# Sequential Properties over Negative Pauli Pascal Table

Authors: Eunmi Choi; Jiin Jo
Journal: International Journal of Mathematics and Mathematical Sciences
 (2020)

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## Abstract

For polynomialsx+yi and x+y−i satisfying the noncommutative multiplication yx=−xy, let C−1 and N−1 be the arithmetic tables, respectively. We investigate sequential properties of various diagonal sums over the tables C−1 and N−1 and prove that they are types of interlocked Fibonacci sequence and Padovan sequence.


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## Body

## 1. Introduction

The Pascal tableC1=ci,j1 is an arithmetic table of a polynomial x+yi=∑j=0ici,j1xi−jyj,i≥0, in which yx=xy is assumed tactically. The Pauli Pascal table C−1=ci,j−1 is an arithmetic table of x+yi=∑j=0ici,j−1xi−jyj with noncommuting variables x,y such that yx=−xy. The ci,j1 and ci,j−1 satisfy the following (see [1]):(1)ci+1,j+11=ci,j1+ci,j+11,ci+1,j+1−1=ci,j−1+−1j+1ci,j+1−1,i,j≥0. It is known that diagonal sums over C1 give a Fibonacci sequence f1,f2,f3,…,, while those over C−1 yield a sequence f1,f0,f2,f1,f3,f2,…, that is interlocked by two Fibonacci sequences [2].Consider a polynomialx+y−i with negative exponent satisfying either yx=xy or yx=−xy and denote the corresponding arithmetic table by either N1=ni,j1 or N−1=ni,j−1 with x+y−i=∑j=0∞ni,j±1x−i−jyj, respectively.By at/s-slope diagonal over a table (t,s≥1), we mean a generalized diagonal moving s steps along x-axis and t steps along y-axis. Over C±1, let dt/s,i±1 denote the t/s-slope diagonal set starting from ci,0±1i≥0 toward northeast direction, and Dt/s,i±1 be the sum of elements in dt/s,i±1. We call it the t/s-slope ith diagonal sum. Similarly, over N±1, let gt/s,j±1 be the t/s-slope diagonal set starting from n1,j±1j≥0 toward southwest direction, and Gt/s,j±1 be the t/s-slope jth diagonal sum. When s=1, we simply say t-slope diagonal sums Dt,i±1 and Gt,j±1.A purpose of the work is to study arithmetic tablesC±1 and N±1. We investigate sequential properties of generalized diagonal sums Dt/s,i±1 and Gt/s,j±1 and find their interrelationships. We particularly give attention to Gt/s,j−1 of N−1 and prove Gt/s,j−1 is a type of interlocked Padovan sequence. The results of the work provide interesting connections of sequences over the arithmetic tables of x+y±i having either commutative yx=xy or noncommutative yx=−xy rules.


## 2. Arithmetic Table and Its Diagonal Sum

The arithmetic tableN1=ni,j1 of x+y−i with yx=xy can be obtained by Taylor series expansion. Every element ni,j1 (i≥1,j≥0) of N1 in Table 1 satisfies a recurrence rule ni+1,j+11=ni,j+11−ni+1,j1.Table 1
N1=ni,j1.
11−11−11−11−1⋯21−23−45−67−8⋯31−36−1015−2128−36⋯41−410−2035−5684−120⋯51−515−3570−126210−330⋯By flippingN1 and passing it over C1, the pile-up table  follows the Pascal rule (1). For example, we get x+y−4=x−41−4x−1y+10x−2y2−20x−3y3+35x−4y4+,…,. Some recurrence rules of t-slope diagonal sum Dt,i±1 over C±1 and 1/t-slope diagonal sum G1/t,j1 over N1 were studied as follows.Lemma 1 (see [3, 4]).
Dt,i1+Dt,i+t1=Dt,i+t+11 with t+1 initials 1,…,1. And G1/t,i1−G1/t,i+t−11=G1/t,i+t1 with t initials 1,−1,1,−1,…,. Moreover, Dt,i−1+Dt,i+2t−1=Dt,i+2t+2−1 (i≥2t+1) with initials 1,…,1︸t+1,2,1,2,1,2,1,…,︸t+1.
The proof of Lemma1 is mainly due to(2)ni,j1=−1jci+j−1,j1,ni,01=ci,01=1andni,11=−ci,11=−i.
LetD1,j−1e and D1,j−1o be subsequences having only eventh and oddth terms, respectively, in D1,i−1|i≥0. From Table 2, we observe(3)D1,i−1|i≥0=1,2,3,5,8,13,21,…,∪1,1,2,3,5,8,13,…,=D1,i−1e|j≥0∪D1,j−1o|j≥0.
