On the Smarandache-Pascal derived sequences of generalized Tribonacci numbers

Abstract

For any sequence recurrence formula, the Smarandache-Pascal derived sequence $\{T_n\}$ of $\{b_n\}$ is defined by $T_{n+1}={\sum }_{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)\cdot b_{k+1}$ for all $n\ge 2$, where $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$ denotes the combination number. The recurrence formula of $\{T_n\}$ is obtained by the properties of the third-order linear recurrence sequence.

1 Introduction

For any sequence $\{b_n\}$, a new sequence $\{T_n\}$ is defined by the following method: $T_1=b_1$, $T_2=b_1+b_2$, $T_3=b_1+2b_2+b_3$, generally, $T_{n+1}={\sum }_{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)\cdot b_{k+1}$ for all $n\ge 2$, where $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$ is the combination number. This sequence is called the Smarandache-Pascal derived sequence of $\{b_n\}$. It was introduced by professor Smarandache in [1] and studied by some authors. For example, Murthy and Ashbacher [2] proposed a series of conjectures related to Fibonacci numbers and the Smarandache-Pascal derived sequence, one of them is as follows.

Conjecture Let $\{b_n\}=\{F_{8n+1}\}=\{F_1,F_9,F_{17},F_{25},\dots \}$, $\{T_n\}$ be the Smarandache-Pascal derived sequence of $\{b_n\}$, then we have the recurrence formula

$$T_{n+1}=49\cdot (T_n-T_{n-1}),n\ge 2.$$

Li and Han [3] studied these problems and proved a generalized conclusion as follows.

Proposition Let $\{X_n\}$ be a second-order linear recurrence sequence with $X_0=u$, $X_1=v$, $X_{n+1}=a{X}_n+b{X}_{n-1}$ for all $n\ge 1$, where $a^2+4b>0$. For any positive integer $d\ge 2$, we define the Smarandache-Pascal derived sequence of $\{X_{dn+1}\}$ as

$$T_{n+1}=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)\cdot X_{dk+1}.$$

Then we have the recurrence formula

$$T_{n+1}=(2+A_d+b\cdot A_{d-2})\cdot T_n-(1+A_d+b\cdot A_{d-2}+{(-b)}^d)\cdot T_{n-1},$$

where the sequence $\{A_n\}$ is defined as $A_0=1$, $A_1=a$, $A_{n+1}=a\cdot A_n+b\cdot A_{n-1}$ for all $n\ge 1$.

It is clear that if we take $b=1$, then $X_n$ is the Fibonacci polynomials, see [4–7].

The main purpose of this paper is, using the elementary method and the properties of the third-order linear recurrence sequence, to unify the above results by proving the following theorem.

Theorem Let $\{X_n\}$ be a third-order linear recurrence sequence $X_{n+3}=a\cdot X_{n+2}+b\cdot X_{n+1}+c\cdot X_n$ with the initial values $X_0=u$, $X_1=v$ and $X_2=w$ for all $n\ge 1$, where a, b and c are positive integers. For any positive integer $d\ge 2$, we define the Smarandache-Pascal derived sequence of $\{X_{dn+1}\}$ as

$$T_{n+1}=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)\cdot X_{dk+1}.$$

Then we have the recurrence formula

$$T_{n+1}=\frac{g_1{g}_3-g_1{g}_6+g_7}{g_3-g_6}\cdot T_n+\frac{g_3{g}_4-g_2{g}_6-g_1{g}_7}{g_3-g_6}\cdot T_{n-1}+\frac{g_3{g}_5-g_2{g}_7}{g_3-g_6}\cdot T_{n-2},$$

where

$$\begin{array}{c}{g}_1=f_1+f_2+c\cdot A_d{f}_5,g_2=f_3{A}_{d+1}-f_1{f}_2+c\cdot A_d{f}_6,\hfill \\ g_3=c\cdot A_{d+1}^2-c\cdot A_d{f}_2+c\cdot A_d{f}_4,g_4=f_3{A}_{d+1}-f_1{f}_2+c\cdot A_d(f_5+f_6),\hfill \\ g_5=c\cdot A_d{f}_6,g_6=c\cdot (A_{d+1}-A_d{f}_2+A_d{f}_4-A_d),g_7=c\cdot A_d{f}_4\hfill \end{array}$$

