THE ORDER OF APPEARANCE OF PRODUCT OF CONSECUTIVE FIBONACCI NUMBERS

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1 THE ORDER OF APPEARANCE OF PRODUCT OF CONSECUTIVE FIBONACCI NUMBERS DIEGO MARQUES Abstract. Let F n be the nth Fibonacci number. The order of appearance z(n) of a natural number n is defined as the smallest natural number k such that n divides F k. For instance, z(f n) = n, for all n 3. In this paper, among other things, we prove that z(f nf n+1f n+) = n(n+1)(n+), for all even positive integers n. 1. Introduction Let (F n ) n 0 be the Fibonacci sequence given by F n+ = F n+1 +F n, for n 0, where F 0 = 0 and F 1 = 1. These numbers are well-known for possessing amazing properties (consult [4] together with its very extensive annotated bibliography for additional references and history). In 193, the Fibonacci Association was created to provide enthusiasts an opportunity to share ideas about these intriguing numbers and their applications. We cannot go very far in the lore of Fibonacci numbers without encountering its companion Lucas sequence (L n ) n 0 which follows the same recursive pattern as the Fibonacci numbers, but with initial values L 0 = and L 1 = 1. The study of the divisibility properties of Fibonacci numbers has always been a popular area of research. Let n be a positive integer number, the order (or rank) of appearance of n in the Fibonacci sequence, denoted by z(n), is defined as the smallest positive integer k, such that n F k (some authors also call it order of apparition, or Fibonacci entry point). There are several results about z(n) in the literature. For instance, z(n) < for all n 1. The proof of this fact is an immediate consequence of the Théorème Fondamental of Section XXVI in [10, p. 300]. Indeed, z(m) < m 1, for all m > (see [15, Theorem, p. 5]) and in the case of a prime number p, one has the better upper bound z(p) p+1, which is a consequence of the known congruence F p ( p 5 ) 0 (mod p), for p,5, where ( a q ) denotes the Legendre symbol of a with respect to a prime q >. In recent papers, the author [, 7, 8] found explicit formulas for the order of appearance of integers related to Fibonacci numbers, such as F m ±1,F mk /F k and Fn. k In particular, it was proved that z(f 4m ±1) = 8m, if m > 1, and z(fn k+1 ) = nfn, k for all non-negative integers k and n > 3 with n 3 (mod ). In this paper, we continue this program and study the order of appearance of product of consecutive Fibonacci numbers. Our main results are the following. Theorem 1.1. We have (i) For n 3, z(f n F n+1 ) = n(n+1). Research supported in part by FAP-DF, FEMAT-Brazil and CNPq-Brazil. 13 VOLUME 50, NUMBER

2 (ii) For n, (iii) For n 1, ORDER OF APPEARANCE OF PRODUCTS OF FIBONACCI NUMBERS { n(n+1)(n+), if n 1 (mod ), z(f n F n+1 F n+ ) = n(n+1)(n+), if n 0 (mod ). z(f n F n+1 F n+ F n+3 ) = n(n+1)(n+)(n+3), if n 0 (mod 3), n(n+1)(n+)(n+3) 3, if n 0,9 (mod 1), n(n+1)(n+)(n+3), if n 3, (mod 1). We recall that thefibonacci factorial ofn(also called Fibonorial), denoted byn F!, is defined as the product of the first n nonzero Fibonacci numbers (sequence A003 in OEIS [13]). In the search for z(n F!), we found that n F! F n! (and thus z(n F!) n!) for n = 1,...,10 (see Table 1). It is therefore reasonable to conjecture that z(n F!) n! and so F 1 F n F n!, for all positive integers n. However, this is not true, because z(110 F!) 110! In fact, ν 11 (F 1 F 110 ) = 1 > 11 = ν 11 (F 110! ). Hence, motivated by this fact, we prove that n ! 4! 5!! 7! 8! 9! z(n F!) 1! 1! ! 88 Table 1. The order of appearance of n F!, for 1 n 10. Theorem 1.. For all p {,3,5,7} and n 1, we have ν p (F 1 F n ) ν p (F n! ). Here ν p (n) denotes the p-adic order of n which, as usual, is the exponent of the highest power of p dividing n. We organize the paper as follows. In Section, we will recall some useful properties of Fibonacci numbers such as d Ocagne s identity and a result concerning the p-adic order of F n. The last two sections will be devoted to the proof of theorems.. Auxiliary Results Before proceeding further, we recall some facts on Fibonacci numbers for the convenience of the reader. Lemma.1. We have (a) F n F m if and only if n m. (b) gcd(f n,f m ) = F gcd(n,m). (c) F n F n, for all n 0 (mod 3). (d) (d Ocagne s identity) ( 1) n F m n = F m F n+1 F n F m+1. (e) F p ( p 5 ) 0 (mod p), for all primes p. Most of the previous items can be proved by using induction together with the well-known Binet s formula: F n = αn β n, for n 0, α β where α = (1+ 5)/ and β = (1 5)/. We refer the reader to [1, 3, 4, 11] for more details and additional bibliography. MAY

