# Some New Codes on thek–Fibonacci Sequence

Authors: M. Hashemi; E. Mehraban
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

In this paper, we define and study thek–Fibonacci sequence matrix for k≥3. By these obtained results, we introduce some new codes on k–Fibonacci sequence.


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## Body

## 1. Introduction

The Fibonacci sequenceFn2 is defined by the recurrence relation F02=0,F12=1, and Fn2=Fn−12+Fn−22,n≥2. The Fibonacci sequence and its generalization sequence (k–Fibonacci sequence number and k–Pell sequence) are famous sequences in mathematics. For information about these sequences, see [1–4].Definition 1.
Fork≥3, the k–Fibonacci sequence, denoted by Fnk−∞∞, is defined by(1)Fnk=Fn−1k+Fn−2k+⋯+Fn−kk,n≥0,Fn−1k=Fn+k−1k−Fn+k−2k+⋯+Fnk,n<0,with initial conditions Fik=0,0≤i≤k−2, and Fk−1k=1 (see [5]).
For example, letk=3, we have(2)Fn3=Fn−13+Fn−23+Fn−3k,n≥0,Fn−13=Fn+23−Fn+13+Fn3,n<0.
So,Fn3−∞∞=⋯,−1,1,0,0,1,1,2,4,7,⋯.
By the definition of thek–Fibonacci sequence, the proof of the following lemma is trivial.Lemma 1.
Fork≥3, the following relations are satisfied about the k–Fibonacci sequence:(i)
Fik=0,i∈ℕ,−k≤i≤−2(ii)
F−1k=1(iii)
F−k−1k=−1
Apostolic and Fraenkle introduced the Fibonacci code which is used in source coding as well as in cryptography (see [6]). For information on the Fibonacci code, see in [7, 8]. The Fibonacci Qp-matrices were introduced (see [9]), and in [10], for p=1, a coding theory is obtained on the Fibonacci Qp-matrices. Also, a Fibonacci coding method is introduced by using Fibonacci polynomials and Fibonacci sequence (see [8, 11–14]). Here, we introduce some coding methods on k–Fibonacci sequence.
In Section2, we define the k–Fibonacci sequence matrix and calculate the determinant of it. Section 3 and Section 4 are devoted to obtain some codes on k–Fibonacci sequence.


