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

Abstract

In 2021, Mork and Ulness studied the Mandelbrot and Julia sets for a generalization of the well-explored function ${\eta }_{\lambda }\left(z\right)=z^2+\lambda$. Their generalization was based on the composition of ${\eta }_{\lambda }$ with the Möbius transformation $\mu \left(z\right)=\frac{1}{z}$ at each iteration step. Furthermore, they posed a conjecture providing a relation between the coefficients of (each order) iterated series of $\mu \left({\eta }_{\lambda }\left(z\right)\right)$ (at $z=0$) and the Catalan numbers. In this paper, in particular, we prove this conjecture in a more precise (quantitative) formulation.

1. Introduction

Let $\eta :\mathbb{C}\to \mathbb{C}$ be a monic complex polynomial of degree $d\ge 2$. We denote by ${\eta }^j$ the j-th iterate of $\eta$, that is,

$${\eta }^j\left(z\right)=\stackrel{j-\text{times}}{\overbrace{\eta \left(\eta \right(\eta (\cdots \eta }}\left(z\right)\cdots \left)\right)).$$

The filled-in Julia set of $\eta$ is defined as

$$K\left(\eta \right)=\{z\in \mathbb{C}:{\eta }^j\left(z\right)\text{does} \text{not} \text{diverge}\}$$

and the Julia set $J\left(\eta \right)$ of the function $\eta$ is defined to be the boundary of the set $K\left(\eta \right)$, i.e., $J\left(\eta \right)=\partial K\left(\eta \right)$ (see, e.g., [1]).

In this work, we are interested in a modified version of the “classical” filled-in Julia set $K\left({\eta }_{\lambda }\right)$ and the Julia set $J\left({\eta }_{\lambda }\right)$ of functions in the quadratic family ${\left({\eta }_{\lambda }\left(z\right)\right)}_{\lambda \in \mathbb{C}}={(z^2+\lambda )}_{\lambda \in \mathbb{C}}$. We observe that the Mandelbrot set $M\left({\eta }_{\lambda }\right)$ is the fractal defined as

$$M\left({\eta }_{\lambda }\right)=\{\lambda \in \mathbb{C}:J\left({\eta }_{\lambda }\right)\text{is} \text{connected}\}.$$

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):

$$\lambda \in M\left({\eta }_{\lambda }\right)\iff {\eta }_{\lambda }^j\left(0\right)\text{does} \text{not} \text{diverge}.$$

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 a lacunary function ${\eta }_{N,k}\left(z\right)={\sum }_{n=1}^N{z}^{P_k\left(n\right)}$, where ${\left(P_n\left(k\right)\right)}_n={\left(\frac{1}{2}(k{n}^2-kn-2)\right)}_n$ is the sequence of centered k-gonal numbers and k 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 $\eta \left(z\right)={\sum }_{n=1}^N{z}^{P_k\left(n\right)}$ with a fixed Möbius transformation $\mathcal{M}\left(z\right)=e^{i\theta }\frac{z-a}{\overline{a}z-1}$ (with $(\theta ,a)\in \mathbb{R}\times \mathbb{D}$, where 𝔻 denotes the the unit disc) at each iteration step. More precisely

$$h^j(z;a,k,N,\theta )=\stackrel{j-\text{times}}{\overbrace{\mathcal{M}({\eta }_{N,k}\left(\mathcal{M}\right({\eta }_{N,k}(\cdots \mathcal{M}({\eta }_{N,k}}}\left(z\right)\cdots \left)\right)\left)\right)).$$

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 ${\eta }_{\lambda }$ and the Möbius transformation ${\mu }_A\left(z\right)=\frac{az+b}{cz+d}$, where $A=(a,b,c,d)\in {\mathbb{C}}^4$. Thus,

$$h^j(z;A)=\stackrel{j-\text{times}}{\overbrace{{\mu }_A({\eta }_{\lambda }({\mu }_A({\eta }_{\lambda }(\cdots {\mu }_A({\eta }_{\lambda }}}\left(z\right)\cdots \left)\right)\left)\right)).$$

