Fibonacci and Lucas Identities the Golden Way

Fibonacci Lucas Identities the Golden Way Kunle Adegoe adegoe00@gmail.com arxiv:1810.12115v1 [math.nt] 25 Oct 2018 Department of Physics Engineering Physics, Obafemi Awolowo University, 220005 Ile-Ife, Nigeria Abstract By expressing Fibonacci Lucas numbers in terms of the powers of the golden ratio, α = (1 + 5)/2 its inverse, β = 1/α = (1 5)/2, a multitude of Fibonacci Lucas identities have been developed in the literature. In this paper, we follow the reverse course: we derive numerous Fibonacci Lucas identities by maing use of the well-nown expressions for the powers of α β in terms of Fibonacci Lucas numbers. 1 Introduction The Fibonacci numbers, F n, the Lucas numbers, L n, are defined, for n Z, through the recurrence relations with F n = F n 1 +F n 2, (n 2), F 0 = 0, F 1 = 1; (1.1) L n = L n 1 +L n 2, (n 2), L 0 = 2, L 1 = 1; (1.2) F n = ( 1) n 1 F n, L n = ( 1) n L n. (1.3) Throughout this paper, we denote the golden ratio, (1 + 5)/2, by α write β = (1 5)/2 = 1/α, so that αβ = 1 α+β = 1. The following well-nown algebraic properties of α β can be proved directly from Binet s formula for the nth Fibonacci number or by induction: α n = α n 1 +α n 2, (1.4) β n = β n 1 +β n 2, (1.5) α n = αf n +F n 1, (1.6) α n 5 = α n (α β) = αl n +L n 1, (1.7) β n = βf n +F n 1, (1.8) β n 5 = β n (α β) = βl n L n 1, (1.9) β n = αf n +F n+1, (1.10) 1

β n 5 = αl n L n+1, (1.11) α n = ( 1) n 1 αf n +( 1) n F n+1 (1.12) β n = ( 1) n αf n +( 1) n F n 1. (1.13) Hoggatt et. al. [4] derived, among other results, the identity F +t = α F t +β t F, which, upon multiplication by α s, can be put in the form α s F +t = α s+ F t +( 1) t α s t F. (1.14) Identity (1.14) is unchanged under interchange of t, s interchange of t s s t; we therefore have three additional identities: As Koshy [7, p.79] noted, the two Binet formulas α s F +t = α s+t F +( 1) α s F t, (1.15) α F s+t = α s+ F t +( 1) t α t F s (1.16) α s t F = α t F s +( 1) s α t F s. (1.17) F n = αn β n α β, L n = α n +β n, (1.18) expressing F n L n in terms of α n β n, can be used in tem to derive an array of identities. Our aim in writing this paper is to derive numerous Fibonacci Lucas identities by emphasizing identities (1.6) (1.11), expressing α n β n in terms of F n L n. Our method relies on the fact that α β are irrational numbers. We will mae frequent use of the fact that if a, b, c d are rational numbers γ is an irrational number, then aγ +b = cγ +d implies that a = c b = d; an observation that was used to advantage by Griffiths [3]. As a quic illustration of our method, tae x = αf p y = F p 1 in the binomial identity to obtain x y n = (x+y) n, which, by multiplying both sides by α q, can be written α Fp Fn p 1 = αnp, (1.19) α +q Fp Fn p 1 = α np+q, (1.20) 2

