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):

λ \in M (η_{λ}) \Leftrightarrow η_{λ}^{j} (0) does not diverge .

Some other recent results related to the Mandelbrot set can be found for example in [3,4,5,6,7,8,9,10].

In 2019, Mork et al. [11] constructed filled-in Julia sets for alacunary function

η_{N, k} (z) = \sum_{n = 1}^{N} z^{P_{k} (n)}

, where

{(P_{n} (k))}_{n} = {(\frac{1}{2} (k n^{2} - k n - 2))}_{n}

is the sequence ofcentered k-gonal numbers andk is any positive integer (for more facts and history of lacunary functions see, e.g., [12,13,14]).

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) = \sum_{n = 1}^{N} z^{P_{k} (n)}

with a fixed Möbius transformation

M (z) = e^{i θ} \frac{z - a}{\bar{a} z - 1}

(with

(θ, a) \in R \times D

, where 𝔻 denotes the the unit disc) at each iteration step. More precisely

h^{j} (z; a, k, N, θ) = \overset{j - times}{\overset{︷}{M (η_{N, k} (M (η_{N, k} (\dots M (η_{N, k}}} (z) \dots))))) .

Very recently, Mork and Ulness [16] continued the previous line of research by dealing with the so-calledj-averaged Mandelbrot set which is a set generated by iterating a function obtained by composing the function

η_{λ}

and the Möbius transformation

μ_{A} (z) = \frac{a z + b}{c z + d}

, where

A = (a, b, c, d) \in C^{4}

. Thus,

h^{j} (z; A) = \overset{j - times}{\overset{︷}{μ_{A} (η_{λ} (μ_{A} (η_{λ} (\dots μ_{A} (η_{λ}}} (z) \dots))))) .

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 \leq n \leq 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]), seeFigure 1,

\begin{matrix} {(H_{n})}_{0 \leq n \leq j} & = {h^{0} (0; A), h^{1} (0; A), \dots, h^{j} (0; A)} \\ = {0, μ_{A} (η_{λ} (0)), \dots, (\overset{j - t i m e s}{\overset{︷}{μ_{A} (η_{λ} (μ_{A} (η_{λ} (\dots μ_{A} (η_{λ}}} (0) \dots))))))} . \end{matrix}

Figure 1. Thej-averaged Mandelbrot sets for

A = (0, 0.5, 1, 0)

λ = x + i y

, with

x \in [- 1.2, 0.8]

y \in [- 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 softwareMathematica

^{®}

(see [17]) that are defined in Appendix 1 of [16].

Mork and Ulness ([16] Theorem 1) proved that thej-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 \in D

be a non-zero complex number. Define the function

H (z, λ)

H (z, λ) : = μ_{A} (η_{λ} (z))

, with

A = (0, 1, 1, 0)

. Therefore,

H (z, λ) = \frac{1}{z^{2} + λ} .

Observe that then-th iteration ofH at

z = 0

is a function of

λ

, say

h_{n} (λ)

, which satisfies the relations:

h_{0} (λ) = 0, h_{1} (λ) = H (0, λ) = \frac{1}{λ} and h_{n + 1} (λ) = H (h_{n} (λ), λ), for n \geq 1 .

(1)

The sequence

{(C_{n})}_{n \geq 0}

of theCatalan numbers, which is called the sequence A000108 in theOEIS [18], is often defined with the help of the central binomial coefficient

(\binom{2 n}{n})

C_{n} = \frac{1}{n + 1} (\binom{2 n}{n}),

thus, its first terms are inTable 1.

Table 1. Values of

C_{n}

forn from 0 to 14.

Table 1. Values of

C_{n}

forn from 0 to 14.

n	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14
$C_{n}$	1	1	2	5	14	42	132	429	1430	4862	16,796	58,786	208,012	742,900	2,674,440

which can lead us to the following recurrence relation (it was first discovered by Euler in 1761; for more facts, see [19])

C_{n} = \frac{4 n - 2}{n + 1} C_{n - 1}, for n \geq 1,

with the initial condition

C_{0} = 1

. Sometimes the sequence

{(C_{n})}_{n \geq 0}

is defined on the basis of the generating function

(1 - \sqrt{1 - 4 x}) / (2 x)

, as the following holds (for

| x | < 1 / 4

)

\sum_{n = 0}^{\infty} \frac{(\binom{2 n}{n})}{n + 1} x^{n} = \frac{2}{1 + \sqrt{1 - 4 x}} .

