# On Partial Sum of Tribonacci Numbers

Authors: Eunmi Choi; Jiin Jo
Journal: International Journal of Mathematics and Mathematical Sciences
 (2015)

---

## Abstract

We study the sums
t
(
k
,
r
)
=
∑
i
=
0
t
T
k
i
+
r of k step apart Tribonacci numbers for any 1
≤
r
≤
k. We prove that s
t
(
k
,
r
) satisfies certain Tribonacci rule s
t
(
k
,
r
)
=
a
k
s
t
-
1
(
k
,
r
)
+
b
k
s
t
-
2
(
k
,
r
)
+
s
t
-
3
(
k
,
r
)
+
λ with integers a
k
,
b
k
,
c
k, and λ.


---

## Body

## 1. Introduction

A Tribonacci sequence{
T
n
}
n
≥
0, which is a generalized Fibonacci sequence {
F
n
}, is defined by the Tribonacci rule T
n
=
T
n
-
1
+
T
n
-
2
+
T
n
-
3 with T
2
=
T
1
=
1 and T
0
=
0. The sequence can be extended to negative subscript n; hence few terms of the sequence are …
,
-
3,2
,
0
,
-
1,1
,
0,0
,
1,1
,
2,4
,
7
,
…. Each term in {
T
n
} is called the Tribonacci number.The sum of Fibonacci numbers is well expressed by∑
i
=
0
n
‍
F
i
=
F
n
+
2
-
1, and moreover the sum of reciprocal Fibonacci numbers was studied intensively in [1–3]. For the sum S
n
=
∑
i
=
0
n
‍
T
i of Tribonacci numbers, there are some researches including [4–7]. In particular Kilic [6] proved the identity S
n
=
1
/
2
(
T
n
+
2
+
T
n
-
1
) by means of generating matrix calculations. And Irmak and Alp [5] proved an identity about the k subscripted Tribonacci sum S
n
,
k
=
∑
i
=
0
n
‍
T
k
i using the three roots of x
3
-
x
2
-
x
-
1
=
0.This paper is devoted to studying the sum of Tribonacci numbers as well as the sum∑
i
=
0
t
‍
T
k
i
+
r of k step apart Tribonacci numbers for any 1
≤
r
≤
k. Our method here is to employ the k-step Tribonacci rule T
n
=
a
k
T
n
-
k
+
b
k
T
n
-
2
k
+
c
k
T
n
-
3
k (a
k
,
b
k
,
c
k
∈
Z) that is a linear combination of k distance Tribonacci numbers in [8]. For this purpose we will display all Tribonacci numbers in rectangle form with k columns, called the k-tribo table:(1)
T
1
T
2
⋯
T
r
⋯
T
k
T
k
+
1
T
k
+
2
⋯
T
k
+
r
⋯
T
2
k
T
2
k
+
1
T
2
k
+
2
⋯
T
2
k
+
r
⋱Then ∑
i
=
0
t
‍
T
k
i
+
r can be regarded as a partial sum of t
+
1 entries in rth column of the table. We denote it by s
t
(
k
,
r
) for 1
≤
r
≤
k.