## 2.k–Fibonacci Sequence Matrix

The2−Fibonacci sequence matrix (the Fibonacci Q1-matrices) is defined in [15]. The aim of this section is to define the k–Fibonacci sequence matrix and get to the determinant of this matrix. First, we define the k–Fibonacci sequence matrix for k=3 and obtain the determinant of it.Definition 2.
The3−Fibonacci sequence matrix, denoted by Qn,3, is defined:(3)Qn,3=Fn+23Fn+13Fn3Fn+13Fn3Fn−13Fn3Fn−13Fn−23,where Fn3 is the element of the 3−Fibonacci sequence.
For example,n=10, and we have(4)Q10,3=F123F113F103F113F103F93F103F93F83=274149811498144814424.
Now, we get the determinant of the3−Fibonacci sequence matrix. By the definition of Fn3, we have(5)Fn−33=Fn3−Fn−13+Fn−23.Theorem 1.
The determinant ofQn,3 is equal to −1.Proof.
Forn=3m+i where 0≤i≤2, we have(6)Q3m+i,3=F3m+i+23F3m+i+13F3m+i3F3m+i+13F3m+i3F3m+i−13F3m+i3F3m+i−13F3m+i−23.
We can perform specific column operations to reduceQ3m+i,3 to Qi,3, where 0≤i≤2. The operations that we use in this process are as follows:(i)
Add column 2 and column 3, then subtract this summation from column 1, and replace it to the result column1c1−c2+c3⟶c1. So, by relation (5), we have(7)Q3m+i,3⟶F3m−1+i+23F3m+i+13F3m+i3F3m−1+i+13F3m+i3F3m+i−13F3m−1+i3F3m+i−13F3m+i−23.(ii)
Add column 1 and column 3, then subtract it from column 2, and replace to the result column 2. Then, by relation (5), we get(8)F3m−1+i+23F3m+i+13F3m+i3F3m−1+i+13F3m+i3F3m+i−13F3m−1+i3F3m+i−13F3m+i−23⟶F3m−1+i+23F3m−1+i+13F3m+i3F3m−1+i+13F3m−1+i3F3m+i−13F3m−1+i3F3m−1+i−13F3m+i−23.(iii)
Add column 1 and column 2, then subtract this summation from column 3, and replace to it. Thus, by relation (5), we have(9)F3m−1+i+23F3m−1+i+13F3m+i3F3m−1+i+13F3m−1+i3F3m+i−13F3m−1+i3F3m−1+i−13F3m+i−23⟶F3m−1+i+23F3m−1+i+13F3m−1+i3F3m−1+i+13F3m−1+i3F3m−1+i−13F3m−1+i3F3m−1+i−13F3m−1+i−23.
Then, we have(10)Q3m+i,3⟶Q3m−1+i,3
Continuing this processm−1 steps on Q3m−1+i,3, we obtain the following:(11)Q3m−1+i,3⟶Qi,3=Fi+23Fi+13Fi3Fi+13Fi3Fi−13Fi3Fi−13Fi−23.
Therefore,detQn,3=detQi,3. It is sufficient that we get the determinant of Qi,3 where 0≤i≤2. If i=0, then by Lemma 1, we obtain(12)Q0,3=F23F13F03F13F03F−13F03F−13F−23=100001010.
So,detQ0,3=−1. If i=1, then by Lemma 1, we have(13)Q1,3=F33F23F13F23F13F03F13F03F−13=110100001.
Therefore,detQ1,3=−1.