3 THE FIBONACCI QUARTERLY The second lemma is a consequence of the previous one. Lemma.. We have (a) If F n m, then n z(m). (b) If n F m, then z(n) m. Proof. For (a), since F n m F z(m), by Lemma.1 (a), we get n z(m). In order to prove (b), we write m = z(n)q +r, where q and r are integers, with 0 r < z(n). So, by Lemma.1 (d), we obtain ( 1) z(n)q F r = F m F z(n)q+1 F z(n)q F m+1. Since n divides both F m and F z(n)q, then it also divides F r implying r = 0 (keep in mind the range of r). Thus, z(n) m. The p-adic order of Fibonacci and Lucas numbers was completely characterized, see [, 9, 1, 14]. For instance, from the main results of Lengyel [9], we extract the following result. Lemma.3. For n 1, we have 0, if n 1, (mod 3); 1, if n 3 (mod ); ν (F n ) = 3, if n (mod 1); ν (n)+, if n 0 (mod 1), ν 5 (F n ) = ν 5 (n), and if p is prime or 5, then { νp (n)+ν ν p (F n ) = p (F z(p) ), if n 0 (mod z(p)); 0, if n 0 (mod z(p)). A proof of this result can be found in [9]. Lemma.4. For any integer k 1 and p prime, we have k logk p 1 1 ν p (k!) k 1 logp p 1, (.1) where, as usual, x denotes the largest integer less than or equal to x. Proof. Recall the well-known Legendre s formula [5]: ν p (k!) = k s p(k), (.) p 1 where s p (k) is the sum of digits of k in base p. Since k has logk/logp +1 digits in base p, and each digit is at most p 1, we get ( ) logk 1 s p (k) (p 1) +1. (.3) logp Therefore, the inequality in (.1) follows from (.) and (.3). Now, we are ready to deal with the proof of theorems. 134 VOLUME 50, NUMBER

4 ORDER OF APPEARANCE OF PRODUCTS OF FIBONACCI NUMBERS 3. The Proof of Theorem Proof of (i). For ɛ {0,1}, one has that F n+ɛ F n F n+1 and so Lemma. (a) yields n + ɛ z(f n F n+1 ). But, gcd(n,n + 1) = 1 and therefore n(n + 1) z(f n F n+1 ). On the other hand, F n+ɛ F n(n+1) (Lemma.1 (a)) and hence, F n F n+1 F n(n+1), since gcd(f n,f n+1 ) = 1. Now, by using Lemma. (b), we conclude that z(f n F n+1 ) n(n+1). In conclusion, we have z(f n F n+1 ) = n(n+1). We remark that the same idea can be used to prove that if m 1,...,m k are positive integers, such that gcd(m i,m j ) = 1 or, for all i j, then z(f m1 F mk ) m 1 m k. If m 1,...,m k are pairwise relatively prime, then z(f m1 F mk ) = m 1 m k. 3.. Proof of (ii). For ɛ {0,1,}, we have F n+ɛ F n(n+1)(n+). By Lemma.1 (b), the numbers F n,f n+1,f n+ are pairwise coprime. In fact, if ɛ 1,ɛ {0,1,} are distinct, then gcd(n+ɛ 1,n+ɛ ) = 1 or. In any case, we get Thus, F n F n+1 F n+ F n(n+1)(n+) and so gcd(f n+ɛ1,f n+ɛ ) = F gcd(n+ɛ1,n+ɛ ) = 1. z(f n F n+1 F n+ ) n(n+1)(n+). (3.1) Now, we use F n+ɛ F n F n+1 F n+, to conclude that n+ɛ divides z(f n F n+1 F n+ ). So, the proof splits in two cases: Case 1: If n is odd, then n,n + 1,n + are pairwise coprime. Therefore, n(n + 1)(n + ) z(f n F n+1 F n+ ). This fact, together with (3.1), yields the result in this case. Case : For n even, we have that F n Fn(n+1)(n+) and so z(f n F n+1 F n+ ) divides n(n + 1)(n + )/. We already know that n + ɛ z(f n F n+1 F n+ ) and gcd(n,n + ) =. If n 0 (mod 4), then n,n + 1,(n + )/ are pairwise coprime. In the case of n (mod 4), the numbers n/,n + 1,n + are pairwise coprime. Thus, in any case, we have n(n + 1)(n + )/ z(f n F n+1 F n+ ) and the desired result is proved Proof of (iii). By the same arguments as before, we conclude that n+ɛ z(f n F n+1 F n+ F n+3 ), for ɛ {0,1,,3}. (3.) Assume first that n 0 (mod 3). Then there exists only one pair among (n,n + ) and (n+1,n+3) whose greatest common divisor is. Without loss of generality, we suppose that gcd(n,n+) =. Again, as in the previous item, we can see that n/ a,n+1,(n+)/ b,n+3 are pairwise coprime, for distinct a,b {0,1} suitably chosen depending on the class of n modulo 4. Thus, n(n+1)(n+)(n+3) = n(n+1)(n+)(n+3) a+b z(f n F n+1 F n+ F n+3 ). Sincetherearetwoevennumbersamongn,n+1,n+,n+3,wehavethatF n+ɛ Fn(n+1)(n+)(n+3). However, gcd(f n,f n+3 ) = F gcd(n,n+3) = 1, because 3 n. Thus, F n,f n+1,f n+,f n+3 are pairwise coprime yielding that F n F n+1 F n+ F n+3 Fn(n+1)(n+)(n+3). We apply Lemma. to get z(f n F n+1 F n+ F n+3 ) n(n+1)(n+)(n+3). MAY

