Weighted Random Averages and Recursive Interpolation in Fibonacci Sequences

Abstract

We investigate the multifractal geometry of irregular sets arising from weighted averages of random variables, where the weights $\left(w_n\right)$ form a positive sequence with exponential growth. Our analysis applies in particular to sequences generated by linear recurrence relations of Fibonacci type, including higher-order generalizations such as the Tetranacci sequence $\left({\mathsf{T}}_n\right)$. Using a Cantor-type construction built from alternating free and forced blocks, we show that the associated exceptional sets may attain full Hausdorff and packing dimension, independently of the precise form of the recurrence. We further develop a probabilistic interpretation of $\left({\mathsf{T}}_n\right)$ through an appropriate Markov representation that encodes its combinatorial evolution and yields sharp asymptotic behavior. Finally, given $n+1$ consecutive terms of a Fibonacci-type sequence, one may construct a polynomial $P_n\left(x\right)$ of degree at most n via Lagrange interpolation; we show that this polynomial admits an implicit recursive representation consistent with the underlying recurrence.

1. Introduction and Main Results

In recent years, numerous researchers have shown interest in studying number sequences and recurrence relations. The Fibonacci sequence ${\left(F_n\right)}_n$ is the most prominent example of recursive sequences characterized by rich and remarkable structural properties. Originating from the pioneering work of Leonardo Fibonacci, it is defined by the series of numbers generated by the pattern

$$1,1,2,3,5,8,13,21,34,55,89,\dots$$

Each subsequent term is computed as the sum of its two preceding terms. The pervasiveness of these numerical entities in natural phenomena enhances their allure. Furthermore, they hold significant importance in diverse academic disciplines including but not limited to algebra, geometry, and number theory. Many researchers have investigated the properties and applications of this sequence in diverse fields. Moreover, it was proved that the sequence ${\left(F_n\right)}_{n\ge 0}$ is closely related to the golden ratio $\sigma =(1+\sqrt{5})/2=1.61803398\dots$.

Numerous authors have undertaken the task of generalizing the Fibonacci sequence by either keeping the original recurrence relation while altering the initial terms or by maintaining the initial terms while slightly modifying the recurrence relation (see, for instance, [1,2,3]). The reader can consult also [4,5] for the study of the m-Fibonacci-like sequence for a positive integer m. Furthermore, one of the most important variations of the Fibonacci sequence is the k-step Fibonacci sequence, commonly denoted by ${\left({\mathsf{T}}_{k,n}\right)}_{n\ge 0}$ (or simply by ${\left({\mathsf{T}}_n\right)}_n$ if context allows clarity), in which each number is the sum of the preceding k-numbers [6,7,8,9]. In particular, when $k=3$, one has the Tribonacci sequence, originally introduced by Feinberg in 1963 [10], this topic has since inspired numerous generalizations, each exhibiting fascinating properties such as Binet formulas, summation formulas and generating functions [11].

Motivated by these developments, we investigate how such recursive sequences behave when used as weights in probabilistic or dynamical constructions. Recent works show that weighted sums involving Fibonacci-type weights may display unexpected oscillatory and fractal features, particularly when the weights grow exponentially [12]. This motivates the study of the multifractal structure of weighted averages driven by general recursive sequences. In this paper, we adopt a probabilistic viewpoint and consider weighted sums of independent random variables. To formalize the setting, let $(\Omega ,\mathcal{F},\mathbb{P})$ be a probability space, and let ${\left(X_n\right)}_{n\ge 1}$ be a sequence of independent and identically distributed (i.i.d.) real-valued random variables with non-degenerate law. Given a sequence of positive weights ${\left(w_n\right)}_{n\ge 1}$, we consider the $W_n={\sum }_{k=1}^n{w}_k$ and the weighted averages

$$Y_n={\displaystyle \frac{{\sum }_{k=1}^n{w}_k{X}_k}{W_n}},n\ge 1.$$

(1)

Such quantities arise naturally in probability theory, statistics, ergodic theory, and the study of non-uniform random walks. A fundamental question concerns the almost sure convergence of $Y_n$. When the weights are sufficiently “diffuse”, in the sense that

$$\underset{1\le k\le n}{\max }{\displaystyle \frac{w_k}{W_n}}⟶0,$$

(2)

the behavior of $\left(Y_n\right)$ is well understood. Classical results of Chow [13,14], Stout [15,16], Padgett–Taylor [17], Rosalsky–Sreehari [18], Bai–Cheng [19], and Cuzick [20], among others, show that $Y_n$ converges almost surely to $\mathbb{E}\left[X_1\right]$ under mild moment assumptions. Further extensions to various forms of dependence were later obtained by Etemadi [21], Shen [22], and related works.

A completely different behavior emerges when the weights grow exponentially, that is,

$${\displaystyle \frac{w_{n+1}}{w_n}}⟶\phi >1,$$

(3)

see Lemma 1. In this regime, the contribution of the last weight $w_n$ does not vanish in the normalization $W_n$; in fact, it converges to a strictly positive proportion of the total mass. Consequently, the diffuseness condition underlying the classical weighted law of large numbers fails, and almost-sure convergence of $\left(Y_n\right)$ cannot hold whenever $X_1$ is non-degenerate; see, for instance, ([12], Theorem 2). Exponentially growing weights arise naturally in recursive random walks whose increments satisfy Fibonacci or more generally k–bonacci-type recursions. Recent works [6,12,23] show that such walks display persistent oscillations and generate full-dimensional fractal sets of exceptional trajectories. Beyond the classical law of large numbers for i.i.d. sequences and the convergence theory for Birkhoff sums [15,24], recent research has focused on the fractal geometry of the irregular sets where such averages fail to converge, revealing that these exceptional sets often have large or even full dimension [25,26].

Motivated by these phenomena, we study in this paper the weighted averages $Y_n$ defined by (1) in the general setting where the random variables $X_k$ take values in a finite alphabet

$$\mathcal{B}=\{x_0<x_1<\dots <x_r\},r\ge 1,$$

(4)

and where the weights satisfy the sole assumption (3). In this regime, the averages typically oscillate rather than converge. For any pair $a,b\in \mathcal{B}$ with $a<b$, we introduce the set

$$\mathcal{C}(a,b)=\left\{\omega \in \Omega :\underset{n\to \infty }{\lim \inf }{Y}_n\left(\omega \right)=a,\underset{n\to \infty }{\lim \sup }{Y}_n\left(\omega \right)=b\right\},$$

which describes trajectories whose oscillation interval is prescribed. The analysis of these sets is partly motivated by earlier work on branching random walks on supercritical Galton–Watson trees, where quantities analogous to $Y_n$ have been investigated [27,28]. A related form of the oscillation set $\mathcal{C}(a,b)$ was also examined in [29], where the normalized sums were shown to oscillate between two prescribed limits a and b. In that setting, the proofs rely on the construction of a Mandelbrot-type multiplicative cascade supported on the boundary of the tree, which provides probabilistic control over the Hausdorff and packing dimensions of the corresponding exceptional sets [30,31]. These techniques, however, fundamentally depend on independence and on the use of homogeneous weights $(a_k=1)$ and thus cannot be extended to non-independent scenarios or to non-stationary weight sequences such as the Fibonacci case. Furthermore, the study of the global irregular set I is naturally linked to earlier developments in multifractal analysis and ergodic theory.

A central result of this paper is Theorem 1, which establishes that exponential growth of the weights alone is sufficient to generate irregular sets of maximal fractal size. More precisely, we show that for any prescribed oscillation interval $[a,b]$ contained in the alphabet, the corresponding oscillation set $\mathcal{C}(a,b)$ has full Hausdorff and packing dimension. In the special case $X_k\in \{0,1\}$ and $w_n=F_n$, this was previously proved ([12], Theroem 2) using the identity

$$\sum _{i=1}^n{F}_i=F_{n+2}-1,$$

which allows exact control of weighted blocks. We show here that no such exact formula is needed: the conclusion remains valid for any weight sequence with geometric growth.

Theorem 1. 

Assume that the weights satisfy (3) and let $x_0\le a<b\le x_r$. Then, the set $\mathcal{C}(a,b)$ is non-empty, and for every $\epsilon >0$, it contains a compact Cantor set K such that

$${\dim }_{\mathcal{H}}K\ge 1-\epsilon .$$

Consequently,

$${\dim }_{\mathcal{H}}\mathcal{C}(a,b)={\dim }_{\mathcal{P}}\mathcal{C}(a,b)=1,$$

where ${\dim }_{\mathcal{H}}$ and ${\dim }_{\mathcal{P}}$ denote the Hausdorff and packing dimension, respectively.