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 ${\left({\mathcal{H}}_n\right)}_{0\le n\le j}$, 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,

$$\begin{array}{cc}\hfill {\left({\mathcal{H}}_n\right)}_{0\le n\le j}& =\{h^0(0;A),h^1(0;A),\cdots ,h^j(0;A)\}\hfill \\ \hfill & =\{0,{\mu }_A\left({\eta }_{\lambda }\left(0\right)\right),\cdots ,(\stackrel{j-times}{\overbrace{{\mu }_A({\eta }_{\lambda }({\mu }_A({\eta }_{\lambda }(\cdots {\mu }_A({\eta }_{\lambda }}}\left(0\right)\cdots ))))))\}.\hfill \end{array}$$

Figure 1. The j-averaged Mandelbrot sets for $A=(0,0.5,1,0)$, $\lambda =x+iy$, 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 software Mathematica ${}^{\circledR }$ (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 ${\mu }_A$ 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 $\lambda \in \mathbb{D}$ be a non-zero complex number. Define the function $H(z,\lambda )$ by $H(z,\lambda ):={\mu }_A\left({\eta }_{\lambda }\left(z\right)\right)$, with $A=(0,1,1,0)$. Therefore,

$$H(z,\lambda )={\displaystyle \frac{1}{z^2+\lambda }}.$$

Observe that the n-th iteration of H at $z=0$ is a function of $\lambda$, say $h_n\left(\lambda \right)$, which satisfies the relations:

$$h_0\left(\lambda \right)=0,h_1\left(\lambda \right)=H(0,\lambda )=\frac{1}{\lambda }\mathrm{and}{h}_{n+1}\left(\lambda \right)=H(h_n\left(\lambda \right),\lambda ), \mathrm{for} n\ge 1.$$

(1)

The sequence ${\left(C_n\right)}_{n\ge 0}$ 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 $\left(\genfrac{}{}{0pt}{}{2n}{n}\right)$ by

$$C_n=\frac{1}{n+1}\left(\genfrac{}{}{0pt}{}{2n}{n}\right),$$

thus, its first terms are in Table 1.

Table 1. Values of $C_n$ for n from 0 to 14.

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

Table 1. Values of $C_n$ for n from 0 to 14.

n	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14
${\mathit{C}}_{\mathit{n}}$	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])

$$C_n=\frac{4n-2}{n+1}{C}_{n-1},\mathrm{for}n\ge 1,$$

with the initial condition $C_0=1$. Sometimes the sequence ${\left(C_n\right)}_{n\ge 0}$ is defined on the basis of the generating function $(1-\sqrt{1-4x})/\left(2x\right)$, as the following holds (for $\left|x\right|<1/4$)

$$\sum _{n=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}{n+1}{x}^n=\frac{2}{1+\sqrt{1-4x}}.$$

The aim of this paper is to obtain a (quantitative) result for the coefficients of the power series of $h_n\left(\lambda \right)$ which implies the Mork–Ulness’ conjecture (qualitative version). More precisely,

Theorem 1.

For all $n\ge 1$, we have

$$h_n\left(\lambda \right)=\frac{1-{(-1)}^n}{2\lambda }+{(-1)}^n\sum _{i=1}^{\lfloor n/2\rfloor }{C}_{i-1}{\lambda }^{3i-1}+O\left({\lambda }^{3\lfloor n/2\rfloor +2}\right),$$

(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 $h_{\infty }^{\left(1\right)}\left(\lambda \right)$ and $h_{\infty }^{\left(2\right)}\left(\lambda \right)$ as

$${\displaystyle h_{\infty }^{\left(1\right)}\left(\lambda \right):=\underset{n\to \infty }{lim}{h}_{2n+1}\left(\lambda \right)=\frac{1}{\lambda }-\sum _{i\ge 0}{C}_i{\lambda }^{3i+2}}$$

and

$${\displaystyle h_{\infty }^{\left(2\right)}\left(\lambda \right):=\underset{n\to \infty }{lim}{h}_{2n}\left(\lambda \right)=\sum _{i\ge 0}{C}_i{\lambda }^{3i+2}.}$$