Similarly, fori=2, we get detQ2,3=−1.
Consequently, the theorem is proven.Example 1.
detQ10,3=−1. Since n=3×3+1, we have(14)Q10,3=F123F113F103F113F103F93F103F93F83=274149811498144814424.
We get the following:Step 1:(15)Q10,3⟶Q7,3=F93F83F73F83F73F63F73F63F53=442413241371374.Step 2:(16)Q7,3⟶Q4,3=F63F53F43F53F43F33F43F33F23=7482421211.Step 3:(17)Q4,3⟶Q1,3=F33F23F13F23F13F03F13F03F−13=110100001.
Therefore, detdetQ10,3=detQ1,3=−1.
Now, we are ready to generalize the idea of the3−Fibonacci sequence matrix to the k–Fibonacci sequence matrices k>3.Definition 3.
Thek–Fibonacci sequence matrices of size k×kk≥4, denoted by Qn,k, are defined as follows:(18)Qn,k=Fn+k−1kFn+k−2k⋯FnkFn+k−2kFn+k−3k⋯Fn−1k⋮⋮⋮⋮FnkFn−1k⋯Fn−k−1k,where Fik is the element of the k–Fibonacci sequence.Example 3.
By the definition ofQn,k, we have(19)Q8,4=F114F104F94F84F104F94F84F74F94F84F74F64F84F74F64F54=108562915562915829158415842.
In Theorem1, we get the determinant of the 3−Fibonacci sequence matrix. Now, we obtain the determinant of the k–Fibonacci sequence matrices for k≥4. By the definition of Fnk, we have(20)Fn−kk=Fnk−Fn−1k+Fn−2k+⋯+Fn−k−1k.Theorem 2.
detQn,k is equal to –1.Proof.
Forn=mk+i,0≤i≤k−1, we have(21)Qmk+i,k=Fmk+i+k−1kFmk+i+k−2kldotsFmk+ikFmk+i+k−2kFmk+i+k−3kldotsFmk+i−1k⋮vdotsvdotsvdotsFmk+ikFmk+i−1kldotsFmk+i−k−1k.
We can perform specific column operations to reduceQmk+i,k to Qi,k, where 0≤i≤k−1. The operations that we use in this process are as follows.
Add column 2 until columnk, then subtract this summation from column 1, and replace it to the result column 1c1−c2+c3+⋯+ck⟶c1. Then, add column 1 until column k except column 2, then subtract it from column 2, and replace it to c2−c1+c3+⋯+ck⟶c2. Continuing the process until column k and by relation (20), we have(22)Qmk+i,k⟶Qm−1k+i,k=Fm−1k+i+k−1kFm−1k+i+k−2k⋯Fm−1k+ikFm−1k+i+k−2kFm−1k+i+k−3k⋯Fm−1k+i−1k⋮⋮⋮⋮Fm−1k+ikFm−1k+i−1k⋯Fm−1k+i−k−1k.
Thus, performing above the operations untilm−1 steps, we get(23)Qm−1k+i,k⟶Qi,k=Fi+k−1kFi+k−2k⋯FikFi+k−2kFi+k−3k⋯Fi−1k⋮⋮⋮⋮FikFi−1k⋯Fi−k−1k.
Therefore,detQn,k=detQi,k,0≤i≤k−1.
Leti=0, by Lemma 1, we have F−1k=Fk−1k=−1 and Fik=0,−k+1≤i≤k−2. So, we get(24)Q0,k=Fk−1kFk−2k⋯F0kFk−2kFk−3k⋯F−1k⋮⋮⋮⋮F0kF−1k⋯F−k+1k=10⋯0000⋯0100⋯10⋮⋮⋮⋮⋮01⋯00.
By elementary matrix operations, we getdetQ0,k=−1.
The other cases are proven similarly.