## 2. Tribonacci Tables

T
n satisfies a Tribonacci rule a
1
T
n
-
1
+
b
1
T
n
-
2
+
c
1
T
n
-
3 with (
a
1
,
b
1
,
c
1
)
=
(
1,1
,
1
) and a 2-step Tribonacci rule a
2
T
n
-
2
+
b
2
T
n
-
4
+
c
2
T
n
-
6 with (
a
2
,
b
2
,
c
2
)
=
(
3,1
,
1
). Moreover the k-step Tribonacci rules for T
n were proved in [8].Lemma 1 (see [8]).
ConsiderT
n
=
a
k
T
n
-
k
+
b
k
T
n
-
2
k
+
T
n
-
3
k with a
k
=
3
T
k
-
T
k
-
6 and b
k
=
-
a
-
k for any 1
≤
k
<
n. The {
a
k
} and {
b
k
} satisfy a
k
+
3
=
a
k
+
2
+
a
k
+
1
+
a
k and b
k
=
b
k
+
1
+
b
k
+
2
+
b
k
+
3 with a
1
=
1, a
2
=
3, a
3
=
7, and b
1
=
b
2
=
1, b
3
=
-
5.The recurrencea
k
+
3
=
a
k
+
2
+
a
k
+
1
+
a
k implies that {
a
k
} is a Tribonacci type sequence. By extending the subscript k to negative integers, we have a
0
=
3, a
-
1
=
-
1, a
-
2
=
-
1, and so forth; thus {
a
k
}
=
{
…
,
-
1
,
-
5,5
,
-
1
,
-
1,3
,
1,3
,
7,11
,
…
}. In particular if 1
≤
k
≤
10 then  the coefficients (
a
k
,
b
k
) are(2)
k
a
k
,
b
k
1
1
1
1,1
1
1
2
3,1
3
7
,
-
5
4
11,5
5
21,1
6
39
,
-
11
7
71,15
8
131
,
-
3
9
241
,
-
23
10
443,41For example, T
31 is expressed by the 10-step Tribonacci rule T
31
=
443
T
21
+
41
T
11
+
T
1
=
443
(
121415
)
+
41
(
274
)
+
1
=
53798080. It shows that T
31 is a combination of three entries at 1st column of 10-tribo table. Similarly, by taking k
=
11, we have a
11
=
815 and b
11
=
-
21, so T
31
=
815
T
20
-
21
T
9
+
T
-
2
=
53798080 which is a combination of three entries at 9th column of 11-tribo table.Thus ifn
=
k
t
+
r (0
≤
t, 1
≤
r
≤
k) then T
n is located at (
t
+
1
)th row and rth column in the k-tribo table and is a combination of three entries at t, t
-
1, and (
t
-
2
)th row of rth column. Now for the partial sum s
t
(
k
,
r
)
=
∑
i
=
0
t
‍
T
k
i
+
r of rth column in k-tribo table, let us begin with k
=
3 and 4.Theorem 2.
Whenk
=
3 or 4, the partial sum s
t
(
k
,
r
) holds as follows:(
1
)
(3)
s
t
3
,
r
=
7
s
t
-
1
3
,
r
-
5
s
t
-
2
3
,
r
+
s
t
-
3
3
,
r
-
1
i
f
r
=
1
7
s
t
-
1
3
,
r
-
5
s
t
-
2
3
,
r
+
s
t
-
3
3
,
r
+
1
i
f
r
≠
1
f
o
r
t
≥
3
=
1
T
3
T
3
t
+
1
+
r
-
4
T
3
t
+
r
+
T
3
t
-
1
+
r
+
1
i
f
r
=
1
1
T
3
T
3
t
+
1
+
r
-
4
T
3
t
+
r
+
T
3
t
-
1
+
r
-
1
i
f
r
≠
1
f
o
r
t
≥
1
;
(
2
)
(4)
s
t
4
,
r
=
11
s
t
-
1
4
,
r
+
5
s
t
-
2
4
,
r
+
s
t
-
3
4
,
r
-
4
i
f
r
=
1
11
s
t
-
1
4
,
r
+
5
s
t
-
2
4
,
r
+
s
t
-
3
4
,
r
+
4
i
f
r
≠
1
f
o
r
t
≥
3
=
1
T
4
2
T
4
t
+
1
+
r
+
6
T
4
t
+
r
+
T
4
t
-
1
+
r
+
4
i
f
r
=
1
1
T
4
2
T
4
t
+
1
+
r
+
6
T
4
t
+
r
+
T
4
t
-
1
+
r
-
4
i
f
r
≠
1
f
o
r
t
≥
1
.Proof.
The 3-tribo table produces a table ofs