A key feature of Theorem 1 is that the structure of the irregular sets $\mathcal{C}(a,b)$ is governed exclusively by the exponential growth of the weights. In particular, no further arithmetic or combinatorial assumption on the weights is required. The only assumption needed is (3), which guarantees that the last part of the weighted sum carries a fixed positive proportion of the total mass, thereby violating the classical diffuseness condition underlying weighted laws of large numbers. This property is precisely what allows us to force the weighted averages to oscillate between prescribed values and ultimately to construct Cantor subsets of full the Hausdorff and packing dimension. The exponential ratio property $w_{n+1}/w_n\to \phi >1$ is well known to hold for the Fibonacci sequence and, more generally, for all k-step Fibonacci-type recurrences through their Binet-type representations. Thus, classical linear recurrences naturally generate exponentially growing weights satisfying the assumptions of Theorem 1. In the present work, we investigate this phenomenon for the Tetranacci sequence, whose structure admits both an algebraic (Binet-type) description and a probabilistic interpretation. We consider the recurrence

$${\mathsf{T}}_{n+4}={\mathsf{T}}_n+{\mathsf{T}}_{n+1}+{\mathsf{T}}_{n+2}+{\mathsf{T}}_{n+3},n\ge 0,$$

(5)

with initial values ${\mathsf{T}}_0=0$, ${\mathsf{T}}_1=1$, ${\mathsf{T}}_2=2$, and ${\mathsf{T}}_3=4$. As shown in [32], ${\mathsf{T}}_n$ admits a closed Binet-type formula in terms of the roots of the characteristic polynomial

$$x^4-x^3-x^2-x-1=0,$$

which possesses a unique real root $\mathsf{p}\in (0,1)$. This yields algebraic asymptotics of the Tetranacci sequence ${\left({\mathsf{T}}_n\right)}_n$. We develop a probabilistic viewpoint based on a color/Markov model (Section 3 and Section 4) that encodes the combinatorial structure of the recurrence. This method yields both an exact identity and an asymptotic expansion.

Theorem 2. 

Let ${\left({\mathsf{T}}_n\right)}_{n\ge 0}$ be the Tetranacci sequence defined by (5). Then

$$\sum _{n=2}^{\infty }{\displaystyle \frac{{\mathsf{T}}_n}{4^n}}={\displaystyle \frac{399}{684}}.$$

Moreover, letting $\varsigma ={\left(1+{\mathsf{p}}^2+2{\mathsf{p}}^3+2{\mathsf{p}}^4\right)}^{-1},$ then we have the asymptotic estimate

$${\mathsf{T}}_n\sim \varsigma {\mathsf{p}}^{-n},n\to \infty ,$$

and in particular

$$\underset{n\to \infty }{\lim }{\displaystyle \frac{{\mathsf{T}}_{n+1}}{{\mathsf{T}}_n}}={\mathsf{p}}^{-1}>1.$$

(6)

Theorem 2 shows in particular that ${\mathsf{T}}_n=o\left(4^n\right)$, and the probabilistic model leads to a second asymptotic characterization in Section 4. Following the approach of [33,34], we encode the sequence into a Markov chain and extract the limiting distribution, obtaining the following refinement.

Theorem 3. 

Let $\left({\mathsf{T}}_n\right)$ be the Tetranacci sequence with initial values ${\mathsf{T}}_0={\mathsf{T}}_1=0$ and ${\mathsf{T}}_2={\mathsf{T}}_3=1$. Then

$$\underset{n\to \infty }{\lim }{\mathsf{p}}^{n-2}{\mathsf{T}}_{n-1}={\displaystyle \frac{1}{2}}{\displaystyle \frac{\mathsf{p}-{\mathsf{p}}^2-{\mathsf{p}}^3}{{(1-\mathsf{p})}^2}},$$

where $\mathsf{p}$ is the real root of $1-\mathsf{p}-{\mathsf{p}}^2-{\mathsf{p}}^3-{\mathsf{p}}^4=0$. Consequently, (6) holds.

Theorem 3 confirms that the Tetranacci sequence grows exponentially according to

$${\mathsf{T}}_n=O\left({\beta }^n\right),\beta ={\mathsf{p}}^{-1}=1.92756\dots ,$$

in agreement with the algebraic asymptotics obtained in [32], but now derived through a fully probabilistic mechanism. Such exponential behavior mirrors the classical Fibonacci case, where the ratio $F_{n+1}/F_n$ converges to the Golden Ratio $\sigma =(1+\sqrt{5})/2$, a phenomenon with deep geometric and analytic implications (see, e.g., [35,36]). More generally, this convergence property extends to the broad family of k–Fibonacci and k-Fibonacci-like sequences, whose successive ratios converge to the real dominant root of the corresponding characteristic polynomial.

The final part of this paper is devoted to a complementary and somewhat independent perspective: the polynomial reconstruction of Fibonacci sequences. Given $n+1$ consecutive values of a k-step Fibonacci-like sequence, we show in Section 5 that one may construct a polynomial ${\mathbb{P}}_n\left(x\right)$ of degree at most n via Lagrange interpolation, which recovers these values exactly. We prove that this interpolating polynomial admits both an implicit and a recursive description; more precisely,

$${\mathbb{P}}_{n+1}\left(x\right){\mathbb{P}}_n\left(x\right)+{\alpha }_{n}x(x-1)\dots (x-n),$$

where ${\alpha }_n$ denotes the leading coefficient of ${\mathbb{P}}_{n+1}$.

The article is organized around three complementary themes, unified by the role played by Fibonacci–type recursions and exponential growth. In Section 2, we first investigate exponentially weighted random averages and show that the sole assumption of geometric growth of the weights is sufficient to generate large irregular sets, culminating in the proof of Theorem 1. Section 3 and Section 4 are then devoted to a detailed probabilistic analysis of the Tetranacci sequence: using color models and Markov chain representations, we derive sharp asymptotic estimates and limit formulas, without resorting to explicit Binet-type formulas or the resolution of Vandermonde systems, thereby illustrating how classical linear recurrences naturally produce the exponential regimes required in Section 2. Finally, Section 5 explores a complementary perspective by studying polynomial reconstruction associated with Fibonacci–type sequences, showing that consecutive values of such recursions admit an implicit and recursive Lagrange interpolation scheme.

2. Irregular Sets via Weighted Random Sequences

In this section we analyse the emergence of irregular behaviour in exponentially weighted averages and construct Cantor-type subsets of the oscillation sets $\mathcal{C}(a,b)$. The key mechanism is that geometric growth of the weights forces the most recent symbols to carry a fixed positive proportion of the total weight. This phenomenon leads to almost-sure oscillations of the averages between the extreme values of the alphabet, and ultimately allows the construction of large fractal subsets inside the irregular set $\mathcal{C}(a,b)$. Before turning to the Cantor construction, we first show that the classical diffuseness condition fails in the exponentially growing regime, the following result will be proved in Appendix A.

Lemma 1. 

Let ${\left(X_k\right)}_{k\ge 1}$ be i.i.d. real-valued random variables taking values in a finite alphabet $\mathcal{B}$ defined by (4). Let ${\left(w_n\right)}_{n\ge 1}$ be a sequence of positive weights satisfying (3).

1. 

The diffuseness condition (2) does not hold. More precisely,

$$\underset{1\le k\le n}{\max }{\displaystyle \frac{w_k}{{\sum }_{i=1}^n{w}_i}}={\displaystyle \frac{w_n}{{\sum }_{i=1}^n{w}_i}}⟶{\displaystyle \frac{\phi -1}{\phi }}>0.$$

2. 

The exponentially weighted averages $Y_n$ do not converge almost surely. More precisely,

$$\underset{n\to \infty }{\lim \sup } Y_n=x_r,\underset{n\to \infty }{\lim \inf } Y_n=x_0,\mathit{a}.\mathit{s}..$$

The proof of Theorem 1 is based on the fact that exponential growth of the weights implies that the last block of terms contributes a fixed positive proportion of the total weight. This allows us to force the weighted average $Y_n$ to approach prescribed values and to construct a Cantor-type subset of $\mathcal{C}(a,b)$ with full Hausdorff and packing dimensions. We will start by considering the case when $a,b\in \mathcal{B}$ in the following result.

Proposition 1. 

Assume that the weights satisfy (3). Let $a<b$ be elements of $\mathcal{B}$. Then, the set $\mathcal{C}(a,b)$ is non-empty, and for every $\epsilon >0$ it contains a compact Cantor set K such that

$${\dim }_{\mathcal{H}}K\ge 1-\epsilon .$$

Consequently,

$${\dim }_{\mathcal{H}}\mathcal{C}(a,b)={\dim }_{\mathcal{P}}\mathcal{C}(a,b)=1.$$

Proof. 