## 3. A Code on thek–Fibonacci Sequence

In this section, we introduce a coding method by using thek–Fibonacci sequence and obtain error detection and correction of it. First, we need the following definition.Definition 4.
Fork=2,n≥4 or k≥3,n≥k+1, let Fnk and Fn+1k be two consecutive elements of the k–Fibonacci sequence. The sequence Ts1Fn+2k is defined as follows:(25)Ts≡sFn+1k+1modFn+2k
Foru=Fn+2k and 1≤i,j≤u, by using definition Ts, we define the matrix Tu=tiju×u as follows:(26)tij=1,ifi=Tj,0.otherwise.Example 3.
Supposek=3 and n=4. Since, F53=4 and u=F63=7, we have(27)T1≡1F53+1=1×5≡5mod7,T2≡2F53+1=2×5≡3mod7,T3≡3F53+1=3×5≡1mod7,T4≡4F53+1=4×5≡6mod7,T5≡5F53+1=5×5≡4mod7,T6≡6F53+1=6×5≡2mod7,T7≡7F53+1=7×5≡7mod7.
Then,(28)tij=1,ifi,j=5,1,3,2,1,3,6,4,4,5,2,6,7,7,0,otherwise.
That is,(29)T7=0010000000001001000000000100100000000010000000001.
Now, by using above notations, we can explain a method coding, which is namedTu− code.
Foru=Fn+2k,1≤i≤u2 and a message matrix M of size u×u,(30)M=m1m2…mu⋮⋮⋮mu2−u+1mu2−u+2…mu2,where mi≥0. We consider a transformation E=M×Tu as the Tu− coding method. Then, M=E×Tu−1 is the Tu− decoding method.Example 4.
Supposek=2andn=3, we have F42=3,F52=5 and T1≡4mod5,T2≡3mod5,T3≡2mod5,T4≡1mod5,andT5≡5mod5. Then,(31)T5=0001000100010001000000001.
Let “mathematics is beautiful” be a message. Then, by Table1 and 0≔blank space, we have(32)M=mathematics0is0beautiful0=F172F52F242F122F92F172F52F242F132F72F232F312F132F232F312F62F92F52F252F242F132F242F252F162F312=1597546368144341597546368233132865713462692332865713462698345750254636823346368750259871346269.
Hence, we get(33)E=M×T5=15975463681443415975463682331328657134626923328657134626983457502546368233463687502598713462690001000100010001000000001=1444636851597342334636851597132865723313462692865713462697502553484636898775025463682331346269.
So, “htameiamcsi s uaebtluti” is the code matrixE.
Note thatM=E×T5−1 is the T5−decoding method.
By using the elementary operations on matrixTu, one can prove the following Lemma.Table 1
English alphabet.
abcdefghijklmnopqrstuvwxyz0Fn+1kFn+2kFn+3kFn+4kFn+5kFn+6kFn+7kFn+8kFn+9kFn+10kFn+11kFn+12kFn+13kFn+14kFn+15kFn+16kFn+17kFn+18kFn+19kFn+20kFn+21kFn+22kFn+23kFn+24kFn+25kFn+26kFn+27kLemma 2.
Foru≥4,detTu=±1. In particular, detT5=1 and detT4=detT7=−1.Note 1.
This lemma follows that forE=M×Tu, we have(34)detE=detM×Tu=detM×detTu=±detM.
In particular, whenu=4,7, we get detE=−detM, and for u=5, we have detE=detM.
By using the above facts, we find the error detection and correction forTu−code where u=4,5,7.
The code matrixE may contain “single,” “double,”…,“u2− fold” errors. Thus, we get(35)1Cu2+2Cu2+⋯+u2Cu2=2u2−1.where rCs=s!/r!×s−r!.
It is clear that the “single” errors in the code matrixE are as follows:(36)1x1e12⋯e1ue21e22⋯e2u⋮⋮⋮eu1eu2⋯euu,2e11x2⋯e1ue21e22⋯e2u⋮⋮⋮eu1eu2⋯euu,…,u2e11e12⋯e1ue21e22⋯e2u⋮⋮⋮eu1eu2⋯xu2,where xi,1≤i≤u2 are possible “destroyed” elements. In this case, there are u2 errors in the matrix code E.
Supposeu=4 and 7. For correction, by using the algebraic equations and the relation detE=−detM, we have(37)x1dete22…e2u⋮⋮eu2…euu−e12dete21…e2u⋮⋮eu2…euu+⋯+−1ue1udete21…e2u−1⋮⋮eu2…euu−1=−detM,e11dete22…e2u⋮⋮eu2…euu−x2dete21…e2u⋮⋮eu2…euu+⋯+−1ue1udete21…e2u−1⋮⋮eu2…euu−1=−detM,⋮eu1dete12…e1u⋮⋮eu−12…eu−1u−1+⋯+−1u2−1eu2−1dete11…e1u⋮⋮eu−11…eu−1u+−1u2xu2dete11…e1u−1⋮⋮eu−11…eu−1u=−detM.
By above relations, we can obtain “single” correction. Using an analogous argument, we can find the error correction forT5−code.


