Weighted generalized Fibonacci sums

Transcription

1 Weighted generalized Fibonacci um Kunle Adegoke arxiv: v1 [math.ca] 1 May 2018 Department of Phyic Engineering Phyic, Obafemi Awolowo Univerity, Ile-Ife, Nigeria Abtract We derive weighted um, including binomial double binomial um, for the generalized Fibonacci equence {G m } where for m 2, G m = G m 1 +G m 2 with initial value G 0 G 1. 1 Introduction The generalized Fibonacci equence {G m } i given, for m 2, by G m = G m 1 + G m 2 with initial value G 0 G 1. In particular, when G 0 = 0 G 1 = 1, we have the Fibonacci equence {F m }, when G 0 = 2 G 1 = 1, we have the equence of Luca number, {L m }. Extenion to negative index i provided through G m = 1 m F m+1 G 0 F m G 1, Vada [9], identity 9. Whenever an identity contain number from two generalized Fibonacci equence, the number from one will be denoted by G while the number from the other will be denoted by H, with the appropriate ubcript. The following identitie connecting the Fibonacci number, the Luca number the generalized Fibonacci number are well known, Vada [9], identitie 8, 10a, 10b, 18: G m+n = 1 G m + G m+1, 1.1 G m+n + 1 n G m n = L n G m, 1.2 G m+n 1 n G m n = G m 1 +G m+1, 1.3 G n+r H m+n G n H m+n+r = 1 n G r H m G 0 H m+r, 1.4 Howard [5], Theorem 3.1, Corrolary 3.5: 1 r G m = +r G m+r F r G m+n+r. 1.5 Note, by the way, that identity 1.1 can be obtained from identity 1.5 through the et of tranformation m m+1, n n r n 1. Our purpoe i to dicover the weighted ummation identitie aociated with the identitie 1.3, ince thoe aociated with identitie are already contained a pecial cae, p = 1 = q, of the reult obtained in an earlier paper [1]. AMS Claification: 11B37, 11B39, 65B10 adegoke00@gmail.com 1

2 2 Weighted um Lemma 1 [1], Lemma 2. Let {X m } be any arbitrary equence, where X m, m Z, atifie a econd order recurrence relation X m = X m a + X m b, where are arbitrary non-vanihing complex function, not dependent on m, a b are integer. Then, for k any integer. f 1 X m b a = X m f k+1 1 X m k+1a, 2.1 f 2 X m a b = X m f k+1 2 X m k+1b, 2.2 X m+a b a = f / 1 X m + / X m k+1b a 2.3 X m+b a b = f / 2 X m + / X m k+1a b, 2.4 Theorem 1. The following identitie hold for integer m, n, r k: +1 F r 1 r +r G m+n+r+r = 1r +r G m+k+1r 1 r G m, n 0, r 1 r+1 G m+r+n+r = 1 r G m + 1 r+1k G m+k+1n+r, n 0 +r G m r+n = 1 r+r G m 1 rf k r G m+k+1n, n+r 0. +r Proof. In identity 1.5, identify = 1 r +r /, = 1 r F r /, a = r, b = n r ue thee in Lemma 1 with X m = G m. 3 Weighted binomial um The following identitie of Luca [8]: F + = F +2k L + = L +2k, 3.1 are the Fibonacci Luca pecialization of the generalized Fibonacci um: G + = G +2k, 3.2 2

3 which i itelf a particular cae of the following identity Theorem 2, identity 3.7: G m rk+n+r = 1 rk +r G m, n 0, 1 r being an evaluation at r = 2, n = 1 m = +2k. In thi ection, we will derive three general binomial ummation identitie for the G equence. Binomial um for Fibonacci or Fibonacci-like equence are alo dicued in reference [6, 3, 7, 4]. Firt we tate a required Lemma. Lemma 2 [1], Lemma 3. Let {X m } be any arbitrary equence. Let X m, m Z, atify a econd order recurrence relation X m = X m a + X m b, where are non-vanihing complex function, not dependent on m, a b are integer. Then, X m ak+a b = X m k, 3.3 X m+ak b = f k 1 X m 3.4 X m+bk a = f k 2 X m, 3.5 for k a non-negative integer. Theorem 2. The following identitie hold for integer n, m r non-negative integer k: G m+rk+n = 1 rk G m, n+r 0, G m rk+n+r = 1 rk +r G m, n 0, r +r G m n+rk+r = 1 1 rk G m, n r Proof. Ue, in Lemma 2, the value of a, b, found in the proof of Theorem 1, with X m = G m. A particular intance of identity 3.6 i G n = F rk+1g 0 F rkg 1,