First observe, using (A2), that, for $m\ge 1$, the tail block $\{n-m+1,\dots ,n\}$ carries a fixed positive proportion of the total weight. Forcing m consecutive symbols equal to $x_r=b$ makes the average $Y_n$ exponentially close to b, and similarly for $x_0=a$. In the following, we construct inductively a compact set $K\subset \Omega$ consisting of trajectories whose weighted averages oscillate between a and b. To this end, for each $j\ge 1$, we choose:

• a free block of length $m_j$, where $X_k$ may take any value in $\mathcal{B}$;
• a forced block of length $t_j$, where all symbols are fixed, equal either to $x_0=a$ or to $x_r=b$.

These two blocks are represented schematically as:

$$\underset{\mathrm{all}\mathrm{choices}\mathrm{allowed}}{\underbrace{\begin{array}{|c|}\hline \mathrm{free}\mathrm{block}\mathrm{of}\mathrm{length}{m}_j\\ \hline\end{array}}}⟶\underset{X_k=x_0\mathrm{or}{x}_r}{\underbrace{\begin{array}{|c|}\hline \mathrm{forced}\mathrm{block}\mathrm{of}\mathrm{length}{t}_j\\ \hline\end{array}}}.$$

The sequences $\left(m_j\right)$ and $\left(t_j\right)$ are chosen so that

$$m_j\to \infty ,t_j\to \infty ,{\displaystyle \frac{t_j}{m_j}}\to 0.$$

(7)

Thus, the free blocks generate the combinatorial branching, while the forced blocks control the oscillation of $Y_n$. Define,

$$M_J=\sum _{j=1}^J{m}_j,T_J=\sum _{j=1}^J{t}_j,L_J=M_J+T_J.$$

(8)

At stage J, the construction produces exactly $2^{M_J}$ cylinders of length $L_J$. For each j, we define the admissible blocks:

$$B_j^a=\left(\mathrm{free}\mathrm{block}\mathrm{of}\mathrm{length}{m}_j\right)\mathrm{followed}\mathrm{by}{t}_j\mathrm{copies}\mathrm{of}{x}_0=a,$$

$$B_j^b=\left(\mathrm{free}\mathrm{block}\mathrm{of}\mathrm{length}{m}_j\right)\mathrm{followed}\mathrm{by}{t}_j\mathrm{copies}\mathrm{of}{x}_r=b.$$

We then define

$$K=\left\{B_1^{{\sigma }_1}{B}_2^{{\sigma }_2}{B}_3^{{\sigma }_3}\cdots :{\sigma }_j\in \{a,b\}\right\}.$$

Let $N=L_j$ be the endpoint of the jth forced block. If ${\sigma }_j=b$, then the last $t_j$ symbols are all equal to $x_r=b$, and by (A2),

$$Y_N={\displaystyle \frac{{\sum }_{k=1}^{N-t_j}{w}_k{X}_k+b{\sum }_{k=N-t_j+1}^N{w}_k}{W_N}}=b+O\left({\phi }^{-t_j}\right).$$

Similarly, if ${\sigma }_j=a$, then

$$Y_N=a+O\left({\phi }^{-t_j}\right).$$

Since $t_j\to \infty$, the errors decay exponentially. Thus, every $\omega \in K$ satisfies

$$\underset{n\to \infty }{\lim \inf } Y_n\left(\omega \right)=a,\underset{n\to \infty }{\mathrm{lim\; sup}} Y_n\left(\omega \right)=b.$$

Therefore,

$$K\subset \mathcal{C}(a,b).$$

The Hausdorff dimension of K is computed using the standard formula for inhomogeneous Cantor constructions ([37], Chapter 4):

$${\dim }_{\mathcal{H}}K=\underset{J\to \infty }{\lim \inf }{\displaystyle \frac{\log \left(2^{M_J}\right)}{\log \left(2^{L_J}\right)}}=\underset{J\to \infty }{\lim \inf }{\displaystyle \frac{M_J}{L_J}}.$$

Using (7), we obtain

$${\displaystyle \frac{M_J}{L_J}}={\displaystyle \frac{M_J}{M_J+T_J}}⟶1.$$

Thus, ${\dim }_{\mathcal{H}}K\ge 1-\epsilon$ for all $\epsilon >0$. On the other hand, since the packing dimension satisfies ${\dim }_{\mathcal{P}}\left(\mathcal{C}(a,b)\right)\le 1$, it follows that

$${\dim }_{\mathcal{H}}\mathcal{C}(a,b)={\dim }_P\mathcal{C}(a,b)=1.$$

□

Proof of Theorem 1. 

Fix $x_0\le a<b\le x_r$ and $\epsilon >0$. The proof relies on constructing a Cantor-type subset $K\subset \mathcal{C}(a,b)$ using free blocks (which generate combinatorial branching) and forced blocks (which push the weighted average close to a or to b). To this end, choose sequences $\left(m_j\right)$ and $\left(t_j\right)$ satisfying (7). Block j consists of:

• a free part of length $m_j$ (all symbols in $\mathcal{B}$ allowed);
• a forced tail of length $t_j$ (symbols fixed, chosen to approximate a or b).

Recall (8) and consider the forced block of length $t_j$ ending at $L_j$:

$$\{L_j-t_j+1,\dots ,L_j\}.$$

For integers $m=0,\dots ,t_j$, define normalized partial tail sums

$$R_j\left(m\right)={\displaystyle \frac{{\sum }_{n=L_j-t_j+1}^{L_j-t_j+m}{w}_n}{{\sum }_{n=L_j-t_j+1}^{L_j}{w}_n}}.$$

Since $w_{n-k}/w_n\sim {\phi }^{-k}$ and the weights decay geometrically along the tail, the increments satisfy

$$R_j(m+1)-R_j\left(m\right)={\displaystyle \frac{w_{L_j-t_j+m+1}}{\sum w_n}}\sim (\phi -1){\phi }^{-(t_j-m)}⟶0$$

uniformly in m as $t_j\to \infty$. Thus ${\left(R_j\left(m\right)\right)}_{m=0}^{t_j}$ becomes asymptotically dense in $[0,1]$. Hence, for any $p\in [0,1]$ and any $\delta >0$, if $t_j$ is large enough, there exists $m_j\left(p\right)\in \{0,\dots ,t_j\}$ such that

$$|R_j\left(m_j\left(p\right)\right)-p|<\delta .$$

(9)

Let

$$p_j=\left\{\begin{array}{cc}a\hfill & \mathrm{if}\mathrm{block}j\mathrm{is}\mathrm{of}\mathrm{type}a,\hfill \\ b\hfill & \mathrm{if}\mathrm{block}j\mathrm{is}\mathrm{of}\mathrm{type}b.\hfill \end{array}\right.$$

Choose ${\delta }_j=1/\left(3j\right)$ in (9). Then, the forced tail of block j is chosen so that

$$\left|{\displaystyle \frac{{\sum }_{n=L_j-t_j+1}^{L_j}{w}_n{X}_n}{{\sum }_{n=L_j-t_j+1}^{L_j}{w}_n}}-p_j\right|<{\displaystyle \frac{1}{3j}}.$$

(10)

Meanwhile, the free part of length $m_j$ contributes

$${\displaystyle \frac{{\sum }_{n=L_{j-1}+1}^{L_j-t_j}{w}_n{X}_n}{{\sum }_{n=L_j-t_j+1}^{L_j}{w}_n}}⟶0,$$

(11)

because its total weight is $O\left({\phi }^{-t_j}\right)$ times the tail weight (by (A2) with $m=t_j$). Combining (10) and (11) gives

$$|Y_{L_j}-p_j|<{\displaystyle \frac{1}{j}}\mathrm{for}\mathrm{all}\mathrm{sufficiently}\mathrm{large}j.$$

(12)

Define

$$K=\left\{B_1^{{\sigma }_1}{B}_2^{{\sigma }_2}{B}_3^{{\sigma }_3}\cdots :{\sigma }_j\in \{a,b\}\right\}.$$

Using (12), we get

$$\underset{n\to \infty }{\lim \inf } Y_n\left(\omega \right)=a,\underset{n\to \infty }{\lim \sup }{Y}_n \left(\omega \right)=b,\forall \omega \in K,$$

so $K\subset \mathcal{C}(a,b)$. Finally, the computation of the Hausdorff and packing dimensions proceeds exactly as in the proof of Proposition 1. □

3. Approximation of ${\mathsf{T}}_{\mathbf{n}}$: Color Model

Beyond their classical algebraic and combinatorial features, Fibonacci sequences and their higher-order extensions naturally arise in probabilistic and number–theoretic settings. From a number–theoretic perspective, probabilistic constructions offer an effective framework for modeling Fibonacci-type linear recurrences, in which the arithmetic nature of the coefficients influences both the growth behavior and the statistical distribution of the sequences. Such an approach allows one to recover key arithmetic properties, including Binet-type formulas, the influence of dominant algebraic roots, and sharp asymptotic estimates, without relying on the explicit resolution of Vandermonde systems. Within this framework, the sequence (5) admits a natural interpretation in terms of stochastic processes, and in particular through connections with Markov chains, which have been widely explored in the literature [6,7,34,38].