which, by identity (1.6), evaluates to α F pfp 1F n +q + Comparing the coefficients of α in (1.21), we find F pfp 1F n +q 1 = αf np+q +F np+q 1. (1.21) Fp Fn p 1 F +q = F np+q ; (1.22) valid for non-negative integer n arbitrary integers p q. Identity (1.22) contains many nown identities as special cases. If we write (1.20) as α +q 5F pf n p 1 = α np+q 5 (1.23) apply identity (1.7), we obtain the Lucas version of (1.22), namely, F pfp 1L n +q = L np+q. (1.24) Lemma 1. The following properties hold for a, b, c d rational numbers: aα+b = cα+d a = c, b = d, P1 1 cα+d = 1 cβ +d = aβ +b = cβ +d a = c, b = d, P2 ( ) ( ) c c+d α, P3 c 2 d 2 cd c 2 d 2 cd ( ) ( ) c c+d β, P4 c 2 d 2 cd c 2 d 2 cd aα+b cα+d = cb da ca db cb α+ c 2 d 2 cd c 2 cd d 2 aβ +b cβ +d = cb da c 2 d 2 cd β + ca db cb c 2 cd d. 2 Properties P3 to P6 follow from properties P1 P2. The rest of this section is devoted to using our method to re-discover nown results or to discover results that may be easily deduced from nown ones. In establishing some of the identities we require the fundamental relations F 2n = F n L n, L n = F n 1 + F n+1 5F n = L n 1 +L n+1. Presumably new results will be developed in section 2. P5 P6 3

Fibonacci Lucas addition formulas To derive the Fibonacci addition formula, use (1.6) to write the identity α p+q = α p α q (1.25) as αf p+q +F p+q 1 = (αf p +F p 1 )(αf q +F q 1 ), from which, by multiplying out the right side, maing use of (1.6) again, we find αf p+q +F p+q 1 = α(f p F q +F p F q 1 +F p 1 F q )+F p F q +F p 1 F q 1 = α(f p F q+1 +F p 1 F q )+F p F q +F p 1 F q 1. (1.26) Equating coefficients of α (property P1) from both sides of (1.26) establishes the well-nown Fibonacci addition formula: F p+q = F p F q+1 +F p 1 F q. (1.27) A similar calculation using the identity produces the subtraction formula α p β q = ( 1) q α p q (1.28) ( 1) q F p q = F p F q 1 F p 1 F q, (1.29) which may, of course, be obtained from (1.27) by changing q to q. The Lucas counterpart of identity (1.27) is obtained by applying (1.6) (1.7) to the identity α p+q 5 = (α p 5)α q, (1.30) proceeding as in the Fibonacci case, giving L p+q = F p L q+1 +F p 1 L q. (1.31) Application of (1.6) to the right side (1.7) to the left side of the identity 5α p+q = (α p 5)(α q 5) (1.32) produces 5F p+q = L p L q+1 +L p 1 L q. (1.33) Fibonacci Lucas multiplication formulas Subtracting identity (1.14) from identity (1.15) gives ( F t α s+ ( 1) α s ) ( = F α s+t ( 1) t α s t). (1.34) Applying identity (1.6) to identity (1.34) equating coefficients of α, we obtain ( ) ( ) F t Fs+ ( 1) F s = F Fs+t ( 1) t F s t, (1.35) which, upon setting = 1, gives F t L s = F s+t ( 1) t F s t. (1.36) 4