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,

Theorem 1.

For all

n \geq 1

, we have

h_{n} (λ) = \frac{1 - {(- 1)}^{n}}{2 λ} + {(- 1)}^{n} \sum_{i = 1}^{⌊ n / 2 ⌋} C_{i - 1} λ^{3 i - 1} + O (λ^{3 ⌊ n / 2 ⌋ + 2}),

(2)

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_{\infty}^{(1)} (λ)

and

h_{\infty}^{(2)} (λ)

h_{\infty}^{(1)} (λ) : = lim_{n \to \infty} h_{2 n + 1} (λ) = \frac{1}{λ} - \sum_{i \geq 0} C_{i} λ^{3 i + 2}

and

h_{\infty}^{(2)} (λ) : = lim_{n \to \infty} h_{2 n} (λ) = \sum_{i \geq 0} C_{i} λ^{3 i + 2} .

They also asserted that these functions should converge in the whole unit disk (or the punctured one for

h_{\infty}^{(1)} (λ)

). However, this is not true (this is expected because of the exponential nature of Catalan numbers). For example, the simple bound

(\binom{2 n}{n}) \geq 4^{n} / (2 n + 1)

, which comes from the fact that

4^{n} = {(1 + 1)}^{2 n} = \sum_{k = 0}^{2 n} (\binom{2 n}{k})

, implies that

C_{n} > 4^{n} / (3 n^{2})

(some other bounds can be found in ([19] Chapter 2) and [20]) and so if

| λ | > 1 / \sqrt[3]{2} \approx 0.793

then

| C_{n} λ^{3 n + 2} | \geq 3^{- 1} {| λ |}^{2} (2^{n} / n^{2}) (\sqrt[3]{2} {| λ |)}^{3 n} > {| λ |}^{2} / 3

(for

n \geq 4

) yielding the divergence of

h_{\infty}^{(2)} (λ)

. In order to compute the radius of convergence, say r, of

h_{\infty}^{(2)} (λ)

, one can write this function as

h_{\infty}^{(2)} (λ) = \sum_{n \geq 0} a_{n} λ^{n}

, where

a_{n} = \{\begin{matrix} C_{(n - 2) / 3}, & i f n \equiv 2 (mod 3), \\ 0, & i f n ≢ 2 (mod 3) . \end{matrix}

Thus,

1 / r = lim {sup}_{n \to \infty} \sqrt[n]{a_{n}}

and, by using

C_{n} \approx 4^{n} / (n^{3 / 2} \sqrt{π})

(which comes from the Stirling formula

n! \approx \sqrt{2 π n} {(n / e)}^{n}

), we obtain

\begin{matrix} \frac{1}{r} & = & lim sup_{n \to \infty} \sqrt[n]{a_{n}} \\ = & lim sup_{n \to \infty} \sqrt[3 n + 2]{C_{n}} \\ = & lim sup_{n \to \infty} \sqrt[3 n + 2]{\frac{4^{n}}{n^{3 / 2} \sqrt{π}}} \\ = & \sqrt[3]{4} . \end{matrix}

Therefore,

B (0, 1 / \sqrt[3]{4})

is the disk of convergence of

h_{\infty}^{(2)} (λ)

(observe that

r = 1 / \sqrt[3]{4} \approx 0.6299

2. 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)

Lemma 1.

For any

n \geq 1

, there exists a power series

P_{n} (λ)

such that

h_{n} (λ) = \{\begin{matrix} λ^{2} + λ^{5} P_{n} (λ^{3}), & i f n i s e v e n; \\ \frac{1}{λ} + λ^{2} P_{n} (λ^{3}), & i f n i s o d d . \end{matrix}

Proof.