and

$$\begin{array}{c}{f}_1=b\cdot A_d+c\cdot A_{d-1}+1,f_2=1+A_{d+2},\hfill \\ f_3=b\cdot A_{d-1}+c\cdot A_d,f_4=1+c\cdot A_{d-1}-\frac{c\cdot A_d^2}{A_{d+1}},\hfill \\ f_5=\frac{A_d}{A_{d+1}},f_6=b\cdot A_{d-1}+c\cdot A_{d-2}-\frac{b\cdot A_d^2+c\cdot A_{d-1}{A}_d+A_d}{A_{d+1}},\hfill \end{array}$$

the sequence $\{A_n\}$ is defined by $A_{n+3}=a\cdot A_{n+2}+b\cdot A_{n+1}+c\cdot A_n$ with the initial values $A_1=0$, $A_2=1$ and $A_3=a$ for all $n\ge 1$.

From our theorem we know that if $\{b_n\}$ is a third-order linear recurrence sequence, then its Smarandache-Pascal derived sequence $\{T_n\}$ is also a third-order linear recurrence sequence.

2 Proof of the theorem

To complete the proof of our theorem, we need the following lemma.

Lemma Let integers $m\ge 0$ and $n\ge 3$. If the sequence $\{X_n\}$ satisfies the recurrence relations $X_{n+3}=a\cdot X_{n+2}+b\cdot X_{n+1}+c\cdot X_n$, $n\ge 0$, then we have the identity

$$X_{m+n}=A_n\cdot X_{m+2}+(b\cdot A_{n-1}+c\cdot A_{n-2})\cdot X_{m+1}+c\cdot A_{n-1}\cdot X_m,$$

where $A_n$ is defined by $A_{n+3}=a\cdot A_{n+2}+b\cdot A_{n+1}+c\cdot A_n$ with the initial values $A_1=0$, $A_2=1$ and $A_3=a$ for all $n\ge 1$.

Proof Now we prove this lemma by mathematical induction. Note that the recurrence formula $X_{m+3}=a\cdot X_{m+2}+b\cdot X_{m+1}+c\cdot X_m=A_3\cdot X_{m+2}+(b\cdot A_2+c\cdot A_1)\cdot X_{m+1}+c\cdot A_2\cdot X_m$ for all $n\ge 1$. That is, the lemma holds for $n=3$ since

$$\begin{array}{rl}{X}_{m+4}& =a\cdot (a\cdot X_{m+2}+b\cdot X_{m+1}+c\cdot X_m)+b\cdot X_{m+2}+c\cdot X_{m+1}\\ =(a^2+b)\cdot X_{m+2}+(ab+c)\cdot X_{m+1}+ac\cdot X_m\\ =A_4\cdot X_{m+2}+(b\cdot A_3+c\cdot A_2)\cdot X_{m+1}+c\cdot A_3\cdot X_m.\end{array}$$

That is, the lemma holds for $n=4$. Suppose that for all integers $2\le n\le k$, we have $X_{m+n}=A_n\cdot X_{m+2}+(b\cdot A_{n-1}+c\cdot A_{n-2})\cdot X_{m+1}+c\cdot A_{n-1}\cdot X_m$. Then, for $n=k+1$, from the recurrence relations for $X_m$ and the inductive hypothesis, we have

$$\begin{array}{rcl}{X}_{m+k+1}& =& a\cdot X_{m+k}+b\cdot X_{m+k-1}+c\cdot X_{m+k-2}\\ =& a\cdot (A_k\cdot X_{m+2}+(b\cdot A_{k-1}+c\cdot A_{k-2})\cdot X_{m+1}+c\cdot A_{k-1}\cdot X_m)\\ +b\cdot (A_{k-1}\cdot X_{m+2}+(b\cdot A_{k-2}+c\cdot A_{k-3})\cdot X_{m+1}+c\cdot A_{k-2}\cdot X_m)\\ +c\cdot (A_{k-2}\cdot X_{m+2}+(b\cdot A_{k-3}+c\cdot A_{k-4})\cdot X_{m+1}+c\cdot A_{k-3}\cdot X_m)\\ =& (a\cdot A_k+b\cdot A_{k-1}+c\cdot A_{k-2})\cdot X_{m+2}+(ab\cdot A_{k-1}+(ac+b^2)\cdot A_{k-2}\\ +2bc\cdot A_{k-3}+c^2\cdot A_{k-4})\cdot X_{m+1}+c(a\cdot A_{k-1}+b\cdot A_{k-2}+c\cdot A_{k-3})\cdot X_m\\ =& A_{k+1}\cdot X_{m+2}+(ab\cdot A_{k-1}+b^2\cdot A_{k-2}+bc\cdot A_{k-3}+c\cdot A_{k-1})\cdot X_{m+1}+c\cdot A_k\cdot X_m\\ =& A_{k+1}\cdot X_{m+2}+(b\cdot A_k+c\cdot A_{k-1})\cdot X_{m+1}+c\cdot A_k\cdot X_m.\end{array}$$