Writing identity (1.34) as F t (α s+ 5 ( 1) α s ) ( 5 = F α s+t 5 ( 1) t α s t ) 5, (1.37) applying identity (1.7) equating coefficients of α produces which, upon setting = 1, gives F t ( Ls+ ( 1) L s ) = F ( Ls+t ( 1) t L s t ), (1.38) 5F t F s = L s+t ( 1) t L s t. (1.39) Adding identity (1.14) identity (1.15), maing use of identity (1.6) equating the coefficients of α, we obtain which, at = t reduces to 2F +t F s = F t ( Fs+ +( 1) F s ) +F ( Fs+t +( 1) t F s t ), (1.40) L t F s = F s+t +( 1) t F s t. (1.41) Similarly, adding identity (1.14) identity (1.15), multiplying through by 5, maing use of identity (1.7) equating the coefficients of α, we have 2F +t L s = F t ( Ls+ +( 1) L s ) +F ( Ls+t +( 1) t L s t ), (1.42) which, at = t reduces to L t L s = L s+t +( 1) t L s t. (1.43) Cassini s identity Since α n β n = (αβ) n = ( 1) n ; (1.44) applying identities (1.6) (1.10) to the left h side of the above identity gives Thus, according to (1.44), we have α n β n = (αf n +F n 1 )( αf n +F n+1 ) = α 2 F 2 n +α(f n F n+1 F n F n 1 )+F n 1 F n+1 = α( F 2 n +F2 n ) F2 n +F n 1F n+1. α( F 2 n +F2 n ) F2 n +F n 1F n+1 = ( 1) n. Comparing coefficients of α 0 from both sides gives Cassini s identity: To derive the Lucas version of (1.45), write F n 1 F n+1 = F 2 n +( 1)n. (1.45) (α n 5)(β n 5) = ( 1) n 5; (1.46) apply (1.7) (1.11) to the left h side, multiply out equate coefficients, obtaining L n 1 L n+1 L 2 n = ( 1)n 1 5. (1.47) 5

General Fibonacci Lucas addition formulas Catalan s identity From identity (1.14), we can derive an addition formula that includes identity (1.27) as a particular case. Using identity (1.6) to write the left h side (lhs) the right h side (rhs) of identity (1.14), we have lhs of (1.14) = αf s F +t +F s 1 F +t (1.48) rhs of (1.14) = αf s+ F t +F s+ 1 F t +α( 1) t F s t F +( 1) t F s t 1 F = α(f s+ F t +( 1) t F s t F )+(F s+ 1 F t +( 1) t F s t 1 F ). (1.49) Comparing the coefficients of α from (1.48) (1.49), we find of which identity (1.27) is a particular case. Setting t = s in (1.50) produces Catalan s identity: F s F +t = F s+ F t +( 1) t F s t F, (1.50) F 2 s = F s+f s +( 1) s+ F 2. (1.51) Multiplying through identity (1.14) by 5 performing similar calculations to above produces L s F +t = L s+ F t +( 1) t L s t F, (1.52) which at t = s gives F 2s = L s+ F s +( 1) s+ F 2. (1.53) Sums of Fibonacci Lucas numbers with subscripts in arithmetic progression Setting x = α p in the geometric sum identity x = 1 xn+1 1 x (1.54) multiplying through by α q gives Thus, we have α p+q = αq α pn+p+q 1 α p. (1.55) α p+q = α(f pn+p+q F q )+(F pn+p+q 1 F q 1 ) αf p +(F p 1 1) from which, with the use of identity (1.6) property P5, we find, (1.56) F p+q = F p(f pn+p+q 1 F q 1 ) (F p 1 1)(F pn+p+q F q ) L p 1+( 1) p 1, (1.57) valid for all integers p, q n. The derivation here is considerably simpler than in the direct use of Binet s formula as done, for example, in Koshy [7, p. 86] by Freitag [2]. 6

Multiplying through identity (1.55) by 5 gives α p+q 5 = αq 5 α pn+p+q 5 1 α p, (1.58) from which, by identities (1.7) (1.6), properties P5 P1, we find L p+q = F p(l pn+p+q 1 L q 1 ) (F p 1 1)(L pn+p+q L q ) L p 1+( 1) p 1. (1.59) Generating functions of Fibonacci Lucas numbers with indices in arithmetic progression Setting x = yα p in the identity multiplying through by α q gives x = 1 1 x (1.60) α p+q y = αq 1 α p y = F q α+f q 1 yf p α+1 yf p 1. (1.61) Application of identity (1.6) properties P5 P1 then produces To find the corresponding Lucas result, we write F p+q y = F q +( 1) q F p q y 1 L p y +( 1) p y 2. (1.62) α p+q 5y = αq 5 1 α p y (1.63) use identity (1.7) properties P5 P1, obtaining L p+q y = L q ( 1) q L p q y 1 L p y +( 1) p y 2. (1.64) Identity (1.62), but not (1.64), was reported in Koshy [7, identity 18, p.230]. The case q = 0 in (1.62) was first obtained by Hoggatt [5] while the case p = 1 in (1.64) is also found in Koshy [7, identity (19.2), p.231]. 2 Main results 2.1 Recurrence relations Theorem 1. The following identities hold for integers p, q r: F p+q+r = F q 1 F r F p 1 +(F q+1 F r +F q 1 F r 1 )F p +F q F r+1 F p+1, (2.1) 7