By definition in (1),

h_{n} (λ)

satisfies the following recurrence relation

h_{n + 1} (λ) = \frac{1}{{(h_{n} (λ))}^{2} + λ},

with

h_{0} (λ) = 0

(since

h_{1} (λ) = H (0, λ) = 1 / λ

). Now, by defining

f_{n} (λ) : = λ h_{n} (λ)

and using the previous recurrence, we obtain

f_{n + 1} (λ) / λ = \frac{1}{{(f_{n} (λ) / λ)}^{2} + λ}

and so

f_{n + 1} (λ) = \frac{λ^{3}}{{(f_{n} (λ))}^{2} + λ^{3}},

(3)

with

f_{0} (λ) = 0

. We claim that

f_{n} (λ) = g_{n} (λ^{3})

for some rational function

g_{n} (λ)

, wheren is any positive integer. Indeed, we can proceed by induction onn. For

n = 1

, we can take

g_{1} (λ) = 1

. Suppose (by induction hypothesis) that

f_{n} (λ) = g_{n} (λ^{3})

, for some formal power series

g_{n} (λ)

, then, by (3), we have

f_{n + 1} (λ) = \frac{λ^{3}}{{(g_{n} (λ^{3}))}^{2} + λ^{3}} = g_{n + 1} (λ^{3}),

where

g_{n + 1} (λ)

can be chosen by satisfying the recurrence

g_{n + 1} (λ) = \frac{λ}{{(g_{n} (λ))}^{2} + λ},

with

g_{0} (λ) = 0

. The inductive process is finished. Observe that, since

h_{n} (λ) = λ^{- 1} g_{n} (λ^{3})

, then it suffices to prove that

g_{n} (λ) = \{\begin{matrix} λ + O (λ^{2}), & if n is even; \\ 1 + O (λ), & if n is odd . \end{matrix}

(4)

The proof is also by induction onn (more precisely, a double induction). For the basis cases, we have

g_{1} (λ) = 1 = 1 + O (λ)

and

g_{2} (λ) = \frac{λ}{1 + λ} = λ (1 - λ + λ^{2} - \dots) = λ + O (λ^{2}),

where we used that for

| λ | < 1

, one has

{(1 + λ)}^{- 1} = \sum_{k \geq 0} {(- λ)}^{n}

(in general, it holds that

{(1 + O (1))}^{- 1} = 1 + O (1)

). Suppose that (4) is valid for all

n \in [1, 2 k]

. Then,

\begin{matrix} g_{2 k + 1} (λ) & = & \frac{λ}{{(g_{2 k} (λ))}^{2} + λ} \\ = & \frac{λ}{{(λ + O (λ^{2}))}^{2} + λ} \\ = & \frac{λ}{λ + O (λ^{2})} \\ = & \frac{1}{1 + O (λ)} = 1 + O (λ), \end{matrix}

where we used

O (λ^{r}) + O (λ^{s}) = O (λ^{min {r, s}})

, since

| λ | < 1

Now, we use the previous fact

\begin{matrix} g_{2 k + 2} (λ) & = & \frac{λ}{{(g_{2 k + 1} (λ))}^{2} + λ} \\ = & \frac{λ}{{(1 + O (λ))}^{2} + λ} \\ = & \frac{λ}{1 + O (λ)} \\ = & λ (1 + O (λ)) = λ + O (λ^{2}) . \end{matrix}

This completes the induction proof of (4).

Therefore, since

| λ | < 1

, we can write

g_{n} (λ) = \{\begin{matrix} λ + \sum_{i \geq 2} c_{i} λ^{i}, & if n is even; \\ 1 + \sum_{i \geq 1} c_{i} λ^{i}, & if n is odd \end{matrix}

and so

h_{n} (λ) = \frac{1}{λ} g_{n} (λ^{3}) = \{\begin{matrix} λ^{2} + \sum_{i \geq 2} c_{i} λ^{3 i - 1}, & if n is even; \\ \frac{1}{λ} + \sum_{i \geq 1} c_{i} λ^{3 i - 1}, & if n is odd . \end{matrix}

This completes the proof. □

Remark 2.

