Weighted Tribonacci sums

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1 Weighted Tribonacci um Kunle Adegoke arxiv: v1 [math.ca] 16 Apr 018 Department of Phyic Engineering Phyic, Obafemi Awolowo Univerity, 0005 Ile-Ife, Nigeria Abtract We derive variou weighted ummation identitie, including binomial double binomial identitie, for Tribonacci number. Our reult contain ome previouly known reult a pecial cae. 1 Introduction For m, the Tribonacci number are defined by T m = T m 1 +T m +T m, T 0 = 0, T 1 = T = By writing T m 1 = T m + T m + T m 4 eliminating T m T m between thi recurrence relation the recurrence relation 1.1, a ueful alternative recurrence relation i obtained for m 4: T m = T m 1 T m 4, T 0 = 0, T 1 = T = 1, T =. 1. Extenion of the definition oft m to negative ubcript i provided by writing the recurrence relation 1. a T m = T m+ T m Anantakitpaial Kuhapatanakul [] proved that T m = T m 1 T m T m. 1.4 The following identity Feng [], equation.; Shah [7], ii i readily etablihed by the principle of mathematical induction: T m+r = T r T m +T r 1 +T r T m 1 +T r+1 T m. 1.5 Irmak Alp [5] derived the following identity for Tribonacci number with indice in arithmetic progreion: T tm+r = λ 1 tt tm 1+r +λ tt tm +r +λ tt tm +r, 1.6 AMS Claification: 11B7, 11B9, 65B10 adegoke00@gmail.com 1

2 where, λ 1 t = α t +β t +γ t, λ t = αβ t αγ t βγ t, λ t = αβγ t, where α, β γ are the root of the characteritic polynomial of the Tribonacci equence x x x 1. Thu, α = , β = 1 1+ω 19+ +ω 19 γ = 1 1+ω 19+ +ω 19, where ω = expiπ/ i a primitive cube root of unity. Note that λ 1 t, λ t λ t are integer for any poitive integer t [5]; in particular, λ 1 1 = 1 = λ 1 = λ 1. Weighted um Lemma 1 [1], Lemma. Let {X m } be any arbitrary equence, where X m, m Z, atifie a econd order recurrence relation X m = f 1 X m a + X m b, where f 1 are arbitrary non-vanihing complex function, not dependent on m, a b are integer. Then, X m ka b+a f 1 = X m f k 1 f 1 X m k+1a,.1 f 1 for k a non-negative integer. X m kb a+b f X m b ak+a+b a /f 1 = X m f k X m k+1b. = f 1X m /f 1 k +X m k+1b a.. Theorem 1. The following identitie hold for any integer m k: T m k 4+ = T m k 1 k T m,.4 1 T m 4k 1+4 = 1 k T m +T m 4k 4.5 T m k+1+ = k+1 T m T m k..6

3 Proof. From the recurrence relation 1., make the identification f 1 =, = 1, a = 1 b = 4 ue thee in Lemma 1 with X = T. Particular intance of identitie.4.6 are the following identitie: T = 4 k T k+4,.7 giving, T = 4,.8 1 T 4 = 1 k T 4k T = k+1 T k Lemma Partial um of ann th order equence. Let {X } be any arbitrary equence, where X, Z, atifie a n th order recurrence relation X = f 1 X c1 + X c + +f n X cn = n f mx cm, where f 1,,..., f n are arbitrary non-vanihing complex function, not dependent on, c 1, c,..., c n are fixed integer. Then, the following ummation identity hold for arbitrary x non-negative integer k : n cm {x cm f m x =1 x X } k =k c m+1 x X X = 1 n. f xcm m Proof. Recurrence relation: X = n f m X cm. We multiply both ide by x um over to obtain n n k c m x X = f m x X cm = x cm f m x X, = c m after hifting the ummation index. Splitting the inner um, we can write n 1 k c m x X = x cm f m x X + x X + x X. = c m =k+1 Since 1 = c m x X c m =1 x X k c m =k+1 x X =k c m+1 x X,