They also asserted that these functions should converge in the whole unit disk (or the punctured one for $h_{\infty }^{\left(1\right)}\left(\lambda \right)$). However, this is not true (this is expected because of the exponential nature of Catalan numbers). For example, the simple bound $\left(\genfrac{}{}{0pt}{}{2n}{n}\right)\ge 4^n/(2n+1)$, which comes from the fact that $4^n={(1+1)}^{2n}={\sum }_{k=0}^{2n}\left(\genfrac{}{}{0pt}{}{2n}{k}\right)$, implies that $C_n>4^n/\left(3n^2\right)$ (some other bounds can be found in ([19] Chapter 2) and [20]) and so if $\left|\lambda \right|>1/\sqrt[3]{2}\approx 0.793$ then $|C_n{\lambda }^{3n+2}|\ge 3^{-1}{\left|\lambda \right|}^2(2^n/n^2)(\sqrt[3]{2}{\left|\lambda \right|)}^{3n}>{\left|\lambda \right|}^2/3$ (for $n\ge 4$) yielding the divergence of $h_{\infty }^{\left(2\right)}\left(\lambda \right)$. In order to compute the radius of convergence, say r, of $h_{\infty }^{\left(2\right)}\left(\lambda \right)$, one can write this function as $h_{\infty }^{\left(2\right)}\left(\lambda \right)={\sum }_{n\ge 0}{a}_n{\lambda }^n$, where

$$a_n=\left\{\begin{array}{cc}{C}_{(n-2)/3},\hfill & ifn\equiv 2(mod3),\hfill \\ 0,\hfill & ifn\not\equiv 2(mod3).\hfill \end{array}\right.$$

Thus, $1/r=lim{sup}_{n\to \infty }\sqrt[n]{a_n}$ and, by using $C_n\approx 4^n/\left(n^{3/2}\sqrt{\pi }\right)$ (which comes from the Stirling formula $n!\approx \sqrt{2\pi n}{(n/e)}^n$), we obtain

$$\begin{array}{ccc}\hfill \frac{1}{r}& =& lim\underset{n\to \infty }{sup}\sqrt[n]{a_n}\hfill \\ & =& lim\underset{n\to \infty }{sup}\sqrt[3n+2]{C_n}\hfill \\ & =& lim\underset{n\to \infty }{sup}\sqrt[3n+2]{\frac{4^n}{n^{3/2}\sqrt{\pi }}}\hfill \\ & =& \sqrt[3]{4}.\hfill \end{array}$$

Therefore, $B(0,1/\sqrt[3]{4})$ is the disk of convergence of $h_{\infty }^{\left(2\right)}\left(\lambda \right)$ (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 $h_n\left(\lambda \right)$.

Lemma 1.

For any $n\ge 1$, there exists a power series $P_n\left(\lambda \right)$ such that

$h_n\left(\lambda \right)=\left\{\begin{array}{cc}{\lambda }^2+{\lambda }^5{P}_n\left({\lambda }^3\right),\hfill & ifniseven;\hfill \\ \frac{1}{\lambda }+{\lambda }^2{P}_n\left({\lambda }^3\right),\hfill & ifnisodd.\hfill \end{array}\right.$

Proof. 

By definition in (1), $h_n\left(\lambda \right)$ satisfies the following recurrence relation

$$h_{n+1}\left(\lambda \right)={\displaystyle \frac{1}{{\left(h_n\left(\lambda \right)\right)}^2+\lambda }},$$

with $h_0\left(\lambda \right)=0$ (since $h_1\left(\lambda \right)=H(0,\lambda )=1/\lambda$). Now, by defining $f_n\left(\lambda \right):=\lambda h_n\left(\lambda \right)$ and using the previous recurrence, we obtain

$$f_{n+1}\left(\lambda \right)/\lambda ={\displaystyle \frac{1}{{(f_n\left(\lambda \right)/\lambda )}^2+\lambda }}$$

and so

$$f_{n+1}\left(\lambda \right)={\displaystyle \frac{{\lambda }^3}{{\left(f_n\left(\lambda \right)\right)}^2+{\lambda }^3}},$$