Note that, by using Lemma 1, we can write

h_{n} (λ) = α_{- 1, n} λ^{- 1} + α_{0, n} λ^{2} + α_{1, n} λ^{5} + \dots = \sum_{k = - 1}^{\infty} α_{k, n} λ^{3 k + 2} \in R [[λ]],

(5)

where

α_{- 1, n} = (1 - {(- 1)}^{n}) / 2

, i.e.,

α_{- 1, n}

is 1if 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

λ \to 0

, for any

k \equiv 0

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

H_{λ}^{(2 n + 1)} (0) = \frac{(2 n + 1)!}{2 π i} \int_{γ_{R}} \frac{H_{λ} (ω)}{ω^{2 n + 2}} d ω = \frac{(2 n + 1)!}{2 π i} \int_{γ_{R}} \frac{1}{(ω^{2} + λ) ω^{2 n + 2}} d ω,

where

γ_{R}

is the circle

γ (t) : = R e^{i t}

, for

t \in [0, 2 π]

and

0 < R < | λ |

. Now, we can use the partial fraction decomposition to deduce that

\frac{1}{(ω^{2} + λ) ω^{2 n + 2}} = \frac{A}{ω + \sqrt{| λ |}} - \frac{A}{ω - \sqrt{| λ |}} + \frac{B}{ω^{2 n + 2}},

for computable constants A and B. Hence, again by the Cauchy integral formula, we have

\int_{γ_{R}} \frac{1}{(ω^{2} + λ) ω^{2 n + 2}} d ω = 2 A π i f (0) - 2 A π i f (0) + 2 B π i f^{(2 n + 1)} (0),

where

f (z) = 1

, for all z. Thus,

H_{λ}^{(2 n + 1)} (0)

is equal to zero as claimed.

Now we show the important connection of the sequence

{(α_{k, n})}_{k \geq 0}

to the Catalan numbers. For the simplicity of notation, we use the following notation in the rest of the text:

α_{k, n} = \{\begin{matrix} d_{k}, & for odd n; \\ e_{k}, & for even n . \end{matrix}

Lemma 2.

Let

{(C_{k})}_{k \geq 0}

be the Catalan sequence. We have

(i): If ${(d_{k})}_{k \geq 0}$ is defined by the recurrence,
$d_{k + 1} = - (C_{1} d_{k} + \dots + C_{k + 1} d_{0}) - C_{k + 2},$
with $d_{0} = - C_{0}$ , then $d_{k} = - C_{k}$ , for all $k \geq 0$ .
(ii): If ${(e_{k})}_{k \geq 1}$ is defined by the recurrence,
$e_{k + 1} = C_{0} e_{k} + \dots + C_{k - 1} e_{1} + C_{k},$
with $e_{1} = C_{1}$ , then $e_{k} = C_{k}$ , for all $k \geq 1$ .

Proof.

Let us recall that Catalan numbers satisfy the Segner recurrence relation (see, e.g., [19], p. 117)

C_{i + 1} = \sum_{j = 0}^{n} C_{j} C_{i - j},

(6)

with

C_{0} = 1

(i). We shall proceed by induction onk. For

k = 0

, one has

d_{0} = - C_{0}

(by definition). Suppose

d_{t} = C_{t}

, for all

t \in [0, k]

. Then,

\begin{matrix} d_{k + 1} & = & - (C_{1} d_{k} + \dots + C_{k + 1} d_{0}) - C_{k + 2} \\ = & C_{1} C_{k} + \dots + C_{k + 1} C_{0} - C_{k + 2} \\ = & \underset{C_{k + 2} - C_{k + 1}}{\underset{︸}{C_{1} C_{k} + \dots + C_{k + 1} C_{0}}} - C_{k + 2} \\ = - C_{k + 1} \end{matrix}

which completes the proof (where we used (6)).

(ii). Again by induction onk, the basis case

e_{1} = C_{1}

follows by definition. Assume now that

e_{t} = C_{t}

, for all

t \in [1, k]

. Then, by the recurrence for

{(e_{k})}_{k}

together with the induction hypothesis, we obtain

\begin{matrix} e_{k + 1} & = & C_{0} e_{k} + \dots + C_{k - 1} e_{1} + C_{k} \\ = & C_{0} C_{k} + \dots + C_{k - 1} C_{1} + C_{k} \\ = & \underset{C_{k + 1} - C_{k}}{\underset{︸}{C_{0} C_{k} + \dots + C_{k - 1} C_{1}}} + C_{k} \\ = C_{k + 1} \end{matrix}

which finishes the proof (where we used again (6)). □

The next lemma gives a helpful recurrence for

C_{n}

, depending on the parity ofn. The proof follows by induction together with (6) (we leave the details to the readers).

Lemma 3.