5 THE FIBONACCI QUARTERLY This finishes the proof in this case. Now suppose that n 0 (mod 3). If 9 n, then gcd(n,(n+ 3)/3) = 1, while gcd(n/3, n + 3) = 1 when 9 n. In any case, for a suitable choice of a,b,c,d,e,f {0,1}, where a b and only one among c,d,e,f is 1, we obtain that n n+1 c 3a, d, n+ e, n+3 f 3 b are pairwise coprime. Here the sets {a,b} and {c,d,e,f} depend on the class of n modulo 4 and 9, respectively. Hence, by (3.), we get n(n+1)(n+)(n+3) = n(n+1)(n+)(n+3) c+d+e+f 3 a+b z(f n F n+1 F n+ F n+3 ), (3.3) since a+b = c+d+e+f = 1. Note that there are even numbers among n,n + 1,n +,n + 3 and also 3 divides both n and n + 3. Thus F n+ɛ Fn(n+1)(n+)(n+3), for ɛ {0,1,,3}. Since gcd(f n,f n+3 ) = and gcd(f n+3,f n+1 F n+ ) = 1, then gcd(f n F n+1 F n+,f n+3 ) =. Now, we use that F n,f n+1,f n+ are pairwise coprime to ensure that F n F n+1 F n+ Fn(n+1)(n+)(n+3). Since F n+3 also divides Fn(n+1)(n+)(n+3), we get F n F n+1 F n+ F n+3 Fn(n+1)(n+)(n+3) where we used Lemma.1 (c). Thus, Lemma. (b) yields F n(n+1)(n+)(n+3), 3 z(f n F n+1 F n+ F n+3 ) n(n+1)(n+)(n+3). (3.4) 3 Combining (3.3) and (3.4), we get { n(n+1)(n+)(n+3) z(f n F n+1 F n+ F n+3 ), n(n+1)(n+)(n+3) } (3.5) 3 holds for all positive integers n 0 (mod 3). In order to complete the proof, it suffices to prove that F n F n+1 F n+ F n+3 Fn(n+1)(n+)(n+3), for all n 0,9 (mod 1) (3.) and F n F n+1 F n+ F n+3 Fn(n+1)(n+)(n+3), for all n 3, (mod 1). (3.7) We claim that (3.) is true. In fact, if n 0 (mod 1), then n+3 3 (mod ). On the one hand, by setting n = 1l and by Lemma.3, we have ν (F n F n+1 F n+ F n+3 ) = ν (F n )+ν (F n+3 ) = ν (n)+3 = ν (l)+5. On the other, n(n+1)(n+)(n+3) = 1l(1l+1)(l+1)(4l+1) and so ν (Fn(n+1)(n+)(n+3)) = ν (1l(1l +1)(l+1)(4l+1))+ = ν (l)+4. This means that ν (F n F n+1 F n+ F n+3 ) > ν (Fn(n+1)(n+)(n+3)) which is enough to prove (3.). A similar argument holds for the case n 9 (mod 1). Now to prove (3.7), we shall show that ν p (F n F n+1 F n+ F n+3 ) ν p (Fn(n+1)(n+)(n+3)), for all primes p. In fact, 13 VOLUME 50, NUMBER