t
(
3
,
r
) as follows:(5)
3
-
t
r
i
b
o
t
a
b
l
e
1
1
1
1
2
1
1
4
7
13
24
44
81
149
274
504
927
1705
⋯
t
s
t
3,1
s
t
3,2
s
t
3,3
1
1
0
1
1
2
1
1
1
5
8
15
2
29
52
96
3
178
326
600
4
1105
2031
3736
Ift
=
4 then it can be observed that(6)
s
4
3,1
=
1105
=
7
178
-
5
29
+
5
-
1
=
7
s
3
3,1
-
5
s
2
3,1
+
s
1
3,1
-
1
,
s
4
3,2
=
2031
=
7
326
-
5
52
+
8
+
1
=
7
s
3
3,2
-
5
s
2
3,2
+
s
1
3,2
+
1
,
s
4
3,3
=
3736
=
7
600
-
5
96
+
15
+
1
=
7
s
3
3,3
-
5
s
2
3,3
+
s
1
3,3
+
1
.For some t
>
4, we assume s
t
(
3
,
r
)
=
7
s
t
-
1
(
3
,
r
)
-
5
s
t
-
2
(
3
,
r
)
+
s
t
-
3
(
3
,
r
)
±
1 with minus sign if r
=
1, otherwise plus sign. Then the next partial sum s
t
+
1
(
3
,
r
) satisfies(7)
s
t
+
1
3
,
r
=
s
t
3
,
r
+
T
3
t
+
1
+
r
=
7
s
t
-
1
3
,
r
-
5
s
t
-
2
3
,
r
+
s
t
-
3
3
,
r
±
1
+
7
T
3
t
+
r
-
5
T
3
t
-
1
+
r
+
T
3
t
-
2
+
r
=
7
s
t
3
,
r
-
5
s
t
-
1
3
,
r
+
s
t
-
2
3
,
r
±
1due to Lemma 1. And we also notice that(8)
T
10
-
4
T
7
+
T
4
+
1
=
149
-
4
24
+
4
+
1
=
2
29
=
2
s
2
3,1
=
T
3
s
2
3,1
,
T
11
-
4
T
8
+
T
5
-
1
=
274
-
4
44
+
7
-
1
=
2
52
=
2
s
2
3,2
=
T
3
s
2
3,2
,
T
12
-
4
T
9
+
T
6
-
1
=
504
-
4
81
+
13
-
1
=
2
96
=
2
s
2
3,3
=
T
3
s
2
3,3
,so it shows T
3
s
2
(
3
,
r
)
=
T
3
(
3
)
+
r
-
4
T
3
(
2
)
+
r
+
T
3
+
r
±
1 with minus sign if r
=
1, otherwise plus. Hence if we assume T
3
s
t
(
3
,
r
)
=
T
3
(
t
+
1
)
+
r
-
4
T
3
t
+
r
+
T
3
(
t
-
1
)
+
r
±
1 for some t then it follows from Lemma 1 that(9)
T
3
s
t
+
1
3
,
r
=
T
3
s
t
3
,
r
+
T
3
t
+
1
+
r
=
T
3
t
+
1
+
r
-
4
T
3
t
+
r
+
T
3
t
-
1
+
r
±
1
+
T
3
T
3
t
+
1
+
r
=
7
T
3
t
+
1
+
r
-
5
T
3
t
+
r
+
T
3
t
-
1
+
r
-
4
T
3
t
+
1
+
r
+
T
3
t
+
r
±
1
=
T
3
t
+
2
+
r
-
4
T
3
t
+
1
+
r
+
T
3
t
+
r
±
1
.
Similarly the 4-tribo table makes the table ofs
t
(
4
,
r
): (10)
4
-
t
r
i
b
o
t
a
b
l
e
1
1
1
1
2
4
1
1
7
13
24
44
81
149
274
504
927
1705
3136
5768
t
s
t
4,1
s
t
4,2
s
t
4,3
s
t
4,4
1
1
0
1
1
2
4
1
1
1
8
14
26
48
2
89
163
300
552
3
1016
1868
3436
6320
Whent
=
3, experimental observations show that(11)
s
3
4,1
=
1016
=
11
89
+
5
8
+
1
-
4
=
11
s
2
4,1
+
5
s
1
4,1
+
s
0
4,1
-
4
,
s
3
4,2
=
1868
=
11
163
+
5
14
+
5
=
11
s
2
4,2
+
5
s
1
4,2
+
s
0
4,2
+
4
,and so on. Hence if we assume s
t
(
4
,
r
)
=
11
s
t
-
1
(
4
,
r
)
+
5
s
t
-
2
(
4
,
r
)
+
s
t
-
3
(
4
,
r
)
±
4 with − sign if r
=
1, otherwise + sign for some t
>
3, then s
t
+
1
(
4
,
r
) is equal to(12)
s
t
+
1
4
,
r
=
s
t
4
,
r
+
T
4
t
+
1
+
r
=
11
s
t
-
1
4
,
r
+
5
s
t
-
2
4
,
r
+
s
t
-
3
4
,
r
±
4
+
11
T
4
t
+
r
+
5
T
4
t
-
1
+
r
+
T
4
t
-
2
+
r
=
11
s
t
4
