International Journal of Mathematics and Mathematical Sciences

Volume 2015, Article ID 301814, 8 pages

http://dx.doi.org/10.1155/2015/301814

## On Partial Sum of Tribonacci Numbers

Department of Mathematics, Hannam University, Daejeon 306-791, Republic of Korea

Received 30 March 2015; Accepted 28 May 2015

Academic Editor: Nawab Hussain

Copyright © 2015 Eunmi Choi and Jiin Jo. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We study the sum of step apart Tribonacci numbers for any . We prove that satisfies certain Tribonacci rule with integers , and .

#### 1. Introduction

A Tribonacci sequence , which is a generalized Fibonacci sequence , is defined by the Tribonacci rule with and . The sequence can be extended to negative subscript ; hence few terms of the sequence are . Each term in is called the Tribonacci number.

The sum of Fibonacci numbers is well expressed by , and moreover the sum of reciprocal Fibonacci numbers was studied intensively in [1–3]. For the sum of Tribonacci numbers, there are some researches including [4–7]. In particular Kilic [6] proved the identity by means of generating matrix calculations. And Irmak and Alp [5] proved an identity about the subscripted Tribonacci sum using the three roots of .

This paper is devoted to studying the sum of Tribonacci numbers as well as the sum of step apart Tribonacci numbers for any . Our method here is to employ the -step Tribonacci rule () that is a linear combination of distance Tribonacci numbers in [8]. For this purpose we will display all Tribonacci numbers in rectangle form with columns, called the -tribo table:Then can be regarded as a partial sum of entries in th column of the table. We denote it by for .

#### 2. Tribonacci Tables

satisfies a Tribonacci rule with and a 2-step Tribonacci rule with . Moreover the -step Tribonacci rules for were proved in [8].

Lemma 1 (see [8]). *Consider with and for any . The and satisfy and with , , , and , .*

The recurrence implies that is a Tribonacci type sequence. By extending the subscript to negative integers, we have , , , and so forth; thus . In particular if then the coefficients areFor example, is expressed by the -step Tribonacci rule . It shows that is a combination of three entries at 1st column of 10-tribo table. Similarly, by taking , we have and , so which is a combination of three entries at 9th column of 11-tribo table.

Thus if (, ) then is located at th row and th column in the -tribo table and is a combination of three entries at , , and th row of th column. Now for the partial sum of th column in -tribo table, let us begin with and 4.

Theorem 2. *When or 4, the partial sum holds as follows:* * *

*Proof. *The 3-tribo table produces a table of as follows:If then it can be observed thatFor some , we assume with minus sign if , otherwise plus sign. Then the next partial sum satisfiesdue to Lemma 1. And we also notice thatso it shows with minus sign if , otherwise plus. Hence if we assume for some then it follows from Lemma 1 thatSimilarly the 4-tribo table makes the table of : When , experimental observations show thatand so on. Hence if we assume with − sign if , otherwise + sign for some , then is equal toOn the other hand, it is easy to see thatSo by assuming for some , we have

*We remark that if in Theorem 2 then we have Since , our result is equal to Theorem 5 in [6] which was proven by means of generating matrix.*

*The expression of has tails according to or not and equals multiplied by . Similarly has tails according to or not and equals multiplied by . However, when the expressions of quite differ from the case of .*

*Theorem 3. Let and . Then one has the following: *

*Proof. *The 5-tribo table yields a table of such that Then by inspecting the table, we find thatso we have for and . Now if we assume the identity is true for some , then the partial sum followsfrom Lemma 1. Moreover we notice thatso we have . Hence, by assuming for some , we haveNow, for , we construct the table of from 6-tribo table:It is not hard to observeso it proves when . Thus if we assume the identity holds for some then Lemma 1 shows thatFinally we also notice . Indeed,So if we assume then

*3. The Tails of the Partial Sum *

*3. The Tails of the Partial Sum*

*With the use of coefficients satisfying the -step Tribonacci rule in Lemma 1, the identities in Theorem 3 can be restated such thatIn this sense, we are able to recast Theorem 2 asWe will call the tail set of .*

*Theorem 4. With in Lemma 1, one has the following:(1),(2),where the tails are defined as follows:*

*Proof. *When , is due to Theorems 2 and 3. If then is the sum of all numbers from to ; henceSo for any , the identity with can be proved by induction.

If , making use of the table of , it is easy to see thatThus it can be generalized to for any and , since .

Now when , can be observed. For instance, the table of shows that These observations together with mathematical induction implyThe rest follows similarly.

Since the coefficients and satisfy and in Lemma 1, can be expressed by , , and with such thatSo also follows immediately.

*In particular if then and ; Theorem 4 shows But since , we have ; this is Lemma 1 in [6]. Moreover Theorem 4 implies It means , that is, Theorem 2 in [6].*

*4. Cyclic Rule for Partial Sums*

*4. Cyclic Rule for Partial Sums**The partial sum and its tail were discussed when . We investigate them for all by showing certain cyclic rules.*

*Theorem 5. For any , satisfies the following cyclic rules:(1) for all ;(2);(3), and .*

*Proof. *When , is clear fromAnd also follows fromsince . Moreoverbecause . Finally, we also have

*Next theorem is about the cyclic rule of the tail satisfying .*

*Theorem 6. For any , satisfies the following cyclic rules:(1) for all ;(2);(3), and .*

*Proof. *When , Theorem 4 shows that satisfies . For instance, if , the set implies .

Suppose . Then Theorem 5 gives rise toNow, for , in Lemma 1 and in Theorem 4 showwhich proves if . Now for any , we havedue to Theorem 5. Similarly (3) also follows from Theorem 5 such thatAnd the rest follows analogously.

*The identity implies that does not depend on . Hence without loss of generality if we assume thenBut since , , , and , we haveTherefore we are able to have tails for all ; for instance,*

*We remark that a special case was proved in [5] by making use of the three roots of . Considering the difficulty of finding roots of the cubic polynomial, the identity in Theorem 4 seems a little bit simple and easy.*

*Conflict of Interests*

*Conflict of Interests**The authors declare that there is no conflict of interests regarding the publication of this paper.*

*References*

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