In this section, using a probabilistic approach, we will prove Theorem 2 below, which provides an asymptotic approximation of $T_n$ for sufficiently large n. Let ${\left\{s_n\right\}}_{n\ge 1}$ denote the number of sequences of ones, twos, threes, and fours that sum to n. In particular, one has $s_1={\mathsf{T}}_1=1$, $s_2={\mathsf{T}}_2=2$, $s_3={\mathsf{T}}_3=4$, and $s_4={\mathsf{T}}_4=7$. Additionally, note that

$$s_n=s_{n-4}+s_{n-3}+s_{n-2}+s_{n-1},n\ge 5$$

which implies first that $s_n={\mathsf{T}}_n$ and then ${\mathsf{T}}_n$ can be combinatorially interpreted as the number of ways to tile a board of length $n-1$ using tiles of size 1, 2, 3 and 4 cells. To illustrate this, consider an infinite board with cells labeled $1,2,3,\dots$, where each cell is independently colored black (B), white (W), gray (G), or yellow (Y) with probability $1/4$ as shown in Figure 1. Moreover, any coloring of the first n cells has a probability of ${(1/4)}^n$.

An infinite tiling can be described as a repeating pattern of sequences composed of black, white, gray, and yellow cells, each of varying length. For example, the tiling in Figure 1 begins with a gray sequence of length 3, followed by a black sequence of length 2, another gray sequence of length 3, then a white sequence of length 1, a yellow sequence of length 4, a black sequence of length 2, and so on.

We define the random variable X given the position of the end of the first black string that is not a multiple of two or the first gray string that is not a multiple of three or the first yellow string that is not a multiple of four. For example, in Figure 1, we have $X=16$. Let $n\ge 1$, note that $\{X=n\}$ occurs if and only if:

1.

Cell n is covered by B, and cell $n+1$ is covered by $\overline{B}$ (a white, gray, or yellow cell), or cell n is covered by G, and cell $n+1$ is covered by $\overline{G}$ or Cell n is covered by Y, and cell $n+1$ is covered by $\overline{Y}$.

2.

Cells 1 through $n-1$ are covered using any combination of white cells, black double cells, gray triple cells, and yellow quadruple cells.

Let $n_G$ (resp. $n_Y$) denotes the number of last gray (resp; Yellow) tiles. It follows that

(i)

If tile number n is black, we will have ${\mathsf{T}}_{n-1}$ ways from 1 to $n-1.$

(ii)

If tile number n is gray and $n_G=3m+2$, for some $m\in \mathbb{N}$, we will have ${\mathsf{T}}_{n-2}$ ways from 1 to $n-2.$ However, if $n_G=3m+1$, we will have ${\mathsf{T}}_{n-1}$ ways from 1 to $n-1.$

(iii)

If tile number n is yellow and $n_Y=4m+3$, for some $m\in \mathbb{N}$, we will have ${\mathsf{T}}_{n-3}$ ways from 1 to $n-3.$ If $n_Y=4m+2$, we will have ${\mathsf{T}}_{n-2}$ ways from 1 to $n-2.$ However, if $n_Y=4m+1$, we will have ${\mathsf{T}}_{n-1}$ ways from 1 to $n-1.$

Since each irregular block occurs with strictly positive probability and the colorings are independent and identically distributed, the probability that no irregular block ever appears is zero. Consequently, $X<\infty$ almost surely. Moreover, $\mathbb{P}(X=1)={\displaystyle \frac{9}{16}}$, $\mathbb{P}(X=2)={\displaystyle \frac{15}{64}}$, $\mathbb{P}(X=3)={\displaystyle \frac{27}{256}}$, $\mathbb{P}(X=4)={\displaystyle \frac{51}{1024}}$, and for all $n\ge 5$ one has

$$\begin{array}{ccc}\hfill \mathbb{P}(X=n)& =& {\displaystyle \frac{{\mathsf{T}}_{n-1}}{4^{n-1}}}{\displaystyle \frac{1}{4}}{\displaystyle \frac{3}{4}}+{\displaystyle \frac{{\mathsf{T}}_{n-2}}{4^{n-2}}}{\displaystyle \frac{1}{4^2}}{\displaystyle \frac{3}{4}}+{\displaystyle \frac{{\mathsf{T}}_{n-1}}{4^{n-1}}}{\displaystyle \frac{1}{4}}{\displaystyle \frac{3}{4}}+{\displaystyle \frac{{\mathsf{T}}_{n-3}}{4^{n-3}}}{\displaystyle \frac{1}{4^3}}{\displaystyle \frac{3}{4}}+{\displaystyle \frac{{\mathsf{T}}_{n-2}}{4^{n-2}}}{\displaystyle \frac{1}{4^2}}{\displaystyle \frac{3}{4}}+{\displaystyle \frac{{\mathsf{T}}_{n-1}}{4^{n-1}}}{\displaystyle \frac{1}{4}}{\displaystyle \frac{3}{4}}\hfill \\ & =& {\displaystyle \frac{9{\mathsf{T}}_{n-1}+6{\mathsf{T}}_{n-2}+3{\mathsf{T}}_{n-3}}{4^{n+1}}}.\hfill \end{array}$$

It follows, since ${\sum }_{n\ge 3}\mathbb{P}(X=n)=1$, that

$$\begin{array}{ccc}\hfill \sum _{n=5}^{+\infty }\mathbb{P}(X=n)& =& \sum _{n=5}^{+\infty }{\displaystyle \frac{9{\mathsf{T}}_{n-1}+6{\mathsf{T}}_{n-2}+3{\mathsf{T}}_{n-3}}{4^{n+1}}}\hfill \\ & =& 1-\sum _{n=1}^4\mathbb{P}(X=n)={\displaystyle \frac{49}{1024}}.\hfill \end{array}$$

This directly yields Theorem 2.

Now, we tile an infinite board by independently placing cells, double cells, triple cells, and quadruple cells with probability $\mathsf{p},$${\mathsf{p}}^2$, ${\mathsf{p}}^3$, and ${\mathsf{p}}^4$, respectively, where $\mathsf{p}$ is defined in Theorem 3. In this model, the probability that a tiling begins with any particular length n of cells is ${\sqrt{\mathsf{p}}}^n.$ Let ${\varsigma }_n$ be the probability that a random tiling is breakable at cell $n,$ i.e, that a cell or double cell, triple cell or quadruple cells begins at cell $n.$ The example in Figure 1 is breakable at cells

$$3,5,8,9,13,16,\dots .$$

Since there are ${\mathsf{T}}_{n-1}$ different ways to tile the first $n-1$ cells,

$${\varsigma }_n={\mathsf{T}}_{n-1}{\mathsf{p}}^{n-1}.$$

(13)

For a tilling to be unbreakable at $n,$ it must be

(i)

breakable at $n-1$ followed by a double, triple or quadruple cell.

(i)

breakable at $n-2$ followed by a triple cell.

(iii)

breakable at $n-3$ followed by a quadruple cell.

Thus, for $n\ge 5,$ one has

$$1-{\varsigma }_n={\varsigma }_{n-1}{\mathsf{p}}^2+{\varsigma }_{n-1}{\mathsf{p}}^3+{\varsigma }_{n-1}{\mathsf{p}}^4+{\varsigma }_{n-2}{\mathsf{p}}^3+{\varsigma }_{n-3}{\mathsf{p}}^4.$$

(14)

Let $\varsigma ={\lim }_{n\to +\infty }{\varsigma }_n.$ Taking a limit in (14) gives

$$1-\varsigma =\varsigma {\mathsf{p}}^2+2\varsigma {\mathsf{p}}^3+2\varsigma {\mathsf{p}}^4.$$

Then, $\varsigma ={\left(1+{\mathsf{p}}^2+2{\mathsf{p}}^3+2{\mathsf{p}}^4\right)}^{-1}$ and ${\mathsf{T}}_n\sim \varsigma {\mathsf{p}}^{-n}.$

Remark 1. 

