open regions Q i , i = 1, 2, 3, 4 in the same figure.For example, Q 1 is the exterior of the circles C 0 , C 1 , surrounded by the lines iτ 2 and 1/2 + iτ 2 ; it is a hyperbolic quadrangle with angles 0, π 2 , π 2 , 0.
Proof.This follows from the formulae (6).Indeed, +1 is the inverse of T 2 .Therefore by (6) In the same way one proves that, for i = 1, 2, 3, Observe that the function −r 2 /τ interchanges conformally Q 1 with Q 2 and Q 3 with Q 4 .As f (τ ) is a Hauptmodul for the group generated by T i then f (− r2 τ ) must be a Hauptmodul also.The lemma follows by matching the values at the cusps.Some explicit known examples are the following: , where f 0 (τ Here n 5 is the Legendre symbol.From now on we write to denote the derivative with respect to τ .

Definition. Define t(τ
From this definition it is seen that We denote by τ * the point of intersection of the circles C 0 , C 1 .Recall that the point i √ r 2 belongs to C 0 .See figure 2.
Lemma 12.The function t(τ ) maps Q 1 univalently onto the upper half plane.One has the mirror symmetry t(τ ) = t(−τ ) and t(i∞ where 0 < ρ min < 1 < ρ max .See figure 3. Also, This last fact and the definition t(τ ) = 1 ρ f (τ ) f (τ )−ρ f (τ )−1 which gives that t(τ ) is a 2 :1 map, yield that t(τ ) maps Q 1 univalently onto the upper half plane.(Hint: If τ moves anticlockwise on the boundary of Q 1 then t(τ ) must move on the real line from −∞ to +∞, without "bouncing back" for, otherwise, a real point would have three preimages at least; thus the derivative of t(τ ) on the line iτ 2 must be purely complex.The image t(Q 1) is open, it must contain a point from the upper half plane and using the mirror symmetry, t(Q 1 ) = t(Q 4 ).Therefore t(Q 1 ) can not touch the real line for, otherwise, again a real point would have three preimages at least.This yields that t(Q 1 ) must be the upper half plane.)Finally observe that the point i √ r 2 , which belongs to C 0 , goes to itself by the transformation , where the last equality follows because f (τ ) takes real values on C 1 .Then, by Lemma 11, and f (τ * ) are the roots of the equation ) should be the smallest root).The lemma follows from these values and the fact that 0 < ρ < 1.
From the definition of t(τ ) one calculates that in a neighbourhood of q = 0 t = e 0 q + e 0 (ρ − 1)e 0 ρ − e 1 q 2 + . . .COMBINATORIAL AND MODULAR SOLUTIONS OF SOME SEQUENCES . . .405 and therefore the local inverse in a neighbourhood of t = 0 is Thus one has t 0 (τ ) = −ρ + e 0 (2 − ρ)q + e 0 {(1 − ρ)e 0 + (ρ − 2)e 1 } q 2 + . . .and Putting (9) into this last equation we get that in a neighbourhood of t = 0 As with E(τ ) we may look at the expression of E(τ )F 0 (τ ) as a function of t.A calculation gives that in a neighbourhood of t = 0 We finally connect our construction with the sequences at the beginning of the paper.
Proof.i) Our aim is to show that the function E(τ ), viewed as a function of the variable t, satisfies where L 1 , L are the operators defined in section 3, see formula (5).If this is so, then using Lemma 8 and Lemma 9 one gets that the coefficients of E(τ ) = 1 + a 1 t + a 2 t 2 . . .satisfy the recurrence (2).To ease the proof we write (resp.˙) for the derivative with respect to τ (resp.t) and E(τ ) = √ E. Thus for a generic function f we have, for example, ḟ = f t .Note: In the space of modular forms (under the group that we have) the function E(τ ) is a 1-form and by a theorem of P. Stiller it satisfies a differential equation of second order in the variable t where t(τ ) is the Hauptmodul for that group, namely equation (12).We give here a direct and self contained proof of this fact adapted from the third proof of Proposition 21 of [17]; the reader may recognize the coefficients A, B below as certain Rankin-Cohen brackets whose definitions we do not need.For a more general point of view the reader may consult [17].
Firstly observe that one trivially has We will see that, up to a factor, this is equation (12).We calculate explicitly the factors A, B as functions of t.Recall that from the definitions 2πiE(τ ) = g(t(τ ))t 0 (τ ), where The relationship between t(τ ) and t 0 (τ ) can be read from the definitions and is given in a neighbourhood of t 0, that is iτ = i∞, by We write for short Thus (dropping the variables) one has 2πiE = ġt t 0 + gt 0 , t 0 = ḣt and t 0 = ḧt 2 + t ḣ.Therefore