That is, the lemma also holds for $n=k+1$. This completes the proof of our lemma by mathematical induction. □

Now we use this lemma to complete the proof of our theorem. From the properties of the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, we have

$$\begin{array}{rcl}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)+\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)& =& \frac{(n-1)!}{k!(n-1-k)!}+\frac{(n-1)!}{(k-1)!(n-k)!}\\ =& \frac{(n-1)!}{(k-1)!(n-k-1)!}(\frac{1}{k}+\frac{1}{n-k})=\left(\genfrac{}{}{0ex}{}{n}{k}\right).\end{array}$$

(1)

For any positive integer d, from the lemma we have $X_{dk+d+1}=A_{d+1}\cdot X_{dk+2}+(b\cdot A_d+c\cdot A_{d-1})\cdot X_{dk+1}+c\cdot A_d\cdot X_{dk}$. By the definition of $T_n$, we may deduce that

$$\begin{array}{rcl}{T}_{n+1}& =& \sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)\cdot X_{dk+1}\\ =& X_1+X_{dn+1}+\sum _{k=1}^{n-1}(\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)+\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right))\cdot X_{dk+1}\\ =& \sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk+1}+\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk+d+1}+X_{dn+1}\\ =& T_n+\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk+d+1}\\ =& T_n+\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot (A_{d+1}\cdot X_{dk+2}+(b\cdot A_d+c\cdot A_{d-1})\cdot X_{dk+1}+c\cdot A_d\cdot X_{dk})\\ =& (b\cdot A_d+c\cdot A_{d-1}+1)\cdot T_n+A_{d+1}\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk+2}\\ +c\cdot A_d\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk}.\end{array}$$

For convenience, we let $f_1(A_k)=b\cdot A_d+c\cdot A_{d-1}+1$ (briefly $f_1$), then the above identity implies that

$$T_{n+1}=f_1\cdot T_n+A_{d+1}\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk+2}+c\cdot A_d\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk}.$$

(2)

From this identity, we can also deduce

$$T_n=f_1\cdot T_{n-1}+A_{d+1}\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk+2}+c\cdot A_d\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk}$$

and

$$T_{n-1}=f_1\cdot T_{n-2}+A_{d+1}\sum _{k=0}^{n-3}\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)\cdot X_{dk+2}+c\cdot A_d\sum _{k=0}^{n-3}\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)\cdot X_{dk}.$$

They are equivalent to

$$\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk+2}=\frac{1}{A_{d+1}}(T_n-f_1\cdot T_{n-1}-c\cdot A_d\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk})$$

(3)

and

$$\sum _{k=0}^{n-3}\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)\cdot X_{dk+2}=\frac{1}{A_{d+1}}(T_{n-1}-f_1\cdot T_{n-2}-c\cdot A_d\sum _{k=0}^{n-3}\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)\cdot X_{dk}).$$

(4)

On the other hand, from the lemma we also deduce $X_{dk+d+2}=A_{d+2}\cdot X_{dk+2}+(b\cdot A_{d+1}+c\cdot A_d)\cdot X_{dk+1}+c\cdot A_{d+1}\cdot X_{dk}$. Then we have

$$\begin{array}{r}\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk+2}\\ =X_2+X_{dn-d+2}+\sum _{k=1}^{n-2}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk+2}\\ =X_2+X_{dn-d+2}+\sum _{k=1}^{n-2}(\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)+\left(\genfrac{}{}{0ex}{}{n-2}{k-1}\right))\cdot X_{dk+2}\\ =\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk+d+2}+\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk+2}\\ =\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot (A_{d+2}\cdot X_{dk+2}+(b\cdot A_{d+1}+c\cdot A_d)\cdot X_{dk+1}+c\cdot A_{d+1}\cdot X_{dk})\\ +\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk+2}\\ =(1+A_{d+2})\cdot \sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk+2}+c\cdot A_{d+1}\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk}\\ +(b\cdot A_{d+1}+c\cdot A_d)\cdot T_{n-1}.\end{array}$$

(5)