Since bothD1,j−1e and D1,j−1o are Fibonacci sequences, we say D1,i−1 is an interlocked Fibonacci sequence. For any t≥1, a sequence fn satisfying fn+fn+t=fn+t+1 with t+1 initials is called a Fibo t-sequence [5].Table 2
G1/t,i1andDt,i−1.
t/i12345678912345678910000000001121325382−12−35−813−2134−551112122333−110−12−211−31111212134−11−12−34−57−101111121215−11−110−12−34Theorem 1.
Dt,i−1 is an interlocked Fibo t-sequence for t≥1.Proof.
LetDt,j−1e and Dt,j−1o be subsequences consisting of eventh or oddth terms, respectively, in Dt,i−1|i≥0. When t=1, D1,i−1 is clearly an interlocked Fibo 1-sequence. When t=2,3, Table 2 shows that(4)D2,i−1|i≥0=1,1,1,2,3,4,6,9,13,19,…,∪1,2,2,3,5,7,10,15,…,=D2,j−1e|j≥0∪D2,j−1o|j≥0,where D2,j−1e and D2,j−1o are Fibo 2-sequences with initials 1,1,1 and 1,2,2. Similarly, D3,i−1|i≥0=D3,j−1e|j≥0∪D3,j−1o|j≥0 shows D3,j−1e=1,1,2,2,3,4,6,8,11,…, and D3,j−1o=1,1,1,1,2,3,4,5,7,10,…, are Fibo 3-sequences having initials 1,1,2,2 and 1,1,1,1. So, Dt,i−1 with t=2,3 is an interlocked Fibo t-sequence.
In general, for anyt≥1, Lemma 1 shows that the sequence(5)…,Dt,i−1,Dt,i+1−1,…,Dt,i+2t−1,Dt,i+2t+1−1,Dt,i+2t+2−1,…,holds Dt,i−1+Dt,i+2t−1=Dt,i+2t+2−1 (i≥2t+1) with initials 1,…,1︸t+1,2,1,2,1,…,︸t+1. Thus, Dt,j−1e|j≥0 and Dt,j−1o|j≥0 satisfy recurrences Dt,j−1e+Dt,j+t−1e=Dt,j+t+1−1e and Dt,j−1o+Dt,j+t−1o=Dt,j+t+1−1o. So, they are Fibo t-sequences having initials 1,…,1t+1tuples2|t1,…,1︸t+1/2,2,…,2︸t+1/22∤t and 1,…,1︸t/2,2,…,2︸t/2+12|t1,…,1t+1tuples2∤t, respectively. Hence, Dt,i−1 is an interlocked Fibo t-sequence.
Now, in order to have the tableN−1=ni,j−1 of x+y−i with yx=−xy, look at the piled-up table  satisfying the Pauli rule (1).
Then, by flipping the upper part upside down, we getN−1 (Table 3) holding(6)ni+1,j+1−1=−1j+1ni,j+1−1−ni+1,j−1fori≥1,j≥0.
Thus,C−1 and N−1 yield expansions of x+y±i with yx=−xy, for instance, x+y−5=x−51+x−1y−3x−2y2−3x−3y3+6x−4y4+6x−5y5+,⋯,.Table 3
N−1=ni,j−1.
012345678910⋯111−1−111−1−111−1⋯210−1010−1010−1⋯311−2−233−4−455−6⋯410−2030−4050−6⋯511−3−366−10−101515−21⋯Theorem 2.
ni,j−1=ci+j−1,j−1j≡0,1mod4−ci+j−1,j−1j≡2,3mod4. In particular, n2i,2j+1−1=0 and n2i−1,2j−1=n2i−1,2j+1−1=n2i,2j−1 for any i,j≥1.Proof.
Sincen5,4−1=6c8,4−1=6n5,5−1=6c9,5−1=6n5,6−1=−10c10,6−1=10n5,7−1=−10c11,7−1=10n5,8−1=15c12,8−1=15, we may assume ni,j−1=ci+j−1,j−1,j≡0,1mod4−ci+j−1,j−1,j≡2,3mod4 for some i,j.
Ifj≡0mod4, then recurrence (6) implies(7)ni+1,j−1=−1jni,j−1−ni+1,j−1−1=ni,j−1−ni+1,j−1−1=ci+j−1,j−1−−1ci+1,j−1−1=−1jci+j−1,j−1+ci+1,j−1−1=ci+j,j−1.
