Sums of Tribonacci and Tribonacci-Lucas Numbers

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1 International Journal of Mathematical Analysis Vol. 1, 018, no. 1, 19-4 HIKARI Ltd, Sums of Tribonacci Tribonacci-Lucas Numbers Robert Frontczak 1 Lesbank Baden-Wuerttemberg (LBBW) Am Hauptbahnhof, Stuttgart, Germany Copyright c 018 Robert Frontczak. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, reproduction in any medium, provided the original work is properly cited. Abstract In this short article, we consider finite nonalternating alternating sums of Tribonacci Tribonacci-Lucas numbers. By applying an elementary telescoping argument, we obtain new identities for these sums. Mathematics Subject Classification: 11B39, 11B37 Keywords: Tribonacci number, Tribonacci-Lucas number, Sum 1 Introduction Preliminaries For n, the Tribonacci numbers (T n ) n 0 (sequence A in The On-Line Encyclopedia of Integer Sequences [10]) the Tribonacci-Lucas numbers (K n ) n 0 (sequence A in [10]) are defined, respectively, by T n+1 = T n + T n 1 + T n, T 0 = 0, T 1 = T = 1, (1) K n+1 = K n + K n 1 + K n, K 0 = 3, K 1 = 1, K = 3. () The first few terms of the sequence (T n ) n 0 are 0, 1, 1,, 4, 7, 13, 4, 44, 81, 149, 74, 504, 97, 1705, 1 Disclaimer: Statements conclusions made in this article are entirely those of the author. They do not necessarily reflect the views of LBBW.

2 0 Robert Frontczak whereas for (K n ) n 0 we have 3, 1, 3, 7, 11, 1, 39, 71, 131, 41, 443, 815, 1499, 757, Both sequences must be regarded as generalizations of the famous Fibonacci numbers. The name Tribonacci numbers was given by Feinberg [5]. Tribonacci as well as Tribonacci-Lucas numbers are members of the general Tribonacci recurrence. The properties of these numbers are studied in many articles. The most recent articles include but are not limited to [1], [], [3], [4], [7], [8] [9]. The Binet formulas are given by T n = α n+1 (α β)(α γ) + β n+1 (β α)(β γ) + γ n+1 (γ α)(γ β), (3) K n = α n + β n + γ n, (4) where α, β γ are roots of the cubic equation x 3 x x 1 = 0, i.e., α = , 3 β = 1 + ω ω , 3 γ = ω ω , 3 where ω = 1+i 3 is a primitive cube root of unity. In what follows, we will also need to define the Tribonacci numbers the Tribonacci-Lucas numbers for negative indices. This can be done as follows (see [9]): We set T n = B n K n = C n where B n = B n 1 B n + B n 3, B 1 = 1, B 0 = B 1 = 0, (5) C n = C n 1 C n + C n 3, C 1 = 1, C 0 = 3, C 1 = 1. (6) We are interested in finding expressions for the sums n T mk, n ( 1)k 1 T mk, n K mk, n ( 1)k 1 K mk, where m is an integer. The first examples of these evaluations, which may be proved by induction on n, are T k = 1 (T n+ + T n 1) (7)

3 Sums of Tribonacci Tribonacci-Lucas numbers 1 T k = 1 (T n+1 + T n 1). (8) Equation (7) appears as Theorem in [7]. In the same paper, the author also states the following expression for the sum of the 4k subscripted Tribonacci numbers ([7], Theorem 5): T 4k = 1 (T T4 4n+4 + 6T 4n + T 4n 4 T 4 ). (9) In this note we show that the sum evaluations from above are special cases of more general sum identities that we will present in the next section. Results We state our results in two separate theorems. which we will use in the proofs of our results. We first provide a lemma, Lemma.1 The following identities hold for Tribonacci Tribonacci- Lucas numbers: T k+n = T k K n T k n C n + T k n, (10) K k+n = K k K n K k n C n + C n k, (11) where the sequence C n may be expressed as C n = α n β n + α n γ n + β n γ n. PROOF: Both identities follow essentially from the Binet forms (3) (4). See [9] [] for details. Now we are ready to state the main results of this paper. Proofs will follow in the next section. Theorem. Let m be a positive integer. Then T mk = T m(n+1) + (1 C m )T mn + T m(n 1) T m B m K m C m, (1) ( 1) k 1 T mk = ( 1)n+1 (T m(n+1) + (1 + C m )T mn T m(n 1) ) + T m B m. K m + C m + (13)