Similarly, applying formula (1) and identity (3), we have

$$\begin{array}{r}\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk}\\ =X_0+X_{dn-d}+\sum _{k=1}^{n-2}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk}\\ =X_0+X_{dn-d}+\sum _{k=1}^{n-2}(\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)+\left(\genfrac{}{}{0ex}{}{n-2}{k-1}\right))\cdot X_{dk}\\ =\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot (A_d\cdot X_{dk+2}+(b\cdot A_{d-1}+c\cdot A_{d-2})\cdot X_{dk+1}+c\cdot A_{d-1}\cdot X_{dk})\\ +\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk}\\ =A_d\cdot \sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk+2}+(1+c\cdot A_{d-1})\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk}\\ +(b\cdot A_{d-1}+c\cdot A_{d-2})\cdot T_{n-1}\\ =\frac{A_d}{A_{d+1}}(T_n-(b\cdot A_d+c\cdot A_{d-1}+1)\cdot T_{n-1}-c\cdot A_d\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk})\\ +(1+c\cdot A_{d-1})\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk}\\ +(b\cdot A_{d-1}+c\cdot A_{d-2})\cdot T_{n-1}\\ =\frac{A_d}{A_{d+1}}\cdot T_n+(b\cdot A_{d-1}+c\cdot A_{d-2}-\frac{b\cdot A_d^2+c\cdot A_{d-1}{A}_d+A_d}{A_{d+1}})\cdot T_{n-1}\\ +(1+c\cdot A_{d-1}-\frac{c\cdot A_d^2}{A_{d+1}})\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk}.\end{array}$$

(6)

For convenience, we let

$$\begin{array}{r}{f}_2=1+A_{d+2},f_3=b\cdot A_{d+1}+c\cdot A_d,f_4=1+c\cdot A_{d-1}-\frac{c\cdot A_d^2}{A_{d+1}},\\ f_5=\frac{A_d}{A_{d+1}},f_6=b\cdot A_{d-1}+c\cdot A_{d-2}-\frac{b\cdot A_d^2+c\cdot A_{d-1}{A}_d+A_d}{A_{d+1}},\end{array}$$

then identities (5) and (6) imply that

$$\begin{array}{c}\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk+2}\hfill \\ =f_2\cdot \sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk+2}+c\cdot A_{d+1}\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk}+f_3\cdot T_{n-1},\hfill \end{array}$$

(7)

$$\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk}=f_4\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk}+f_5{T}_n+f_6{T}_{n-1}.$$

(8)

Combining (2), (3), (7) and (8), we deduce

$$\begin{array}{rcl}{T}_{n+1}& =& (f_1+f_2+c\cdot A_d{f}_5)\cdot T_n+(f_3{A}_{d+1}-f_1{f}_2+c\cdot A_d{f}_6)\cdot T_{n-1}\\ +(c\cdot A_{d+1}^2-c\cdot A_d{f}_2+c\cdot A_d{f}_4)\cdot \sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk}.\end{array}$$

(9)

Applying formula (1), we deduce

$$\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)+\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)=\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)+\left(\genfrac{}{}{0ex}{}{n-2}{k-1}\right)+\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right).$$

(10)

From this and identities (3) and (4), note that $X_{dk+d}=A_d\cdot X_{dk+2}+(b\cdot A_{d-1}+c\cdot A_{d-2})\cdot X_{dk+1}+c\cdot A_{d-1}\cdot X_{dk}$, we have