F p+q+r = F p 1 F r F q 1 +(F p+1 F r +F p 1 F r 1 )F q +F p F r+1 F q+1, (2.2) F p+q+r = F q 1 F p F r 1 +(F q+1 F p +F q 1 F p 1 )F r +F q F p+1 F r+1, (2.3) F p+q+r = F r 1 F q F p 1 +(F r+1 F q +F r 1 F q 1 )F p +F r F q+1 F p+1, (2.4) L p+q+r = F q 1 F r L p 1 +(F q+1 F r +F q 1 F r 1 )L p +F q F r+1 L p+1, (2.5) L p+q+r = F p 1 F r L q 1 +(F p+1 F r +F p 1 F r 1 )L q +F p F r+1 L q+1, (2.6) L p+q+r = F q 1 F p L r 1 +(F q+1 F p +F q 1 F p 1 )L r +F q F p+1 L r+1, (2.7) L p+q+r = F r 1 F q L p 1 +(F r+1 F q +F r 1 F q 1 )L p +F r F q+1 L p+1, (2.8) 5F p+q+r = L q 1 L r F p 1 +(L q+1 L r +L q 1 L r 1 )F p +L q L r+1 F p+1, (2.9) 5F p+q+r = L p 1 L r F q 1 +(L p+1 L r +L p 1 L r 1 )F q +L p L r+1 F q+1, (2.10) 5F p+q+r = L q 1 L p F r 1 +(L q+1 L p +L q 1 L p 1 )F r +L q L p+1 F r+1, (2.11) 5F p+q+r = L r 1 L q F p 1 +(L r+1 L q +L r 1 L q 1 )F p +L r L q+1 F p+1, (2.12) 5L p+q+r = L q 1 L r L p 1 +(L q+1 L r +L q 1 L r 1 )L p +L q L r+1 L p+1, (2.13) 5L p+q+r = L p 1 L r L q 1 +(L p+1 L r +L p 1 L r 1 )L q +L p L r+1 L q+1, (2.14) 5L p+q+r = L q 1 L p L r 1 +(L q+1 L p +L q 1 L p 1 )L r +L q L p+1 L r+1, (2.15) 5L p+q+r = L r 1 L q L p 1 +(L r+1 L q +L r 1 L q 1 )L p +L r L q+1 L p+1. (2.16) Proof. Identities (2.1) (2.16) are derived by applying (1.6) (1.7) to the following identities: α p+q+r = α p α q α r, α p+q+r 5 = (α p 5)α q α r, 5α p+q+r = (α p 5)(α q 5)α r 5α p+q+r 5 = (α p 5)(α q 5)(α r 5). Note that identities (2.2) (2.4) are obtained from identity (2.1) by interchanging two indices from p, q r. Theorem 2. The following identities hold for integers p, q, r, s t: F p+q r F t s+r +F p+q r 1 F t s+r 1 = F p s F t+q +F p s 1 F t+q 1, (2.17) F p+q r L t s+r +F p+q r 1 L t s+r 1 = F p s L t+q +F p s 1 L t+q 1, (2.18) L p+q r F t s+r +L p+q r 1 F t s+r 1 = L p s F t+q +L p s 1 F t+q 1 (2.19) L p+q r L t s+r +L p+q r 1 L t s+r 1 = L p s L t+q +L p s 1 L t+q 1. (2.20) 8