Similarly, ifj≡1mod4, then(8)ni+1,j−1=−ci+j−1,j−1−ci+1,j−1−1=−1jci+j−1,j−1+ci+1,j−1−1=ci+j,j−1.
The other casesj≡2,3mod4 can be proved analogously. In particular, −1n2i−1,2j+1−1−n2i,2j−1=n2i,2j+1−1=0, so n2i,2j−1=n2i−1,2j+1−1.
Theorem2 can be compared to ni,j1=ci+j−1,j1j≡0mod2−ci+j−1,j1j≡1mod2 in (2).


## 3. Diagonal Sum overN−1

We will discusst-slope diagonal gt,i−1 and its sum Gt,i−1 on N−1.Theorem 3.
Whent=1, G1,0−1=1, G1,1−1=2, and G1,i−1=0 for i≥2.
Whent=1/2, −1iG1/2,i−1−G1/2,i+1−1=G1/2,i+2−1 for i≥1.Proof.
The first few 1-slope diagonal sets and sums inN−1 are as follows:(9)
In general, by (6) and n2i,2j+1−1=0 for all i,j (Theorem 2), we have(10)G1,2l−1=n1,2l−1+n2,2l−1−1+n3,2l−2−1+,…,+n2l,1−1+n2l+1,0−1=n1,2l−1+n3,2l−2−1+n5,2l−4−1+n7,2l−6−1+,…,+n2l−1,2−1+n2l+1,0−1=n1,2l−1+n2,2l−2−1−n3,2l−3−1+,…,+n2l−2,2−1−n2l−1,1−1+n2l+1,0−1=n1,2l−1+n2,2l−2−1−,…,−n2l−3,3−1−n2l−2,2−1−n2l−1,1−1+n2l+1,0−1=0,because n2k−1,2l+1−1=n2k,2l−1, n1,2l−1=−n1,2l−1−1=−n2,2l−2−1 and n2l−1,1−1=n2l,0−1=n2l+1,0−1 in Theorem 2.
Similarly,n2k−1,2l+1−1=n2k,2l−1=n2k−1,2l−1 and n2k,2l+1−1=0 also show(11)G1,2l+1−1=n1,2l+1−1+n2,2l−1+n3,2l−1−1+n4,2l−2−1+,⋯,+n2l+1,1−1+n2l+2,0−1=2n2,2l−1+n4,2l−2−1+n6,2l−4−1+,⋯,+n2l,2−1+n2l+2,0−1=2n1,2l−1+n3,2l−2−1+n5,2l−4−1+,⋯,+n2l−1,2−1+n2l+1,0−1=2n1,2l−1+n2,2l−1−1+n3,2l−2−1+,⋯,+n2l−1,2−1+n2l,1−1+n2l+1,0−1=2G1,2l−1=0.
Therefore, we haveG1,i−1|i≥0=1,2,0,0,0,0,0,…,.
The1/2-slope diagonals sets and their sums G1/2,i−1 in N−1 are as follows:(12)
Thus,G1/2,i−1|i≥1=1,0,−1,1,2,−1,−3,2,5,−3,…, satisfies(13)G1/2,5−1−G1/2,6−1=−G1/2,7−1,G1/2,6−1−G1/2,7−1=G1/2,8−1.
So, we haveG1/2,i−1−G1/2,i+1−1=−1iG1/2,i+2−1 for 1≤i≤7.
Now, wheni=2l+1, we have(14)G1/2,2l+1−1=n1,2l+1−1+n2,2l−1−1+n3,2l−3−1+,…,+nl,3−1+nl+1,1−1=n1,2l+1−1+−1n1,2l−1−1−n2,2l−2−1+−1n2,2l−3−1−n3,2l−4−1+−1n3,2l−5−1−n4,2l−6−1+,…,+−1nl−1,3−1−nl,2−1+nl+1,1−1=−n1,2l−1−1+n2,2l−3−1+n3,2l−5−1+,…,+nl−1,3−1+nl+1,1−1+n1,2l−1+n2,2l−2−1+n3,2l−4−1+n4,2l−6−1+,…,+nl,2−1=−G1/2,2l−1−1+G1/2,2l−1,for n1,2l+1−1=n1,2l−1. Hence, G1/2,2l−1−1−G1/2,2l−1=−G1/2,2l+1−1.