4 the preceding identity can be written x X = Thu, we have where S = n cm x cm f m x X + =1 x X n cm x cm f m x X +S Removing bracket, we have S = =1 S = S k x = n cm x cm f m x X =1 x X. =k c m+1 from which the reult follow by grouping the S term. =k c m+1 x X +S =k c m+1 x X x X, n x cm f m, Lemma Generating function. Under the condition of Lemma, if additionally x k X k vanihe in the limit a k approache infinity, then n x cm f cm m S x = x =1 x X X = 1 n, f xcm m o that S x i a generating function for the equence {X }. Theorem Sum of Tribonacci number with indice in arithmetic progreion. For arbitrary x, any integer t r any non-negative integer k, the following identity hold: where, 1 λ1 tx λ tx λ tx k x T t+r = T r +xλ t+x λ tt r t +xλ tt r t x k+1 T k+1t+r x k+ λ t+xλ tt kt+r x k+ λ tt k 1t+r, λ 1 t = α t +β t +γ t, λ t = αβ t αγ t βγ t, λ t = αβγ t, where α, β γ are the root of the characteritic polynomial of the Tribonacci equence x x x 1. Proof. Write identity 1.6 a X = f 1 X 1 + X + X identify the equence {X } = {T t+r } the contant c 1 = 1, c =, c = the function f 1 = λ 1 t, = λ t, = λ t, ue thee in Lemma.. 4

5 Corollary Generating function of the Tribonacci number with indice in arithmetic progreion. For any integer t r, any non-negative integer k arbitrary x for which x k T k vanihe a k approache infinity, the following identity hold: x T t+r = T r +xλ +x λ T r t +xλ T r t 1 λ 1 x λ x λ x, where, λ 1 = α t +β t +γ t, λ = αβ t αγ t βγ t, λ = αβγ t, where α, β γ are the root of the characteritic polynomial of the Tribonacci equence x x x 1. Many intance of Theorem may be explored. In particular, we have λ 1 t+λ t+λ t 1 T t+r = T r λ t+λ tt r t λ tt r t +T k+1t+r +λ t+λ tt kt+r +λ tt k 1t+r,.11 which at r = 0 give λ 1 t+λ t+λ t 1 T t = λ t+λ tt t 1 T t T t λ tt t 1 T t T t +T k+1t +λ t+λ tt kt +λ tt k 1t ;.1 1+λ 1 t λ t+λ t 1 T t+r = T r +λ t λ tt r t λ tt r t + 1 k T k+1t+r k λ t λ tt kt+r 1 k λ tt k 1t+r, which at r = 0 give 1+λ 1 t λ t+λ t 1 T t = λ t λ tt t 1 T t T t λ tt t 1 T t T t + 1 k T k+1t + 1 k λ t λ tt kt 1 k λ tt k 1t..14 Many previouly known reult are particular intance of the identitie For example, Theorem 5 of [6] i obtained from identity.1 by etting t = 4. Sum of 5

6 Tribonacci number with indice in arithmetic progreion are alo dicued in reference [4, 5, 6] reference therein, uing variou technique. Weighted um of the form k p T t+r, where p i a non-negative integer, may be evaluated by etting x = e y in the identity of Theorem, differentiating both ide p time with repect to y then etting y = 0. The implet example in thi category are the following: T +r = T r +T r+1 +k 1T k+r 1 +k 1T k+r +k T k+r+1.15 T +r = T r 1 5T r 6T r+1 +k k +T k+r 1 +k k +5T k+r +k 4k +6T k+r+1,.16 with the particular cae T = +k 1T k 1 +k 1T k +k T k+1.17 T = 6+k k +T k+r 1 +k k +5T k +k 4k +6T k Weighted binomial um Lemma 4 [1], Lemma. Let {X m } be any arbitrary equence. Let X m, m Z, atify a econd order recurrence relation X m = f 1 X m a + X m b, where f 1 are non-vanihing complex function, not dependent on m, a b are integer. Then, for k a non-negative integer. X m bk+b a = X m f k,.1 Xm+a bk+b = f k 1 X m. Xm+b ak+a = f k X f 1 m,. f 1 6