,
r
+
5
s
t
-
1
4
,
r
+
s
t
-
2
4
,
r
±
4
.
On the other hand, it is easy to see that(13)
T
17
+
6
T
13
+
T
9
+
T
4
=
10609
+
6
927
+
81
+
4
=
4
2
1016
=
T
4
2
s
3
4,1
,
T
18
+
6
T
14
+
T
10
-
T
4
=
19513
+
6
1705
+
149
-
4
=
4
2
1868
=
T
4
2
s
3
4,2
,
⋮
T
20
+
6
T
16
+
T
12
-
T
4
=
66012
+
6
5768
+
504
-
4
=
4
2
6320
=
T
4
2
s
3
4,4
.So by assuming T
4
2
s
t
-
1
(
4
,
r
)
=
T
4
t
+
r
+
6
T
4
(
t
-
1
)
+
r
+
T
4
(
t
-
2
)
+
r
±
T
4 for some t, we have(14)
T
4
t
+
1
+
r
+
6
T
4
t
+
r
+
T
4
t
-
1
+
r
±
T
4
=
11
T
4
t
+
r
+
5
T
4
t
-
1
+
r
+
T
4
t
-
2
+
r
+
6
T
4
t
+
r
+
T
4
t
-
1
+
r
±
T
4
=
T
4
t
+
r
+
6
T
4
t
-
1
+
r
+
T
4
t
-
2
+
r
±
T
4
+
16
T
4
t
+
r
=
T
4
2
s
t
-
1
4
,
r
+
T
4
2
T
4
t
+
r
=
T
4
2
s
t
-
1
4
,
r
+
T
4
t
+
r
=
T
4
2
s
t
4
,
r
.We remark that ifr
=
4 in Theorem 2 then we have (15)
1
T
4
2
T
4
t
+
2
+
6
T
4
t
+
1
+
T
4
t
-
4
=
s
t
4,4
=
∑
i
=
0
t
T
4
i
+
4
=
∑
i
=
0
t
+
1
T
4
i
.Since T
4
=
4, our result is equal to Theorem 5 in [6] which was proven by means of generating matrix.The expression ofs
t
(
4
,
r
) has tails ±
4 according to r
=
1 or not and equals T
4
(
t
+
1
)
+
r
+
6
T
4
t
+
r
+
T
4
(
t
-
1
)
+
r
±
4 multiplied by 1
/
T
4
2. Similarly s
t
(
3
,
r
) has tails ±
1 according to r
=
1 or not and equals T
3
(
t
+
1
)
+
r
-
4
T
3
t
+
r
+
T
3
(
t
-
1
)
+
r
±
1 multiplied by 1
/
T
3. However, when k
≥
5 the expressions of s
t
(
k
,
r
) quite differ from the case of k
=
3,4.Theorem 3.
Let{
λ
(
r
)
}
r
=
1
5
=
{
-
7,3
,
5,1
,
9
} and {
λ
(
r
)
}
r
=
1
6
=
{
-
12,6
,
4
,
-
2,8
,
10
}. Then one has the following:(
1
)
(16)
s
t
5
,
r
=
21
s
t
-
1
5
,
r
+
s
t
-
2
5
,
r
+
s
t
-
3
5
,
r
+
λ
r
f
o
r
t
≥
3
=
1
22
T
5
t
+
1
+
r
+
2
T
5
t
+
r
+
T
5
t
-
1
+
r
-
λ
r
f
o
r
t
≥
1
;
(
2
)
(17)
s
t
6
,
r
=
39
s
t
-
1
6
,
r
-
11
s
t
-
2
6
,
r
+
s
t
-
3
6
,
r
+
λ
r
f
o
r
t
≥
3
=
1
28
T
6
t
+
1
+
r
-
10
T
6
t
+
r
+
T
6
t
-
1
+
r
-
λ
r
f
o
r
t
≥
1
.Proof.
The 5-tribo table yields a table ofs
t
(
5
,
r
) such that (18)
t
s
t
5,1
s
t
5,2
s
t
5,3
s
t
5,4
s
t
5,5
1
1
0
1
1
2
4
7
1
1
1
1
1
1
1
14
25
46
85
156
2
288
529
973
1790
3292
3
6056
11138
20486
37680
⋯Then by inspecting the table, we find that(19)
s
3
5,1
=
21
288
+
14
+
1
-
7
=
21
s
2
5,1
+
s
1
5,1
+
s
0
5,1
-
7
,
s
3
5,2
=
21
529
+
25
+
1
+
3
=
21
s
2
5,2
+
s
1
5,2
+
s
0
5,2
+
3
,
⋮
s
3
5,5
=
21
3292
+
156
+
7
+
9
=
21
s
2
5,5
+
s
1
5,5
+
s
0
5,5
+
9
,so we have s
3
(
t
,
r
)
=
21
s
2
(
t
,
r
)
+
s
1
(
t
,
r
)
+
s
t
-
3
(
t
,
r
)
+
λ
(
r
) for t
=
5 and 1
≤
r
≤
t. Now if we assume the identity is true for some t, then the t