Let

{(C_{n})}_{n \geq 0}

be the Catalan sequence. Then,

C_{2 n} = 2 \sum_{j = 1}^{n} C_{j - 1} C_{2 n - j}

and

C_{2 n + 1} = C_{n}^{2} + 2 \sum_{j = 1}^{n} C_{j - 1} C_{2 n - j + 1}

for all

n \geq 0

(with

C_{0} = 1

Now, we are ready to deal with the proof.

3. The Proof of the Theorem 1

First, observe that (2) can be rewritten for any as

h_{2 j} (λ) = C_{0} λ^{2} + \dots + C_{j - 1} λ^{3 j - 1} + O (λ^{3 j + 2})

(7)

and

h_{2 j + 1} (λ) = \frac{1}{λ} - (C_{0} λ^{2} + \dots + C_{j - 1} λ^{3 j - 1}) + O (λ^{3 j + 2}),

(8)

where we adopt the convention that

C_{0} λ^{2} + \dots + C_{j - 1} λ^{3 j - 1} = 0

for

j = 0

Now, we want to prove the following fact:

Claim. It holds that

{(C_{0} λ^{2} + \dots + C_{n - 1} λ^{3 n - 1})}^{2} = C_{1} λ^{4} + \dots + C_{n} λ^{3 n + 1} + O (λ^{3 n + 2}),

(9)

for a non-negative integern.

Proof.

The proof is by induction onn. The identity is true for

n = 0

, since

C_{1} = C_{0}^{2}

. Suppose that (9) holds, then one has

\begin{matrix} {(C_{0} λ^{2} + \dots + C_{n - 1} λ^{3 n - 1} + C_{n} λ^{3 n + 2})}^{2} & = & {(C_{0} λ^{2} + \dots + C_{n - 1} λ^{3 n - 1})}^{2} \\ + 2 (C_{0} λ^{2} + \dots + C_{n - 1} λ^{3 n - 1}) C_{n} λ^{3 n + 2} + {(C_{n} λ^{3 n + 2})}^{2} . \end{matrix}

Since we desire to evaluate the identity up to

O (λ^{3 n + 5})

, then

2 (C_{0} λ^{2} + \dots + C_{n - 1} λ^{3 n - 1}) C_{n} λ^{3 n + 2} + {(C_{n} λ^{3 n + 2})}^{2} = 2 C_{0} C_{n} λ^{3 n + 4} + O (λ^{3 n + 7}) .

(10)

On the other hand, in the induction hypothesis

{(C_{0} λ^{2} + \dots + C_{n - 1} λ^{3 n - 1})}^{2} = C_{1} λ^{4} + \dots + C_{n} λ^{3 n + 1} + O (λ^{3 n + 2}),

the terms of order

λ^{3 n + 4}

were neglected (since we were interested in

O (λ^{3 n + 2})

). Thus, we can improve the previous identity by considering these terms (note that this procedure does not affect the induction hypothesis). Additionally, since the sum of two numbers, which are congruent to 2 modulo 3, is congruent to 1 modulo 3, there is no term of magnitude

λ^{3 n + 3}

{(C_{0} λ^{2} + \dots + C_{n - 1} λ^{3 n - 1})}^{2}

. Let us also suppose thatn is odd (the even case is carried out along the same lines). We then have

\begin{matrix} {(C_{0} λ^{2} + \dots + C_{n - 1} λ^{3 n - 1})}^{2} & = & C_{1} λ^{4} + \dots + C_{n} λ^{3 n + 1} \\ + 2 (C_{1} C_{n - 1} + \dots + C_{(n - 1) / 2} C_{(n + 1) / 2}) λ^{3 n + 4} + O (λ^{3 n + 5}) . \end{matrix}

(11)

Now, we combine (10) and (11) together with Lemma 3 to arrive at

\begin{matrix} {(C_{0} λ^{2} + \dots + C_{n - 1} λ^{3 n - 1} + C_{n} λ^{3 n + 2})}^{2} & = & {(C_{0} λ^{2} + \dots + C_{n - 1} λ^{3 n - 1})}^{2} \\ + 2 (C_{0} λ^{2} + \dots + C_{n - 1} λ^{3 n - 1}) C_{n} λ^{3 n + 2} + {(C_{n} λ^{3 n + 2})}^{2} \\ = & C_{1} λ^{4} + \dots + C_{n} λ^{3 n + 1} \\ + 2 (C_{1} C_{n - 1} + \dots + C_{(n - 1) / 2} C_{(n + 1) / 2}) λ^{3 n + 4} + O (λ^{3 n + 5}) \\ + 2 C_{0} C_{n} λ^{3 n + 4} + O (λ^{3 n + 7}) \\ = & C_{1} λ^{4} + \dots + C_{n + 1} λ^{3 n + 4} + O (λ^{3 n + 5}) \end{matrix}

which finishes the proof of the claim.