Proof. Identity (2.17) is proved by applying identity (1.6) to the identity α p+q r α t s+r = α p s α t+q, multiplying out the products applying property P1. Identities (2.18) (2.19) are derived by writing α p+q r (α t s+r 5) = α p s (α t+q 5) (α p+q r 5)α t s+r = (α p s 5)α t+q, applying identities (1.6) (1.7) property P1. Finally identity (2.20) is derived from (α p+q r 5)(α t s+r 5) = (α p s 5)(α t+q 5). 2.2 Summation identities 2.2.1 Binomial summation identities Lemma 2. The following identities hold for positive integer n arbitrary x y: y x n = (x+y) n, (2.21) ( 1) (x+y) x n = ( 1) n y n, (2.22) ( 1) y (x+y) n = x n, (2.23) y 1 x n = n(x+y) n 1, (2.24) ( 1) (x+y) 1 x n = ( 1) n ny n 1 (2.25) ( 1) 1 y 1 (x+y) n = nx n 1. (2.26) =1 Proof. Setting z = 0 in the binomial identity y e z x n = (x+ye z ) n (2.27) 9

gives identity (2.21). Identities (2.22) (2.23) are obtained from identity (2.21) by obvious transformations. Identity (2.24) is obtained by differentiating identity (2.27) with respect to z then setting z to zero. More generally, ( n ) r y x n = dr ) n dz r(x+yez. (2.28) z=0 Identities (2.25) (2.26) are obtained from identity (2.24) by transformations. Theorem 3. The following identities hold for integers, t, s positive integer n: ( 1) t F Fn t F (s+)n (t+) = Ft+ n F sn, (2.29) =1 ( 1) t F Fn t L (s+)n (t+) = Ft+L n sn, (2.30) ( 1) F +t Fn t F (s+)n = ( 1) n(t+1) FF n n(s t), (2.31) ( 1) F +t Fn t L (s+)n = ( 1) n(t+1) FL n n(s t), (2.32) ( 1) (t+1) F Fn +t F sn t = Ft n F n(s+), (2.33) ( 1) (t+1) F Fn +t L sn t = Ft n L n(s+), (2.34) ( 1) t ( 1) t =1 ( 1) t ( 1) t =1 ( 1) F 1 ( 1) F 1 =1 +t Fn t +t Fn t F 1 F 1 F n t F n t F (+s)n+t s (+t) = nf n 1 +t F s(n 1), (2.35) L (+s)n+t s (+t) = nf n 1 +t L s(n 1), (2.36) F (+s)n s = ( 1) n(t+1)+t nf n 1 F (s t)(n 1), (2.37) L (+s)n s = ( 1) n(t+1)+t nf n 1 L (s t)(n 1), (2.38) ( 1) t+1 ( 1) (t+1) F 1 F n +t F s(n 1)+t t = nft n 1 F (s+)(n 1) (2.39) =1 ( 1) t+1 ( 1) (t+1) F 1 F n +t L s(n 1)+t t = nft n 1 L (s+)(n 1). (2.40) =1 10