7 Theorem 4. The following identitie hold for any integer m any non-negative integer k: 1 T m 4k+ = 1 k T m,.4 T m k+4 = k T m.5 1 T m+k+ = k T m..6 Proof. Identify X = T in Lemma 4 ue the f 1,, a b value found in the proof of Theorem 1. Particular cae o.4,.5.6 are the following identitie: 1 T = 1 k T 4k,.7 T 4 = k T k.8 1 T = k Tk 1 T k T k..9 4 Weighted double binomial um Lemma 5. Let {X m } be any arbitrary equence, X m atifying a third order recurrence relation X m = f 1 X m a + X m b + X m c, where f 1, are arbitrary nonvanihing function a, b c are integer. Then, the following identitie hold: =0 =0 =0 =0 =0 f f f f f 1 f X m ck+c b+b a = X m k, 4.1 X m bk+b c+c a = X m k, 4. X m ak+a c+c b = X m f 1 k, 4. 1 X m c ak+c b+b = f k 1 X m, X m c bk+c a+a = f k X m, 4.5 f 1 7

8 =0 1 X m b ck+b a+a = f k X m. 4.6 f 1 Proof. Only identity 4.1 need to be proved a identitie are obtained from 4.1 by re-arranging the recurrence relation. The proof of 4.1 i by induction on k, imilar to the proof of Lemma of [1]. Theorem 5. The following identitie hold for non-negative integer k, integer m integer r / { 17, 4, 1,0}: =0 T r 1 +T r T r+1 Tr T m r+k++ = T m, 4.7 Tr k =0 =0 =0 1 1 =0 1 =0 T r Tr+1 T r 1 +T r T m r+1k + = T r 1T r +T r 1 Tr 1 +T r T r T r+1 T r T r T m T r 1 +T r k, 4.8 T m r 1k + = T m T k r, 4.9 T m k++r+1 = 1 k Tr+1 T m, 4.10 T r T m k++r = 1 k Tr 1 +T r T m 4.11 T r T r+1 T r 1 +T r T m+k++r = 1 k T r T r 1 +T r k T m. 4.1 Proof. Write the identity 1.5 a T m = T r T m r +T r 1 +T r T m r 1 +T r+1 T m r, identify f 1 = T r, = T r 1 +T r, = T r+1, a = r +, b = r +1, c = r ue thee in Lemma 5 with X = T. Reference [1] K. Adegoke, Weighted um of ome econd-order equence, arxiv: [math.nt] 018. [] P. Anantakitpaial K. Kuhapatanakul, Reciprocal um of the Tribonacci number, Journal of Integer equence , 1 9. [] J. Feng, More identitie on the Tribonacci number, Ar Combinatorial C 011, [4] R. Frontczak, Sum of Tribonacci Tribonacci-Luca number, International Journal of Mathematical Analyi 1:1 018,

9 [5] N. Irmak M. Alp, Tribonacci number with indice in arithmetic progreion their um, Mikolc Mathematical Note 14:1 01, [6] E. Kilic, Tribonacci equence with certain indice their um, Ar Combinatorial , 1. [7] D. V. Shah, Some Tribonacci identitie, Mathematic Today 7 011,

Weighted generalized Fibonacci sums

Weighted generalized Fibonacci um Kunle Adegoke arxiv:1805.01538v1 [math.ca] 1 May 2018 Department of Phyic Engineering Phyic, Obafemi Awolowo Univerity, 220005 Ile-Ife, Nigeria Abtract We derive weighted