4 which at r = n give 1 Gn L = k+1g 0 k G 1, 3.10 n L k n ince F 2n = L n. In particular, we have 1 G = F k+1 G 0 F k G Similarly, from identitie , we have G n+r = 1 rk +r G rk, n 0, r +r G r = 1 1 rk G n+rk, n 0, r which at r = n give In particular, we have 1 n G 2n = 1 nk L k n G nk n L n G n = 1 1 nk G 2nk G 2 = 1 G k 3.16 G = G 2k Weighted double binomial um Lemma 3 [2], Lemma 5. Let {X m } be any arbitrary equence, X m atifying a third order recurrence relation X m = X m a + X m b + X m c, where, are arbitrary non-vanihing function a, b c are integer. Let k be a non-negative integer. Then, the following identitie hold: =0 X m ck+c b+b a = X m k, 4.1 4

5 =0 =0 =0 =0 =0 =0 k X m bk+b c+c a = X m k, 4.2 X m ak+a c+c b = X m k, X m c ak+c b+b = f 1 X m, X m c bk+c a+a = f 2 X m X m b ck+b a+a = f 3 X m. 4.6 Note that each of identitie can be written in the following repective equivalent form: k k k X m ak c a b a = X m, 4.7 k k =0 k =0 k =0 k =0 k =0 k k k k k X m ak b a c a = X m f k 1 X m bk a b c b = X m f k 2, 4.8, f 3 X m+ak c b = f k 1 X m, f 3f 1X m+bk c a = f k 2X m f 2 X m+ck b a = f k 3 X m Theorem 3. The following identitie hold for integer n, m, r non-negative integer k: k k 1 n+ + G m 2nk+n+1+n 1 = 1 nk G m, 4.13 =0 k k 1 n+ + G m 2nk+n 1+n+1 = 1 nk G m, 4.14 =0 k k k 1 n =0 Gm n+1k n F n = G m, n 0, 4.15 k

6 1 Gm+n+1k 2+n+1 =0 =0 =0 1 n++ 1 n++ Gm+2k n+1+2n Gm 2k n 1+2n = 1 n+1kg m, n 0, 4.16 k = 1 G m, n 0, 4.17 = 1 G m, n Proof. Write identity 1.3 a G m = 1 n G m 2n + G m n 1 + G m n+1, identify a = 2n, b = n+1,c=n 1, = 1 n, = = ue thee in Lemma 3 withx m = G m. Theorem 4. The following identitie hold for integer n, m r non-negative integer k: =0 1 n+ 1 n++ =0 1 n++ =0 G G G n+rg n G H m+rk+n r+r = 0 Gr G 0 0 G n G H m+nk+r n+n = 1 nk Gr G n+r n+r H m, G 0 0, G n+r G H m+n+rk n+n r = 1 n+1k Gr n G n H m, G n+r 0, 4.20 H m, G n 0, 4.21 k k G 1 n++ 0 G n+r G + H m n+rk+r+n = 1 n+1k Gn H m, G r 0, =0 r G r 4.22 k k G 1 n+ 0 G n G + H m nk+r+n+r = 1 nk Gn+r H m, G r 0, =0 r G r 4.23 k k G 1 n++ n+rg k n G0 G + H m rk+n+n+r = H m, G r 0. =0 r G r 4.24 Proof. Write identity 1.4 a G r H m = 1 n G n H m+n+r + 1 n G n+r H m+n + G 0 H m+r, identify a = n r, b = n, c = r, = 1 n G n /G, = 1 n G n+r /G r, = G 0 /G r ue thee in Lemma 3 with X m = H m. Reference [1] K. Adegoke, Weighted um of ome econd-order equence, arxiv: [math.nt]

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