Proof. Choosing x = α s+ F t y = ( 1) t α s t F in identity (2.21) taing note of identity (1.14), we have ( 1) t F Fn t α (s+)n (t+) = F+tα n ns. (2.41) Application of identity (1.6) property P1 to (2.41) produces identity (2.29). To prove (2.30), multiply through identity (2.41) by 5 to obtain ( 1) t F Fn t {α (s+)n (t+) 5} = F+t{α n ns 5}. (2.42) Use of identity (1.7) property P1 in identity (2.42) gives identity (2.30). Identities (2.31) (2.40) are derived in a similar fashion. Lemma 3. The following identities hold true for integer r, non-negative integer n, arbitrary x y: ( ) 2n+1 n+ +1 (xy) r (x y) 2+1 = x r+n+1 y r n y r+n+1 x r n, (2.43) n+ +1 2 +1 ( ) 2n+1 n+ +1 (x(x y)) r y 2+1 = x r+n+1 (x y) r n (x y) r+n+1 x r n (2.44) n+ +1 2 +1 2n+1 n+ +1 + +1 (y(y x)) r x 2+1 = y r+n+1 (y x) r n (y x) r+n+1 y r n. 2 +1 (2.45) Proof. Jennings [6, Lemma (i)] derived an identity equivalent to the following: ( )( ) 2n+1 n+ +1 z 2 2 1 = z2 z 2n z 2n. (2.46) n+ +1 2 +1 z z 2 1 Setting z 2 = x/y in the above identity clearing fractions gives identity (2.43). Identity (2.44) is obtained by replacing y with x y in identity (2.43). Identity (2.45) is obtained by interchanging x with y in identity (2.44). Theorem 4. The following identities hold for non-negative integer n integers s,, r t: ( ) ( 1) t 2n+1 n+ +1 (F +t F t ) r F 2+1 F r(2s+)+s t (2t+) n+ +1 2 +1 (2.47) = F r+n+1 +t Ft r n F s(r+n+1)+(s+)(r n) F r+n+1 t F r n +t F (s+)(r+n+1)+s(r n), ( 1) t 2n+1 n+ +1 = F r+n+1 +t + +1 2 +1 F r n (F +t F t ) r F 2+1 L r(2s+)+s t (2t+) L s(r+n+1)+(s+)(r n) Ft r+n+1 F r n +t L (s+)(r+n+1)+s(r n), (2.48) 11

( 1) t 2n+1 n+ +1 = ( 1) tn F r+n+1 +t + +1 2 +1 F r n (F +t F t ) r F 2+1 t F r(2s t)+s++(2+t) F s(2r+1) t(r n) ( 1) tn+t F r+n+1 F r n +t F s(2r+1) t(r+n+1), (2.49) ( 1) t 2n+1 n+ +1 = ( 1) tn F r+n+1 +t + +1 2 +1 F r n (F +t F t ) r F 2+1 t L r(2s t)+s++(2+t) L s(2r+1) t(r n) ( 1) tn+t F r+n+1 F r n +t L s(2r+1) t(r+n+1), (2.50) ( ) ( 1) (t 1) 2n+1 n+ +1 (F t F ) r F 2+1 +t F s(2r+1)+( t)r ( t) n+ +1 2 +1 = ( 1) (t 1)n Ft r+n+1 F r n ( 1) (t 1)(n+1) F r+n+1 F s(2r+1)+(r+n+1)+t(n r) Ft r n F s(2r+1) (n r) t(r+n+1) ( ) ( 1) (t 1) 2n+1 n+ +1 (F t F ) r F 2+1 +t L s(2r+1)+( t)r ( t) n+ +1 2 +1 = ( 1) (t 1)n Ft r+n+1 F r n L s(2r+1)+(r+n+1)+t(n r) ( 1) (t 1)(n+1) F r+n+1 Ft r n L s(2r+1) (n r) t(r+n+1). (2.51) (2.52) Proof. Each of identities (2.47) (2.52) is proved by settingx = α s+ F t y = ( 1) t α s t F in the identities of Lemma 3 taing note of identity (1.14) while maing use of identities (1.6) (1.7) property P1. 2.2.2 Summation identities not involving binomial coefficients Lemma 4 ([1, Lemma 1]). Let (X t ) (Y t ) be any two sequences such that X t Y t, t Z, are connected by a three-term recurrence relation hx t = f 1 X t a +f 2 Y t b, where h, f 1 f 2 are arbitrary non-vanishing complex functions, not dependent on t, a b are integers. Then, the following identity holds for integer n: f 2 f n 1 h Y t na b+a = h n+1 X t f n+1 1 X t (n+1)a. Theorem 5. The following identities hold for integers n,, s t: ( 1) n+t 1 F ( 1) F n(s )+s t+2 = F t F (s+)(n+1) F t+(n+1) F s(n+1), (2.53) 12

