The Proof of a Conjecture Relating Catalan Numbers to an Averaged Mandelbrot-Möbius Iterated Function

: In 2021, Mork and Ulness studied the Mandelbrot and Julia sets for a generalization of the well-explored function η λ ( z ) = z 2 + λ . Their generalization was based on the composition of η λ with the Möbius transformation µ ( z ) = 1 z at each iteration step. Furthermore, they posed a conjecture providing a relation between the coefﬁcients of (each order) iterated series of µ ( η λ ( z )) (at z = 0) and the Catalan numbers. In this paper, in particular, we prove this conjecture in a more precise (quantitative) formulation.

In this work, we are interested in a modified version of the "classical" filled-in Julia set K(η λ ) and the Julia set J(η λ ) of functions in the quadratic family (η λ (z)) λ∈C = (z 2 + λ) λ∈C . We observe that the Mandelbrot set M(η λ ) is the fractal defined as M(η λ ) = {λ ∈ C : J(η λ ) is connected}.
We point out that there is a more "workable" way of considering the Mandelbrot set (we refer to [2], Theorem 14.14) for a proof of the usually referred fundamental theorem of the Mandelbrot set): Some other recent results related to the Mandelbrot set can be found for example in [3][4][5][6][7][8][9][10].
In 2021, Mork et al. [15] followed up on the aforementioned article and considered a generalization of the filled-in Julia sets and their corresponding Mandelbrot sets by composing the lacunary function η(z) = ∑ N n=1 z P k (n) with a fixed Möbius transformation M(z) = e iθ z−a az−1 (with (θ, a) ∈ R × D, where D denotes the the unit disc) at each iteration step. More precisely h j (z; a, k, N, θ) = j−times M(η N,k (M(η N,k (· · · M(η N,k (z) · · · ))))).
Very recently, Mork and Ulness [16] continued the previous line of research by dealing with the so-called j-averaged Mandelbrot set which is a set generated by iterating a function obtained by composing the function η λ and the Möbius transformation )))).
The name "j-averaged" is used here since the points of the resulting fractal are colored according to the total number of members of the following sequence of iterations (H n ) 0≤n≤j , that escaped from the circle with radius 2 (the concrete algorithm for coloring of points of this fractal you can find in Appendix 1 of [16]), see Figure 1, )))))}.  1], j = 1, 2, 3, 4 (the first row from the left to the right) and for and j = 5, 7, 10, 100 (the second row from the left to the right). We used functions in the software Mathematica ® (see [17]) that are defined in Appendix 1 of [16].
Mork and Ulness ( [16] Theorem 1) proved that the j-averaged Mandelbrot set for the Möbius transformation µ A with A = (0, 1, 1, 0) has threefold rotational symmetry and dihedral mirror symmetry. Additionally, they raised a conjecture (see [16], Conjecture 2)) concerning the coefficients of these iterations. Before stating their conjecture, we introduce some basic notations.
Let λ ∈ D be a non-zero complex number. Define the function H(z, λ) by H(z, λ) := µ A (η λ (z)), with A = (0, 1, 1, 0). Therefore, Observe that the n-th iteration of H at z = 0 is a function of λ, say h n (λ), which satisfies the relations: The sequence (C n ) n≥0 of the Catalan numbers, which is called the sequence A000108 in the OEIS [18], is often defined with the help of the central binomial coefficient ( 2n n ) by thus, its first terms are in Table 1. which can lead us to the following recurrence relation (it was first discovered by Euler in 1761; for more facts, see [19]) with the initial condition C 0 = 1. Sometimes the sequence (C n ) n≥0 is defined on the basis of the generating function (1 − √ 1 − 4x)/(2x), as the following holds (for |x| < 1/4) The aim of this paper is to obtain a (quantitative) result for the coefficients of the power series of h n (λ) which implies the Mork-Ulness' conjecture (qualitative version). More precisely, where C n is the n-th Catalan number.

