aa r X i v : . [ m a t h . N T ] J u l ON THE L.C.M. OF SHIFTED FIBONACCI NUMBERS
CARLO SANNA † Abstract.
Let ( F n ) n ≥ be the sequence of Fibonacci numbers. Guy and Matiyasevich provedthat log lcm( F , F , . . . , F n ) ∼ απ · n as n → + ∞ , where lcm is the least common multiple and α := (cid:0) √ / s = ( s n ) n ≥ in {− , +1 } there exists an effectivelycomputable rational number C s > F + s , F + s , . . . , F n + s n ) ∼ απ · C s · n , as n → + ∞ . Moreover, we show that if ( s n ) n ≥ is a sequence of independent uniformly distributed randomvariables in {− , +1 } then E (cid:2) log lcm( F + s , F + s , . . . , F n + s n ) (cid:3) ∼ απ ·
15 Li (1 / · n , as n → + ∞ , where Li is the dilogarithm function. Introduction
Let ( F n ) n ≥ be the sequence of Fibonacci numbers, defined recursively by F = F = 1 and F n +2 = F n +1 + F n , for every integer n ≥
1. Guy and Matiyasevich [8] proved that, as n → + ∞ ,(1) log lcm( F , F , . . . , F n ) ∼ απ · n , where lcm denotes the least common multiple and α := (cid:0) √ / F k in (1) is replaced by a shifted Fibonaccinumber F k ±
1, for various choices of signs. Arithmetic properties of shifted Fibonacci havebeen studied before. For example, Bugeaud, Luca, Mignotte, and Siksek [5] determined all theshifted Fibonacci numbers that are perfect powers; Marques [7] gave formulas for the orderof appearance of shifted Fibonacci numbers; and Pongsriiam [10] found all shifted Fibonaccinumbers that are products of Fibonacci numbers.Our first result concerns periodic sequences of signs.
Theorem 1.1.
For every periodic sequence s = ( s n ) n ≥ in {− , +1 } , there exists an effectivelycomputable rational number C s > such that log lcm( F + s , F + s , . . . , F n + s n ) ∼ απ · C s · n , as n → + ∞ . (The least common multiple starts from F + s to avoid zero terms.) We computed the constant C s for periodic sequences s with short period. We found that C s = 1 / C s = 1 / s with C s = 1 / Mathematics Subject Classification.
Primary: 11B39, Secondary: 11B37, 11N37.
Key words and phrases. asymptotic formula; Fibonacci number; least common multiple. † C. Sanna is a member of GNSAGA of INdAM and of CrypTO, the group of Cryptography andNumber Theory of Politecnico di Torino. s C s s C s s C s s C s ----+ / -+--+ / +---- / +-+++ / ---+- / -+-+- / +--+- / ++-+- / --+-- / -+-++ / +-+-- / ++-++ / --+-+ / -++-+ / +-+-+ / +++-+ / -+--- / -++++ / +-++- / ++++- / Table 1.
All period-5 sequences s such that C s = 1 / s C s s C s s C s s C s -----+ / --++-+ / +-+-++ / ++--++ / ----++ / --++++ / +-++-- / ++-+-- / ---+-- / -+---- / +-+++- / +++-+- / ---+-+ / -+---+ / +-++++ / +++-++ / --+-++ / -+--++ / ++---- / ++++-- / --++-- / -+-+-- / ++--+- / +++++- / Table 2.
All period-6 sequences s such that C s = 1 / Theorem 1.2.
Let ( s n ) n ≥ be a sequence of independently uniformly distributed random vari-ables in {− , +1 } . Then E (cid:2) log lcm( F + s , F + s , . . . , F n + s n ) (cid:3) ∼ απ ·
15 Li (1 / · n , as n → + ∞ , where Li ( z ) := P ∞ n = 1 z n /n denotes the dilogarithm. Using the methods of the proofs of Theorem 1.1 and Theorem 1.2, it should be possible toprove similar results, where the sequence of Fibonacci numbers is replaced by the sequence ofLucas numbers or by a sequence of integers powers ( a n ) n ≥ , with a ≥ s n ) n ≥ .We leave these as problems for the interested reader. Notation.