Most classical proofs of Tetranacci or Fibonacci sums use generating functions, recurrences, or combinatorial arguments. In particular, consider the generating function

$$G\left(x\right)=\sum _{n=0}^{\infty }{T}_n{x}^n={\displaystyle \frac{T_0+(T_1-T_0)x+(T_2-T_1-T_0)x^2+(T_3-T_2-T_1-T_0)x^3}{1-x-x^2-x^3-x^4}}.$$

We want to compute

$$\sum _{n=5}^{\infty }\mathbb{P}(X=n)=\sum _{n=5}^{\infty }{\displaystyle \frac{9T_{n-1}+6T_{n-2}+3T_{n-3}}{4^{n+1}}}={\displaystyle \frac{1}{4}}\sum _{n=5}^{\infty }{\displaystyle \frac{9T_{n-1}+6T_{n-2}+3T_{n-3}}{4^n}}.$$

Using the function G with $x=1/4$, shifting indices, and substituting $G\left(x\right)$, we obtain

$$\sum _{n=5}^{\infty }\mathbb{P}(X=n)={\displaystyle \frac{49}{1024}}.$$

4. Approximation of ${\mathsf{T}}_{\mathbf{n}}$: Graph Model

4.1. Markov Chain

A family of random variables $({\mathsf{X}}_0,{\mathsf{X}}_1,\dots )$ is referred to as a Markov chain if, given a specific value $x_{n_0}$, the variables $({\mathsf{X}}_0,{\mathsf{X}}_1,\dots ,{\mathsf{X}}_{n_0-1})$ and $({\mathsf{X}}_{n_0+1},\dots ,)$ are independent of each other. Consequently, a straightforward way to characterize a Markov chain is to define the distribution of ${\mathsf{X}}_{n-1}$ conditioned on the value of ${\mathsf{X}}_n$, for each n. Let us consider the following Markov chain:

1.

$B_1=B_2=B_3=B_4=0.$

2.

$\mathbb{P}(B_{i+1}=1|B_i=0)=p_1$, $\mathbb{P}(B_{i+1}=2|B_i=0)=p_2$, $\mathbb{P}(B_{i+1}=3|B_i=0)=p_3$ and $\mathbb{P}(B_{i+1}=0|B_i=0)=1-p_1-p_2-p_3$.

3.

$\mathbb{P}(B_{i+1}=0|B_i=1)=\mathbb{P}(B_{i+1}=1|B_i=2)=\mathbb{P}(B_{i+1}=2|B_i=3)=1$.

Specifically, we have the transitive probabilities shown in

Table 1. Calculation of $\mathbb{P}(B_i=b_i|B_{i-1}=b_{i-1})$.

${\mathit{b}}_{\mathit{i}-1}$	0	0	0	0	1	2	3
$b_i$	0	1	2	3	0	1	2
$\mathbb{P}(B_i=b_i|B_{i-1}=b_{i-1})$	$1-p_1-p_2-p_3$	$p_1$	$p_2$	$p_3$	1	1	1

, which can also be illustrated visually with a transition graph (Figure 2).

Notice that each Markov chain is irreducible (it can move from any state to any other state with positive probability) and aperiodic (as is any chain that contains a state with positive probability of moving back to itself) will have a limiting distribution. To this end, we have ${\displaystyle \underset{n\to \infty }{\lim }\mathbb{P}(B_n=k)}$ exists for all $k=0,1,2,3$. Moreover, one has

$$\left\{\begin{array}{cc}\mathbb{P}(B_{n+1}=1)\hfill & =p_1\mathbb{P}(B_n=0)+\mathbb{P}(B_n=2)\hfill \\ \hfill & \\ \mathbb{P}(B_{n+1}=0)\hfill & =(1-p_1-p_2-p_3)\mathbb{P}(B_n=0)+\mathbb{P}(B_n=1)\hfill \\ \hfill & \\ \mathbb{P}(B_{n+1}=2)\hfill & =p_2\mathbb{P}(B_n=0)+1-\mathbb{P}(B_n=0)-\mathbb{P}(B_n=1)-\mathbb{P}(B_n=2).\hfill \end{array}\right.$$

(15)

We denote ${\displaystyle \underset{n\to \infty }{\lim }\mathbb{P}(B_n=0)={\alpha }_0}$, ${\displaystyle \underset{n\to \infty }{\lim }\mathbb{P}(B_n=1)={\alpha }_1}$, and ${\displaystyle \underset{n\to \infty }{\lim }\mathbb{P}(B_n=2)={\alpha }_2}$; then, using (15), we get

$$\left\{\begin{array}{cc}{\alpha }_1\hfill & =p_1{\alpha }_0+{\alpha }_2\hfill \\ \hfill & \\ {\alpha }_0\hfill & =(1-p_1-p_2-p_3){\alpha }_0+{\alpha }_1\hfill \\ \hfill & \\ {\alpha }_2\hfill & =p_2{\alpha }_0+1-{\alpha }_0-(p_1{\alpha }_0+{\alpha }_2)-{\alpha }_2\hfill \end{array}\right.\mathrm{and}\mathrm{then}\left\{\begin{array}{cc}{\alpha }_1\hfill & =p_1{S}_{123}^{-1}{\alpha }_1+{\alpha }_2\hfill \\ \hfill & \\ {\alpha }_0\hfill & =S_{123}^{-1}{\alpha }_1\hfill \\ \hfill & \\ 3{\alpha }_2\hfill & =1+(p_2-p_1-1){\alpha }_0.\hfill \end{array}\right.$$

where $S_{123}=p_1+p_2+p_3$. It follows that

$$\left\{\begin{array}{cc}{\alpha }_1\hfill & =p_1{S}_{123}^{-1}{\alpha }_1+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}(p_2-p_1-1)S_{123}^{-1}{\alpha }_1\hfill \\ \hfill & \\ {\alpha }_0\hfill & =S_{123}^{-1}{\alpha }_1\hfill \\ \hfill & \\ 3{\alpha }_2\hfill & =1+(p_2-p_1-1)S_{123}^{-1}{\alpha }_1\hfill \end{array}\right.$$

$$\left\{\begin{array}{cc}{\alpha }_1\hfill & ={\displaystyle \frac{1}{3}}{\left(1-{\displaystyle \frac{2}{3}}{p}_1{S}_{123}^{-1}-{\displaystyle \frac{1}{3}}(p_2-1)S_{123}^{-1}\right)}^{-1}\hfill \\ \hfill & \\ {\alpha }_0\hfill & =S_{123}^{-1}{\alpha }_1\hfill \\ \hfill & \\ {\alpha }_2\hfill & ={\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}(p_2-p_1-1)S_{123}^{-1}{\alpha }_1.\hfill \end{array}\right.$$

(16)

We define the event $A_n=\left\{B_{n+1}=B_{n+2}=0\right\}$. Then,

$$\begin{array}{cc}\hfill \mathbb{P}\left(A_n\right)& =\mathbb{P}(B_{n+2}=0|B_{n+1}=0)\mathbb{P}(B_{n+1}=0)\hfill \\ \hfill & =(1-S_{123})\left((1-S_{123})\mathbb{P}(B_n=0)+\mathbb{P}(B_n=1)\right)\hfill \\ \hfill & ={(1-S_{123})}^2\mathbb{P}(B_n=0)+(1-S_{123})\mathbb{P}(B_n=1),\hfill \end{array}$$

(17)

which implies that ${\lim }_n\mathbb{P}\left(A_n\right)$ exists and depends on $p_1$, $p_2$, and $p_3$. More precisely, using (16), we have

$$\begin{array}{cc}\hfill \underset{n\to \infty }{\lim }\mathbb{P}\left(A_n\right)& ={(1-S_{123})}^2{\alpha }_0+(1-S_{123}){\alpha }_1.\hfill \end{array}$$

(18)

Example 1. 

Take $p_1=p_2=0.25$ and $p_3=0.125$ then, $S_{123}=0.625$ and from (17), we obtain ${\alpha }_1=0.2941,$${\alpha }_0=0.4705$, and ${\alpha }_2=0.2353$, which implies that

$$\mathbb{P}\left(A_n\right)=0.1406\mathbb{P}(B_n=0)+0.375\mathbb{P}(B_n=1).$$

Moreover, it follows, using (15), that

$$\left\{\begin{array}{cc}\mathbb{P}(B_{n+1}=0)\hfill & =0.375\mathbb{P}(B_n=0)+\mathbb{P}(B_n=1)\hfill \\ \hfill & \\ \mathbb{P}(B_{n+1}=1)\hfill & =0.25\mathbb{P}(B_n=0)+\mathbb{P}(B_n=2)\hfill \\ \hfill & \\ \mathbb{P}(B_{n+1}=2)\hfill & =1-0.75\mathbb{P}(B_n=0)-\mathbb{P}(B_n=1)-\mathbb{P}(B_n=2).\hfill \end{array}\right.$$

In Table 2, we will calculate the first few probabilities of the event $A_n$.