On the other hand, wheni=2l, we have(15)G1/2,2l−1=n1,2l−1+n1,2l−2−1−n2,2l−3−1+,⋯,+nl−1,2−1−nl,0−1+nl+1,0−1=n1,2l−2−1+n2,2l−4−1+⋯+nl+1,0−1−n1,2l−1−1+n2,2l−3−1+⋯+nl,0−1=G1/2,2l−2−1−G1/2,2l−1−1,for n1,2l−1=−n1,2l−1−1. Thus, G1/2,2l−2−1−G1/2,2l−1−1=G1/2,2l−1.
Let us continue to work with1/3-slope diagonals in N−1.Theorem 4.
Fori≥1, G1/3,i−1+−1iG1/3,i+2−1=G1/3,i+3−1.Proof.
ObserveG1/3,5−1−G1/3,7−1=G1/3,8−1 and G1/3,6−1+G1/3,8−1=G1/3,9−1 from(16)
Let2∤i. Since n1,i+3−1=−n1,i+2−1 and np,q−1=0 for 2|p, 2∤q, we have(17)G1/3,i−1+−1iG1/3,i+2−1=n1,i−1+n2,i−3−1+n3,i−6−1+⋯−n1,i+2−1+n2,i−1−1+n3,i−4−1+⋯=−n1,i+2−1+n1,i−1−n2,i−1−1+n2,i−3−1−n3,i−4−1+n3,i−6−1−n4,i−7−1+⋯=−n1,i+2−1+−1in2,i−1+n3,i−3−1+−1in4,i−6−1+n5,i−9−1+⋯=n1,i+3−1+n2,i−1+n3,i−3−1+n4,i−6−1+n5,i−9−1+⋯=G1/3,i+3−1.
Now, let2|i. Again with np,q−1=0 (2|p, 2∤q), we also have(18)G1/3,i−1+−1iG1/3,i+2−1=n1,i−1+n2,i−3−1+n3,i−6−1+⋯−n1,i+2−1+n2,i−1−1+n3,i−4−1+⋯=−n1,i+2−1+n1,i−1−n2,i−1−1+n2,i−3−1−n3,i−4−1+n3,i−6−1−n4,i−7−1+⋯=n1,i+3−1+n2,i−1+n3,i−3−1+n4,i−6−1+n5,i−9−1+⋯=G1/3,i+3−1.
We now have recurrence rules of1/t-slope diagonal sums over N−1.Theorem 5.
G1/t,i−1−G1/t,i+t−1−1=−1iG1/t,i+t−1 with even t>1.
AndG1/t,i−1+−1iG1/t,i+t−1−1=G1/t,i+t−1 with odd t>1.Proof.
Let2|t. Since n1,i+t−1−1=−1in1,i+t−1, recurrence (6) yields(19)G1/t,i−1−G1/t,i+t−1−1=n1,i−1+n2,i−t−1+n3,i−2t−1+n4,i−3t−1+⋯+ni/t,i−i/t−1t−1−n1,i+t−1−1+n2,i−1−1+n3,i−t−1−1+⋯+ni/t+1,i−i/t−1t−1−1=−n1,i+t−1−1+n1,i−1−n2,i−1−1+n2,i−t−1−n3,i−t−1−1+n3,i−2t−1−n4,i−2t−1−1+⋯+ni/t,i−i/t−1t−1−ni/t+1,i−i/t−1t−1−1=−1in1,i+t−1+−1in2,i−1+−1in3,i−t−1+⋯+−1ini/t+1,i−i/t−1t−1=−1iG1/t,i+t−1.
Now, assume2∤t and 2∣i. Then, n2,i−1−1=n2,i−t−1=n4,i−2t−1−1=n4,i−3t−1=,⋯,=0 for 2∤i−t. However, since n1,i+t−1−1=n1,i+t−1, we have(20)G1/t,i−1+−1iG1/t,i+t−1−1=n1,i−1+n2,i−t−1+n3,i−2t−1+⋯−n1,i+t−1−1+n2,i−1−1+n3,i−t−1−1+⋯=n1,i+t−1−1+n1,i−1+n2,i−1−1+n2,i−t−1+n3,i−t−1−1+⋯=n1,i+t−1−1+n1,i−1−n2,i−1−1−n2,i−t−1−n3,i−t−1−1+⋯=n1,i+t−1+−1in2,i−1−−1i−tn3,i−t−1+−1in4,i−2t−1+⋯=n1,i+t−1+n2,i−1+n3,i−t−1+n4,i−2t−1+n5,i−3t−1+⋯=G1/t,i+t−1.