We employ the LandauBachmann “Big Oh” notation O with its usual meaning.Any dependence of the implied constants is indicated with subscripts. We let ϕ denote theEuler’s totient function. We reserve the letter p for prime numbers.2. Preliminaries on Fibonacci and Lucas Numbers
Let ( L n ) n ≥ be the sequence of Lucas numbers, defined recursively by L = 1, L = 3, and L n +2 = L n +1 + L n , for every integer n ≥
1. It is well known that the Binet’s formulas(2) F n = α n − β n α − β and L n = α n + β n , hold for every integer n ≥
1, where α := (cid:0) √ / β := (cid:0) − √ /
2. It is useful (proof ofLemma 2.3 later) to extend the sequences of Fibonacci and Lucas numbers to negative indicesusing (2). Let us define(3) Φ n := Y ≤ k ≤ n gcd( n,k ) = 1 (cid:16) α − e π i kn β (cid:17) , N THE L.C.M. OF SHIFTED FIBONACCI NUMBERS 3 for each integer n ≥
2, and put Φ := 1. It can be proved that each Φ n is an integer [13,p. 428]. Moreover, from (2) and (3) it follows that(4) F n = Y n ∈ D ( n ) Φ d and L n = Y n ∈ D ′ ( n ) Φ d , for every integer n ≥
1, where D ( n ) := { d ∈ N : d | n } and D ′ ( n ) := D (2 n ) \D ( n ). In particular,using (4) one can prove by induction that Φ n > n ≥ n . Lemma 2.1.
For all integers m > n ≥ we have gcd(Φ m , Φ n ) | m .Proof. For m ≥ m = 6 ,
12, and n ≥
3, it is known [13, Lemma 7] that gcd(Φ m , Φ n ) dividesthe greatest prime factor of m/ gcd(3 , m ), and consequently it divides m . The remaining casesfollow easily since Φ = Φ = 1, Φ = 2, Φ = 3, Φ = 5, Φ = 4, and Φ = 6. (cid:3) Lemma 2.2.
For all integers n ≥ , we have log Φ n = ϕ ( n ) log α + O (1) .Proof. See, e.g., [11, Lemma 2.1(iii)]. (cid:3)
The next lemma belongs to the folklore and provides a way to write shifted Fibonaccinumbers as products of Fibonacci and Lucas numbers.
Lemma 2.3.
For every integer k , we have F k +1 − F k L k +1 , F k +1 + 1 = F k +1 L k ,F k +2 − F k L k +2 , F k +2 + 1 = F k +2 L k ,F k +3 − F k +2 L k +1 , F k +3 + 1 = F k +1 L k +2 ,F k +4 − F k +3 L k +1 , F k +4 + 1 = F k +1 L k +3 . Proof.
Employing (2) and αβ = −
1, a quick algebraic manipulation yields(5) F a + b + ( − b F a − b = F a L b , for all integers a, b . Each of the eight identities corresponds to a particular choice of a, b in (5),noting that F − = 1 and F − = − (cid:3) Finally, we need a lemma about the greatest common divisor of a Fibonacci number and aLucas number.
Lemma 2.4.
For all integers m, n , we have that gcd( F m , L n ) is equal to , , or L gcd( m,n ) .Proof. See [9]. (cid:3) Further preliminaries
For every sequence s = ( s n ) n ≥ in {− , +1 } and for every integer n ≥
5, define ℓ s ( n ) = lcm( F + s , . . . , F n + s n ) . (Starting from F instead of F does not affect the asymptotic and simplifies a bit the nextarguments.) Furthermore, define the sets F s ( n ) := (cid:8) h ∈ [2 , n ] : s h − = ( − h ∨ s h − = ( − h +1 ∨ s h +1 = ( − h +1 ∨ s h +2 = ( − h +1 (cid:9) , L s ( n ) := (cid:8) h ∈ [2 , n ] : s h − = ( − h +1 ∨ s h − = ( − h ∨ s h +1 = ( − h ∨ s h +2 = ( − h (cid:9) , and M s ( n ) := [ h ∈ F s ( n ) D ( h ) ∪ [ h ∈ L s ( n ) D ′ ( h ) . The next lemma is the key to the proofs of Theorem 1.1 and Theorem 1.2.
C. SANNA
Lemma 3.1. As n → + ∞ , we have log ℓ s ( n ) = X d ∈ M s ( n ) ϕ ( d ) log α + O (cid:18) n log n (cid:19) . Proof.