(3)

with $f_0\left(\lambda \right)=0$. We claim that $f_n\left(\lambda \right)=g_n\left({\lambda }^3\right)$ for some rational function $g_n\left(\lambda \right)$, where n is any positive integer. Indeed, we can proceed by induction on n. For $n=1$, we can take $g_1\left(\lambda \right)=1$. Suppose (by induction hypothesis) that $f_n\left(\lambda \right)=g_n\left({\lambda }^3\right)$, for some formal power series $g_n\left(\lambda \right)$, then, by (3), we have

$$f_{n+1}\left(\lambda \right)={\displaystyle \frac{{\lambda }^3}{{\left(g_n\left({\lambda }^3\right)\right)}^2+{\lambda }^3}}=g_{n+1}\left({\lambda }^3\right),$$

where $g_{n+1}\left(\lambda \right)$ can be chosen by satisfying the recurrence

$$g_{n+1}\left(\lambda \right)={\displaystyle \frac{\lambda }{{\left(g_n\left(\lambda \right)\right)}^2+\lambda }},$$

with $g_0\left(\lambda \right)=0$. The inductive process is finished. Observe that, since $h_n\left(\lambda \right)={\lambda }^{-1}{g}_n\left({\lambda }^3\right)$, then it suffices to prove that

$$g_n\left(\lambda \right)=\left\{\begin{array}{cc}\lambda +O\left({\lambda }^2\right),\hfill & \text{if}n\text{is} \text{even};\hfill \\ 1+O\left(\lambda \right),\hfill & \text{if}n\text{is} \text{odd}.\hfill \end{array}\right.$$

(4)

The proof is also by induction on n (more precisely, a double induction). For the basis cases, we have $g_1\left(\lambda \right)=1=1+O\left(\lambda \right)$ and

$$g_2\left(\lambda \right)={\displaystyle \frac{\lambda }{1+\lambda }}=\lambda (1-\lambda +{\lambda }^2-\cdots )=\lambda +O\left({\lambda }^2\right),$$

where we used that for $\left|\lambda \right|<1$, one has ${(1+\lambda )}^{-1}={\sum }_{k\ge 0}{(-\lambda )}^n$ (in general, it holds that ${(1+O\left(1\right))}^{-1}=1+O\left(1\right)$). Suppose that (4) is valid for all $n\in [1,2k]$. Then,

$$\begin{array}{ccc}\hfill g_{2k+1}\left(\lambda \right)& =& {\displaystyle \frac{\lambda }{{\left(g_{2k}\left(\lambda \right)\right)}^2+\lambda }}\hfill \\ & =& {\displaystyle \frac{\lambda }{{(\lambda +O\left({\lambda }^2\right))}^2+\lambda }}\hfill \\ & =& {\displaystyle \frac{\lambda }{\lambda +O\left({\lambda }^2\right)}}\hfill \\ & =& {\displaystyle \frac{1}{1+O\left(\lambda \right)}}=1+O\left(\lambda \right),\hfill \end{array}$$

where we used $O\left({\lambda }^r\right)+O\left({\lambda }^s\right)=O\left({\lambda }^{min\{r,s\}}\right)$, since $\left|\lambda \right|<1$.

Now, we use the previous fact

$$\begin{array}{ccc}\hfill g_{2k+2}\left(\lambda \right)& =& {\displaystyle \frac{\lambda }{{\left(g_{2k+1}\left(\lambda \right)\right)}^2+\lambda }}\hfill \\ & =& {\displaystyle \frac{\lambda }{{(1+O\left(\lambda \right))}^2+\lambda }}\hfill \\ & =& {\displaystyle \frac{\lambda }{1+O\left(\lambda \right)}}\hfill \\ & =& \lambda (1+O\left(\lambda \right))=\lambda +O\left({\lambda }^2\right).\hfill \end{array}$$

This completes the induction proof of (4).

Therefore, since $\left|\lambda \right|<1$, we can write

$$g_n\left(\lambda \right)=\left\{\begin{array}{cc}\lambda +\sum _{i\ge 2}{c}_i{\lambda }^i,\hfill & \text{if}n\text{is} \text{even};\hfill \\ 1+\sum _{i\ge 1}{c}_i{\lambda }^i,\hfill & \text{if}n\text{is} \text{odd}\hfill \end{array}\right.$$

and so

$$h_n\left(\lambda \right)=\frac{1}{\lambda }{g}_n\left({\lambda }^3\right)=\left\{\begin{array}{cc}{\lambda }^2+\sum _{i\ge 2}{c}_i{\lambda }^{3i-1},\hfill & \text{if}n\text{is} \text{even};\hfill \\ {\displaystyle \frac{1}{\lambda }}+\sum _{i\ge 1}{c}_i{\lambda }^{3i-1},\hfill & \text{if}n\text{is} \text{odd}.\hfill \end{array}\right.$$

This completes the proof. □

Remark 2.