Similarly, if2∤t and 2∤i, then n2,i−1=n4,i−2t−1=n6,i−4t−1=,⋯,=0, so(21)G1/t,i−1+−1iG1/t,i+t−1−1=n1,i−1+n2,i−t−1+n3,i−2t−1+⋯−n1,i+t−1−1+n2,i−1−1+n3,i−t−1−1+⋯=−n1,i+t−1−1+n1,i−1−n2,i−1−1+n2,i−t−1−n3,i−t−1−1+⋯=−n1,i+t−1+−1in2,i−1−−1i−tn3,i−t−1+−1in4,i−2t−1+⋯=n1,i+t−1+n2,i−1+n3,i−t−1+n4,i−2t−1+n5,i−3t−1+⋯=G1/t,i+t−1.
A more explicit relation of the diagonal setsdt,i−1 and g1/t,i−1 is as follows.Theorem 6.
Letdt,i,k and g1/t,i,k be the kth elements of the sets dt,i−1 and g1/t,i−1, respectively. Then, g1/2,i,k=−1kd1,i,kifi≡0,1mod4−1k+1d1,i,kifi≡2,3mod4 and g1/3,i,k=d2,i,kifi+k≡0,1mod4−d2,i,kifi+k≡2,3mod4. In general, any kth elements g1/t+1,i,k in g1/t+1,i−1 and dt,i,k in dt,i−1 are the same, except for signs.Proof.
Notedt,i−1=dt,i,k|k≥0=ci,0−1,ci−t,1−1,…,ci−kt,k−1,…, of C−1 and g1/t,i−1=g1/t,i,k|k≥0=n1,i−1,n2,i−t−1,…,nk+1,i−kt−1,…, of N−1. Theorem 2 and the symmetricity of C−1 imply(22)g1/2,i,0=n1,i−1=ci,i−1=ci,0−1=d1,i,0,i≡0,1mod4,−ci,i−1=−ci,0−1=−d1,i,0,i≡2,3mod4,g1/2,i,1=n2,i−2−1=ci−1,1−1=d1,i,1,i≡2,3mod4,−ci−1,1−1=−d1,i,1,i≡0,1mod4,g1/2,i,2=n3,i−4−1=ci−2,2−1=d1,i,2,i≡0,1mod4,−ci−2,2−1=−d1,i,2,i≡2,3mod4,g1/2,i,3=n4,i−6−1=ci−3,3−1=d1,i,3,i≡2,3mod4,−ci−3,3−1=−d1,i,3,i≡0,1mod4,which shows g1/2,i,k=−1kd1,i,k,i≡0,1mod4−1k+1d1,i,k,i≡2,3mod4 for 0≤k≤3.
Now, for anykth element in the diagonal set g1/2,i−1, we note that(23)g1/2,i,k=nk+1,i−2k−1=ci−k,i−2k−1=ci−k,k−1i−2k≡0,1mod4,−ci−k,i−2k−1=−ci−k,k−1i−2k≡2,3mod4.
By mod 4, ifi−2k≡0, then i≡0 or 2 according to k≡0,2 or k≡1,3. Thus, i−2k≡1 implies i≡1 or 3 according to k≡0,2 or k≡1,3. Similarly, i−2k≡2 means i≡0 (if k≡1,3) or i≡2 (if k≡0,2). And i−2k≡3 says i≡1 (if k≡1,3) or i≡3 (if k≡0,2).
Thus, we have the following 4 cases (all congruences are by mod 4):(i)
Leti≡0. If k≡0,2 then i−2k≡0, so g1/2,i,k=ci−k,k−1=−1kci−k,k−1=−1kd1,i,k. If k≡1,3 then i−2k≡2, so g1/2,i,k=−1kci−k,k−1=−1kd1,i,k. Thus, g1/2,i,k=−1kd1,i,k for any k.(ii)
Leti≡1. If k≡0,2 then i−2k≡1, so g1/2,i,k=−1kci−k,k−1=−1kd1,i,k. If k≡1,3 then i−2k≡3, so g1/2,i,k=−ci−k,k−1=−1kd1,i,k. Thus, g1/2,i,k=−1kd1,i,k for any k.(iii)
Leti≡2. If k≡1,3 then i−2k≡0, so g1/2,i,k=−1k+1ci−k,k−1=−1k+1d1,i,k. If k≡0,2 then i−2k≡2, so g1/2,i,k=−ci−k,k−1=−1k+1d1,i,k.(iv)
Leti≡3. If k≡1,3, then i−2k≡1, so g1/2,i,k=ci−k,k−1=−1k+1d1,i,k. If k≡0,2, then i−2k≡3, so g1/2,i,k=−ci−k,k−1=−1k+1d1,i,k.
