New Types of Distance Padovan Sequences via Decomposition Technique

Abstract

In this paper, we introduce new kinds of generalized Padovan sequences and study their properties using number decomposition techniques. In particular, we consider three types of generalized Padovan sequences defined by the same recurrence equation with distinct initial conditions which follows from special number decomposition. Using the number decomposition method, we give their mutual relations and direct binomial formulas for considered sequences. Moreover, we give some combinatorial properties of these sequences and also define their matrix generators.

1. Introduction and Preliminary Results

Let $k\ge 1$ be an integer. Sequences defined by the k-th order linear homogeneous recurrence equation of the form

$$a_n=b_1{a}_{n-1}+b_2{a}_{n-2}+\dots +b_k{a}_{n-k},\mathrm{for}n\ge k,$$

(1)

where coefficients $b_j\ge 0$ for $j\in \{1,\dots ,k-1\}$ and $b_k>0$ are integers, with fixed integers $a_i,i\in \{0,\dots ,k-1\}$, are called Fibonacci-type (or Fibonacci-like) sequences, and their elements are called numbers of the Fibonacci type (or Fibonacci-like), respectively. We denote by $\mathcal{F}$ the family of Fibonacci-type sequences defined by the Equation (1) and by $F(a_i;b_j)$ for $i\in \{0,\dots ,k-1\}$ and $j\in \{1,\dots ,k\}$ the sequences belonging to $\mathcal{F}$. The family $\mathcal{F}$ includes a number of well-known sequences that have been intensively studied in the literature. Below, we list some classical Fibonacci-type sequences defined by the second- and the third-order linear recursions. The identification number of each sequence coming from the On-line Encyclopedia of Integer Sequences, which is an online database of integer sequences [1], is given in brackets.

• $F(1,1;1,1)$—Fibonacci sequence (A000045).
• $F(2,1;1,1)$—Lucas sequence (A000032).
• $F(0,1;2,1)$—Pell sequence (A000129).
• $F(2,2;2,1)$—Pell–Lucas sequence (A0002203).
• $F(0,1;1,2)$—Jacobsthal sequence (A001045).
• $F(2,1;1,2)$—Jacobsthal–Lucas sequence (A014551).
• $F(1,1,1;1,0,1)$—Narayana sequence (A000930).
• $F(1,1,1;0,1,1)$—Padovan sequence (A000931).
• $F(3,0,2;0,1,1)$—Perrin sequence (A001608).

For other well-known sequences from $\mathcal{F}$ defined by the second- or the third-order recurrence equation, see, for example, [2]. Sequences from the family $\mathcal{F}$ are widely studied in many areas of modern research from different points of view, for example, in medicine [3], theoretical physics [4], particularly physics of the high energy [5,6,7], engineering, computer science [8,9], architecture, nature, arts, [10,11], and others. For historical background issues of the particular cases of Fibonacci-type sequences, see [12]. The family $\mathcal{F}$ includes a special subfamily of distance Fibonacci-type sequences of the form $F(a_i;b_j)$, where $i\in \{1,\dots ,k-1\}$, $j\in \{1,\dots ,k\}$, $b_k\ne 0$, $b_p\ne 0$ for a fixed p such that $1\le p\le k-1$, and $b_j=0$ for $j\in \{1,\dots ,k\}\backslash \{p,k\}$. In other words, the subfamily of distance Fibonacci-type sequences consists of sequences that are defined by the k-th-order recurrence equation, and each term of such a sequence is the sum of two terms, such that one of them is at the distance k from n, for an arbitrary $k\ge 2$. For convenience, we will write $F(a_i;b_p=n_p,b_k=n_k)$ for this subfamily. In this subfamily, the most interesting are sequences that generalize the well-known classical sequences. These sequences are intensively studied in different contexts, for example, for their combinatorial properties, graph interpretations, or hypercomplex number theory. Below, we recall some distance Fibonacci-type sequences related to the main topic of this paper.

• $F(a_i=i+1;b_1=b_k=1)$, $i\in \{0,\dots ,k-1\}$, $k\ge 2$ (M.Kwaśnik et al. [13]).
• $F(a_i=1;b_{k-1}=b_k=1)$, $i\in \{0,\dots ,k-1\}$, $k\ge 2$ (U.Bednarz et al. [14]).
• $F(a_i=1;b_2=b_k=1)$, $i\in \{0,\dots ,k-1\}$, $k\ge 1$ (I.Włoch et al. [15]).
• $F(a_i=1;b_3=b_k=1)$, $i\in \{0,\dots ,k-1\}$, $k\ge 1$ (E.Özkan et al. [16]).

The above sequences generalize Fibonacci sequence, Narayana sequence, Padovan sequence, and others.

In this paper, being a sequel of the paper [16], we introduce a subfamily $F(a_i;b_3=b_k=1)$, where $i\in \{0,\dots k-1\}$, generated by special decompositions of a number n. Sequences of this subfamily generalize the classical Padovan sequence $F(1,1,1;0,1,1)$ and Narayana sequence $F(1,1,1;1,0,1)$ simultaneously. It is worth mentioning that terms of Padovan sequence called Padovan numbers, usually denoted by $Pv\left(n\right)$, were discovered in 1924 by a French student of architecture, G. Cordonnier, and independently by a Dutch Benedictine monk and architect, D.H. van der Laan. The name Padovan numbers was used by I. Steward in honor of the contemporary architect Richard Padovan, who studied and popularized the plastic number well known as the following limit:

$$\underset{n\to \infty }{lim}\frac{Pv(n+1)}{Pv\left(n\right)}=1,3247\dots .$$

(2)

For historical details, see, for example, [17]. In the literature, we can find many generalizations of Padovan numbers in different contexts. We can meet them, for example, in the graph theory, as a result of counting sets of different types [18,19], in combinatorics [20], in the theory of polynomials [21], and in cryptography, where properties of such sequences and new matrix generations used in encoding and decoding algorithms are particularly desired; see, for example, [22,23]. In this paper, we consider three types of generalized distance Padovan numbers based on the same recurrence equation with distinct initial conditions that follow from special number decompositions. We derive direct formulas, prove some properties of the considered numbers, reveal their connection with Pascal’s triangle, and define new matrix generators for them.