Assume n ≥ n = 4 K + 4, for some real number K ≥
1. Using Lemma 2.3, wecan write each F i + s i ( i = 5 , . . . , n ) as a product of a Fibonacci number and a Lucas number,which, in light of Lemma 2.4, have a greatest common divisor not exceeding 3. Therefore,(6) log ℓ s ( n ) = log lcm (cid:18) lcm i ∈ F ′ s ( n ) F i , lcm j ∈ L ′ s ( n ) L j (cid:19) + O (1) , where F ′ s ( n ) , L ′ s ( n ) ⊆ [2 , K + 3] ∩ Z are defined by2 k ∈ F ′ s ( n ) ⇐⇒ (cid:0) (1 ≤ k ≤ K ) ∧ ( s k +1 = − ∨ s k +2 = − (cid:1) (7) ∨ (cid:0) (2 ≤ k ≤ K + 1) ∧ ( s k − = − ∨ s k − = +1) (cid:1) , k + 1 ∈ F ′ s ( n ) ⇐⇒ (cid:0) (1 ≤ k ≤ K ) ∧ ( s k +1 = +1 ∨ s k +3 = +1 ∨ s k +4 = +1) (cid:1) ∨ (cid:0) (2 ≤ k ≤ K + 1) ∧ s k = − (cid:1) , k ∈ L ′ s ( n ) ⇐⇒ (cid:0) (1 ≤ k ≤ K ) ∧ ( s k +1 = +1 ∨ s k +2 = +1) (cid:1) ∨ (cid:0) (2 ≤ k ≤ K + 1) ∧ ( s k − = − ∨ s k − = +1) (cid:1) , k + 1 ∈ L ′ s ( n ) ⇐⇒ (cid:0) (1 ≤ k ≤ K ) ∧ ( s k +1 = − ∨ s k +3 = − ∨ s k +4 = − (cid:1) ∨ (cid:0) (2 ≤ k ≤ K + 1) ∧ s k = +1 (cid:1) , for every integer k ∈ [1 , K + 1]. Since F i , L i ≤ i for every integer i ≥
1, replacing all thebounds on k in (7) with 2 ≤ k ≤ n/ O ( n ) in (6), that is,(8) log ℓ s ( n ) = log lcm (cid:18) lcm i ∈ F s ( n ) F i , lcm j ∈ L s ( n ) L j (cid:19) + O ( n ) . Suppose that p v || ℓ s ( n ), for some prime number p ≤ n and some integer v ≥
1. Then p v | F i + s i for some integer i ∈ [5 , n ], and consequently p v ≤ F n + 1 ≤ n . Hence,(9) log (cid:16) Y p v || ℓ n p ≤ n p v (cid:17) ≤ log (cid:16) Y p v || ℓ n p ≤ n n (cid:17) ≤ { p : p ≤ n } · n · log 2 = O (cid:18) n log n (cid:19) , since the number of primes not exceeding x is O ( x/ log x ).Writing each F i , L j in (8) as a product of Φ d ’s using (4), and taking into account Lemma 2.1and (9), we obtain that log ℓ s ( n ) = log Y d ∈ M s ( n ) Φ d + O (cid:18) n log n (cid:19) . Hence, by Lemma 2.2, we getlog ℓ s ( n ) = X d ∈ M s ( n ) log Φ d + O (cid:18) n log n (cid:19) = X d ∈ M s ( n ) ϕ ( d ) log α + O (cid:18) n log n (cid:19) , since M s ( n ) ⊆ [2 , n ] and consequently M s ( n ) ≤ n . (cid:3) For all integers r ≥ m ≥
1, and for every x ≥
1, let us define A r,m ( x ) := { n ≤ x : n ≡ r (mod m ) } . We need two lemmas about unions of D ( n ), respectively D ′ ( n ), with n ∈ A r,m ( x ). N THE L.C.M. OF SHIFTED FIBONACCI NUMBERS 5
Lemma 3.2.
Let r, m be positive integers and let S be the set of s ∈ { , . . . , m } such that thereexists an integer t ≥ satisfying st ≡ r (mod m ) . For each s ∈ S , let t ( s ) be the minimal t .Then, for all x ≥ , we have [ n ∈ A r,m ( x ) D ( n ) = [ s ∈ S A s,m (cid:18) xt ( s ) (cid:19) . Proof.