Note that, by using Lemma 1, we can write

$$h_n\left(\lambda \right)={\alpha }_{-1,n}{\lambda }^{-1}+{\alpha }_{0,n}{\lambda }^2+{\alpha }_{1,n}{\lambda }^5+\cdots =\sum _{k=-1}^{\infty }{\alpha }_{k,n}{\lambda }^{3k+2}\in \mathbb{R}\left[\left[\lambda \right]\right],$$

(5)

where ${\alpha }_{-1,n}=(1-{(-1)}^n)/2$, i.e., ${\alpha }_{-1,n}$ is 1 if n is odd and 0 if n is even. In particular, $h_n\left(\lambda \right)$ is an analytic function in some neighborhood of $\lambda =0$, when n is even, and for n odd, $h_n\left(\lambda \right)$ 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 $h_n\left(\lambda \right)=0$ as $\lambda \to 0$, for any $k\equiv 0$ or $1(mod3)$. 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 $H_{\lambda }\left(z\right):=H(z,\lambda )$ vanish (for fixed λ) at $z=0$. This last assertion follows from Cauchy’s integral formula. Indeed, we have

$$H_{\lambda }^{(2n+1)}\left(0\right)={\displaystyle \frac{(2n+1)!}{2\pi i}}{\int }_{{\gamma }_R}{\displaystyle \frac{H_{\lambda }\left(\omega \right)}{{\omega }^{2n+2}}}d\omega ={\displaystyle \frac{(2n+1)!}{2\pi i}}{\int }_{{\gamma }_R}{\displaystyle \frac{1}{({\omega }^2+\lambda ){\omega }^{2n+2}}}d\omega ,$$

where ${\gamma }_R$ is the circle $\gamma \left(t\right):=R{e}^{it}$, for $t\in [0,2\pi ]$ and $0<R<\left|\lambda \right|$. Now, we can use the partial fraction decomposition to deduce that

$${\displaystyle \frac{1}{({\omega }^2+\lambda ){\omega }^{2n+2}}}={\displaystyle \frac{A}{\omega +\sqrt{\left|\lambda \right|}}}-{\displaystyle \frac{A}{\omega -\sqrt{\left|\lambda \right|}}}+{\displaystyle \frac{B}{{\omega }^{2n+2}}},$$

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

$${\int }_{{\gamma }_R}{\displaystyle \frac{1}{({\omega }^2+\lambda ){\omega }^{2n+2}}}d\omega =2A\pi if\left(0\right)-2A\pi if\left(0\right)+2B\pi i{f}^{(2n+1)}\left(0\right),$$

where $f\left(z\right)=1$, for all z. Thus, $H_{\lambda }^{(2n+1)}\left(0\right)$ is equal to zero as claimed.

Now we show the important connection of the sequence ${\left({\alpha }_{k,n}\right)}_{k\ge 0}$ to the Catalan numbers. For the simplicity of notation, we use the following notation in the rest of the text:

$${\alpha }_{k,n}=\left\{\begin{array}{cc}{d}_k,\hfill & \text{for} \text{odd}n;\hfill \\ e_k,\hfill & \text{for} \text{even}n.\hfill \end{array}\right.$$

Lemma 2.

Let ${\left(C_k\right)}_{k\ge 0}$ be the Catalan sequence. We have

(i) 

If ${\left(d_k\right)}_{k\ge 0}$ is defined by the recurrence,

$$d_{k+1}=-(C_1{d}_k+\cdots +C_{k+1}{d}_0)-C_{k+2},$$

with $d_0=-C_0$, then $d_k=-C_k$, for all $k\ge 0$.

(ii) 

If ${\left(e_k\right)}_{k\ge 1}$ is defined by the recurrence,

$$e_{k+1}=C_0{e}_k+\cdots +C_{k-1}{e}_1+C_k,$$

with $e_1=C_1$, then $e_k=C_k$, for all $k\ge 1$.