Now, we return to the proof of (2). Again, the proof is by induction onn. For the basis case, we have

h_{1} (λ) = \frac{1}{λ}

and, by Lemma 1,

h_{2} (λ) = λ^{2} + O (λ^{5}) .

Suppose that (2) is true for

h_{n} (λ)

with

n \in [1, 2 j]

. Then, by the recurrence relation for

{(h_{n} (λ))}_{n}

together with the induction hypothesis, we infer that

h_{2 j + 1} (λ) = \frac{1}{{(h_{2 j} (λ))}^{2} + λ} = \frac{1}{{(C_{0} λ^{2} + \dots + C_{j - 1} λ^{3 j - 1} + O (λ^{3 j + 2}))}^{2} + λ} .

However, we can use (9) to write

{(C_{0} λ^{2} + \dots + C_{j - 1} λ^{3 j - 1} + O (λ^{3 j + 2}))}^{2} + λ = C_{0} λ + C_{1} λ^{4} + \dots + C_{j} λ^{3 j + 1} + O (λ^{3 j + 2}) .

(12)

From Lemma 1 and Remark 2, one has

h_{2 j + 1} (λ) = \frac{1}{λ} + d_{0} λ^{2} + d_{1} λ^{5} + \dots + d_{j - 1} λ^{3 j - 1} + O (λ^{3 j + 2}) .

Thus, the coefficients

d_{0}

d_{1}

, ...,

d_{j - 1}

satisfy the following equality

1 \equiv (C_{0} λ + C_{1} λ^{4} + \dots + C_{k} λ^{3 j + 1} + O (λ^{3 j + 2})) (\frac{1}{λ} + d_{0} λ^{2} + d_{1} λ^{5} + \dots + d_{j - 1} λ^{3 j - 1} + O (λ^{3 j + 2}))

and so

1 \equiv 1 + λ (\sum_{i = 0}^{j - 1} d_{i} λ^{3 i + 2}) + C_{1} λ^{4} (\frac{1}{λ} + \sum_{i = 0}^{j - 2} d_{i} λ^{3 i + 2}) + \dots + C_{k} λ^{3 j + 1} (\frac{1}{λ}) + O (λ^{3 j + 1}) .

By reordering this sum, we obtain

0 \equiv (d_{0} + C_{1}) λ^{3} + (d_{1} + C_{1} d_{0} + C_{2}) λ^{6} + \dots + (d_{j - 1} + C_{1} d_{j - 2} + \dots + C_{j}) λ^{3 j} + O (λ^{3 j + 1}) .

Therefore,

d_{0} = - C_{1} = - C_{0}

and

d_{t} = - (C_{1} d_{t - 1} + \dots + C_{t} d_{0}) - C_{t},

for all

t \in [1, j - 1]

. By Lemma 3 (i), we conclude that

d_{t} = - C_{t}

, for all

t \in [1, j - 1]

which yields that

h_{2 j + 1} (λ) = \frac{1}{λ} - (C_{0} λ^{2} + C_{1} λ^{5} + \dots + C_{k - 1} λ^{3 j - 1}) + O (λ^{3 j + 2})

as desired.

Thus, we determine that (2) holds for

h_{n} (λ)

for all

n \in [1, 2 j + 1]

. To finish the proof, we must prove that (2) is also true for

n = 2 j + 2

. First, one has that

h_{2 j + 2} (λ) = \frac{1}{{(h_{2 j + 1} (λ))}^{2} + λ} = \frac{1}{{((1 / λ) - (C_{0} λ^{2} \dots + C_{j - 1} λ^{3 j - 1}) + O (λ^{3 j + 2}))}^{2} + λ} .