In [16], Özkan et al. introduced and studied distance Padovan numbers $F_3(k,n)$ defined for integers $k\ge 1$ and $n\ge 0$ in the following way:

$$F_3(k,n)=F_3(k,n-3)+F_3(k,n-k),\mathrm{for}n\ge max\{3,k\}$$

(3)

with $F_3(k,i)=1$, for $i\in \{0,\dots ,max\{3,k\}-1\}$.

Based on this idea we introduce other kinds of distance Padovan numbers as follows.

Let $k\ge 1$, $n\ge 0$ be integers. Then, for a fixed $j\in \{1,2,3\}$, distance Padovan numbers $F_3^{\left(j\right)}(k,n)$ of the j-th kind are given by the recursion

$$F_3^{\left(j\right)}(k,n)=F_3^{\left(j\right)}(k,n-3)+F_3^{\left(j\right)}(k,n-k)\mathrm{for}n\ge max\{3,k\}$$

(4)

with the following initial conditions $F_3^{\left(j\right)}(k,i)=a_i^{\left(j\right)}$ for $i\in \{0,1,\dots ,max\{3,k\}-1\}$

$$F_3^{\left(1\right)}(k,i)=a_i^{\left(1\right)}=\left\{\begin{array}{cc}1\hfill & \mathrm{for}i=3p,p\ge 0\mathrm{or}k=1\mathrm{or}k=i=2,\hfill \\ 0\hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

(5)

$$F_3^{\left(2\right)}(k,i)=a_i^{\left(2\right)}=1,$$

(6)

$$F_3^{\left(3\right)}(k,i)=a_i^{\left(3\right)}=\left\{\begin{array}{cc}0\hfill & \mathrm{for}i=3p,p\ge 0\mathrm{or}k=i=2,\hfill \\ 1\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

(7)

Clearly, $F_3^{\left(2\right)}(k,n)=F_3(k,n)$. Moreover, $F_3^{\left(1\right)}(1,n)=F_3^{\left(2\right)}(1,n)=F_3^{\left(3\right)}(1,n)$ is Narayana number and for $k=2$ we have $F_3^{\left(1\right)}(2,n+2)=F_3^{\left(2\right)}(2,n)=F_3^{\left(3\right)}(2,n+3)=Pv\left(n\right)$.

Table 1, Table 2 and Table 3 present numbers $F_3^{\left(j\right)}(k,n)$, where $j\in \{1,2,3\}$ for some initial values of k and n.

Table 1. Numbers $F_3^{\left(1\right)}(k,n)$ for $k\in \{1,2,\dots ,6\}$.

n	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
$F_3^{\left(1\right)}(1,n)$	1	1	1	2	3	4	6	9	13	19	28	41	60	88	129	189
$F_3^{\left(1\right)}(2,n)$	1	0	1	1	1	2	2	3	4	5	7	9	12	16	21	28
$F_3^{\left(1\right)}(3,n)$	1	0	0	2	0	0	4	0	0	8	0	0	16	0	0	32
$F_3^{\left(1\right)}(4,n)$	1	0	0	1	1	0	1	2	1	1	3	3	2	4	6	5
$F_3^{\left(1\right)}(5,n)$	1	0	0	1	0	1	1	0	2	1	1	3	1	3	4	2
$F_3^{\left(1\right)}(6,n)$	1	0	0	1	0	0	2	0	0	3	0	0	5	0	0	8

Table 2. Numbers $F_3^{\left(2\right)}(k,n)$ for $k\in \{1,2,\dots ,6\}$.

n	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
$F_3^{\left(2\right)}(1,n)$	1	1	1	2	3	4	6	9	13	19	28	41	60	88	129	189
$F_3^{\left(2\right)}(2,n)$	1	1	1	2	2	3	4	5	7	9	12	16	21	28	37	49
$F_3^{\left(2\right)}(3,n)$	1	1	1	2	2	2	4	4	4	8	8	8	16	16	16	32
$F_3^{\left(2\right)}(4,n)$	1	1	1	1	2	2	2	3	4	4	5	7	8	9	12	15
$F_3^{\left(2\right)}(5,n)$	1	1	1	1	1	2	2	2	3	3	4	5	5	7	8	9
$F_3^{\left(2\right)}(6,n)$	1	1	1	1	1	1	2	2	2	3	3	3	5	5	5	8

Table 3. Numbers $F_3^{\left(3\right)}(k,n)$ for $k\in \{1,2,\dots ,6\}$.

n	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
$F_3^{\left(3\right)}(1,n)$	1	1	1	2	3	4	6	9	13	19	28	41	60	88	129	189
$F_3^{\left(3\right)}(2,n)$	0	1	0	1	1	1	2	2	3	4	5	7	9	12	16	21
$F_3^{\left(3\right)}(3,n)$	0	1	1	0	2	2	0	4	4	0	8	8	0	16	16	0
$F_3^{\left(3\right)}(4,n)$	0	1	1	0	1	2	1	1	3	3	2	4	6	5	6	10
$F_3^{\left(3\right)}(5,n)$	0	1	1	0	1	1	1	2	1	2	3	2	4	4	4	7
$F_3^{\left(3\right)}(6,n)$	0	1	1	0	1	1	0	2	2	0	3	3	0	5	5	0

Now we give generating functions ${\sum }_{n=0}^{\infty }{F}_3^{\left(j\right)}(k,n)x^n$ of the sequence $\left\{F_3^{\left(j\right)}(k,n)\right\}$. Note that the generating function for $\left\{F_3^{\left(2\right)}(k,n)\right\}$ was given by Özkan et al. in [16].

Theorem 1.

[16] Let $k\ge 1$, $n\ge 0$ be integers. The generating function of $\left\{F_3^{\left(2\right)}(k,n)\right\}$ has the following form:

$$g^{\left(2\right)}(k,x)=\frac{1+t}{1-x^3-x^k},\mathit{where}t=\left\{\begin{array}{cc}0\hfill & \mathit{for}k=1,\hfill \\ x\hfill & \mathit{for}k=2,\hfill \\ x+x^2\hfill & \mathit{for}k\ge 3.\hfill \end{array}\right.$$

(8)

Theorem 2.