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 on k. For $k=0$, one has $d_0=-C_0$ (by definition). Suppose $d_t=C_t$, for all $t\in [0,k]$. Then,

$$\begin{array}{ccc}\hfill d_{k+1}& =& -(C_1{d}_k+\cdots +C_{k+1}{d}_0)-C_{k+2}\hfill \\ & =& C_1{C}_k+\cdots +C_{k+1}{C}_0-C_{k+2}\hfill \\ & =& \underset{C_{k+2}-C_{k+1}}{\underbrace{C_1{C}_k+\cdots +C_{k+1}{C}_0}}-C_{k+2}\hfill \\ & & =-C_{k+1}\hfill \end{array}$$

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 $t\in [1,k]$. Then, by the recurrence for ${\left(e_k\right)}_k$ together with the induction hypothesis, we obtain

$$\begin{array}{ccc}\hfill e_{k+1}& =& C_0{e}_k+\cdots +C_{k-1}{e}_1+C_k\hfill \\ & =& C_0{C}_k+\cdots +C_{k-1}{C}_1+C_k\hfill \\ & =& \underset{C_{k+1}-C_k}{\underbrace{C_0{C}_k+\cdots +C_{k-1}{C}_1}}+C_k\hfill \\ & & =C_{k+1}\hfill \end{array}$$

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).

Lemma 3.

Let ${\left(C_n\right)}_{n\ge 0}$ be the Catalan sequence. Then,

${\displaystyle C_{2n}=2\sum _{j=1}^n{C}_{j-1}{C}_{2n-j}}$ and ${\displaystyle C_{2n+1}=C_n^2+2\sum _{j=1}^n{C}_{j-1}{C}_{2n-j+1}}$,

for all $n\ge 0$ (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

$$h_{2j}\left(\lambda \right)=C_0{\lambda }^2+\cdots +C_{j-1}{\lambda }^{3j-1}+O\left({\lambda }^{3j+2}\right)$$

(7)

and

$$h_{2j+1}\left(\lambda \right)=\frac{1}{\lambda }-(C_0{\lambda }^2+\cdots +C_{j-1}{\lambda }^{3j-1})+O\left({\lambda }^{3j+2}\right),$$

(8)

where we adopt the convention that $C_0{\lambda }^2+\cdots +C_{j-1}{\lambda }^{3j-1}=0$ for $j=0$.

Now, we want to prove the following fact:

Claim. It holds that

$${(C_0{\lambda }^2+\cdots +C_{n-1}{\lambda }^{3n-1})}^2=C_1{\lambda }^4+\cdots +C_n{\lambda }^{3n+1}+O\left({\lambda }^{3n+2}\right),$$

(9)

for a non-negative integer n.

Proof. 

The proof is by induction on n. The identity is true for $n=0$, since $C_1=C_0^2$. Suppose that (9) holds, then one has

$$\begin{array}{ccc}\hfill {(C_0{\lambda }^2+\cdots +C_{n-1}{\lambda }^{3n-1}+C_n{\lambda }^{3n+2})}^2& =& {(C_0{\lambda }^2+\cdots +C_{n-1}{\lambda }^{3n-1})}^2\hfill \\ & & +2(C_0{\lambda }^2+\cdots +C_{n-1}{\lambda }^{3n-1})C_n{\lambda }^{3n+2}+{\left(C_n{\lambda }^{3n+2}\right)}^2.\hfill \end{array}$$

Since we desire to evaluate the identity up to $O\left({\lambda }^{3n+5}\right)$, then

$$2(C_0{\lambda }^2+\cdots +C_{n-1}{\lambda }^{3n-1})C_n{\lambda }^{3n+2}+{\left(C_n{\lambda }^{3n+2}\right)}^2=2C_0{C}_n{\lambda }^{3n+4}+O\left({\lambda }^{3n+7}\right).$$

(10)

On the other hand, in the induction hypothesis

$${(C_0{\lambda }^2+\cdots +C_{n-1}{\lambda }^{3n-1})}^2=C_1{\lambda }^4+\cdots +C_n{\lambda }^{3n+1}+O\left({\lambda }^{3n+2}\right),$$

the terms of order ${\lambda }^{3n+4}$ were neglected (since we were interested in $O\left({\lambda }^{3n+2}\right)$). 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 ${\lambda }^{3n+3}$ in ${(C_0{\lambda }^2+\cdots +C_{n-1}{\lambda }^{3n-1})}^2$. Let us also suppose that n is odd (the even case is carried out along the same lines). We then have

$$\begin{array}{ccc}\hfill {(C_0{\lambda }^2+\cdots +C_{n-1}{\lambda }^{3n-1})}^2& =& C_1{\lambda }^4+\cdots +C_n{\lambda }^{3n+1}\hfill \\ & & +2(C_1{C}_{n-1}+\cdots +C_{(n-1)/2}{C}_{(n+1)/2}){\lambda }^{3n+4}+O\left({\lambda }^{3n+5}\right).\hfill \end{array}$$