Therefore, we haveg1/2,i,k=−1kd1,i,kifi≡0,1−1k+1d1,i,kifi≡2,3.
Similarly, in the setg1/3,i−1, the kth element g1/3,i,k (0≤k≤3) are(24)g1/3,i,0=n1,i−1=ci,0−1=d2,i,0,i≡0,1,−ci,0−1=−d2,i,0,i≡2,3,g1/3,i,1=n2,i−3−1=ci−2,1−1=d2,i,1,i≡0,3,−ci−2,1−1=−d2,i,1,i≡1,2,g1/3,i,2=d2,i,2,i≡2,3,−d2,i,2,i≡0,1,g1/3,i,3=d2,i,3,i≡1,2,−d2,i,3,i≡0,3,where all congruences are by mod 4. Thus, we generally have(25)g1/3,i,k=nk+1,i−3k−1=ci−2k,k−1,i−3k≡0,1−ci−2k,k−1,i−3k≡2,3=d2,i,k,i+k≡0,1,−d2,i,k,i+k≡2,3.
Now, the next table showsdt,i−1 and g1/t+1,i−1 for the first few t.(26)
And in general by Theorem2, we have (congruences are by mod 4)(27)g1/t+1,i,k=nk+1,i−kt+1−1=ci−tk,k−1=dt,i,k,i−kt+1≡0,1,−ci−tk,k−1=−dt,i,k,i−kt+1≡2,3.
We are now ready to obtain a recurrence onG1/t,i−1.Theorem 7.
G1/t,i−1−G1/t,i+2t−1−1=G1/t,i+2t−1 for any t,i>0.Proof.
Theorem3 implies G1/2,i−1−G1/2,i+1−1=−1iG1/2,i+2−1 and G1/2,i+1−1−G1/2,i+2−1=−1i+1G1/2,i+2−1. So, we have(28)G1/2,i−1−G1/2,i+2−1=−1iG1/2,i+2−1−G1/2,i+3−1=−1i−1i+2G1/2,i+4−1=G1/2,i+4−1.
Similarly, Theorem4 shows G1/3,i−1+−1iG1/3,i+2−1=G1/3,i+3−1 and G1/3,i+2−1+−1i+2G1/3,i+4−1=G1/3,i+5−1. Thus, if 2|i, then(29)G1/3,i−1−G1/3,i+4−1=G1/3,i+3−1−G1/3,i+5−1=G1/3,i+3−1+−1i+3G1/3,i+5−1=G1/3,i+6−1,
while if2∤i, then(30)G1/3,i−1−G1/3,i+4−1=G1/3,i+3−1−G1/3,i+5−1=G1/3,i+6−1.
Now, consider anyt≥1. If 2|t, then G1/t,i−1−G1/t,i+t−1−1=−1iG1/t,i+t−1 and G1/t,i+t−1−1−G1/t,i+2t−1−1=−1i+t−1G1/t,i+2t−1−1 by Theorem 5. So,(31)G1/t,i−1−G1/t,i+2t−1−1=−1iG1/t,i+t−1−G1/t,i+2t−1−1=−1i−1i+tG1/t,i+2t−1=G1/t,i+2t−1.
On the other hand, if2∤t, then G1/t,i−1+−1iG1/t,i+t−1−1=G1/t,i+t−1 and G1/t,i+t−1−1+−1i+t−1G1/t,i+2t−1−1=G1/t,i+2t−1−1. However, since the latter identity equals −−1iG1/t,i+t−1−1−G1/t,i+2t−1−1=−1i+tG1/t,i+2t−1−1, and the sum of the two identities yields G1/t,i−1−G1/t,i+2t−1−1=G1/t,i+2t−1.


## 4. Extended Sequences ofDt,i±1 and G1/t,i±1

Lemma1 shows that Dt,i1|i>0 is a Fibo t-sequence. By extending subscripts i backward up to all integers, we have a sequence Dt,i′1|i∈ℤ satisfying Dt,i′1+Dt,i+t′1=Dt,i+t′1, which is also a Fibo t-sequence. On the contrary, the sequence G1/t,i1|i>0 satisfies G1/t,i1−G1/t,i+t−11=G1/t,i+t+11. By extending i to all integers, we get a sequence G1/t,i′1|i∈ℤ in [3] satisfying(32)G1/t,i′1+G1/t,i+1′1=D1/t,i+t′1.In fact, fromG1/3,i1=−1,1,0,−1,2,−2,1,1,−3,…,, we have(33)G′1/3,i1=…,1,1,−2,2,−1,0,1,−1︸i<0,①,0,0,1,0,1,1,1,2,2,3,4,5,…︸i>0,such that G1/3,i′1+G1/3,i+1′1=G1/3,i+3′1 (see Table 4).Table 4
Dt,i′1|i∈ℤ and G1/t,i′1|i∈ℤ.