Let $k\ge 1$, $n\ge 0$ be integers. The generating function $g^{\left(j\right)}(k,x)$ of the sequence $\left\{F_3^{\left(j\right)}(k,n)\right\}$, for $j\in \{1,3\}$ has the following form:

$$g^{\left(1\right)}(k,x)=\frac{1}{1-x^3-x^k},$$

(9)

$$g^{\left(3\right)}(k,x)=\frac{x+x^2}{1-x^3-x^k}.$$

(10)

Proof. 

Let $g^{\left(j\right)}(k,x)={\displaystyle \sum _{n=0}^{\infty }}{F}_3^{\left(j\right)}(k,n)x^n$. Using the definition (4), we have

$g^{\left(j\right)}(k,x)-x^3{g}^{\left(j\right)}(k,x)-x^k{g}^{\left(j\right)}(k,x)=t$, where $t=\left\{\begin{array}{cc}1\hfill & \mathrm{for}{F}_3^{\left(1\right)}(k,n),\hfill \\ x+x^2\hfill & \mathrm{for}{F}_3^{\left(3\right)}(k,n),k\ge 2.\hfill \end{array}\right.$ Hence $g^{\left(j\right)}(k,x)=\frac{t}{1-x^3-x^k}$, for $j\in \{1,3\}$, which ends the proof. □

2. Number Decompositions

Now we give an interpretation of distance Padovan numbers $F_3^{\left(j\right)}(k,n)$ for $j\in \{1,2,3\}$ and $k\ne 3$ with respect to a special kind of decompositions of a number n. For combinatorial concepts not defined here, see [24]. Recall that for $n\ge 1$ being an integer, a decomposition of the number n is its ordered partition. In other words, a decomposition of n is a t-tuple $(n_1,\dots ,n_t)$, where $1\le n_i\le n$, $i\in \{1,\dots ,t\}$, $t\ge 1$, and ${\displaystyle \sum _{i=1}^t}{n}_i=n.$ For example, $\left(4\right),(3,1),(1,3)(2,2),(1,1,2),(1,2,1),(2,1,1),(1,1,1,1)$ are different decompositions of the number 4.

Let $k\ge 1$, $k\ne 3$, and $n\ge k-1$ be integers. A t-tuple $(n_1,\dots ,n_t)$ is called a $(3,k)$-decomposition of the number n if $n_i\in \{3,k\}$ for $i\in \{1,\dots ,t\}$. We say that $(n_1,\dots ,n_t)$ is a ${(3,k)}_{r^{+}}$-decomposition of the number n if $n_i\in \{3,k\}$ for $i\in \{1,\dots ,t-1\}$, and $n_t=r$, where $1\le r\le min\{3,k\}-1$. Similarly, we say that $(n_1,\dots ,n_t)$ is a ${(3,k)}_{r^{-}}$-decomposition of the number n if $n_1=r$, and $n_i\in \{3,k\}$ for $i\in \{2,\dots ,t\}$.

Let us introduce the following decompositions’ parameters:

• ${\gamma }_0(3,k,n)$ - the number of all $(3,k)$-decompositions;
• ${\gamma }_{r^{-}}(3,k,n)$ - the number of all ${(3,k)}_{r^{-}}$-decompositions;
• ${\gamma }_{r^{+}}(3,k,n)$ - the number of all ${(3,k)}_{r^{+}}$-decompositions;
• ${\gamma }_{\le r^{-}}(3,k,n)$ - the number of all $(3,k)$ and all ${(3,k)}_{s^{-}}$-decompositions,
  where $1\le s\le r$;
• ${\gamma }_{\le r^{+}}(3,k,n)$ - the number of all $(3,k)$ and all ${(3,k)}_{s^{+}}$-decompositions,
  where $1\le s\le r$;
• $\gamma (3,k,n)$ - the number of all $(3,k)$, ${(3,k)}_{r^{-}}$ and ${(3,k)}_{r^{+}}$-decomposition,
  where $1\le r\le min\{3,k\}-1$.

Clearly,

• ${\gamma }_{r^{-}}(3,k,n)={\gamma }_{r^{+}}(3,k,n)$,
• ${\gamma }_{\le r^{-}}(3,k,n)={\gamma }_{\le r^{+}}(3,k,n)$.

Moreover, if $k>3,r=min\{3,k\}-1$, then

• ${\gamma }_{\le r^{-}}(3,k,n)={\gamma }_0(3,k,n)+{\displaystyle \sum _{r=1}^2}{\gamma }_{r^{-}}(3,k,n)$,
• ${\gamma }_{\le r^{+}}(3,k,n)={\gamma }_0(3,k,n)+{\displaystyle \sum _{r=1}^2}{\gamma }_{r^{+}}(3,k,n)$,
• $\gamma (3,k,n)={\gamma }_0(3,k,n)+{\displaystyle \sum _{r=1}^2}({\gamma }_{r^{-}}(3,k,n)+{\gamma }_{r^{+}}(3,k,n))$.

Theorem 3.

Let $k\ge 1$, $k\ne 3$, $n\ge 1$ be integers and $r=min\{3,k\}-1$. Then

(i) 

${\gamma }_0(3,k,n)=F_3^{\left(1\right)}(k,n)$,

(ii) 

${\gamma }_{\le r^{-}}(3,k,n)={\gamma }_{\le r^{+}}(3,k,n)=F_3^{\left(2\right)}(k,n)$,

(iii) 

${\gamma }_{\le r^{-}}(3,k,n)-{\gamma }_0(3,k,n)=F_3^{\left(3\right)}(k,n)$,

(iv) 

$\gamma (3,k,n)=F_3^{\left(1\right)}(k,n)+2F_3^{\left(2\right)}(k,n)$.

Proof. 