(11)

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

$$\begin{array}{ccc}\hfill {(C_0{\lambda }^2+\cdots +C_{n-1}{\lambda }^{3n-1}+C_n{\lambda }^{3n+2})}^2& =& {(C_0{\lambda }^2+\cdots +C_{n-1}{\lambda }^{3n-1})}^2\hfill \\ & & +2(C_0{\lambda }^2+\cdots +C_{n-1}{\lambda }^{3n-1})C_n{\lambda }^{3n+2}+{\left(C_n{\lambda }^{3n+2}\right)}^2\hfill \\ & =& C_1{\lambda }^4+\cdots +C_n{\lambda }^{3n+1}\hfill \\ & & +2(C_1{C}_{n-1}+\cdots +C_{(n-1)/2}{C}_{(n+1)/2}){\lambda }^{3n+4}+O\left({\lambda }^{3n+5}\right)\hfill \\ & & +2C_0{C}_n{\lambda }^{3n+4}+O\left({\lambda }^{3n+7}\right)\hfill \\ & =& C_1{\lambda }^4+\cdots +C_{n+1}{\lambda }^{3n+4}+O\left({\lambda }^{3n+5}\right)\hfill \end{array}$$

which finishes the proof of the claim.

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

$$h_1\left(\lambda \right)=\frac{1}{\lambda }$$

and, by Lemma 1,

$$h_2\left(\lambda \right)={\lambda }^2+O\left({\lambda }^5\right).$$

Suppose that (2) is true for $h_n\left(\lambda \right)$ with $n\in [1,2j]$. Then, by the recurrence relation for ${\left(h_n\left(\lambda \right)\right)}_n$ together with the induction hypothesis, we infer that

$$h_{2j+1}\left(\lambda \right)={\displaystyle \frac{1}{{\left(h_{2j}\left(\lambda \right)\right)}^2+\lambda }}={\displaystyle \frac{1}{{(C_0{\lambda }^2+\cdots +C_{j-1}{\lambda }^{3j-1}+O\left({\lambda }^{3j+2}\right))}^2+\lambda }}.$$

However, we can use (9) to write

$${(C_0{\lambda }^2+\cdots +C_{j-1}{\lambda }^{3j-1}+O\left({\lambda }^{3j+2}\right))}^2+\lambda =C_0\lambda +C_1{\lambda }^4+\cdots +C_j{\lambda }^{3j+1}+O\left({\lambda }^{3j+2}\right).$$

(12)

From Lemma 1 and Remark 2, one has

$$h_{2j+1}\left(\lambda \right)={\displaystyle \frac{1}{\lambda }}+d_0{\lambda }^2+d_1{\lambda }^5+\cdots +d_{j-1}{\lambda }^{3j-1}+O\left({\lambda }^{3j+2}\right).$$

Thus, the coefficients $d_0$, $d_1$, ..., $d_{j-1}$ satisfy the following equality

$$1\equiv (C_0\lambda +C_1{\lambda }^4+\cdots +C_k{\lambda }^{3j+1}+O\left({\lambda }^{3j+2}\right))\left({\displaystyle \frac{1}{\lambda }}+d_0{\lambda }^2+d_1{\lambda }^5+\cdots +d_{j-1}{\lambda }^{3j-1}+O\left({\lambda }^{3j+2}\right)\right)$$

and so

$$1\equiv 1+\lambda \left(\sum _{i=0}^{j-1}{d}_i{\lambda }^{3i+2}\right)+C_1{\lambda }^4\left({\displaystyle \frac{1}{\lambda }}+\sum _{i=0}^{j-2}{d}_i{\lambda }^{3i+2}\right)+\cdots +C_k{\lambda }^{3j+1}\left({\displaystyle \frac{1}{\lambda }}\right)+O\left({\lambda }^{3j+1}\right).$$

By reordering this sum, we obtain

$$0\equiv (d_0+C_1){\lambda }^3+(d_1+C_1{d}_0+C_2){\lambda }^6+\cdots +(d_{j-1}+C_1{d}_{j-2}+\cdots +C_j){\lambda }^{3j}+O\left({\lambda }^{3j+1}\right).$$

Therefore, $d_0=-C_1=-C_0$ and

$$d_t=-(C_1{d}_{t-1}+\cdots +C_t{d}_0)-C_t,$$

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

$$h_{2j+1}\left(\lambda \right)={\displaystyle \frac{1}{\lambda }}-(C_0{\lambda }^2+C_1{\lambda }^5+\cdots +C_{k-1}{\lambda }^{3j-1})+O\left({\lambda }^{3j+2}\right)$$

as desired.