4 Robert Frontczak The corresponding identities for the Tribonacci-Lucas numbers are contained in the next theorem: Theorem.3 Let m be a positive integer. Then K mk = K m(n+1) + (1 C m )K mn + K m(n 1) + C m K m 3, (14) K m C m ( 1) k 1 K mk = ( 1)n+1 (K m(n+1) + (1 + C m )K mn K m(n 1) ) + C m + K m + 3. K m + C m + (15) Before providing proofs, we present some explicit evaluations. The identities in equations (7), (8) (9) follow from the first part of Theorem. for m = 1, 4, respectively. For m = 3 we obtain T 3k = T 3n+3 4T 3n + T 3n 3 1, (16) which may be stated in the equivalent form T 3k = T 3n+ T 3n 1. (17) Moreover, we have ( 1) k 1 T k = ( 1)n+1 (T n+1 T n 1 ) + 1, (18) ( 1) k 1 T k = ( 1)n+1 (T n + T n 1 ), (19) K k = K n+ + K n 6, (0) ( 1) k 1 K k = ( 1)n+1 (K n+1 K n 1 ) +, (1) K k = K n+1 + K n 4, () ( 1) k 1 K k = ( 1)n+1 (K n + K n 1 ) +. (3)

5 Sums of Tribonacci Tribonacci-Lucas numbers 3 3 The Proofs In this section we prove Theorems..3. The method of proof is completely elementary. The idea is to combine Lemma.1 with the following sum identities: Proposition 3.1 Let f(k) be a real sequence m, n j be positive integers. Then ( ) f(m(k + j)) f(m(k j)) = n+j k=n+1 j f(mk) j j f(mk), (4) ) ( 1) (f(m(k k 1 + j)) f(m(k j)) = n+j k=n+1 j ( 1) k+j 1 f(mk) j ( 1) k+j 1 f(mk). (5) j PROOF: These sum relations may be proved straightforwardly by shifting the summation index. See [6] for details first applications. PROOF of the main Theorems: To prove Theorem. we start with equation (10). Replacing n by m k by mk results in T m(k+1) T m(k 1) = K m T mk (1 + C m )T m(k 1) + T m(k ). (6) Set f(k) = T k apply (4) with j = 1 to get T m(n+1) + T mn T m = K m Since, T mk (1 + C m ) T m(k 1) = T m(k ) = T m(k 1) + T mk T mn, T mk T mn T m(n 1) + T m, T m(k ). (7) the first part of Theorem. follows immediately after rearrangement. prove the second part, set f(k) = T k apply (5) with j = 1 to get ( 1) n+1 T m(n+1) + ( 1) n T mn + T m = To

6 4 Robert Frontczak K m ( 1) k 1 T mk (1 + C m ) ( 1) k 1 T m(k 1) + ( 1) k 1 T m(k ). (8) Simplifying gives the stated relation. The proof of Theorem.3 is very similar. Combine (11) with (4) (5) setting f(k) = K k j = 1. As the proof is straightforwardly completed, we omit the details. References [1] P. Anantakitpaisal K. Kuhapatanakul, Reciprocal Sums of the Tribonacci Numbers, Journal of Integer Sequences, 19 (016), Article [] M. Catalani, Identities for Tribonacci-related Sequences, Cornell University Library, (00). [3] M. Elia, Derived Sequences, the Tribonacci Recurrence Cubic Forms, The Fibonacci Quarterly, 39 (001), no., [4] V. Facó D. Marques, Tribonacci Numbers the Brocard- Ramanujan Equation, Journal of Integer Sequences, 19 (016), Article [5] M. Feinberg, Fibonacci-Tribonacci, The Fibonacci Quarterly, 1 (1963), no. 3, [6] R. Frontczak, Sums of Powers of Fibonacci Lucas Numbers: A New Bottom-Up Approach, submitted for publication, October 017. [7] E. Kilic, Tribonacci Sequences with Certain Indices Their Sums, Ars. Comb., 86 (008), [8] T. Komatsu, On the Sum of Reciprocal Tribonacci Numbers, Ars. Comb., 98 (011), [9] N. Yilmaz N. Taskara, Tribonacci Tribonacci-Lucas Numbers via the Determinants of Special Matrices, Applied Mathematical Sciences, 8 (014), no. 39, [10] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences. Received: December 19, 017; Published: January 18, 018