+
1 partial sum follows(20)
s
t
+
1
5
,
r
=
s
t
5
,
r
+
T
5
t
+
1
+
r
=
21
s
t
-
1
5
,
r
+
s
t
-
2
5
,
r
+
s
t
-
3
5
,
r
+
λ
r
+
21
T
5
t
+
r
+
T
5
t
-
1
+
r
+
T
5
t
-
2
+
r
=
21
s
t
5
,
r
+
s
t
-
1
5
,
r
+
s
t
-
2
5
,
r
+
λ
rfrom Lemma 1. Moreover we notice that(21)
T
21
+
2
T
16
+
T
11
+
7
=
121415
+
2
5768
+
274
+
7
=
22
6056
=
22
s
3
5,1
,
T
22
+
2
T
17
+
T
12
-
3
=
223317
+
2
10609
+
504
-
3
=
22
11138
=
22
s
3
5,2
,
⋮
T
25
+
2
T
20
+
T
15
-
9
=
1389537
+
2
66012
+
3136
-
9
=
22
s
3
5,5
,so we have T
5
(
4
)
+
r
+
2
T
5
(
3
)
+
r
+
T
5
(
2
)
+
r
-
λ
(
r
)
=
22
s
3
(
5
,
r
). Hence, by assuming 22
s
t
(
5
,
r
)
=
T
5
(
t
+
1
)
+
r
+
2
T
5
t
+
r
+
T
5
(
t
-
1
)
+
r
-
λ
(
r
) for some t, we have(22)
22
s
t
+
1
5
,
r
=
T
5
t
+
1
+
r
+
2
T
5
t
+
r
+
T
5
t
-
1
+
r
-
λ
r
+
22
T
5
t
+
1
+
r
=
21
T
5
t
+
1
+
r
+
T
5
t
+
r
+
T
5
t
-
1
+
r
+
2
T
5
t
+
1
+
r
+
T
5
t
+
r
-
λ
r
=
T
5
t
+
2
+
r
+
2
T
5
t
+
1
+
r
+
T
5
t
+
r
-
λ
r
.
Now, for(
2
), we construct the table of s
t
(
6
,
r
) from 6-tribo table:(23)
t
s
t
6,1
s
t
6,2
s
t
6,3
s
t
6,4
s
t
6,5
s
t
6,6
1
1
0
1
1
2
4
7
13
1
1
1
25
45
83
153
281
517
2
952
1750
3219
5921
10890
20030
3
36842
67762
124634
229238
421634
⋯It is not hard to observe(24)
s
3
6,1
=
39
952
-
11
25
+
1
-
12
=
39
s
2
6,1
-
11
s
1
6,1
+
s
0
6,1
-
12
,
s
3
6,2
=
39
1750
-
11
45
+
1
+
6
=
39
s
2
6,2
-
11
s
1
6,2
+
s
0
6,2
+
6
,
⋮
s
3
6,6
=
39
20030
-
11
517
+
13
+
10
=
39
s
2
6,6
-
11
s
1
6,6
+
s
0
6,6
+
10
,so it proves s
t
(
6
,
r
)
=
39
s
t
-
1
(
6
,
r
)
-
11
s
t
-
2
(
6
,
r
)
+
s
t
-
3
(
6
,
r
)
+
λ
(
r
) when t
=
3. Thus if we assume the identity holds for some t then Lemma 1 shows that(25)
s
t
+
1
6
,
r
=
s
t
6
,
r
+
T
6
t
+
1
+
r
=
39
s
t
-
1
6
,
r
-
11
s
t
-
2
6
,
r
+
s
t
-
3
6
,
r
+
λ
r
+
39
T
6
t
+
r
-
11
T
6
t
-
1
+
r
+
T
6
t
-
2
+
r
=
39
s
t
6
,
r
-
11
s
t
-
1
6
,
r
+
s
t
-
2
6
,
r
+
λ
r
.
Finally we also notice28
s
2
(
6
,
r
)
=
T
6
(
3
)
+
r
-
10
T
6
(
2
)
+
r
+
T
6
+
r
-
λ
(
r
). Indeed,(26)
T
19
-
10
T
13
+
T
7
+
12
=
35890
-
10
927
+
24
+
12
=
28
952
=
28
s
2
6,1
,
T
20
-
10
T
14
+
T
14
-
6
=
66012
-
10
1705
+
44
-
6
=
28
1750
=
28
s
2
6,2
,
⋮
T
24
-
10
T
18
+
T
18
-
10
=
755476
-
10
19513
+
504
-
10
=
28
s
2
6,6
.So if we assume 28
s
t
-
1
(
6
,
r
)
=
T
6
t
+
r
-
10
T
6
(
t
-
1
)
+
r
+
T
6
(
t
-
2
)
+
r
-
λ
(
r
) then(27)
28
s
t
6
,
r
=
T
6
t
+
r
-
10
T
6
t
-
1
+
r
+
T
6
t
-
2
+
r
-
λ
r
+
28
T
6
t
+
r
=
39
T
6
t
+
r
-
11
T
6
t
-
1
+
r
+
T
6
t
-
2
+
r
-
10
T
6
t
+
r
+
T
6
t
-
1
+
r
-
λ
r
=
T
6
t
+
1
+
r
-
10
T
6
t
+
r
+
T
6
t
-
1
+
r
-
λ
r
.