Thus, we determine that (2) holds for $h_n\left(\lambda \right)$ for all $n\in [1,2j+1]$. To finish the proof, we must prove that (2) is also true for $n=2j+2$. First, one has that

$$h_{2j+2}\left(\lambda \right)={\displaystyle \frac{1}{{\left(h_{2j+1}\left(\lambda \right)\right)}^2+\lambda }}={\displaystyle \frac{1}{{((1/\lambda )-(C_0{\lambda }^2\cdots +C_{j-1}{\lambda }^{3j-1})+O\left({\lambda }^{3j+2}\right))}^2+\lambda }}.$$

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

$${\left(h_{2j+1}\left(\lambda \right)\right)}^2+\lambda ={\displaystyle \frac{1}{{\lambda }^2}}-(C_0\lambda +C_1{\lambda }^4+\cdots +C_{j-1}{\lambda }^{3j-2})+C_k{\lambda }^{3j+1}+O\left({\lambda }^{3j+2}\right).$$

(13)

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

$$h_{2j+2}\left(\lambda \right)=e_0{\lambda }^2+e_1{\lambda }^5+\cdots +e_j{\lambda }^{3j+2}+O\left({\lambda }^{3j+5}\right),$$

where $e_0=1$. Hence,

$$1\equiv \left({\displaystyle \frac{1}{{\lambda }^2}}-(C_0\lambda +\cdots +C_{j-1}{\lambda }^{3j-2})+C_k{\lambda }^{3j+1}+O\left({\lambda }^{3j+2}\right)\right)({\lambda }^2+e_1{\lambda }^5+\cdots +e_j{\lambda }^{3j+2}+O\left({\lambda }^{3j+5}\right)).$$

Thus,

$$1\equiv 1+{\lambda }^{-2}\left(\sum _{i=1}^j{e}_i{\lambda }^{3i+2}\right)-C_0\lambda \left(\sum _{i=0}^j{e}_i{\lambda }^{3i+2}\right)+\cdots +C_{j-1}{\lambda }^{3j-2}·{\lambda }^2+O\left({\lambda }^{3j+3}\right)$$

which can be re-written as

$$0\equiv (e_1-1){\lambda }^3+(e_2-C_0{e}_1-C_1){\lambda }^6+\cdots +(e_j-e_{j-1}-C_1{e}_{j-2}-\cdots -C_{j-2}{e}_1+C_{j-1}){\lambda }^{3j}+O\left({\lambda }^{3j+3}\right).$$

We then deduce that $e_1=1=C_1$ and

$$e_t=C_0{e}_{j-1}+C_1{e}_{j-2}+\cdots +C_{j-2}{e}_1-C_{j-1},$$

for all $t\in [1,j]$. By Lemma 3 (ii), we have $e_t=C_t$, for all $t\in [1,j]$, yielding that

$$h_{2j+2}\left(\lambda \right)=C_0{\lambda }^2+C_1{\lambda }^5+\cdots +C_j{\lambda }^{3j+2}+O\left({\lambda }^{3j+5}\right).$$

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 $h_n\left(\lambda \right)$ (the n-th iteration of $1/(z^2+\lambda )$ at $z=0$) and the Catalan sequence ${\left(C_k\right)}_k$. Indeed, we prove a quantitative version of their conjecture by showing that the sequence ${\left(h_n\left(\lambda \right)-(\frac{1-{(-1)}^n}{2\lambda }+{(-1)}^n{\sum }_{i=1}^{\lfloor n/2\rfloor }{C}_{i-1}{\lambda }^{3i-1})\right)}_n$ tends to zero (with order ${\left|\lambda \right|}^{3\lfloor n/2\rfloor +2}$) as $n\to \infty$.