Putting this into the definition of
The coefficient B is calculated as follows.One has by definition and also , where S(t, τ ) := ( t t ) − 1 2 ( t t ) 2 is the Schwarzian.Here we recall two basic facts about the Schwarzian: −S(t, τ )/t 2 = S(τ, t) (see [8,Exercise 9,p. 377] or use the composition formula for the Schwarzian).Therefore and S(τ, t) can be calculated explicitly as a function of t as in [5, pp. 131-135] (or see [8, Theorem 10.2.1]) because by Lemma 12, the function τ (t), the inverse of t(τ ) defined on that lemma, maps conformally the upper half plane (in the variable t) onto a hyperbolic quadrangle (in the variable τ ).Moreover, by the same lemma, it sends the points 0, ρ min , ρ max , ∞ to the points i∞, i √ r 2 , τ * , 1/2 respectively, and at these last points the quadrangle has angles 0, π/2, π/2, 0. Therefore (see [5] or [8]), for some constants (accessory parameters) β i .Also at a neighbourhood of infinity one has These conditions imply that β 2 , β 3 can be given in terms of β 1 alone.
ii) From the definition of F 0 (τ ), F (τ ) one trivially has Also, Ḟ0 = ) .Therefore with a suitable combination we can make the terms F 0 and F 0 disappear, that is, ...
We will prove that EF 0 , viewed as a function of the variable t, satisfies and this will prove our theorem because this equation is equivalent to the desired recursion by Lemma 9. We do this basically by showing that ( 17) and ( 18) are, up to a factor, equal.
As we already proved that LE = 0 one has that LEF 0 = ...
(here P i = P i (t) are the polynomials defined by L = P 1 ) We calculate the coefficients of the last equation: using Ė E = −A + t t 2 (which is ( 14)) and the fact that In the same way Using (20) one may simplify the first and second terms of the last inner sum obtaining But in part (i), formula (15), we have calculated the Schwarzian S(t, τ )/t 2 = −S(τ, t) = A 2 2 + Ȧ − 2B.Inserting this into the last equation one sees that everything in the inner sum cancels out except the last term.This yields Thus using ( 22) and ( 21) in (19) one gets LEF 0 = EP 1 ... 17) and ( 16) yield where the last equality follows using the definition of E and the derivative of ( 13).This ends our proof.
The following two lemmas complement the last theorem.
Lemma 13.The following hold: COMBINATORIAL AND MODULAR SOLUTIONS OF SOME SEQUENCES . . .409 ii) E(− r2 τ ) = − τ 2 r2 E(τ ) and E(τ ) has, as a function of q, radius of convergence 1. iii) Proof.i) The definition of t 0 (τ ) gives Using this last formula, the definition of t(τ ) and the definition of E(τ ) one gets the desired formula.
ii-iii) By Lemma 10 the function f (τ ) = 0, 1, ρ in the open upper half plane.From the expression (i) one has that E(τ ) has radius of convergence 1 as a function of q.The same happens with both t(τ ) and t 0 (τ ) and therefore with F (τ ).
The transformation formulae follow from ( 7) and its derivative with respect to τ using the definition of E(τ ).
iv 1), the proof being similar to that of Theorem 6.17 on [2, p. 134].Also the function ∞ 0 F (iτ )τ s−1 dτ has an analytic continuation to all the s-complex plane, the later integral being equal to Multiplying this by formula (ii) gives the result.Note: in Proposition 1.2 of [3] it is stated that N = 1/r 2 should be a natural number but this is unnecessary.
The reason for introducing the constant ζ F is the following result.
Lemma 14.The functions E(τ ) and have, as functions of the parameter t (recall formulas (10)  Doing the same with the function E(τ ) one has that now, due to the transformation formula (ii) of Lemma 14, this function has radius of convergence ρ min .

Generating functions
In this section we give some results concerning the combinatorial numbers appearing in Theorem 1.
Remark: The above theorem is related to the Legendre transform.Given a generic sequence β k one may generate another sequence ∆ n called the Legendre transform as One has the inversion formula ( [14]) Proof.Observe that where q = e 2πiτ and q = e 2πiτ , and the last equality follows from (25) and the trivial fact that 1 0 q i dτ = 0 (resp.= 1) if i = 0 (resp.i = 0).Using (23) in the inner sum of the last formula gives the result.
Our final observation is one concerning the solution (p) = (p 1 , p 2 , . . . ) in Theorem 1, where the definitions of d n,k , c k are given.First, observe that from (24) and (25) one has, if q = e 2πiτ , where we have used Lemma 1. Therefore using this last formula, if c k is any sequence and q = e 2πiτ , on has formally that In particular, if c k is the sequence of numbers defined as in Theorem 1 and putting y = 1 in the last equation one gets which is a formal generating function of the solutions of (n + 1) 2 u n+1 − f n u n − n 2 u n−1 = 0, the recurrence given in Theorem 1.

Figure 1 .
Figure 1.Q 0 is mapped conformally onto the upper open half plane by the function f (τ ).

Figure 3 .
Figure 3. Q 1 is mapped conformally onto the upper open half plane by the function t(τ ).

( 1 +
i, k) x k i! = k s(i, k) x ki! and the definition of β i,j,k one has q −1 ) y (1 + q−1 ) x

c
k d n,k y k .
(11)11)), radius of convergence ρ min and greater than or equal to ρ max , respectively.2/τinterchanges conformally Q 1 with Q 2 and Q 3 with Q 4 , and maps the point i/ √ r 2 (in the τ plane) to ρ min (in the t plane).So its radius of convergence is either greater than or equal to ρ max .Note: one may use here Poincaré's theorem (see for example[1, p. 141]) to prove that the radius of convergence is exactly ρ max .
F ) as a multivalued function of the parameter t = t(τ ) we get that this funcion has radius of convergence either ρ min , ρ max or ∞.By the transformation formula (iv) of Lemma 14, looking carefully at what happens around t = ρ min , one gets that the function has no singularity there.For doing this one should recall, as already observed, that the transformation −r COMBINATORIAL AND MODULAR SOLUTIONS OF SOME SEQUENCES . . .411 )Rev.Un.Mat.Argentina, Vol.59, No.2 (2018)