## 3. The Tailsλ
(
k
,
r
) of the Partial Sum s
t
(
k
,
r
)

With the use of coefficients(
a
k
,
b
k
) satisfying the k-step Tribonacci rule T
k
t
+
r
=
a
k
T
k
(
t
-
1
)
+
r
+
b
k
T
k
(
t
-
2
)
+
r
+
T
k
(
t
-
3
)
+
r in Lemma 1, the identities in Theorem 3 can be restated such that(28)
s
t
5
,
r
=
a
5
s
t
-
1
5
,
r
+
b
5
s
t
-
2
5
,
r
+
s
t
-
3
5
,
r
+
λ
r
with
λ
r
r
=
1
5
=
-
7,3
,
5,1
,
9
,
s
t
6
,
r
=
a
6
s
t
-
1
6
,
r
+
b
6
s
t
-
2
6
,
r
+
s
t
-
3
6
,
r
+
λ
r
with
λ
r
r
=
1
6
=
-
12,6
,
4
,
-
2,8
,
10
.In this sense, we are able to recast Theorem 2 as(29)
s
t
3
,
r
=
a
3
s
t
-
1
3
,
r
+
b
3
s
t
-
2
3
,
r
+
s
t
-
3
3
,
r
+
λ
r
with
λ
r
r
=
1
3
=
-
1,1
,
1
,
s
t
4
,
r
=
a
4
s
t
-
1
4
,
r
+
b
4
s
t
-
2
4
,
r
+
s
t
-
3
4
,
r
+
λ
r
with
λ
r
r
=
1
4
=
-
4,4
,
4,4
.We will call {
λ
(
r
)
}
r
=
1
k the tail set of s
t
(
k
,
r
)
=
∑
i
=
0
t
‍
T
k
i
+
r.Theorem 4.
With(
a
k
,
b
k
)
(
1
≤
k
≤
10
) in Lemma 1, one has the following:(1)
s
t
(
k
,
r
)
=
a
k
s
t
-
1
(
k
,
r
)
+
b
k
s
t
-
2
(
k
,
r
)
+
s
t
-
3
(
k
,
r
)
+
λ
(
k
,
r
),
(2)
s
t
(
k
,
r
)
=
1
/
a
k
+
b
k
T
k
t
+
1
+
r
+
b
k
+
1
T
k
t
+
r
+
T
k
t
-
1
+
r
-
λ
k
,
r,where the tails {
λ
(
k
,
r
)
} are defined as follows:(30)
k
λ
k
,
r
r
=
1
k
1
1
1
1
1
1
2
0,2
3
-
1,1
,
1
4
-
4,4
,
4,4
5
-
7,3
,
5,1
,
9
6
-
12,6
,
4
,
-
2,8
,
10
7
-
29,13,9
,
-
7,15,17,25
8
-
48,16,16
,
-
16,16,16,16,48
9
-
87,35,21
,
-
31,25,15,9
,
49,73
10
-
176,66,44
,
-
66,44,22,0
,
66,88,154Proof.
When3
≤
k
≤
6, (
1
) is due to Theorems 2 and 3. If k
=
1 then s
t
(
1,1
)
=
∑
i
=
0
t
‍
T
i
+
1 is the sum of all t
+
1 numbers from T
1 to T
t
+
1; hence(31)
s
1
1,1
+
s
2
1,1
+
s
3
1,1
=
T
1
+
T
2
+
T
1
+
T
2
+
T
3
+
T
1
+
T
2
+
T
3
+
T
4
=
T
1
+
T
2
+
T
3
+
T
4
+
T
5
+
T
1
-
T
3
=
s
4
1,1
-
1
.So for any t
>
0, the identity s
t
(
1,1
)
=
s
t
-
1
(
1,1
)
+
s
t
-
2
(
1,1
)
+
s
t
-
3
(
1,1
)
+
λ
(
1,1
) with λ
(
1,1
)
=
1 can be proved by induction.
Ifk
=
2, making use of the table of s
t
(
2
,
r
), it is easy to see that(32)
s
3
2,1
=
34
=
3
s
2
2,1
+
s
1
2,1
+
s
0
2,1
+
λ
2,1
with
λ
2,1
=
0
,
s
3
2,2
=
62
=
3
s
2
2,2
+
s
1
2,2
+
s
0
2,2
+
λ
2,2
with
λ
2,2
=
2
.Thus it can be generalized to s
t
(
2
,
r
)
=
a
2
s
t
-
1
(
2
,
r
)
+
b
2
s
t
-
2
(
2
,
r
)
+
s
t
-
3
(
2
,
r
)
+
λ
(
k
,
r
) for any t
>
0 and 1
≤
r
≤
2, since (
a
2
,
b
2
)
=
(
3,1
).
Now when7
≤
k
≤
10, (
1
) can be observed. For instance, the table of s
t
(
7
,
r
) shows that (33)
s
3
7,1
=
71
3181
+
15
45
+
1
-
29
=
71
s
2
7,1
+
15
s
1
7,1
+
s
0
7,1
-
29
,
s
3
7,2
=
71
5850
+
15
82
+
1
+
13
=
71
s
2
7,2
+
15
s
1
7,2
+
s
0
7,2
+
13
,
⋮
s
3
7,7
=
71
123144
+
15
1729
+
24
+
25
=
71
s
2
7,7
+
15
s
1
7,7
+
s
0
7,7
+
25
.These observations together with mathematical induction imply(34)
s
t
7
,
r
=
71
s
t
-
1
7
,
r
+
15
s
t
-
2
7
,
r
+
s
t
-
3
7
,
r
+
λ
7
,
r
=
a
7
s
t
-
1
7
,
r
+
b
7
s
t
-
2
7
,
r
+
s
t
-
3
7
,
r
+
λ
7
,
r
.The rest follows similarly.
Since the coefficientsa
k and b
k satisfy a
k
=
3
T
k
-
T
k
-
6 and b
k
=
-
a
-
k in Lemma 1, s
t
(
k
,
r
) can be expressed by T
k
t
+
r, T
k
(
t
-
1
)
+
r, and T
k
(
t
-
2
)
+
r with λ
(
k
,
r
) such that(35)
a
1
+
b
1
s
t
1
,
r
=
2
s
t
1
,
r
=
T
t
+
1
+
r
+
2
T
t
+
r
+
T
t
-
1
+
r
-
λ
1
,
r
,
a
2
+
b
2
s
t
2
,
r
=
4
s
t
2
,
r
=
T
2
t
+
1
+
r
+
2
T
2
t
+
r
+
T
2
t
-
1
+
r
-
λ
2
,
r
,
⋮
a
10
+
b
10
s
t
10
,
r
=
484
s
t
10
,
r
=
T
10
t
+
1
+
r
+
42
T
10
t
+
r
+
T
10
t
-
1
+
r
-
λ
10
,
r
.So (
2
) also follows immediately.In particular ifk
=
1 then r
=
1 and a
1
=
b
1
=
λ
(
1,1
)
=
1; Theorem 4 shows (36)
s
t
1,1
=
s
t
-
1
1,1
+
s
t
-
2
1,1
+
s
t
-
3
1,1
+
1
.But since s
t
(
1,1
)
=
∑
i
=
0
t
+
1
‍
T
i
=
S
t
+
1, we have S
t
+
1
=
S
t
+
S
t
-
1
+
S
t
-
2
+
1; this is Lemma 1 in [6]. Moreover Theorem 4 implies (37)
s
t
1,1
=
1
2
T
t
+
2
+
2
T
t
+
1
+
T
t
-
1
=
1
2
T
t
+
3
+
T
t
+
1
-
1
.It means S
t
+
1
=
1
/
2
(
T
t
+
3
+
T
t
+
1
-
1
), that is, Theorem 2 in [6].