( 1) n+t 1 F ( 1) L n(s )+s t+2 = F t L (s+)(n+1) F t+(n+1) L s(n+1), (2.54) ( 1) nt+ 1 F t ( 1) t F n(s t)+s +2t = F F (s+t)(n+1) F +(n+1)t F s(n+1), (2.55) ( 1) nt+ 1 F t ( 1) t L n(s t)+s +2t = F L (s+t)(n+1) F +(n+1)t L s(n+1), (2.56) ( 1) ns+t 1 F s ( 1) s F n( s)+ t+2s = F t F (+s)(n+1) F t+(n+1)s F (n+1) (2.57) ( 1) ns+t 1 F s ( 1) s L n( s)+ t+2s = F t L (+s)(n+1) F t+(n+1)s L (n+1). (2.58) Proof. Write identity (1.14) as α s+ F t = α s F t+ + ( 1) t 1 α s t F identify h = α s+, f 1 = α s, f 2 = F, a =, b = 0, X t = F t Y t = ( 1) t 1 α s t in Lemma 4. Application of identity (1.6) to the resulting summation identity yields identity (2.53). Identity (2.54) is obtained by multiplying the α sum by 5 using identity (1.7). Identities (2.55) (2.58) are obtained from identities (2.53) (2.54) by interchanging t; s, in turn, since identity (1.14) remains unchanged under these operations. Lemma 5. The following identities hold for integers r n arbitrary x y: (x y) y r x = y r n x n+1 y r+1, (2.59) x y r (x+y) = y r n (x+y) n+1 y r+1 (2.60) (x y) x r y = x r+1 x r n y n+1. (2.61) Proof. Identity (2.59) is obtained by replacing x with x/y in identity (1.54). Identity (2.60) is obtained by replacing x with x+y in identity (2.59). Finally, identity (2.61) is obtained by interchanging x y in identity (2.59). Theorem 6. The following identities hold for integers r, n, s, t: ( 1) t F ( 1) t F F +t Fr t F r(s+)+s t = Ft r n F n+1 +t F s(r+1)+(r n) Ft r+1 F (s+)(r+1), (2.62) F +t Fr t L r(s+)+s t = Ft r n F n+1 +t L s(r+1)+(r n) Ft r+1 L (s+)(r+1), (2.63) 13

( 1) t F ( 1) t F ( 1) rt F t ( 1) t F r F +t F r(s t)+s++t = F r n F n+1 +t F r(s t)+tn+s ( 1) t(r+1) F r+1 F (s t)(r+1), ( 1) rt F t ( 1) t F r F +t L r(s t)+s++t = F r n F n+1 +t L r(s t)+tn+s ( 1) t(r+1) F r+1 L (s t)(r+1), (2.64) (2.65) F r +t F tf s(r+1)+s t+ = F r+1 +t F s(r+1) F r n +t Fn+1 t F s(r+1)+(n+1) (2.66) F r +t F tl s(r+1)+s t+ = F r+1 +t L s(r+1) F r n +t Fn+1 t L s(r+1)+(n+1). (2.67) Proof. Identities (2.62) (2.63) identities (2.66) (2.67) are obtained by setting x = α s F +t y = α s+ F t in identities (2.59) (2.61) while taing note of identity (1.14). Identities (2.64) (2.65) are derived by setting x = α s+ F t y = ( 1) t α s t F in identity (2.60). Theorem 7. The following identities hold for integers p, q n: F p+q = (n+1) F pf p(n+1)+q 1 (F p 1 1)F p(n+1)+q L p 1+( 1) p 1 (F 2p 2F p )(F p(n+2)+q 1 F p+q 1 ) + (F 2p 1 2F p 1 +1)(F 2p+1 2F p+1 +1) (F 2p 2F p ) 2 (F 2p 1 2F p 1 +1)(F p(n+2)+q F p+q ) (F 2p 1 2F p 1 +1)(F 2p+1 2F p+1 +1) (F 2p 2F p ) 2 L p+q = (n+1) F pl p(n+1)+q 1 (F p 1 1)L p(n+1)+q L p 1+( 1) p 1 (F 2p 2F p )(L p(n+2)+q 1 L p+q 1 ) + (F 2p 1 2F p 1 +1)(F 2p+1 2F p+1 +1) (F 2p 2F p ) 2 (F 2p 1 2F p 1 +1)(L p(n+2)+q L p+q ) (F 2p 1 2F p 1 +1)(F 2p+1 2F p+1 +1) (F 2p 2F p ). 2 (2.68) (2.69) Proof. Differentiating identity (1.54) with respect to x multiplying through by x gives x = (n+1) xn+1 x 1 xn+2 x (x 1). (2.70) 2 Setting x = α p multiplying through by α q produces α p+q = (n+1)αp(n+1)+q + αp(n+2)+q α p+q α p 1 2α p α 2p 1, (2.71) from which the results follow through the use of identities (1.6) (1.7) properties P1 P5. 14