After only 14 steps in the chain, the probability $\mathbb{P}\left(A_n\right)$ reaches a point where it remains constant from step to step. However, one can compute this limiting value using (16) and obtain

$${\displaystyle \underset{n\to \infty }{\lim }\mathbb{P}\left(A_n\right)=0.1406{\alpha }_0+0.375{\alpha }_1=0.1764.}$$

Table 2. Calculation of $\mathbb{P}\left(A_n\right)$ for $p_1=p_2=0.25$ and $p_3=0.125$.

n	$\mathbb{P}({\mathit{B}}_{\mathit{n}}=0)$	$\mathbb{P}({\mathit{B}}_{\mathit{n}}=1)$	$\mathbb{P}({\mathit{B}}_{\mathit{n}}=2)$	$\mathbb{P}\left({\mathit{A}}_{\mathit{n}}\right)$
1–4	1	0	0	$0.1406$
5	$0.375$	$0.25$	0	$0.0525$
6	$0.3906$	$0.0934$	$0.4687$	$0.089$
7	$0.2398$	$0.5663$	$0.1449$	$0.246$
8	$0.6562$	$0.2048$	$0.1089$	$0.169$
9	$0.4508$	$0.2729$	$0.1941$	$0.1657$
10	$0.4419$	$0.3068$	$0.1949$	$0.1771$
11	$0.4725$	$0.3053$	$0.1668$	$0.1809$
12	$0.4824$	$0.2849$	$0.1735$	$0.1746$
13	$0.4658$	$0.2941$	$0.1787$	$0.1757$
14	$0.4687$	$0.2951$	$0.1787$	$0.1765$

Table 2. Calculation of $\mathbb{P}\left(A_n\right)$ for $p_1=p_2=0.25$ and $p_3=0.125$.

n	$\mathbb{P}({\mathit{B}}_{\mathit{n}}=0)$	$\mathbb{P}({\mathit{B}}_{\mathit{n}}=1)$	$\mathbb{P}({\mathit{B}}_{\mathit{n}}=2)$	$\mathbb{P}\left({\mathit{A}}_{\mathit{n}}\right)$
1–4	1	0	0	$0.1406$
5	$0.375$	$0.25$	0	$0.0525$
6	$0.3906$	$0.0934$	$0.4687$	$0.089$
7	$0.2398$	$0.5663$	$0.1449$	$0.246$
8	$0.6562$	$0.2048$	$0.1089$	$0.169$
9	$0.4508$	$0.2729$	$0.1941$	$0.1657$
10	$0.4419$	$0.3068$	$0.1949$	$0.1771$
11	$0.4725$	$0.3053$	$0.1668$	$0.1809$
12	$0.4824$	$0.2849$	$0.1735$	$0.1746$
13	$0.4658$	$0.2941$	$0.1787$	$0.1757$
14	$0.4687$	$0.2951$	$0.1787$	$0.1765$

Remark 2. 

We have

$$\mathbb{P}(B_1,\dots ,B_8)=(0,0,0,0,\mathbf{3},2,1,0,0,\mathbf{3},2)=p_3^2(1-p_1-p_2-p_3).$$

1. 

Whenever a 3 appears in the sequence $b_5,\dots ,b_n$, there is a probability factor of $p_3$. Following the 3, a 2 occurs with probability one, then a 1 follows the 2 with probability one, and finally, a 0 follows the 1 with probability one.

2. 

If we observe a 0 that is following another 0, the probability factor is $1-p_1-p_2-p_3$.

Let $s_3$, $s_2$, and $s_1$ represent the number of threes, twos and ones in $(b_5,\dots ,b_n,0,0)$, respectively. Then, the number of 0, denoted by $s_0$, is given by $s_0=n+2-4-(s_1+s_2+s_3)$. Furthermore, applying Remark 2 along with the assumption of the event $A_n:=\{B_{n+1}=B_{n+2}=0\}$, we have

(1) the number of 2 that are preceded by 3 is $s_3$ and then the number of 2 that preceded by 0 is $s_2-s_3$;

(2) the number of 1 that are preceded by a 2 is $s_2$ and then the number of 1 that preceded by 0 is $s_1-s_2$;

(3) the number of 0 that preceded by a 1 is $s_1$ and then the number of 0 that preceded by 0 is $n-2-(s_1+s_2+s_3)-s_1=n-2-s_2-s_3-2s_1.$

4.2. Distribution of $(B_1,B_2,\dots ,B_n)$

In this section, we will study the distribution of $(B_1,B_2,\dots ,B_n)$ conditioned on the event $A_n$. Notice, if $b_1=b_2=b_3=b_4=0$, that

$$\begin{array}{cc}\hfill & \mathbb{P}((B_1,\dots ,B_n)=(b_1,\dots ,b_n)|A_n)\hfill \\ \hfill & ={\displaystyle \frac{\mathbb{P}\left((B_1,\dots ,B_{n+2})=(b_1,\dots ,b_n,0,0)\right)}{\mathbb{P}\left(A_n\right)}}\hfill \\ \hfill & ={\displaystyle \frac{{\prod }_{i=5}^n\mathbb{P}\left(B_i=b_i|B_{i-1}=b_{i-1}\right)\mathbb{P}\left(B_{n+1}=0|B_n=b_n\right)\mathbb{P}\left(B_{n+2}=0|B_{n+1}=0\right)}{\mathbb{P}\left(A_n\right)}}\hfill \\ \hfill & ={\displaystyle \frac{p_3^{s_3}{p}_2^{s_2-s_3}{p}_1^{s_1-s_2}{(1-p_1-p_2-p_3)}^{n-2-s_2-s_3-2s_1}}{\mathbb{P}\left(A_n\right)}}.\hfill \end{array}$$

(19)

Proposition 2. 

Let $\mathsf{p}\in (0,1)$ the golden probability; that is, the unique solution of $g\left(\mathsf{p}\right):=1-\mathsf{p}-{\mathsf{p}}^2-{\mathsf{p}}^3-{\mathsf{p}}^4=0$. Take $p_1={\mathsf{p}}^2$, $p_2={\mathsf{p}}^3$ and $p_3={\mathsf{p}}^4$, we have

1. 

$\mathbb{P}\left(A_{n+4}\right)=\mathsf{p}\mathbb{P}\left(A_{n+3}\right)+{\mathsf{p}}^2\mathbb{P}\left(A_{n+2}\right)+{\mathsf{p}}^3\mathbb{P}\left(A_{n+1}\right)+{\mathsf{p}}^4\mathbb{P}\left(A_n\right)$, for all $n\ge 2.$

2. 

$\mathbb{P}((B_1,\dots ,B_n)=(b_1,\dots ,b_n)|A_n)={\displaystyle \frac{1}{{\mathsf{T}}_{n-1}}}$

and then

$$\underset{n\to \infty }{\lim }{\mathsf{p}}^{n-2}{\mathsf{T}}_{n-1}={\displaystyle \frac{1}{2}}{\displaystyle \frac{\mathsf{p}-{\mathsf{p}}^2-{\mathsf{p}}^3}{{(1-\mathsf{p})}^2}}.$$

Proof. 

First, observe that g has a unique solution in $(0,1)$ (since g is a continuous and decreasing function with $g\left(1\right){\lim }_{x\to 0^{+}}g\left(x\right)<0)$.

1.

Notice, for $k=0,1,\dots$, that

$$\mathbb{P}\left(A_{n+k}\right)=\mathbb{P}(B_{n+k+2}=0|B_{n+k+1}=0)\mathbb{P}(B_{n+k+1}=0)=\mathsf{p}\mathbb{P}(B_{n+k+1}=0).$$

(20)

Then

$$\begin{array}{ccc}\hfill \mathbb{P}\left(A_{n+4}\right)& =& \mathsf{p}\mathbb{P}(B_{n+5}=0)=\mathsf{p}\left(\mathsf{p}\mathbb{P}(B_{n+4}=0)+\mathbb{P}(B_{n+4}=1)\right)\hfill \\ & {=}^{\left(20\right)}& \mathsf{p}\left(\mathbb{P}(A_{n+3}=0)+\mathbb{P}(B_{n+4}=1)\right)\hfill \\ & =& \mathsf{p}\mathbb{P}(A_{n+3}=0)+\mathsf{p}\left({\mathsf{p}}^2\mathbb{P}(B_{n+3}=0)+\mathbb{P}(B_{n+3}=2)\right)\hfill \\ & {=}^{\left(20\right)}& \mathsf{p}\mathbb{P}(A_{n+3}=0)+{\mathsf{p}}^2\mathbb{P}(A_{n+2}=0)+{\mathsf{p}}^3\mathbb{P}(A_{n+1}=0)+{\mathsf{p}}^4\mathbb{P}\left(A_n\right).\hfill \end{array}$$

As a consequence, if we define the sequence ${\left\{{\Gamma }_n\right\}}_n$ by ${\Gamma }_0=1$ and ${\Gamma }_n={\displaystyle \frac{\mathbb{P}\left(A_{n+1}\right)}{{\mathsf{p}}^{n-1}}}$, for all $n\ge 1$. Then,

$$\begin{array}{ccc}\hfill {\Gamma }_{n+4}& =& {\displaystyle \frac{\mathbb{P}\left(A_{n+5}\right)}{{\mathsf{p}}^{n+3}}}={\displaystyle \frac{\mathbb{P}\left(A_{n+4}\right)}{{\mathsf{p}}^{n+2}}}+{\displaystyle \frac{\mathbb{P}\left(A_{n+3}\right)}{{\mathsf{p}}^{n+1}}}+{\displaystyle \frac{\mathbb{P}\left(A_{n+2}\right)}{{\mathsf{p}}^n}}+{\displaystyle \frac{\mathbb{P}\left(A_{n+1}\right)}{{\mathsf{p}}^{n-1}}}\hfill \\ & =& {\Gamma }_{n+3}+{\Gamma }_{n+2}+{\Gamma }_{n+1}+{\Gamma }_n,\hfill \end{array}$$

which implies that ${\left\{{\Gamma }_n\right\}}_n$ is Tetranacci sequence.