## 4. Cyclic Rule for Partial Sums

The partial sums
t
(
k
,
r
) and its tail λ
(
k
,
r
) were discussed when 1
≤
k
≤
10. We investigate them for all k
>
0 by showing certain cyclic rules.Theorem 5.
For anyk
>
0, s
t
(
k
,
r
) satisfies the following cyclic rules:(1)
s
t
(
k
,
r
)
+
s
t
(
k
,
r
+
1
)
+
s
t
(
k
,
r
+
2
)
=
s
t
(
k
,
r
+
3
) for all 1
≤
r
≤
k
-
3;
(2)
s
t
(
k
,
k
-
2
)
+
s
t
(
k
,
k
-
1
)
+
s
t
(
k
,
k
)
=
s
t
+
1
(
k
,
1
)
-
1;
(3)
s
t
(
k
,
k
-
1
)
+
s
t
(
k
,
k
)
+
s
t
+
1
(
k
,
1
)
=
s
t
+
1
(
k
,
2
), and s
t
(
k
,
k
)
+
s
t
+
1
(
k
,
1
)
+
s
t
+
1
(
k
,
2
)
=
s
t
+
1
(
k
,
3
).Proof.
When1
≤
r
≤
k
-
3, (
1
) is clear from(38)
s
t
k
,
r
+
s
t
k
,
r
+
1
+
s
t
k
,
r
+
2
=
∑
i
=
0
t
T
k
i
+
r
+
T
k
i
+
r
+
1
+
T
k
i
+
r
+
2
=
∑
i
=
0
t
T
k
i
+
r
+
3
=
s
t
k
,
r
+
3
.And (
2
) also follows from(39)
s
t
k
,
k
-
2
+
s
t
k
,
k
-
1
+
s
t
k
,
k
=
∑
i
=
0
t
T
k
i
+
k
-
2
+
T
k
i
+
k
-
1
+
T
k
i
+
k
=
∑
i
=
0
t
T
k
i
+
k
+
1
=
∑
j
=
1
t
+
1
T
k
j
+
1
=
∑
j
=
0
t
+
1
T
k
j
+
1
-
T
1
=
s
t
+
1
k
,
1
-
1
,since T
1
=
1. Moreover(40)
s
t
k
,
k
-
1
+
s
t
k
,
k
+
s
t
+
1
k
,
1
=
∑
i
=
0
t
T
k
i
+
k
-
1
+
∑
i
=
0
t
T
k
i
+
k
+
∑
i
=
0
t
+
1
T
k
i
+
1
=
∑
i
=
1
t
+
1
T
k
i
-
1
+
T
k
i
+
T
k
i
+
1
+
T
1
=
∑
i
=
1
t
+
1
T
k
i
+
2
+
T
2
=
s
t
+
1
k
,
2
,because T
1
=
T
2. Finally, we also have(41)
s
t
k
,
k
+
s
t
+
1
k
,
1
+
s
t
+
1
k
,
2
=
∑
i
=
1
t
+
1
T
k
i
+
∑
i
=
1
t
+
1
T
k
i
+
1
+
T
1
+
∑
i
=
1
t
+
1
T
k
i
+
2
+
T
2
=
∑
i
=
1
t
+
1
T
k
i
+
T
k
i
+
1
+
T
k
i
+
2
+
T
1
+
T
2
=
∑
i
=
1
t
+
1
T
k
i
+
3
+
T
0
+
T
1
+
T
2
=
s
t
+
1
k
,
3
.Next theorem is about the cyclic rule of the tail{
λ
(
k
,
r
)
}
r
=
1
k satisfying s
t
(
k
,
r
)
=
a
k
s
t
-
1
(
k
,
r
)
+
b
k
s
t
-
2
(
k
,
r
)
+
s
t
-
3
(
k
,
r
)
+
λ
(
k
,
r
).Theorem 6.
For anyk
>
0, {
λ
(
k
,
r
)
}
r
=
1
k satisfies the following cyclic rules:(1)
λ
(
k
,
r
)
+
λ
(
k
,
r
+
1
)
+
λ
(
k
,
r
+
2
)
=
λ
(
k
,
r
+
3
) for all 1
≤
r
≤
k
-
3;
(2)
λ
(
k
,
k
-
2
)
+
λ
(
k
,
k
-
1
)
+
λ
(
k
,
k
)
=
λ
(
k
,
1
)
+
(
a
k
+
b
k
);
(3)
λ
(
k
,
k
-
1
)
+
λ
(
k
,
k
)
+
λ
(
k
,
1
)
=
λ
(
k
,
2
), and λ
(
k
,
k
)
+
λ
(
k
,
1
)
+
λ
(
k
,
2
)
=
λ
(
k
,
3
).Proof.
When1
≤
k
≤
10, Theorem 4 shows that {
λ
(
k
,
r
)
} satisfies (
1
). For instance, if k
=
10, the set {
λ
(
10
,
r
)
}
=
{
-
176,66
,
44
,
-
66,44,22,0
,
66,88,154
} implies -
176
+
66
+
44
=
-
66
,
…
,
22
+
0
+
66
=
88
,
a
n
d
0
+
66
+
88
=
154.
Suppose1
≤
r
≤
k
-
3. Then Theorem 5 gives rise to(42)
λ
k
,
r
+
λ
k
,
r
+
1
+
λ
k
,
r
+
2
=
s
t
k
,
r
+
s
t
k
,
r
+
1
+
s
t
k
,
r
+
2
-
a
k
s
t
-
1
k
,
r
+
s
t
-
1
k
,
r
+
1
+
s
t
-
1
k
,
r
+
2
-
b
k
s
t
-
2
k
,
r
+
s
t
-
2
k
,
r
+
1
+
s
t
-
2
k
,