On the one hand, let n ∈ A r,m ( x ) and pick d ∈ D ( n ). Clearly, n = dt for some integer t ≥
1. Let s ∈ { , . . . , m } such that d ≡ s (mod m ). Then st ≡ dt ≡ n ≡ r (mod m ) andconsequently s ∈ S and t ≥ t ( s ). Therefore, d = n/t ≤ x/t ( s ), so that d ∈ A s,m ( x/t ( s )).On the other hand, suppose that d ∈ A s,m ( x/t ( s )) for some s ∈ S . Letting n := dt ( s ), wehave n ≡ st ( s ) ≡ r (mod m ) and n ≤ x , that is, n ∈ A r,m ( x ). Finally, d ∈ D ( n ). (cid:3) Lemma 3.3.
Let r, m be positive integers and let S be the set of s ∈ { , . . . , m } such thatthere exists an odd integer t ≥ satisfying st ≡ r (mod m ) . For each s ∈ S , let t ( s ) be theminimal t . Then, for all x ≥ , we have [ n ∈ A r,m ( x ) D ′ ( n ) = [ s ∈ S A s, m (cid:18) xt ( s ) (cid:19) . Proof.
On the one hand, let n ∈ A r,m ( x ) and pick d ∈ D ′ ( n ). Then 2 n = dt for some oddinteger t ≥
1. In particular, d is even. Let s ∈ { , . . . , m } such that d ≡ s (mod m ). Then st ≡ d t ≡ n ≡ r (mod m ), and consequently s ∈ S and t ≥ t ( s ). Therefore, d = 2 n/t ≤ x/t ( s ), so that d ∈ A s, m (2 x/t ( s )).On the other hand, suppose that d ∈ A s, m (2 x/t ( s )) for some s ∈ S . In particular, d iseven and d ≡ s (mod m ). Letting n := d t ( s ), we have n ≡ st ( s ) ≡ r (mod m ) and n ≤ x ,that is, n ∈ A r,m ( x ). Finally, 2 n = dt ( s ) and t ( s ) is odd, so that d ∈ D ′ ( n ). (cid:3) Finally, we need two asymptotic formulas for sums of the Euler’s function over an arithmeticprogression.
Lemma 3.4.
Let r, m be positive integers. Then, for every x ≥ , we have S r,m ( x ) := X n ∈ A r,m ( x ) ϕ ( n ) = 3 π · c r,m x + O r,m ( x log x ) , where c r,m := 1 m Y p | mp | r (cid:18) p (cid:19) − Y p | mp ∤ r (cid:18) − p (cid:19) − . Proof.
This is a special case of the asymptotic formula, given by Shapiro [12, Theorem 5.5A.2],for P n ≤ x ϕ ( f ( n )), where f a polynomial with integers coefficients, no multiple roots, andsatisfying f ( n ) ≥ n ≥ (cid:3) Lemma 3.5.
Let r, m be positive integers and let z ∈ (0 , . Then, for every x ≥ , we have X n ∈ A r,m ( x ) ϕ ( n ) (cid:0) − z ⌊ x/n ⌋ (cid:1) = 3 π · c r,m (1 − z ) Li ( z ) z · x + O r,m (cid:0) x (log x ) (cid:1) . C. SANNA
Proof.