2.

Using (19) and $(p_1,p_2,p_3)=({\mathsf{p}}^2,{\mathsf{p}}^3,{\mathsf{p}}^4)$, we get

$$\begin{array}{ccc}\hfill \mathbb{P}((B_1,\dots ,B_n)& =(b_1,\dots ,b_n)|A_n)={\displaystyle \frac{{(1-p_1-p_2-p_3)}^{n-2}}{\mathbb{P}\left(A_n\right)}}\hfill & \\ \hfill & ·p_3^{s_3}{p}_2^{s_2-s_3}{p}_1^{s_1-s_2}{(1-p_1-p_2-p_3)}^{-s_2-s_3-2s_1}\hfill \\ \hfill & ={\displaystyle \frac{{(1-{\mathsf{p}}^2-{\mathsf{p}}^3-{\mathsf{p}}^4)}^{n-2}}{\mathbb{P}\left(A_n\right)}}{\left({\displaystyle \frac{1-{\mathsf{p}}^2-{\mathsf{p}}^3-{\mathsf{p}}^4}{\mathsf{p}}}\right)}^{-s_2-s_3-2s_1}\hfill \\ \hfill & ={\displaystyle \frac{{\mathsf{p}}^{n-2}}{\mathbb{P}\left(A_n\right)}}{\left({\displaystyle \frac{1-{\mathsf{p}}^2-{\mathsf{p}}^3-{\mathsf{p}}^4}{\mathsf{p}}}\right)}^{-s_2-s_3-2s_1}={\displaystyle \frac{{\mathsf{p}}^{n-2}}{\mathbb{P}\left(A_n\right)}},\hfill \end{array}$$

which does not depends of $s_0$, $s_1,s_2$, and $s_3$. Therefore, each of the elements of ${\Omega }_{n-1}$ with $b_n\notin \{2,3\}$ is equally likely to appear. We calculate the first values of $\mathbb{P}\left(A_n\right)$ in

Table 3. Calculation of $\mathbb{P}\left(A_n\right)$.

n	$\mathbb{P}({\mathit{B}}_{\mathit{n}}=0)$	$\mathbb{P}({\mathit{B}}_{\mathit{n}}=1)$	$\mathbb{P}({\mathit{B}}_{\mathit{n}}=2)$	$\mathbb{P}({\mathit{B}}_{\mathit{n}}=3)$	$\mathbb{P}\left({\mathit{A}}_{\mathit{n}}\right)=\mathsf{p}\mathbb{P}({\mathit{B}}_{\mathit{n}+1}=0)$
4	1	0	0	0	$p^2$
5	$\mathsf{p}$	$p^2$	${\mathsf{p}}^3$	${\mathsf{p}}^4$	$2p^3$
6	$2{\mathsf{p}}^2$	$2{\mathsf{p}}^3$	$2{\mathsf{p}}^4$	${\mathsf{p}}^5$	$4{\mathsf{p}}^4$
7	$4{\mathsf{p}}^3$	$4{\mathsf{p}}^4$	$3{\mathsf{p}}^5$	$2{\mathsf{p}}^6$	$8{\mathsf{p}}^5$
8	$8{\mathsf{p}}^4$	$7{\mathsf{p}}^5$	$6{\mathsf{p}}^6$	$4{\mathsf{p}}^7$	$15{\mathsf{p}}^6$

.

Since ${\mathsf{T}}_0={\mathsf{T}}_1=0$, ${\mathsf{T}}_2={\mathsf{T}}_3=1$, we obtain

$${\displaystyle \frac{\mathbb{P}\left(A_4\right)}{{\mathsf{p}}^2}}=1={\mathsf{T}}_3,{\displaystyle \frac{\mathbb{P}\left(A_5\right)}{{\mathsf{p}}^3}}=2={\mathsf{T}}_4,{\displaystyle \frac{\mathbb{P}\left(A_6\right)}{{\mathsf{p}}^4}}=4={\mathsf{T}}_5,\mathrm{and}{\displaystyle \frac{\mathbb{P}\left(A_7\right)}{{\mathsf{p}}^5}}=8={\mathsf{T}}_6.$$

Therefore

$${\mathsf{T}}_{n-1}={\displaystyle \frac{\mathbb{P}\left(A_n\right)}{{\mathsf{p}}^{n-2}}}.$$

Notice, under our hypothesis on $p_1,p_2$, and $p_3$, that $S_{123}=1-\mathsf{p}$ and the value of ${\alpha }_1$ in (16) is reduced to

$$\begin{array}{ccc}\hfill {\alpha }_1& =& {\displaystyle \frac{1}{3}}{\left[1-{\displaystyle \frac{2}{3}}{\displaystyle \frac{{\mathsf{p}}^2}{1-\mathsf{p}}}-{\displaystyle \frac{1}{3}}{\displaystyle \frac{{\mathsf{p}}^3-1}{1-\mathsf{p}}}\right]}^{-1}={\displaystyle \frac{1}{3}}{\left[-{\displaystyle \frac{2{\mathsf{p}}^2+{\mathsf{p}}^3-1+3\mathsf{p}-3}{3(1-\mathsf{p})}}\right]}^{-1}\hfill \\ & =& {\displaystyle \frac{\mathsf{p}-1}{{\mathsf{p}}^3+2{\mathsf{p}}^2+3\mathsf{p}-4}}.\hfill \end{array}$$

In addition, again by (16), we get

$$\left\{\begin{array}{cc}{\alpha }_1\hfill & ={\displaystyle \frac{\mathsf{p}-1}{{\mathsf{p}}^3+2{\mathsf{p}}^2+3\mathsf{p}-4}}\hfill \\ \hfill & \\ {\alpha }_0\hfill & ={\displaystyle \frac{1}{4-{\mathsf{p}}^3-2{\mathsf{p}}^2-3\mathsf{p}}}\hfill \\ \hfill & \\ {\alpha }_2\hfill & ={\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}{\displaystyle \frac{{\mathsf{p}}^3-{\mathsf{p}}^2-1}{4-{\mathsf{p}}^3-2{\mathsf{p}}^2-3\mathsf{p}}}.\hfill \end{array}\right.$$

Recall (18), we deduce that

$$\underset{n\to \infty }{\lim }{\mathsf{p}}^{n-2}{\mathsf{T}}_{n-1}={\displaystyle \frac{{\mathsf{p}}^2}{4-{\mathsf{p}}^3-2{\mathsf{p}}^2-3\mathsf{p}}}+{\displaystyle \frac{{\mathsf{p}}^2-\mathsf{p}}{{\mathsf{p}}^3+2{\mathsf{p}}^2+3\mathsf{p}-4}}={\displaystyle \frac{\mathsf{p}}{4-{\mathsf{p}}^3-2{\mathsf{p}}^2-3\mathsf{p}}}$$

$$\underset{n}{\lim }{\mathsf{p}}^{n-2}{\mathsf{T}}_{n-1}={\displaystyle \frac{1}{2}}(1-S_{123}){\left(1-p_1{S}_{123}^{-1}\right)}^{-1}{S}_{123}^{-1}={\displaystyle \frac{1}{2}}(S_{123}^{-1}-1){\left(1-p_1{S}_{123}^{-1}\right)}^{-1}.$$

By definition of $\mathsf{p}$, one has $S_{123}=1-\mathsf{p}$, and then

$$\underset{n\to \infty }{\lim }{\mathsf{p}}^{n-2}{\mathsf{T}}_{n-1}={\displaystyle \frac{1}{2}}{\displaystyle \frac{\mathsf{p}-{\mathsf{p}}^2-{\mathsf{p}}^3}{{(1-\mathsf{p})}^2}}.$$

□

Remark 3. 