## 4. Coding and Decoding on thek–Fibonacci Sequence Matrix

Here, we find coding and decoding on thek–Fibonacci sequence matrix Qn,k and get its error detection and correction.For k≥3 and an initial message Mk×k, we will name a transformation E=M×Qn,k as the k–Fibonacci coding and a transformation M=E×Qn,k−1 as the k–Fibonacci decoding. Also, the matrix E is as a code matrix. Now, we explain the above method by an example.Example 4.
Supposek=4, n=5, and(38)M=1245243210213457.
We have(39)Q5,4=F84F74F64F54F74F64F54F44F64F54F44F34F54F44F34F24=15842842142112110.
By the above notations, we have(40)E=M×Q5,4=1245243210213457×1245243210213457=57291787638219251374111573215.
Also, we obtain(41)M=E×Q5,4−1=57291787638219251374111573215×1245243210213457−1=1245243210213457.
The error-correction algorithm for the FibonacciQ2-matrices is described in [16]. Now, we explain error detection and correction for the 3−Fibonacci coding.(42)E=M×Qn,3=m1m2m3m4m5m6m7m8m9Fn+23Fn+13Fn3Fn+13Fn3Fn−13Fn3Fn−13Fn−23=e1e2e3e4e5e6e7e8e9,M=E×Qn,3−1=e1e2e3e4e5e6e7e8e9Fn+23Fn+13Fn3Fn+13Fn3Fn−13Fn3Fn−13Fn−23−1=m1m2m3m4m5m6m7m8m9.
Also,(43)detQn,3=Fn+23Fn3Fn−23−Fn−132−Fn+13Fn+13Fn−23−Fn−13Fn3+Fn3Fn+13Fn−13−Fn32=−1,M=e1e2e3e4e5e6e7e8e9Fn−132−Fn3Fn−23Fn+13Fn−23−Fn3Fn−13Fn32−Fn+13Fn−13Fn+13Fn−23−Fn3Fn−13Fn32−Fn+23Fn−23Fn+23Fn−13−Fn3Fn+13Fn32−Fn+13Fn−13Fn+23Fn−13−Fn+13Fn3Fn+132−Fn+23Fn3.
Sincemi≥0,1≤i≤9, we have(44)m1=e1Fn−132−Fn3Fn−23+e2Fn+13Fn−23−Fn3Fn−13+e3Fn32−Fn+13Fn−13>0,(45)m2=e1Fn+13Fn−23−Fn3Fn−13+e2Fn32−Fn+23Fn−23+e3Fn+23Fn−13−Fn+13Fn3>0,(46)m3=e1Fn32−Fn+13Fn−13+e2Fn+23Fn−13−Fn3Fn+13+e3Fn+132−Fn+23Fn3>0,(47)m4=e4Fn−132−Fn3Fn−23+e5Fn+13Fn−23−Fn3Fn−13+e6Fn32−Fn+13Fn−13>0,m5=e4Fn+13Fn−23−Fn3Fn−13+e5Fn32−Fn+23Fn−23+e6Fn+23Fn−13−Fn+13Fn3>0,m6=e4Fn32−Fn+13Fn−13+e5Fn+23Fn−13−Fn3Fn+13+e6Fn+132−Fn+23Fn3>0,m7=e7Fn−132−Fn3Fn−23+e8Fn+13Fn−23−Fn3Fn−13+e9Fn32−Fn+13Fn−13>0,m8=e7Fn+13Fn−23−Fn3Fn−13+e8Fn32−Fn+23Fn−23+e9Fn+23Fn−13−Fn+13Fn3>0,m9=e7Fn32−Fn+13Fn−13+e8Fn+23Fn−13−Fn3Fn+13+e9Fn+132−Fn+23Fn3>0.
From (44), we get(48)e1Fn−132+e2Fn+13Fn−23+e3Fn32>e1Fn3Fn−23+e2Fn3Fn−13+e3Fn+13Fn−13.
By using (45), we have(49)e1Fn+13Fn−23+e2Fn32+e3Fn+23Fn−13>e1Fn3Fn−13+e2Fn+23Fn−23+e3Fn+13Fn3.
From (46), we get(50)e1Fn32+e2Fn+23Fn−13+e3Fn+132>e1Fn+13Fn−13−e2Fn3Fn+13+e3Fn+23Fn3.
By relations (48)–(50), we have(51)e2Fn3Fn−13−Fn+13Fn−23+Fn3Fn−23−Fn−132<e3Fn32−Fn+13Fn−13,(52)e2Fn+23Fn−23−Fn32+Fn3Fn−13−Fn+13Fn−23<e3Fn+23Fn−13−Fn+13Fn3,(53)e2Fn+23Fn−13−Fn3Fn+13+Fn32−Fn+13Fn−13>e3Fn+23Fn3−Fn+132.
LetA1=Fn32−Fn+13Fn−13,A2=Fn+23Fn−13−Fn+13Fn3, and A3=Fn+23Fn3−Fn+132. Now, there exist 33=27 cases for A1>=<0,A2>=<0,A3>=<0. We calculate two cases and the rest of the cases get in similar ways.Case 1.
LetA1>0,A2>0, and A3>0. By using (51) and (53), we have(54)e2Fn3Fn−13−Fn+13Fn−23e1A1+Fn3Fn−23−Fn−132A1<e3e1,e2Fn+23Fn−13−Fn3Fn+13e1A3+Fn32−Fn+13Fn−13A3>e3e1·