Remark 1.
We remark that Mork and Ulness [16] posed a slightly different conjecture. In fact, we can express their question by defining h They also asserted that these functions should converge in the whole unit disk (or the punctured one for h (1) ∞ (λ)). However, this is not true (this is expected because of the exponential nature of Catalan numbers). For example, the simple bound ( 2n n ) ≥ 4 n /(2n + 1), which comes from the fact that 4 n = (1 + 1) 2n = ∑ 2n k=0 ( 2n k ), implies that C n > 4 n /(3n 2 ) (some other bounds can be found in ( [19] Chapter 2) and [20]) and so if |λ| ∞ (λ). In order to compute the radius of convergence, say r, of h (2) ∞ (λ), one can write this function as h (2) ∞ (λ) = ∑ n≥0 a n λ n , where Thus, 1/r = lim sup n→∞ n √ a n and, by using C n ≈ 4 n /(n 3/2 √ π) (which comes from the Stirling formula n! ≈ √ 2πn(n/e) n ), we obtain Therefore,

Auxiliary Results
Before proceeding further, we shall present some useful tools related to the previous sequences.
Our the first ingredient provides a useful form to the Laurent series of h n (λ).

Remark 2.
Note that, by using Lemma 1, we can write where α −1,n = (1 − (−1) n )/2, i.e., α −1,n is 1 if n is odd and 0 if n is even. In particular, h n (λ) is an analytic function in some neighborhood of λ = 0, when n is even, and for n odd, h n (λ) has a simple pole at origin (with residue equal to 1).

Remark 3.
Another viewpoint of Lemma 1 (and consequently, of Remark 2) is that the k-th derivative of h n (λ) = 0 as λ → 0, for any k ≡ 0 or 1 (mod 3). This fact can also be proved by a harder (but maybe theoretically useful) combination of induction, the generalized Chain Rule (Faà di Bruno's formula) and the fact that all odd order derivatives of H λ (z) := H(z, λ) vanish (for fixed λ) at z = 0. This last assertion follows from Cauchy's integral formula. Indeed, we have where γ R is the circle γ(t) := Re it , for t ∈ [0, 2π] and 0 < R < |λ|. Now, we can use the partial fraction decomposition to deduce that for computable constants A and B. Hence, again by the Cauchy integral formula, we have Now we show the important connection of the sequence (α k,n ) k≥0 to the Catalan numbers. For the simplicity of notation, we use the following notation in the rest of the text: α k,n = d k , for odd n; e k , for even n.
Lemma 2. Let (C k ) k≥0 be the Catalan sequence. We have (i) If (d k ) k≥0 is defined by the recurrence, is defined by the recurrence, e k+1 = C 0 e k + · · · + C k−1 e 1 + C k , with e 1 = C 1 , then e k = C k , for all k ≥ 1.
(i). We shall proceed by induction on k. For k = 0, one has d 0 = −C 0 (by definition). Suppose d t = C t , for all t ∈ [0, k]. Then, which completes the proof (where we used (6)).
(ii). Again by induction on k, the basis case e 1 = C 1 follows by definition. Assume now that e t = C t , for all t ∈ [1, k]. Then, by the recurrence for (e k ) k together with the induction hypothesis, we obtain which finishes the proof (where we used again (6)).
The next lemma gives a helpful recurrence for C n , depending on the parity of n. The proof follows by induction together with (6) (we leave the details to the readers).
for all n ≥ 0 (with C 0 = 1). Now, we are ready to deal with the proof.

The Proof of the Theorem 1
First, observe that (2) can be rewritten for any as and where we adopt the convention that C 0 λ 2 + · · · + C j−1 λ 3j−1 = 0 for j = 0. Now, we want to prove the following fact: Claim. It holds that for a non-negative integer n.
The proof is then complete.
Author Contributions: P.T. and K.V. conceived of and designed the investigation and provided background for the investigation; P.T. applied the Mathematica code to perform the investigation; both authors analyzed the data; K.V. wrote the original draft of manuscript; both authors edited the manuscript. Both authors have read and agreed to the published version of the manuscript.