r
+
2
-
s
t
-
3
k
,
r
+
s
t
-
3
k
,
r
+
1
+
s
t
-
3
k
,
r
+
2
=
s
t
k
,
r
+
3
-
a
k
s
t
-
1
k
,
r
+
3
-
b
k
s
t
-
2
k
,
r
+
3
-
s
t
-
3
k
,
r
+
3
=
λ
k
,
r
+
3
.
Now, for(
2
), (
a
k
,
b
k
) in Lemma 1 and λ
(
k
,
r
) in  Theorem 4 show(43)
k
a
k
+
b
k
λ
k
,
k
-
2
+
λ
k
,
k
-
1
+
λ
k
,
k
-
a
k
+
b
k
λ
k
,
1
3
2
-
1
+
1
+
1
-
2
-
1
4
16
4
+
4
+
4
-
16
-
4
5
22
5
+
1
+
9
-
22
-
7
⋮
10
484
66
+
88
+
154
-
484
-
176which proves (
2
) if 1
≤
k
≤
10. Now for any k
>
0, we have(44)
λ
k
,
k
-
2
+
λ
k
,
k
-
1
+
λ
k
,
k
=
s
t
k
,
k
-
2
+
s
t
k
,
k
-
1
+
s
t
k
,
k
-
a
k
s
t
-
1
k
,
k
-
2
+
s
t
-
1
k
,
k
-
1
+
s
t
-
1
k
,
k
-
b
k
s
t
-
2
k
,
k
-
2
+
s
t
-
2
k
,
k
-
1
+
s
t
-
2
k
,
k
-
s
t
-
3
k
,
k
-
2
+
s
t
-
3
k
,
k
-
1
+
s
t
-
3
k
,
k
=
s
t
+
1
k
,
1
-
1
-
a
k
s
t
k
,
1
-
1
-
b
k
s
t
-
1
k
,
1
-
1
-
s
t
-
2
k
,
1
-
1
=
s
t
+
1
k
,
1
-
a
k
s
t
k
,
1
-
b
k
s
t
-
1
k
,
1
-
s
t
-
2
k
,
1
-
1
+
a
k
+
b
k
+
1
=
λ
k
,
1
+
a
k
+
b
k
,due to Theorem 5. Similarly (3) also follows from Theorem 5 such that(45)
λ
k
,
k
-
1
+
λ
k
,
k
+
λ
k
,
1
=
s
t
k
,
k
-
1
+
s
t
k
,
k
+
s
t
k
,
1
-
a
k
s
t
-
1
k
,
k
-
1
+
s
t
-
1
k
,
k
+
s
t
-
1
k
,
1
-
b
k
s
t
-
2
k
,
k
-
1
+
s
t
-
2
k
,
k
+
s
t
-
2
k
,
1
-
s
t
-
3
k
,
k
-
1
+
s
t
-
3
k
,
k
+
s
t
-
3
k
,
1
=
s
t
+
1
k
,
2
-
a
k
s
t
k
,
2
-
b
k
s
t
-
1
k
,
2
-
s
t
-
2
k
,
2
=
λ
k
,
2
.
And the rest follows analogously.The identityλ
(
k
,
r
)
=
s
t
(
k
,
r
)
-
a
k
s
t
-
1
(
k
,
r
)
-
b
k
s
t
-
2
(
k
,
r
)
-
s
t
-
3
(
k
,
r
) implies that {
λ
(
k
,
r
)
} does not depend on t. Hence without loss of generality if we assume t
=
3 then(46)
λ
k
,
r
=
s
3
k
,
r
-
a
k
s
2
k
,
r
-
b
k
s
1
k
,
r
-
s
0
k
,
r
.But since s
0
(
k
,
r
)
=
T
r, s
1
(
k
,
r
)
=
T
r
+
T
k
+
r, s
2
(
k
,
r
)
=
T
r
+
T
k
+
r
+
T
2
k
+
r, and s
3
(
k
,
r
)
=
T
r
+
T
k
+
r
+
T
2
k
+
r
+
T
3
k
+
r
=
2
T
r
+
(
1
+
b
k
)
T
k
+
r
+
(
1
+
a
k
)
T
2
k
+
r, we have(47)
λ
k
,
r
=
1
-
a
k
-
b
k
T
r
+
1
-
a
k
T
k
+
r
+
T
2
k
+
r
.Therefore we are able to have tails {
λ
(
k
,
r
)
} for all k
>
10; for instance,(48)
λ
11
,
r
=
-
305,105,81
,
-
119,67,29
,
-
23,73,79,129,281
,
λ
12
,
r
=
-
564,212,131
,
-
220,124,36
,
-
60,100,76,116,292,484
,
λ
13
,
r
=
-
1071,387,261
,
-
423,225,63
,
-
135,153,81,99,333,513,945
,
⋮We remark that a special cases
t
-
1
(
k
,
k
)
=
∑
i
=
0
t
‍
T
k
i was proved in [5] by making use of the three roots of x
3
-
x
2
-
x
-
1
=
0. Considering the difficulty of finding roots of the cubic polynomial, the identity s
t
-
1
(
k
,
k
)
=
1
/
a
k
+
b
k
(
T
k
(
t
+
1
)
+
(
b
k
+
1
)
T
k
t
+
T
k
(
t
-
1
)
-
λ
(
k
,
k
)
) in Theorem 4 seems a little bit simple and easy.



---

*Source: 301814-2015-06-15.xml*