(i) (by induction on n) At the beginning, we prove that the equality (i) holds for $n\in \{1,\dots ,max\{3,k\}-1\}$. If $k=1$, then $n\in \{1,2\}$, and consequently we obtain the unique $(3,1)$-decomposition of n of the form $\left(1\right)$ or $(1,1)$, respectively. Thus, ${\gamma }_0(3,1,1)=F_3^{\left(1\right)}(1,1)$, and ${\gamma }_0(3,1,2)=F_3^{\left(1\right)}(1,2).$ If $k=2$, then there is no $(3,2)$-decomposition of $n=1$ and there is the unique $(3,2)$-decomposition of $n=2$ of the form $\left(2\right)$, so ${\gamma }_0(3,2,1)=F_3^{\left(1\right)}(2,1)$, and ${\gamma }_0(3,1,2)=F_3^{\left(1\right)}(2,2)$. Assume now that $k>3$. If $1\le n\le max\{3,k\}-1$, then ${\gamma }_0(3,k,n)=1$ if $n=3p$, $p\ge 1$, otherwise ${\gamma }_0(3,k,n)=0$. Hence, ${\gamma }_0(3,k,n)=F_3^{\left(1\right)}(k,n)$ for initial values $1\le n\le max\{3,k\}-1$.

Let $n\ge max\{3,k\}$ and suppose that $\left(i\right)$ is true for an arbitrary $m<n$. We shall show that ${\gamma }_0(3,k,n)=F_3^{\left(1\right)}(k,n).$ Let us consider a t-tuple $(n_1\dots ,n_t)$, $t\ge 1$ being a $(3,k)$-decomposition of n. If $n_t=k$, then $n={\displaystyle \sum _{i=1}^{t-1}}{n}_i+k$, and so $n-k={\displaystyle \sum _{i=1}^{t-1}}{n}_i.$ By induction hypothesis there are $F_3^{\left(1\right)}(k,n-k)$ distinct $(3,k)$-decompositions of n with $n_t=k$. If $n_t=3$, then by analogy there are $F_3^{\left(1\right)}(k,n-3)$ distinct $(3,k)$-decompositions of n with $n_t=3$. Finally, we obtain ${\gamma }_0(3,k,n)=F_3^{\left(1\right)}(k,n-k)+F_3^{\left(1\right)}(k,n-3)=F_3^{\left(1\right)}(k,n)$, and the result follows.

In the same way, we prove $\left(ii\right)$–$\left(iv\right)$. □

Decomposition interpretation of distance Padovan numbers can be used for finding binomial formulas for $F_3^{\left(j\right)}(k,n),$ where $j\in \{1,2,3\}.$ Note that for $F_3^{\left(2\right)}(k,n)$ the binomial formula was discovered by Özkan et al. in [16].

Theorem 4.

[16] Let $k\ge 1$, $n\ge 0$ be integers. Then

$$F_3^{\left(2\right)}(k,n+t)=\sum _{i=0}^{\lfloor \frac{n}{k}\rfloor }\left(\genfrac{}{}{0pt}{}{i+\lfloor \frac{n-ik}{3}\rfloor }{i}\right),\mathit{where}t=\left\{\begin{array}{cc}2\hfill & \mathit{for}k=1,\hfill \\ 1\hfill & \mathit{for}k=2,\hfill \\ 0\hfill & \mathit{for}k\ge 3.\hfill \end{array}\right.$$

(11)

Below, we give a binomial formula for $F_3^{\left(j\right)}(k,n)$, $j\in \{1,3\}$.

Theorem 5.

If $k\ge 1$, $k\ne 3$, $n\ge 0$, $p\ge 0$, $q\ge 0$ are integers, then

$$F_3^{\left(1\right)}(k,n)=\left\{\begin{array}{cc}{\displaystyle \sum _{{\displaystyle \begin{array}{c}p,t\ge 0\\ n=kp+3q\end{array}}}}\left(\genfrac{}{}{0pt}{}{p+q}{q}\right)\hfill & \mathit{for}n=kp+3q,\hfill \\ 0\hfill & \mathit{otherwise}.\hfill \end{array}\right.$$

(12)

Proof. 

For $n=0$ the statement is obvious. If $n\ge 1$, then by Theorem 3 (i), it follows that $F_3^{\left(1\right)}(k,n)$ is the number of all $(3,k)$-decompositions of n. Such decompositions are possible if $n=kp+3q$, where $p\ge 0$, $q\ge 0$ are integers, and then $n=(n_1,\dots ,n_{p+q})$ is a $(p+q)$-tuple where $n_i\in \{3,k\}$ for $i\in \{1,\dots ,p+q\}$. By the fundamental combinatorial statements we obtain that the total numbers of $(p+q)$-tuples is equal to

$$\sum _{{\displaystyle \begin{array}{c}p,q\ge 0\\ n=kp+3q\end{array}}}{C}_{p+q}^p=\sum _{{\displaystyle \begin{array}{c}p,q\ge 0\\ n=kp+3q\end{array}}}{C}_{p+q}^q=\sum _{{\displaystyle \begin{array}{c}p,q\ge 0\\ n=kp+3q\end{array}}}\left(\genfrac{}{}{0pt}{}{p+q}{p}\right)=\sum _{{\displaystyle \begin{array}{c}p,q\ge 0\\ n=kp+3q\end{array}}}\left(\genfrac{}{}{0pt}{}{p+q}{q}\right),$$

which ends the proof. □

Corollary 1.

If $k\ge 2$, $k\ne 3$, $n\ge 0$, $p\ge 0$, $q\ge 0$ are integers, then

$$F_3^{\left(3\right)}(k,n)=\sum _{i=1}^{min\{3,k\}-1}\sum _{{\displaystyle \begin{array}{c}p,q\ge 0\\ n-i=kp+3q\end{array}}}\left(\genfrac{}{}{0pt}{}{p+q}{q}\right).$$

(13)

3. Connections of $F_3^{\left(j\right)}(k,n)$ with Pascal’s Triangle and Some Identities

Sequences from the family $\mathcal{F}$ have interesting connections with Pascal’s triangle; in particular, these sequences can be generated by elements, rows, or diagonals of Pascal’s triangle. Sequences of the Fibonacci-type and their relations with Pascal’s triangle have been studied in many papers recently; interesting results can be found in [16]. We show some relations between numbers $F_3^{\left(j\right)}(k,n)$ and binomial coefficients which appear in rows of Pascal’s triangle.

Theorem 6.

Let $k\ge 1,k\ne 3$ and $i\ge 1$ be a fixed integer. Then, for $n\ge ki$ and $j\in \{1,2,3\}$ holds

$$F_3^{\left(j\right)}(k,n)=\sum _{p=0}^i\left(\genfrac{}{}{0pt}{}{i}{p}\right)F_3^{\left(j\right)}(k,n-pk-3(i-p)).$$

(14)

Proof. 