However, by (9) and after a straightforward calculation, we arrive at

{(h_{2 j + 1} (λ))}^{2} + λ = \frac{1}{λ^{2}} - (C_{0} λ + C_{1} λ^{4} + \dots + C_{j - 1} λ^{3 j - 2}) + C_{k} λ^{3 j + 1} + O (λ^{3 j + 2}) .

(13)

Now, we use Lemma 1 (and Remark 2) to write

h_{2 j + 2} (λ) = e_{0} λ^{2} + e_{1} λ^{5} + \dots + e_{j} λ^{3 j + 2} + O (λ^{3 j + 5}),

where

e_{0} = 1

. Hence,

1 \equiv (\frac{1}{λ^{2}} - (C_{0} λ + \dots + C_{j - 1} λ^{3 j - 2}) + C_{k} λ^{3 j + 1} + O (λ^{3 j + 2})) (λ^{2} + e_{1} λ^{5} + \dots + e_{j} λ^{3 j + 2} + O (λ^{3 j + 5})) .

Thus,

1 \equiv 1 + λ^{- 2} (\sum_{i = 1}^{j} e_{i} λ^{3 i + 2}) - C_{0} λ (\sum_{i = 0}^{j} e_{i} λ^{3 i + 2}) + \dots + C_{j - 1} λ^{3 j - 2} \cdot λ^{2} + O (λ^{3 j + 3})

which can be re-written as

0 \equiv (e_{1} - 1) λ^{3} + (e_{2} - C_{0} e_{1} - C_{1}) λ^{6} + \dots + (e_{j} - e_{j - 1} - C_{1} e_{j - 2} - \dots - C_{j - 2} e_{1} + C_{j - 1}) λ^{3 j} + O (λ^{3 j + 3}) .

We then deduce that

e_{1} = 1 = C_{1}

and

e_{t} = C_{0} e_{j - 1} + C_{1} e_{j - 2} + \dots + C_{j - 2} e_{1} - C_{j - 1},

for all

t \in [1, j]

. By Lemma 3 (ii), we have

e_{t} = C_{t}

, for all

t \in [1, j]

, yielding that

h_{2 j + 2} (λ) = C_{0} λ^{2} + C_{1} λ^{5} + \dots + C_{j} λ^{3 j + 2} + O (λ^{3 j + 5}) .

The proof is then complete. □

4. Conclusions

This paper is devoted to the proof of a conjecture formulated by Mork and Ulness ([16], Conjecture 4.2). Roughly speaking, they computationally observed the relation between the coefficients of

h_{n} (λ)

(then-th iteration of

1 / (z^{2} + λ)

z = 0

) and the Catalan sequence

{(C_{k})}_{k}

. Indeed, we prove a quantitative version of their conjecture by showing that the sequence

{(h_{n} (λ) - (\frac{1 - {(- 1)}^{n}}{2 λ} + {(- 1)}^{n} \sum_{i = 1}^{⌊ n / 2 ⌋} C_{i - 1} λ^{3 i - 1}))}_{n}

tends to zero (with order

{| λ |}^{3 ⌊ n / 2 ⌋ + 2}

) as

n \to \infty

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.

Funding

The authors was supported by the Project of Specific Research PrF UHK no. 2101/2021, University of Hradec Králové, Czech Republic.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

The authors would like to thank anonymous referees for their careful corrections and their comments that helped to improve the quality of the paper.

Conflicts of Interest

The author declares no conflict of interest.

References

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$Fractalfract 05 00092 g001$

Figure 1. Thej-averaged Mandelbrot sets for

A = (0, 0.5, 1, 0)

λ = x + i y

, with

x \in [- 1.2, 0.8]

y \in [- 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 softwareMathematica

^{®}

(see [17]) that are defined in Appendix 1 of [16].

Figure 1. Thej-averaged Mandelbrot sets for

A = (0, 0.5, 1, 0)

λ = x + i y

, with

x \in [- 1.2, 0.8]

y \in [- 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 softwareMathematica

^{®}

(see [17]) that are defined in Appendix 1 of [16].

$Fractalfract 05 00092 g001$

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

Abstract

1. Introduction

2. Auxiliary Results

3. The Proof of the Theorem 1

4. Conclusions

Author Contributions

Funding

Institutional Review Board Statement

Informed Consent Statement

Data Availability Statement

Acknowledgments

Conflicts of Interest

References

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