For every integer k ≥
1, we have ⌊ x/n ⌋ = k if and only if x/ ( k + 1) < n ≤ x/k . Hence, X n ∈ A r,m ( x ) ϕ ( n ) (cid:0) − z ⌊ x/n ⌋ (cid:1) = X k ≤ x (cid:0) − z k (cid:1) (cid:18) S r,m (cid:16) xk (cid:17) − S r,m (cid:18) xk + 1 (cid:19)(cid:19) = X k ≤ x (cid:16)(cid:0) − z k (cid:1) − (cid:0) − z k − (cid:1)(cid:17) S r,m (cid:16) xk (cid:17) = (1 − z ) X k ≤ x z k − (cid:18) π · c r,m x k + O r,m (cid:18) x log xk (cid:19)(cid:19) = 3 π · c r,m (1 − z ) ∞ X k = 1 z k − k · x + O r,m X k > x x k ! + O r,m X k ≤ x x log xk = 3 π · c r,m (1 − z ) Li ( z ) z · x + O r,m (cid:0) x (log x ) (cid:1) , where we employed Lemma 3.4. (cid:3) Proof of Theorem 1.1
Let s = ( s n ) n ≥ be a periodic sequence in {− , +1 } , and let T ≥ s , it follows that there exist R , R ⊆ { , . . . , m } , where m := 2 T , suchthat F s ( n ) = [ r ∈ R A r,m ( n/
2) and L s ( n ) = [ r ∈ R A r,m ( n/ , for every integer n ≥ R ⊆ { , . . . , m } and positiverational numbers ( θ r ) r ∈R such that M s ( n ) = [ r ∈ R A r, m ( θ r n ) , for every integer n ≥ ℓ s ( n ) = X r ∈ R X d ∈ A r, m ( θ r n ) ϕ ( d ) log α + O (cid:18) n log n (cid:19) = 3 log απ · C s · n + O s (cid:18) n log n (cid:19) , where C s := X r ∈ R c r, m θ r is a positive rational number effectively computable in terms of s , . . . , s T .The proof is complete. 5. Proof of Theorem 1.2
Let s = ( s n ) n ≥ be a sequence of independent and uniformly distributed random variables in {− , +1 } , and let n ≥ k ∈ [2 , n/ k / ∈ F s ( n ), respectively k / ∈ L s ( n ), depends only on s k − , s k − , s k +1 , s k +2 .In particular, if the integers k , k ∈ [2 , n/
2] satisfy | k − k | ≥ (cid:0) k / ∈ F s ( n ) , k / ∈ F s ( n ) (cid:1) , (cid:0) k / ∈ L s ( n ) , k / ∈ L s ( n ) (cid:1) , and (cid:0) k / ∈ F s ( n ) , k / ∈ L s ( n ) (cid:1) are pairs of independent events.Moreover, we have P (cid:2) k / ∈ F s ( n ) (cid:3) = P (cid:2) k / ∈ L s ( n ) (cid:3) = 2 − = 16 − . N THE L.C.M. OF SHIFTED FIBONACCI NUMBERS 7
Therefore, it follows that P (cid:2) d / ∈ M s ( n ) (cid:3) = P h ^ ≤ k ≤ n/ d ∈ D ( k ) (cid:0) k / ∈ F s ( n ) (cid:1) ∧ ^ ≤ h ≤ n/ d ∈ D ′ ( h ) (cid:0) h / ∈ L s ( n ) (cid:1)i = Y ≤ k ≤ n/ d ∈ D ( k ) P (cid:2) k / ∈ F s ( n ) (cid:3) · Y ≤ h ≤ n/ d ∈ D ′ ( h ) P (cid:2) h / ∈ L s ( n ) (cid:3) = 16 −⌊ n/ (2 d ) ⌋ · − (cid:0) ⌊ n gcd(2 ,d ) / (2 d ) ⌋−⌊ n/ (2 d ) ⌋ (cid:1) = 16 −⌊ n gcd(2 ,d ) / (2 d ) ⌋ , for every integer d ≥
6. Consequently, by Lemma 3.1, we get E (cid:2) log ℓ s ( n ) (cid:3) = (log α ) X d ∈ M s ( n ) ϕ ( d ) P [ d ∈ M s ( n )] + O (cid:18) n log n (cid:19) (10) = (log α ) X d ≤ n ϕ ( d ) (cid:16) − −⌊ n gcd(2 ,d ) / (2 d ) ⌋ (cid:17) + O (cid:18) n log n (cid:19) . In turn, by Lemma 3.5, we have X d ≤ n ϕ ( d ) (cid:16) − −⌊ n gcd(2 ,d ) / (2 d ) ⌋ (cid:17) = X d ∈ A , ( n/ ϕ ( d ) (cid:16) − −⌊ n/ (2 d ) ⌋ (cid:17) (11) + X d ∈ A , ( n ) ϕ ( d ) (cid:16) − −⌊ n/d ⌋ (cid:17) = 3 π · (cid:16) c , c , (cid:17)
15 Li (1 / · n + O (cid:0) n (log n ) (cid:1) = 3 π ·
15 Li (1 / · n + O (cid:0) n (log n ) (cid:1) . Finally, putting together (10) and (11), we obtain E (cid:2) log ℓ s ( n ) (cid:3) ∼ απ ·
15 Li (1 / · n , as n → + ∞ .The proof is complete. References
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