tD′t,i1: i<00i>01…,−21,13,−8,5,−3,2,−1,1,0,11,2,3,5,8,13,21,34,55,…2…,−2,3,0,−2,1,1,−1,0,1,0,0,11,1,2,3,4,6,9,13,19,28,41,⋯3…,1,−2,1,0,1,−1,0,0,1,0,0,0,11,1,2,3,4,5,7,10,14,19,26,…4…,1,0,0,1,−1,0,0,0,1,0,0,0,0,11,1,1,1,2,3,4,5,6,8,11,15,…tG′1/t,i1:i<00i>02…,34,−21,13,−8,5,−3,2,−1,10,1,1,2,3,5,8,13,21,34,55,…3…,4,−3,1,1,−2,2,−1,0,1,−1,10,0,1,0,1,1,1,2,2,3,4,5,7,9,12,…4…,10,7,−5,4,−3,2,−1,1,−1,10,0,0,1,0,0,1,1,0,1,2,1,1,2,4,6,…5…,−4,4,−3,2,−1,0,1,−1,1,−1,1,0,0,0,0,1,0,0,0,1,1,0,0,1,0,1,3,…A sequencepn satisfying pn+pn+1=pn+t with t>1 initials is called a Padovan t-sequence [6, 7]. In particular, it is a Fibonacci sequence if t=2. Identity (32) yields the next lemma immediately.Lemma 2 (see [3]).
G1/t,i′1|i∈ℤ is a Padovan t-sequence with initials G′1/t,01=1 and G′1/t,i1=01≤i<t.Now, overC−1 and N−1, we consider extended sequences of Dt,i−1|i≥0 and G1/t,i−1|i≥0, in which subscripts i are extended to all integers. From Dt,i−1+Dt,i+2t−1=Dt,i+2t+2−1 in Lemma 1, we easily have an extended sequence Dt,i′−1|i∈ℤ of Dt,i−1|i≥0 satisfying(34)D′t,i−1+D′t,i+2t−1=D′t,i+2t+2−1,i∈ℤ.So,Dt,i′−1|i∈ℤ is an interlocked Fibo t-sequence as in Theorem 1.On the contrary, letG′1/t,−i−1=G1/t,i−1 for i≥0. Then, G1/t,i−1−G1/t,i+2t−1−1=G1/t,i+2t−1 in Theorem 7 implies G′1/t,−i−1−G′1/t,−i+2t−1−1=G′1/t,−i+2t−1. So, by setting j=−i, we have(35)G′1/t,j−1−G′1/t,j−2t−1−1=G′1/t,j−2t−1.That is,G′1/t,j+2t−1=G′1/t,j+2−1+G′1/t,j−1. It shows that G′1/t,i−1|i∈ℤ is an extended sequence of G1/t,i−1|i≥0 satisfying(36)G′1/t,i−1+G′1/t,i+2−1=G′1/t,i+2t−1,i∈ℤ.Theorem 8.
G1/t,i′−1|i∈ℤ is an interlocked Padovan t-sequence with initials G′1/t,0−1=1 and G′1/t,i−1=01≤i<2t.Proof.
The subsequence having only eventh terms ofG1/t,i′−1|i∈ℤ is(37)G′1/t,i−1e=…,G′1/t,0−1,G′1/t,2−1,…,G′1/t,2k−1,G′1/t,2k+1−1,….
Then,G′1/t,2k−1+G′1/t,2k+1−1=G′1/t,2k+1−1 in (36) implies that the sum of consecutive two eventh terms equals t distanced eventh term. So, G′1/t,i−1e is a Padovan t-sequence. Similarly, the subsequence G′1/t,i−1o of oddth terms of G′1/t,i−1 also satisfies that the sum of consecutive two oddth terms equals t distanced oddth term. Thus, G1/t,i′−1|i∈ℤ is an interlocked Padovan t-sequence.Corollary 1.
The sequencesG′1/2,i1 and G′1/2,i−1 are equal to D′1,i1 and D′1,i−1, respectively. And G′1/3,i−1 is an interlocked Fibo 5-sequence.Proof.