3 Concluding remars A Fibonacci-lie sequence (G ) Z is one whose initial terms G 0 G 1 are given integers, not both zero, G = G 1 +G 2, G = G +2 G +1. (3.1) Identities (1.4) (1.5) show that the sequences (A ) Z (B ) Z, where A = α B = β are Fibonacci-lie, with respective initial terms A 0 = α 0 = 1, A 1 = α B 0 = β 0 = 1, B 1 = β = 1/α. The identity F s t G +m = F m t G +s +( 1) s+t+1 F m s G +t, (3.2) can therefore be written for the golden ration its inverse as F s t α +m = F m t α +s +( 1) s+t+1 F m s α +t (3.3) F s t β +m = F m t β +s +( 1) s+t+1 F m s β +t. (3.4) Proceeding as in previous calculations, many identities can be derived using identities (3.3) (3.4). For example, setting x = F m t α +s y = ( 1) s+t F m s α +t in the binomial identity ( 1) y x n = (x y) n multiplying through by α p produces ( 1) (s+t+1) F m sfm tα n (+t)+(+s)(n )+p = Fs tα n (+m)n+p, from which we find ( 1) (s+t+1) F m sfm tf n (s+)n+p+(t s) = Fs t n F (+m)n+p ( 1) (s+t+1) F m sfm tl n (s+)n+p+(t s) = Fs t n L (+m)n+p. References [1] Kunle Adegoe, Weighted sums of some second-order sequences, The Fibonacci Quarterly 56:3 (2018), 252 262. [2] H. T. Freitag, On summations expansions of Fibonacci numbers, The Fibonacci Quarterly 11:1 (1973), 63 71. [3] M. Griffiths, Extending the domains of definition of some Fibonacci identities, The Fibonacci Quarterly 50:4 (2012), 352 359. 15

[4] V. E. Hoggatt Jr. J. W. Phillips H. T. Leonard Jr., Twenty-four master identities, The Fibonacci Quarterly 9:1 (1971), 1 17. [5] V. E. Hoggatt Jr., Some special Fibonacci Lucas generating functions, The Fibonacci Quarterly 9:2 (1971), 121 133. [6] D. Jennings, Some polynomial identities for the Fibonacci Lucas numbers, The Fibonacci Quarterly 31:2 (1993), 134 137. [7] T. Koshy, Fibonacci Lucas numbers with applications, Wiley-Interscience, (2001). [8] S. Vada, Fibonacci Lucas numbers, the golden section: theory applications, Dover Press, (2008). 2010 Mathematics Subect Classification: Primary 11B39; Secondary 11B37. Keywords: Horadam sequence, Fibonacci number, Lucas number, Fibonacci-lie number, Generating function, Golden ratio. 16