Then, we get(55)e2Fn3Fn−13−Fn+13Fn−23e1A1+Fn3Fn−23−Fn−132A1<e2Fn+23Fn−13−Fn3Fn+13e1A3+Fn32−Fn+13Fn−13A3.
So,(56)A3Fn3Fn−23−Fn−132−A1Fn32−Fn+13Fn−13A1A3<e2A1Fn+23Fn−13−Fn3Fn+13−A3Fn3Fn−13−Fn+13Fn−23e1A1A3·
SinceA1=Fn32−Fn+13Fn−13,A3=Fn+23Fn3−Fn+132 and by using relation (43), we have(57)A3Fn3Fn−23−Fn−132−A1Fn32−Fn+13Fn−13=Fn3detQn,3,A1Fn+23Fn−13−Fn3Fn+13−A3Fn3Fn−13−Fn+13Fn−23=Fn+13detQn,3·
Therefore,(58)e1e2<Fn+13Fn3·
Now, we consider (52) and (53). Then,(59)e2Fn+23Fn−23−Fn32e1A2+Fn3Fn−13−Fn+13Fn−23A2<e3e1,e2Fn+23Fn−13−Fn3Fn+13e1A3+Fn32−Fn+13Fn−13A3>e3e1·
Hence, we have(60)e2Fn+23Fn−23−Fn32e1A2+Fn3Fn−13−Fn+13Fn−23A2>e2Fn+23Fn−13−Fn3Fn+13e1A3+Fn32−Fn+13Fn−13A3,A3Fn3Fn−13−Fn+13Fn−2−A2Fn32−Fn+13Fn−13A2A3<e2A2Fn+23Fn−13−Fn3Fn+13−A3Fn+23Fn−23−Fn32e1A2A3·
We get,(61)A3Fn3Fn−13−Fn+13Fn−23−A2Fn32−Fn+13Fn−13=−Fn+13detQn,3,A2Fn+23Fn−13−Fn3Fn+13−A3Fn+23Fn−23−Fn32=Fn+23detQn,3.
So,(62)e1e2>Fn+23Fn+13.
From (58) and (62), we have(63)Fn+23Fn+13<e1e2<Fn+13Fn3·
Similarly, one can prove that(64)Fn+23Fn+13<e2e3<Fn+13Fn3,Fn+23Fn3<e1e3<Fn+13Fn−13.Case 2.
LetA1<0,A2<0, and A3<0. By using (51) and (53), we have(65)e2Fn3Fn−13−Fn+13Fn−23e1A1+Fn3Fn−23−Fn−132A1>e3e1,e2Fn+23Fn−13−Fn3Fn+13e1A3+Fn32−Fn+13Fn−13A3<e3e1.
Then, we get(66)e2Fn3Fn−13−Fn+13Fn−23e1A1+Fn3Fn−23−Fn−132A1>e2Fn+23Fn−13−Fn3Fn+13e1A3+Fn32−Fn+13Fn−13A3·
So,(67)A3Fn3Fn−23−Fn−132−A1Fn32−Fn+13Fn−13A1A3>e2A1Fn+23Fn−13−Fn3Fn+13−A3Fn3Fn−13−Fn+13Fn−23e1A1A3·
By using (43), we have(68)A3Fn3Fn−23−Fn−132−A1Fn32−Fn+13Fn−13=Fn3detQn,3,A1Fn+23Fn−13−Fn3Fn+13−A3Fn3Fn−13−Fn+13Fn−23=Fn+13detQn,3.
Therefore, we have(69)e1e2>Fn+13Fn3·
In the following, by relations (52) and (53), we get(70)e2Fn+23Fn−23−Fn32e1A2+Fn3Fn−13−Fn+13Fn−23A2>e3e1,e2Fn+23Fn−13−Fn3Fn+13e1A3+Fn32−Fn+13Fn−13A3<e3e1·
Then, we have(71)e2Fn+23Fn−23−Fn32e1A2+Fn3Fn−13−Fn+13Fn−23A2>e2Fn+23Fn−13−Fn3Fn+13e1A3+Fn32−Fn+13Fn−13A3,A3Fn3Fn−13−Fn+13Fn−23−A2Fn32−Fn+13Fn−13A2A3>e2A2Fn+23Fn−13−Fn3Fn+13−A3Fn+23Fn−23−Fn32e1A2A3·
We get(72)A3Fn3Fn−13−Fn+13Fn−23−A2Fn32−Fn+13Fn−13=−Fn+13detQn,3,A2Fn+23Fn−13−Fn3Fn+13−A3Fn+23Fn−23−Fn32=Fn+23detQn,3.
So,(73)e1e2<Fn+23Fn+13·
By using relations (69) and (73), we have(74)Fn+13Fn3<e1e2<Fn+23Fn+13.
Similarly, we get(75)Fn+13Fn3<e2e3<Fn+23Fn+13,Fn+13Fn−13<e1e3<Fn+23Fn3.
Similarly, by the above argument, we get(76)Fn+13Fn3≶e1e2≶Fn+23Fn+13,(77)Fn+13Fn3≶e2e3≶Fn+23Fn+13,(78)Fn+13Fn−13≶e1e3≶Fn+23Fn3.
So, by above facts, we have(79)limn⟶∞Fn+13Fn3=limn⟶∞Fn+23Fn+13=λ,limn⟶∞Fn+13Fn−13=limn⟶∞Fn+23Fn3=λ2,e1e2≈⋋,e2e3≈⋋,e1e3≈⋋2,where ⋋=1.839 is the golden ratio of the 3−Fibonacci sequence.
Hence, we have(80)e4e5≈⋋,e5e6≈⋋,e4e6≈⋋2,e7e8≈⋋,e8e9≈⋋,e7e9≈⋋2.
Now, we are in position that the above results are generalized to thek–Fibonacci coding.