(we prove by induction on n). Suppose that $n=ki$. Using the definition of $F_3^{\left(j\right)}(k,n)$, we obtain the following expression

$$F_3^{\left(j\right)}(k,ki)=F_3^{\left(j\right)}(k,ik-3)+F_3^{\left(j\right)}(k,ki-k).$$

Applying the definition of $F_3^{\left(j\right)}(k,n)$ for each summand we have

$$F_3^{\left(j\right)}(k,ki)=F_3^{\left(j\right)}(k,ki-6)+2F_3^{\left(j\right)}(k,ki-3-k)+F_3^{\left(j\right)}(k,ki-2k).$$

Analogously in the third step we obtain the equality

$$F_3^{\left(j\right)}(k,ki)=F_3^{\left(j\right)}(k,ki-9)+3F_3^{\left(j\right)}(k,ki-k-6)+3F_3^{\left(j\right)}(k,ki-2k-3)+F_3^{\left(j\right)}(k,ki-3k).$$

Repeating it into i steps we obtain the following identity:

$$F_3^{\left(j\right)}(k,ik)=\left(\genfrac{}{}{0pt}{}{i}{0}\right)F_3^{\left(j\right)}(k,ik-3i)\dots \left(\genfrac{}{}{0pt}{}{i}{p}\right)F_3^{\left(j\right)}(k,(i-p)(k-3))\dots \left(\genfrac{}{}{0pt}{}{i}{i}\right)F_3^{\left(j\right)}(k,0)$$

$$=\sum _{p=0}^i\left(\genfrac{}{}{0pt}{}{i}{p}\right)F_3^{\left(j\right)}(k,(i-p)(k-3))=\sum _{p=0}^i\left(\genfrac{}{}{0pt}{}{i}{p}\right)F_3^{\left(j\right)}(k,ik-pk-3(i-p)).$$

Thus, the formula is true for $n=ki$.

Let $n>ki$ and suppose now that for every $t<n$ holds

$$F_3^{\left(j\right)}(k,t)=\sum _{p=0}^i\left(\genfrac{}{}{0pt}{}{i}{p}\right)F_3^{\left(j\right)}(k,t-pk-3(i-p)).$$

We shall show that

$$F_3^{\left(j\right)}(k,n)=\sum _{p=0}^i\left(\genfrac{}{}{0pt}{}{i}{p}\right)F_3^{\left(j\right)}(k,n-pk-3(i-p)).$$

By the definition of $F_3^{\left(j\right)}(k,n)$, and next by induction hypothesis, we obtain that

$$F_3^{\left(j\right)}(k,n)=\sum _{p=0}^i\left(\genfrac{}{}{0pt}{}{i}{p}\right)F_3^{\left(j\right)}(k,n-pk-3(i-p))$$

$$=F_3^{\left(j\right)}(k,n)=\sum _{p=0}^i\left(\genfrac{}{}{0pt}{}{i}{p}\right)(F_3^{\left(j\right)}(k,n-3-pk-3(i-p))+F_3^{\left(j\right)}(k,k,n-k-pk-3(i-p))$$

$$=\sum _{p=0}^i\left(\genfrac{}{}{0pt}{}{i}{p}\right)F_3^{\left(j\right)}(k,n-3-pk-3(i-p))+\sum _{p=0}^i\left(\genfrac{}{}{0pt}{}{i}{p}\right)F_3^{\left(j\right)}(k,n-k-pk-3(i-p))$$

$$=F_3^{\left(j\right)}(k,n-3)+F_3^{\left(j\right)}(k,n-k)=F_3^{\left(j\right)}(k,n),$$

and by the principle of induction the result follows. □

Now we prove some identities for sums of distance Padovan numbers $F_3^{\left(j\right)}(k,n)$.

Theorem 7.

Let $n\ge 0$, $k\ge 2$ be integers. Then the following identities hold:

$\left(i\right)$${\displaystyle \sum _{i=0}^n}{F}_3^{\left(j\right)}(2,2i)=\left\{\begin{array}{cc}{F}_3^{\left(j\right)}(2,2n+3)\hfill & \mathit{for}j=1,\hfill \\ F_3^{\left(j\right)}(2,2n+3)-1\hfill & \mathit{for}j\in \{2,3\},\hfill \end{array}\right.$

$\left(ii\right)$${\displaystyle \sum _{i=0}^n}{F}_3^{\left(j\right)}(3,3i)=F_3^{\left(j\right)}(3,3n+3)-1$ for $j\in \{1,2\}$,

$\left(iii\right)$${\displaystyle \sum _{i=0}^n}{F}_3^{\left(j\right)}(k,ki)=F_3^{\left(j\right)}(k,kn+3)$ for $k\ge 4,j\in \{1,2,3\}.$

Proof. 

(iii) (by induction on n). We give the proof for $k\ge 4$ and $j=1$. For $j\in \{2,3\}$ we prove similarly. For $n=0$ we have $F_3^{\left(1\right)}(k,0)=1=F_3^{\left(1\right)}(k,3)$. Assume the equality (iii) holds for an arbitrary $n\ge 0$. We will prove it for $n+1$. By induction hypothesis and the definition of $F_3^{\left(1\right)}(k,n)$ we obtain

${\displaystyle \sum _{i=0}^{n+1}}{F}_3^{\left(1\right)}(k,ki)={\displaystyle \sum _{i=0}^n}{F}_3^{\left(1\right)}(k,ki)+F_3^{\left(1\right)}(k,kn+k)=F_3^{\left(1\right)}(k,kn+3)+F_3^{\left(1\right)}(k,kn+k)=F_3^{\left(1\right)}(k,kn+k+3)$.

Thus, the identity (iii) is valid.

Analogously we can prove the identities (i) and (ii). □

Theorem 8.