$$\begin{array}{r}\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk}\\ =X_0+X_{dn-d}+\left(\genfrac{}{}{0ex}{}{n-1}{n-2}\right)\cdot X_{dn-2d}+\sum _{k=1}^{n-3}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)\cdot X_{dk}\\ =X_0+X_{dn-d}+(n-1)\cdot X_{dn-2d}+\sum _{k=1}^{n-3}(\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)+\left(\genfrac{}{}{0ex}{}{n-3}{k-1}\right)+\left(\genfrac{}{}{0ex}{}{n-2}{k-1}\right))\cdot X_{dk}\\ =X_0+X_{dn-d}+(n-1)\cdot X_{dn-2d}+\sum _{k=0}^{n-3}\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)\cdot X_{dk}-X_0\\ +\sum _{k=0}^{n-3}\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)\cdot X_{dk+d}-X_{dn-2d}+\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk+d}-(n-2)\cdot X_{dn-2d}-X_{dn-d}\\ =\sum _{k=0}^{n-3}\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)\cdot X_{dk}+\sum _{k=0}^{n-3}\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)\cdot X_{dk+d}+\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk+d}\\ =A_d\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk+2}+A_d\sum _{k=0}^{n-3}\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)\cdot X_{dk+2}+c\cdot A_{d-1}\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk}\\ +(c\cdot A_{d-1}+1)\sum _{k=0}^{n-3}\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)\cdot X_{dk}+(b\cdot A_{d-1}+c\cdot A_{d-2})\cdot (T_{n-1}+T_{n-2})\\ =\frac{A_d}{A_{d+1}}{T}_n+(b\cdot A_{d-1}+c\cdot A_{d-2}-\frac{b\cdot A_d^2+c\cdot A_{d-1}{A}_d}{A_{d+1}})\cdot T_{n-1}\\ +(b\cdot A_{d-1}+c\cdot A_{d-2}-\frac{b\cdot A_d^2+c\cdot A_{d-1}{A}_d+A_d}{A_{d+1}})\cdot T_{n-2}\\ +(c\cdot A_{d-1}-\frac{c\cdot A_d^2}{A_{d+1}})\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk}\\ +(c\cdot A_{d-1}+1-\frac{c\cdot A_d^2}{A_{d+1}})\sum _{k=0}^{n-3}\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)\cdot X_{dk}\\ =f_5{T}_n+(f_5+f_6)T_{n-1}+f_6{T}_{n-2}\\ +(f_4-1)\sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk}+f_4\sum _{k=0}^{n-3}\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)\cdot X_{dk}.\end{array}$$

(11)

Combining (2), (3), (7) and (11), we deduce

$$\begin{array}{rcl}{T}_{n+1}& =& (f_1+f_2+c\cdot A_d{f}_5)\cdot T_n+(f_3{A}_{d+1}-f_1{f}_2+c\cdot A_d(f_5+f_6))\cdot T_{n-1}+c\cdot A_d{f}_6\cdot T_{n-2}\\ +c(A_{d+1}-A_d{f}_2+A_d{f}_4-A_d)\cdot \sum _{k=0}^{n-2}\left(\genfrac{}{}{0ex}{}{n-2}{k}\right)\cdot X_{dk}\\ +c\cdot A_d{f}_4\cdot \sum _{k=0}^{n-3}\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)\cdot X_{dk}.\end{array}$$

(12)

From identity (9) we can also deduce

$$\begin{array}{rcl}{T}_n& =& (f_1+f_2+c\cdot A_d{f}_5)\cdot T_{n-1}+(f_3{A}_{d+1}-f_1{f}_2+c\cdot A_d{f}_6)\cdot T_{n-2}\\ +(c\cdot A_{d+1}^2-c\cdot A_d{f}_2+c\cdot A_d{f}_4)\cdot \sum _{k=0}^{n-3}\left(\genfrac{}{}{0ex}{}{n-3}{k}\right)\cdot X_{dk}.\end{array}$$

(13)

For convenience, we let

$$\begin{array}{c}{g}_1=f_1+f_2+c\cdot A_d{f}_5,g_2=f_3{A}_{d+1}-f_1{f}_2+c\cdot A_d{f}_6,\hfill \\ g_3=c\cdot A_{d+1}^2-c\cdot A_d{f}_2+c\cdot A_d{f}_4,g_4=f_3{A}_{d+1}-f_1{f}_2+c\cdot A_d(f_5+f_6),\hfill \\ g_5=c\cdot A_d{f}_6,g_6=c\cdot (A_{d+1}-A_d{f}_2+A_d{f}_4-A_d),g_7=c\cdot A_d{f}_4.\hfill \end{array}$$

Inserting (9) and (13) into (12), we deduce

$$T_{n+1}=\frac{g_1{g}_3-g_1{g}_6+g_7}{g_3-g_6}\cdot T_n+\frac{g_3{g}_4-g_2{g}_6-g_1{g}_7}{g_3-g_6}\cdot T_{n-1}+\frac{g_3{g}_5-g_2{g}_7}{g_3-g_6}\cdot T_{n-2}.$$

(14)

This completes the proof of our theorem.

Remark In fact, using the above formulas, we can also obtain the recurrence formula of the Smarandache-Pascal derived sequence $\{T_n\}$ of $\{u_n\}$, where $\{u_n\}$ denotes the m th-order linear recursive sequences as follows:

$$u_n=a_1{u}_{n-1}+a_2{u}_{n-2}+\cdots +a_{m-1}{u}_{n-m+1}+a_m{u}_{n-m},$$

with initial values $u_i\in \mathbb{N}$ for $n>m$ and $0\le i<m$.