The parameters $p_1={\mathsf{p}}^2$, $p_2={\mathsf{p}}^3$, and $p_3={\mathsf{p}}^4$ are chosen in accordance with the random tiling model, where the probability that the tiling starts with a tile of length k is proportional to ${\mathsf{p}}^k$. These transition probabilities therefore capture the probabilistic structure governing the tile-length distribution.

5. Recursive Polynomial Interpolation Associated with Fibonacci-Type Sequences

Over the last decade, there have been several developments regarding the Fibonacci sequence $\left(F_n\right)$ with $F_0=F_1=1$. In particular, several new approaches exist to construct polynomials based on the Fibonacci sequence. One can cite, for instance, Fibonacci coefficient polynomials [39], which are generated by putting Fibonacci sequence as the coefficient, and Fibonacci polynomials [40], which are generated recursively, similar to the Fibonacci sequence. A novel approach can be devised to construct a polynomial using interpolation techniques. The proposed methodology involves identifying the Fibonacci points $\left\{(n,F_n),n=0,1,\dots \right\}$ and subsequently devising a polynomial that intersects these points through various interpolation techniques, such as Lagrange and Newton interpolations, as discussed in [41,42]. In this section, we present an independent but complementary contribution to the present work. Specifically, this section explores an algebraic viewpoint, showing how consecutive values of Fibonacci-type sequences can be reconstructed via recursive Lagrange interpolation, thereby highlighting the versatility of these sequences across different mathematical frameworks.

First, observe that in [41], for all $n\ge 1$, we have

$$F_n={\displaystyle \frac{1}{\sqrt{5}}}({\sigma }^{n+1}-{\phi }^{n+1})$$

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where $\sigma ={\displaystyle \frac{1+\sqrt{5}}{2}}$ and $\phi ={\displaystyle \frac{1-\sqrt{5}}{2}}$. Let

$$\{A_n:=(n,{(-1)}^n{F}_n),n=0,1,\dots \},$$

we prove that the Lagrange interpolation can be obtained implicitly and recursively. The introduction of the alternating factor ${(-1)}^n$ in the interpolation points is motivated by two main reasons. First, it yields a particularly simple recursive relation for the polynomials $P_n$, namely

$$P_{n+1}\left(x\right)-P_n\left(x\right)={\alpha }_{n}x(x-1)\dots (x-n).$$

Second, this choice allows the combinatorial identity

$$\sum _{i=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right)F_i=F_{2n}$$

to be used in its most transparent form when computing the leading coefficient. Our first main result in this section is the following.

Theorem 4. 

The Lagrange interpolation of order n of is given

$${\mathbb{P}}_n\left(x\right)={\displaystyle \frac{{(-1)}^n}{n!}}\sum _{i=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right)F_i\prod _{j\ne i}^n(x-j),$$

where $\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right)={\displaystyle \frac{n!}{i!(n-i)!}}.$ Moreover, the leading coefficient ${\alpha }_n$ of ${\mathbb{P}}_n\left(x\right)$ is ${\displaystyle \frac{{(-1)}^n}{n!}}{F}_{2n}$.

Remark 4. 

1. 

Observe, from the proof of Theorem 4, that

$$\sum _{i=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right)F_i=F_{2n},(n\ge 0).$$

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It follows that

$$\begin{array}{cc}\hfill {\mathbb{P}}_n(n+1)& ={\displaystyle \frac{{(-1)}^n}{n!}}\sum _{i=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right)F_i\prod _{j\ne i}^n(n+1-j)\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^n}{n!}}\sum _{i=0}^n{\displaystyle \frac{n!}{i!(n-i)!}}{\displaystyle \frac{(n+1)!}{n+1-i}}{F}_i={(-1)}^n\sum _{i=0}^n{\displaystyle \frac{(n+1)!}{i!(n+1-i)!}}{F}_i\hfill \\ \hfill & ={(-1)}^n\sum _{i=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{i}{n+1}}\right)F_i{=}^{\left(\right)}{(-1)}^n\left(F_{2n+2}-F_{n+1}\right).\hfill \end{array}$$

This implies that

$${\mathbb{P}}_{n+1}(n+1)-{\mathbb{P}}_n(n+1)={(-1)}^{n+1}{F}_{n+1}-{(-1)}^n\left(F_{2n+2}-F_{n+1}\right)={(-1)}^{n+1}{F}_{2n+2}.$$

2. 

Using Binet’s formula, $F_{2n}=({\sigma }^{2n+1}-{\phi }^{2n+1})/\sqrt{5}$. Since, for large n, we have ${\phi }^{2n+1}\approx 0$, we can deduce from the asymptotic behavior that the leading coefficient of ${\mathbb{P}}_n\left(x\right)$ is

$${\alpha }_n\sim {\displaystyle \frac{{(-1)}^n}{\sqrt{5}n!}}{\sigma }^{2n+1}.$$

Thus, the leading coefficient grows essentially like ${\sigma }^{2n+1}/n!$, showing that the golden ratio directly governs the exponential growth rate of the polynomial’s leading term.

Theorem 5. 

There exists, for each $n\in \mathbb{N}$, a polynomial function $Q_n$ of degree n such that

$${\mathbb{P}}_{n+1}\left(x\right)=1+\sum _{i=1}^n{\displaystyle \frac{{(-1)}^{i+1}}{(i+1)!}}{F}_{2i+2}{Q}_i\left(x\right).$$

Proof. 

Observe that ${\mathbb{P}}_{n+1}$ and ${\mathbb{P}}_n$ pass through the $n+1$ same points: $A_0,A_1,\dots ,A_n$, where $A_i=(i,{(-1)}^i{F}_i)$. Consequently, we can write a relation

$${\mathbb{P}}_{n+1}\left(x\right)-{\mathbb{P}}_n\left(x\right)={\alpha }_{n}x(x-1)\dots (x-n):={\alpha }_n{Q}_n\left(x\right),$$

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where ${\alpha }_n={\displaystyle \frac{{(-1)}^{n+1}}{(n+1)!}}{F}_{2n+2}$ is the leading coefficient of ${\mathbb{P}}_{n+1}$. Hence, we may deduce an implicit formula as follows

$$\begin{array}{cc}\hfill {\mathbb{P}}_{n+1}\left(x\right)& ={\mathbb{P}}_n\left(x\right)+{\displaystyle \frac{{(-1)}^{n+1}}{(n+1)!}}{F}_{2n+2}{Q}_n\left(x\right)\hfill \\ \hfill & ={\mathbb{P}}_1\left(x\right)+\sum _{i=1}^n{\displaystyle \frac{{(-1)}^{i+1}}{(i+1)!}}{F}_{2i+2}{Q}_i\left(x\right)\hfill \\ \hfill & =1+\sum _{i=1}^n{\displaystyle \frac{{(-1)}^{i+1}}{(i+1)!}}{F}_{2i+2}{Q}_i\left(x\right).\hfill \end{array}$$

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As an example, from Theorem 5, we can derive $P_n\left(x\right)$ for $n=1,2$ as follows (see Figure 3 and Figure 4).

$$\begin{array}{ccc}{\mathbb{P}}_1\left(x\right)& =& 1-2x\\ {\mathbb{P}}_2\left(x\right)& =& {\displaystyle \frac{5}{2}}{x}^2-{\displaystyle \frac{9}{2}}x+1.\end{array}$$

6. Conclusions

In this work, we investigated probabilistic, geometric, and algebraic aspects of Fibonacci-type sequences. The main contribution of this paper is Theorem 1, which shows that the assumption of exponential growth of the weights suffices to generate irregular sets of full Hausdorff and packing dimension for weighted random averages. We further illustrated this general mechanism through a probabilistic analysis of the Tetranacci sequence. Using color models and Markov chain representations, we derived sharp asymptotic estimates for ${\mathsf{T}}_n$ and established its exponential growth behavior without resorting to explicit Binet-type formulas. Finally, we introduced an independent but complementary algebraic contribution based on recursive Lagrange interpolation, showing that consecutive values of Fibonacci-type sequences admit an implicit polynomial reconstruction. Several questions remain open and deserve further investigation:

1.

A natural problem is to determine whether the methods developed in this work can be extended to more general recursive sequences, such as generalized Lucas sequences or k-step Fibonacci sequence ${\mathsf{T}}_{k,n}$. In particular, it would be of interest to study whether similar interpolation schemes and probabilistic representations can be constructed in this context and whether there are suitable Markov or branching structures to describe their growth and limiting behavior.

2.

The color and Markov models introduced in this work may also be viewed as potential tools for stochastic simulation. In particular, their connection with Fibonacci-type recursions suggests possible applications to Monte Carlo methods and pseudo-random number generation, which we leave for future investigation.