Let $k\ge 1$, $n\ge 0$ be integers. Then

(i) 

${\displaystyle \sum _{i=0}^n}{F}_3^{\left(j\right)}(1,i)=F_3^{\left(j\right)}(1,n+3)-1$ for $j\in \{1,2,3\}$,

(ii) 

${\displaystyle \sum _{i=0}^n}{F}_3^{\left(j\right)}(2,i)=\left\{\begin{array}{cc}{F}_3^{\left(j\right)}(2,n+5)-1\hfill & \mathit{for}j\in \{1,3\},\hfill \\ F_3^{\left(j\right)}(2,n+5)-2\hfill & \mathit{for}j=2,\hfill \end{array}\right.$

(iii) 

${\displaystyle \sum _{i=0}^n}{F}_3^{\left(j\right)}(3,i)=\left\{\begin{array}{cc}{F}_3^{\left(j\right)}(3,n+1)+F_3^{\left(j\right)}(3,n+2)+F_3^j(n+3)-1\hfill & \mathit{for}j=1,\hfill \\ F_3^{\left(j\right)}(3,n+1)+F_3^{\left(j\right)}(3,n+2)+F_3^j(n+3)-3\hfill & \mathit{for}j=2,\hfill \\ F_3^{\left(j\right)}(3,n+1)+F_3^{\left(j\right)}(3,n+2)+F_3^j(n+3)-2\hfill & \mathit{for}j=3,\hfill \end{array}\right.$

(iv) 

${\displaystyle \sum _{i=0}^n}{F}_3^{\left(j\right)}(4,i)=\left\{\begin{array}{cc}{F}_3^{\left(j\right)}(4,n+4)+F_3^{\left(j\right)}(4,n+6)-1\hfill & \mathit{for}j=1,\hfill \\ F_3^{\left(j\right)}(4,n+4)+F_3^{\left(j\right)}(4,n+6)-3\hfill & \mathit{for}j=2,\hfill \\ F_3^{\left(j\right)}(4,n+4)+F_3^{\left(j\right)}(4,n+6)-2\hfill & \mathit{for}j=3,\hfill \end{array}\right.$

(v) 

${\displaystyle \sum _{i=0}^n}{F}_3^{\left(1\right)}(6,i)=F_3^{\left(1\right)}(6,3n+6)-1$.

Proof. 

(by induction on n)

(i)

If $n=0$, then $F_3^{\left(j\right)}(1,0)=1=2-1=F_3^{\left(j\right)}(1,3)-1$.

Assume that the formula (i) holds for n. We will prove it for $n+1$. By induction hypothesis we have

${\displaystyle \sum _{i=0}^{n+1}}{F}_3^{\left(j\right)}(1,i)=F_3^{\left(j\right)}(1,n+3)-1+F_3^{\left(j\right)}(1,n+1)=F_3^{\left(j\right)}(1,n+4)-1$.

(iii) We prove the case $j=2$. If $n=0$, then

$F_3^{\left(2\right)}(3,0)=1=1+1+2-3=F_3^{\left(3\right)}(3,1)+F_3^{\left(2\right)}(3,2)+F_3^{\left(2\right)}(3,3)-3$.

Assume that the formula (iii) holds for n. We will prove it for $n+1$. By induction hypothesis we have

$$\begin{gathered} {\displaystyle \sum _{i=0}^{n+1}}{F}_3^{\left(2\right)}(3,i)=F_3^{\left(2\right)}(3,n+1)+F_3^{\left(2\right)}(3,n+2)+F_3^{\left(2\right)}(3,n+3)-3+F_3^{\left(2\right)}(3,n+1)=F_3^{\left(2\right)}(3,n+2)+ \\ {F}_3^{\left(2\right)}(3,n+3)+F_3^{\left(2\right)}(3,n+4)-3 \end{gathered}$$.

Analogously, we prove the remaining formulas. □

4. Determinant Generators

Matrix generators of Fibonacci-type sequences have been studied by many authors. In a number of papers we can find Padovan generators and related matrices in which terms of Padovan sequences are obtained by Q-matrices, determinants, or permanents; see, for example, results given by Sokhuma [25,26]. In this paper, we give determinant generators for distance Padovan numbers $F_3^{\left(j\right)}(k,n)$, $j\in \{1,2,3\}$, constructed on a basis of a matrix of initial conditions.

By a matrix of initial conditions of the sequence $\left\{F_3^{\left(j\right)}(k,n)\right\}$ we mean a matrix $A_k^{\left(j\right)}$ of size $m\times m,$ where $m=max\{3,k\}$, such that consecutive main corner minors $\det A_{k,i}^{\left(j\right)}$ of the matrix $A_k^{\left(j\right)}$ are equal $F_3^{\left(j\right)}(k,i-1)$, i.e.,

$A_k^{\left(j\right)}={\left[a_{ij}\right]}_{m\times m}$ and $\det A_{k,i}^{\left(j\right)}=\left|\begin{array}{ccc}{a}_{11}& \dots & a_{1i}\\ ⋮& & ⋮\\ a_{i1}& \dots & a_{ii}\end{array}\right|=F_3^{\left(j\right)}(k,i-1)$ for $i\in \{1,\dots ,m\}$.

Clearly, the matrix of initial conditions is not determined uniquely.

At the beginning we discuss cases $k\in \{1,2,3\}$. For $k=1$ we set the following matrices of initial conditions $A_1^{\left(1\right)}=A_1^{\left(2\right)}=A_1^{\left(3\right)}=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 0& 1& 1\end{array}\right]$, for $k=2$ we choose matrices

$$A_2^{\left(1\right)}=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 0\\ 0& 1& 1\end{array}\right],A_2^{\left(2\right)}=\left[\begin{array}{ccc}1& 0& 1\\ 1& 1& 1\\ 0& 1& 1\end{array}\right],A_2^{\left(3\right)}=\left[\begin{array}{ccc}0& -1& -1\\ 1& 1& 1\\ 0& 1& 1\end{array}\right],$$

and for $k=3$ we set

$$A_3^{\left(1\right)}=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 0& 1& 1\end{array}\right],A_3^{\left(2\right)}=\left[\begin{array}{ccc}1& 0& 1\\ 1& 1& 1\\ 0& 1& 1\end{array}\right],A_3^{\left(3\right)}=\left[\begin{array}{ccc}0& -1& 0\\ 1& 1& 1\\ 0& 1& 1\end{array}\right].$$

We construct a determinant generator $A_{k,n}^{\left(j\right)}$ for $F_3^{\left(j\right)}(k,n)$ with $k\in \{1,2,3\}$ and $j\in \{1,2,3\}$ starting from the matrix $A_k^{\left(j\right)}$ by adding rows and columns according to the rule presented below.