The proof is due to Tables4 and 5. And the interlocked Padovan 3-sequence G′1/3,i−1 in Theorem 8 satisfies G′1/3,i−1+G′1/3,i+2−1=G′1/3,i+6−1. Thus,(38)G′1/3,i−1+G′1/3,i+8−1=G′1/3,i−1+G′1/3,i+2−1+G′1/3,i+4−1=G′1/3,i+6−1+G′1/3,i+4−1=G′1/3,i+10−1,which is a recurrence of interlocked Fibo 5-sequence.
We note thatG′1/3,i−1 is interlocked by two Padovan subsequences(39)G′1/3,i−1e=…,0,1,1,1,2,2,3,4,5,7,9,12,16,…=G′1/3,i−1o,with initials 1,1,2. However, G′1/4,i−1 is interlocked by distinct Padovan 4-sequences(40)G′1/4,i−1e=…,0,1,1,0,1,2,1,1,3,3,2,4,6,5,6,10,11,11,16,…,G′1/4,i−1o=…,1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,…,where these two sequences satisfy G′1/4,i−1e+G′1/4,i+2−1e=G′1/4,i+1−1o. Moreover, the sequences are obtained explicitly from Pascal and Pauli tables as follows. Write C1=r01;r11;r21;… and C−1=r0−1;r1−1;r2−1;… by means of ith rows ri1 and ri−1. Let(41)ℭ1=r01;r01;r01︸;r11;r11;r11︸;…,ℭ−1=r0−1;r0−1︸;r1−1;r1−1︸;r2−1;…︸;…,be tables having duplicated rows of C1 and C−1.Table 5
D′t,i−1|i∈ℤandG1/t,i′−1|i∈ℤ, 1≤t≤5.
tD′t,i−1:i<00,i>01…,8,13,5,−8,−3,5,2,−3,−1,2,1,−1,0,1,1⓪,1,1,2,1,3,2,5,…2…,3,−2,−2,1,−1,2,2,−1,0,0,−1,1,1,0,1,0,⓪,1,1,1,2,1,2,2,…,3…,−2,0,2,1,−1,−1,1,0,−1,0,1,1,0,0,1,0,0,⓪,1,1,1,1,2,1,2,…4…,−1,−1,0,1,0,−1,0,0,1,1,0,0,0,1,0,0,0,⓪,1,1,1,1,1,2,1,…tG′1/t,i−1:i<00,i>02…,2,1,−1,0,1,①,0,1,1,2,1,3,2,5,3,8,5,13,8,21,13,34,21,…3…,0,1,0,−1,1,①,0,0,1,0,1,1,1,0,2,1,2,1,3,1,4,2,5,2,7,3,…4…,1,2,−1,−1,1,①,0,0,0,1,0,0,1,1,0,1,0,1,1,1,1,2,0,2,1,2,…5…,2,1,−1,−1,1,①,0,0,0,0,1,0,0,0,1,1,0,0,1,0,1,0,1,1,1,1,…Theorem 9.
Inℭ1, the sequence of 1-slope diagonal sums equals G′1/4,i−1o. And in ℭ−1, the sequence of 1-slope diagonal sums is interlocked by two Padovan 3-sequences P=pi and Q=qi, where P is the ordinary Padovan sequence such that qi+4=pi+pi+3=2pi+pi+1 for all i.Proof.
Fromℭ1=1;1;1︸;1,1;1,1;1,1︸;1,2,1;1,2,1;1,2,1︸;…,, the sequence of 1-slope diagonal sums is 1,1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,…, that corresponds to G′1/4,i−1o. Similarly, the sequence of 1-slope diagonal sums of ℭ−1=1;1︸;1,1;1,1︸;1,0,1;1,0,1︸;1,1,1,1;1,1,1,1︸;1,0,2,0,1;…, is(42)ξi=1,1,1,2,2,1,2,3,3,3,4,4,5,6,7,7,9,10,12,13,16,17,….
That satisfiesξi+ξi+2=ξi+6 for some i. Note(43)ξi=ck,0−1+ck−1,1−1+ck−1,2−1+ck−2,3−1+ck−2,4−1+⋯,ifi=2k,ξi=ck,0−1+ck,1−1+ck−1,2−1+ck−1,3−1+ck−2,4−1+ck−2,5−1+⋯,ifi=2k+1.
Then,ξi+ξi+2=ξi+6 for all i by the Pauli recurrence (1). And ξi is an interlocked Padovan 3-sequence by P=pi=1,1,2,2,3,4,5,7,9,12,16,…, of eventh terms and Q=qi=1,2,1,3,3,4,6,7,10,13,17,…, of oddth terms. Clearly, P is the ordinary Padovan sequence satisfying qi+4=pi+pi+3=2pi+pi+1 for all i.



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