Here, fork>3, we get relations among entries the code matrix E. Similar to k=3, we can obtain the following relations among the first-row entries code matrix E.(81)e1e2≈μk,e2e3≈μk,…,ek−1ek≈μk,e1e3≈μk2,e2e4≈μk2,…,ek−2ek≈μk2,⋮e1ek≈μkk−1,where μk is the golden ratio of the k–Fibonacci sequence.
In general, the following relations among the entries of each row of the code matrixE=eijk×k are obtained by(82)Fn+k−2kFn+k−s−2k≶eijeij+s≶Fn+k−1kFn+k−s−1k,where i,j=1,2,…,k,s=1,2,…,k−1 and 2≤j+s≤k,n>k.
Now, we calculate the determinant of the code matrixE.
For the coding matrixQn,k, E=M×Qn,k and detQn,k=−1, and we have(83)detE=−detM.
Here, we calculate the error detection and correction for thek–Fibonacci coding.
Letk=3. According to the matrix E of the order 3×3, we have “single,” “double,”…, “nine-fold” errors. The first assumption is that there exists only one error in the matrix E received from the communication channel. It is clear that there are nine different cases for it, as follows:(84)1ae2e3e4e5e6e7e8e9,2e1be3e4e5e6e7e8e9,3e1e2ce4e5e6e7e8e9,4e1e2e3de5e6e7e8e9,5e1e2e3e4ee6e7e8e9,6e1e2e3e4e5fe7e8e9,7e1e2e3e4e5e6ge8e9,8e1e2e3e4e5e6e7he9,9e1e2e1e4e5e6e7e8i,where a,b,…,i are possible “destroyed” entries.
FromdetE=−detM, we have(85)1ae5e9−e6e8−e2e4e9−e6e7+e3e4e8−e5e7=−detM,2e1e5e9−e6e8−be4e9−e6e7+e3e4e8−e5e7=−detM,⋮9e1e5i−e6e8−e2e4i−e6e7+e3e4e8−e5e7=−detM.
In a similar way, we will obtain a double error for the matrixE. For example, we consider a bivariate case for matrix E as follows:(86)abe3e4e5e6e7e8e9,,
in which possible cases are2C9=36. Similarly, we obtain “triple,” “four-fold,” …, “nine-fold” errors, which the total number of cases is 1C9+2C9+⋯+9C9=29−1=511 errors. By using detE=−detM and the relations (76)–(78), we can correct up to “single,” “double,” … , “eight” errors except “nine” errors. Therefore, we get that the correctable possibility of the method is equal to 510/511=0.9980=%99.80.
In general, fork>3, by the above method, the code matrix E may contain “single,” “double,”…,“k2− fold” errors. Thus,(87)1Ck2+2Ck2+⋯+k2Ck2=2k2−1.
Therefore, there are2k2−1 errors. By detE=−detM and relation (82), we can give that the correct ability of the k–Fibonacci sequence matrix coding is equal to(88)2k2−22k2−1.
Then, for large value ofk, the correct possibility of this method is(89)2k2−22k2−1≃1=%100.


## 5. Conclusion

In this paper, we give some codes on thek–Fibonacci sequence. These coding methods are the applications of this sequence. Also, we obtain the following results:(1)
Foru=4,5 and 7, we get error detection and correction.(2)
Fork>3, the correct ability of the k–Fibonacci sequence matrix coding is 2k2−2/2k2−1≃%100. In particular, for k=3, the correct ability of this method is equal to 510/511=%99.80.



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*Source: 7660902-2021-08-27.xml*