$A_{1,n}^{\left(j\right)}={\left[\begin{array}{cc}\overline{)\begin{array}{ccc}& & \\ & A_1^{\left(j\right)}& \\ & & \end{array}}& \begin{array}{cc}0& \dots \\ 1& \ddots \\ 0& \ddots \end{array}\\ \begin{array}{ccc}0& 0& 1\\ ⋮& ⋮& \ddots \end{array}& \begin{array}{cc}1& \ddots \\ \ddots & \ddots \end{array}\end{array}\right]}_{n\times n,}$$A_{2,n}^{\left(j\right)}={\left[\begin{array}{cc}\overline{)\begin{array}{ccc}& & \\ & A_2^{\left(j\right)}& \\ & & \end{array}}& \begin{array}{cc}0& \dots \\ 1& \ddots \\ \mathrm{-1}& \ddots \end{array}\\ \begin{array}{ccc}0& 0& 1\\ ⋮& ⋮& \ddots \end{array}& \begin{array}{cc}0& \ddots \\ \ddots & \ddots \end{array}\end{array}\right]}_{n\times n,}$

$A_{3,n}^{\left(j\right)}={\left[\begin{array}{cc}\overline{)\begin{array}{ccc}& & \\ & A_3^{\left(j\right)}& \\ & & \end{array}}& \begin{array}{cc}0& \dots \\ 2& \ddots \\ 0& \ddots \end{array}\\ \begin{array}{ccc}0& 0& 1\\ ⋮& ⋮& \ddots \end{array}& \begin{array}{cc}0& \ddots \\ \ddots & \ddots \end{array}\end{array}\right]}_{n\times n.}$

Theorem 9.

Let $k\in \{1,2\}$, $n\ge 1$ be integers. Then

(i) 

$F_3^{\left(j\right)}(1,n-1)=\mathit{det}{A}_{1,n}^{\left(j\right)}$$forj\in \{1,2\}$,

(ii) 

$F_3^{\left(j\right)}(2,n-1)=\mathit{det}{A}_{2,n}^{\left(j\right)}$$forj\in \{1,2,3\}$,

(iii) 

$F_3^{\left(j\right)}(3,n-1)=\mathit{det}{A}_{3,n}^{\left(j\right)}$$forj\in \{1,2,3\}$.

Proof. 

(by induction on n)

We will prove (i) for $j=1$. The remaining cases can be proved analogously.

If $n\le 3$, then by the definition of a matrix of initial conditions we have

$\det A_{1,n}^{\left(1\right)}=F_3^{\left(1\right)}(1,n-1)$. For $n=4$ we have

$$\begin{gathered} \det A_{1,4}^{\left(1\right)}=\left|\begin{array}{cccc}1& 0& 1& 0\\ 1& 1& 1& 1\\ 0& 1& 1& 0\\ 0& 0& 1& 1\end{array}\right|=\left|\begin{array}{ccc}1& 0& 1\\ 1& 1& 1\\ 0& 1& 1\end{array}\right|+\left|\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right|=\left|\begin{array}{ccc}1& 0& 1\\ 1& 1& 1\\ 0& 1& 1\end{array}\right|+\left|\begin{array}{c}1\end{array}\right|=\det A_{1,3}^{\left(1\right)}+\det A_{1,1}^{\left(1\right)}= \\ {F}_3^{\left(1\right)}(1,2)+F_3^{\left(1\right)}(1,0)=F_3^{\left(1\right)}(1,3) \end{gathered}$$.

Assume that $n>4$ and $\det A_{1,n}^{\left(1\right)}=F_3^{\left(1\right)}(1,n-1)$. We shall show that $\det A_{1,n+1}^{\left(1\right)}=F_3^{\left(1\right)}(1,n)$.

We calculate the determinant of $A_{1,n+1}^{\left(1\right)}$ by the Laplace expansion along the last column. The minor corresponding to $a_{n+1,n+1}$ equals $\det A_{1,n}^{\left(1\right)}$. For the second minor we use expansion along the last row two times, and we have that it equals $\det A_{1,n-2}^{\left(1\right)}$.

By induction hypothesis and the recurrence relation for $F_3^{\left(1\right)}(1,n)$ we have $\det A_{1,n+1}^{\left(1\right)}=\det A_{1,n}^{\left(1\right)}+\det A_{1,n-2}^{\left(1\right)}=F_3^{\left(1\right)}(1,n-1)+F_3^{\left(1\right)}(1,n-3)=F_3^{\left(1\right)}(1,n)$, which completes the proof. □

For $k=4,j=1$ a matrix of initial conditions has the form

$A_4^{\left(1\right)}=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 1& 1& 0\\ 0& 0& 1& 0\end{array}\right]$, and for $k>4$, $j=1$ we can construct a matrix of initial conditions $A_k^{\left(1\right)}$ starting from $A_4^{\left(1\right)}$ as follows:

$$A_k^{\left(1\right)}={\left[\begin{array}{cc}\overline{)\begin{array}{cccc}& & & \\ & & & \\ & & A_4^{\left(1\right)}& \\ & & & \end{array}}& \begin{array}{cc}0& \dots \\ 0& \dots \\ 1& \ddots \\ 0& \ddots \end{array}\\ \begin{array}{cccc}0& 0& 0& 1\\ ⋮& ⋮& \ddots & \ddots \end{array}& \begin{array}{cc}0& \ddots \\ \ddots & \ddots \end{array}\end{array}\right]}_{k\times k.}$$

(15)

The coefficients of the matrix $A_k^{\left(1\right)}={\left[a_{ij}\right]}_{k\times k}$ can be described in the form

$$a_{ij}=\left\{\begin{array}{cc}1\hfill & \mathrm{for}j=i\mathrm{where}i\in \{1,3\},\hfill \\ \hfill & j=i+2\mathrm{where}i\in \{2,\dots ,k-2\},\hfill \\ \hfill & j=i-1\mathrm{where}i\in \{3,\dots ,k\},\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

(16)

Basing on the matrix $A_k^{\left(1\right)}$ we define the matrix $A_{k,n}^{\left(1\right)}={\left[a_{ij}\right]}_{n\times n}$ for an arbitrary $n>k$ as follows:

$$a_{ij}=\left\{\begin{array}{cc}1\hfill & \mathrm{for}j=i\mathrm{where}i\in \{1,3\},\hfill \\ \hfill & j=i+2\mathrm{where}i\in \{2,\dots ,n-2\},\hfill \\ \hfill & j=i-1\mathrm{where}i\in \{3,\dots ,n\},\hfill \\ {(-1)}^{(k+1)}\hfill & \mathrm{for}j=i+k-1\mathrm{where}i\in \{2,\dots ,n-(k-1\left)\right\},\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

(17)

For special values of k we obtain the following matrices.

$$A_{4,n}^{\left(1\right)}={\left[\begin{array}{cc}\overline{)\begin{array}{cccc}& & & \\ & A_4^{\left(j\right)}& & \\ & & & \\ & & & \end{array}}& \begin{array}{ccc}0& \dots & 0\\ \mathrm{-1}& \dots & 0\\ 1& \ddots & 0\\ 0& \ddots & \mathrm{-1}\end{array}\\ \begin{array}{cccc}0& 0& 0& 1\\ ⋮& ⋮& ⋮& \ddots \\ 0& 0& 0& 0\end{array}& \begin{array}{ccc}0& \ddots & 1\\ \ddots & \ddots & 0\\ 0& 1& 0\end{array}\end{array}\right]}_{n\times n.}$$

(18)

$$A_{5,n}^{\left(1\right)}={\left[\begin{array}{cc}\overline{)\begin{array}{cccccc}& & & & & \\ & & & & & \\ & & A_5^{\left(1\right)}& & & \\ & & & & & \\ & & & & & \end{array}}& \begin{array}{ccc}0& \dots & 0\\ 1& \dots & 0\\ 0& \ddots & 0\\ 1& & 1\\ 0& \ddots & 0\end{array}\\ \begin{array}{cccccc}0& 0& 0& & 0& 1\\ ⋮& ⋮& ⋮& & ⋮& \ddots \\ 0& 0& 0& & 0& 0\end{array}& \begin{array}{ccc}0& \ddots & 1\\ \ddots & \ddots & 0\\ 0& 1& 0\end{array}\end{array}\right]}_{n\times n.}$$

(19)

Proving analogously, as in Theorem 9, we obtain the following Theorem.

Theorem 10.

Let $k\ge 4$, $n\ge 1$ be integers. Then $F_3^{\left(1\right)}(k,n-1)=det{A}_{k,n}^{\left(1\right)}$.

For $j=2$ and $k\ge 4$ we define a matrix $A_k^{\left(2\right)}={\left[a_{ij}\right]}_{k\times k}$ as follows:

$$a_{ij}=\left\{\begin{array}{cc}1\hfill & \mathrm{for}j=i\mathrm{where}i\in \{1,\dots ,k\},\hfill \\ \hfill & j=i-1\mathrm{where}i\in \{2,\dots ,k\},\hfill \\ 0\hfill & \mathrm{otherwise}\hfill \end{array}\right.$$

(20)

and for $n>k$ coefficients of a matrix $A_{k,n}^{\left(2\right)}={\left[a_{ij}\right]}_{n\times n}$ have the following form

$$a_{ij}=\left\{\begin{array}{cc}1\hfill & \mathrm{for}j=i\mathrm{where}i\in \{1,\dots ,k\},\hfill \\ \hfill & j=i+2\mathrm{where}i\in \{k-1,\dots ,n-2\},\hfill \\ \hfill & j=i-1\mathrm{where}i\in \{2,\dots ,n\},\hfill \\ {(-1)}^{(k+1)}\hfill & \mathrm{for}j=i+k-1\mathrm{where}i\in \{2,\dots ,n-(k-1\left)\right\},\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

(21)

Theorem 11.

Let $k\ge 4$, $n\ge 1$ be integers. Then $F_3^{\left(2\right)}(k,n-1)=det{A}_{k,n}^{\left(2\right)}$.

For $j=3$ and $k\ge 4$ we define a matrix $A_k^{\left(3\right)}={\left[a_{ij}\right]}_{k\times k}$ as follows:

$$a_{ij}=\left\{\begin{array}{cc}1\hfill & \mathrm{for}i=2,j=1\mathrm{and}i=j=3,\hfill \\ \hfill & j=i+2\mathrm{where}i\in \{2,\dots ,k-2\},\hfill \\ \hfill & j=i-1\mathrm{where}i\in \{5,\dots ,k\},\hfill \\ -1\hfill & \mathrm{for}i=1,j=2,\hfill \\ 0\hfill & \mathrm{otherwise}\hfill \end{array}\right.$$

(22)

and for $n>k$ a matrix $A_{k,n}^{\left(3\right)}={\left[a_{ij}\right]}_{n\times n}$ has the following coefficients

$$a_{ij}=\left\{\begin{array}{cc}1\hfill & \mathrm{for}i=2,j=1\mathrm{and}i=j=3,\hfill \\ \hfill & j=i+2\mathrm{where}i\in \{2,\dots ,k-2\},\hfill \\ \hfill & j=i-1\mathrm{where}i\in \{5,\dots ,k\},\hfill \\ -1\hfill & \mathrm{for}i=1,j=2,\hfill \\ {(-1)}^{(k+1)}\hfill & \mathrm{for}j=i+(k-1)\mathrm{where}i\in \{k-1,\dots ,n-(k-1\left)\right\},\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

(23)

Theorem 12.

Let $k\ge 4$, $n\ge 1$ be integers. Then $F_3^{\left(3\right)}(k,n-1)=det{A}_{k,n}^{\left(3\right)}$.

5. Conclusions

The concept of generalized Padovan sequences is a part of a wide theory of Fibonacci-type sequences, which are intensively studied among others with respect to their distinct interpretations. Such an approach can be used for describing properties of these sequences and we can find it, for example, in the graph theory; for details, see [27,28]. Matrix generators of Fibonacci-like sequences used in encoding–decoding algorithms seem to be very important; see, for example, [29,30]. In this paper, we use number decomposition interpretation to generate and study different kinds of generalized Padovan sequences. The number decomposition technique can probably be useful in further research for other Fibonacci-type sequences. It also seems interesting to extend the concept of distance Padovan numbers to polynomials.