6 Case 1: If p = 5, then ORDER OF APPEARANCE OF PRODUCTS OF FIBONACCI NUMBERS ν 5 (F n F n+1 F n+ F n+3 ) = ν 5 (F n+j ) = ν 5 (n+j) = ν 5 (Fn(n+1)(n+)(n+3)). Case : If p or 5, then p divides at most one among F n,f n+1,f n+,f n+3. Suppose that this is the case (otherwise ν p (F n F n+1 F n+ F n+3 ) = 0 and we are done). Then, let j be the integer belonging to {0,1,,3} such that p F n+j. Thus, On the other hand, ν p (F n F n+1 F n+ F n+3 ) = ν p (F n+j ) = ν p (n+j)+ν p (F z(p) ). ν p (Fn(n+1)(n+)(n+3)) = ν p (n(n+1)(n+)(n+3)) ν p ()+ν p (F z(p) ). If p > 3, ν p () = 0 and the desired inequality follows. In the case of p = 3, we have ν 3 (Fn(n+1)(n+)(n+3)) = ν 3 (n)+ν 3 (n+3) 1+ν p (F z(p) ) v 3 (n+j)+ν 3 (F z(p) ) = ν 3 (F n F n+1 F n+ F n+3 ). Case 3: When p =, we use that n 3, (mod 1). Let us suppose that n 3 (mod 1) (the other case can be handled in the same way). Then n+3 (mod 1) and by Lemma.3, we obtain ν (F n F n+1 F n+ F n+3 ) = ν (F n )+ν (F n+3 ) = 4. However, if n = 1l+3, then ν (Fn(n+1)(n+)(n+3)) = ν (F 1(4l+1)(3l+1)(1l+5)(l+1) ) = ν (1(4l+1)(3l+1)(1l+5)(l+1))+ = 4. The proof is then complete. For p = 5, we have ν 5 (F 1 F n ) = 4. The Proof of Theorem 1. n ν 5 (F j ) = j=1 n ν 5 (j) = ν 5 (n!) = ν 5 (F n! ). In the case of p =, we first use Mathematica to see that the result holds for all 1 n 48. So, we shall assume n 49. Now, we note that ν (F n ) 0 if and only if 3 n. Thus, the only terms with non-zero -adic order among F 1,...,F n are F 3,...,F 3 n/3 and so j=1 ν (F 1 F n ) = [ν (F 3 )+ν (F )+ν (F 9 )]+ν (F 1 ) +[ν (F 15 )+ν (F 18 )+ν (F 1 )]+ν (F 4 ) + +ν (F 1 n/1 )+l, where l {0,1,4,5} and depends on the residue class of n modulo 1. By Lemma.3, each bracketed term in the above sum is 5 and thus, n/1 n n n ν (F 1 F n ) = 5 + ν (F 1j )+l = 9 +ν (!)+l j=1 We now apply Lemma.4 (with p = ) together with the fact that l 5, to get the bound ν (F 1 F n ) 5n +4. (4.1) MAY

7 THE FIBONACCI QUARTERLY On other other hand, since 1 n! (because n > 3), Lemma.3 yields Again, we use Lemma.4 to obtain ν (F n! ) = ν (n!)+. ν (F n! ) n logn +1. (4.) log The proof of this case finishes by noting that the right-hand side of (4.) is greater than 5n/+4, for n 49. When p = 3 or 7, we again use Lemma.3 to get ν p (F 1 F n ) = = n/z(p) j=1 (ν p (z(p)j)+ν p (F z(p) )) ( ) n n ν p (F z(p) z(p) )+ν p!, z(p) where we used that ν p (z(p)) = 0, since by Lemma.1 (e), p F p±1 and so by Lemma. (b), one has that z(p) (p±1). Now, we apply Lemma.4 to obtain n n z(p) 1 ν p (F 1 F n ) ν p (F z(p) z(p) )+. (4.3) p 1 On the other hand, ν p (F n! ) = ν p (n!)+ν p (F z(p) ) and hence, again by Lemma.4, we have ν p (F n! ) n logn p 1 1+ν p (F logp z(p) ). (4.4) By combining (4.3) and (4.4), it suffices to prove for p = 3 that n logn n log3 and for p = 7 that n logn n log7 However, both these inequalities hold for all n 13. For the remaining cases, we use a simple Mathematica routine to check that ν p (F 1 F n ) ν p (F n! ) is also valid for n = 1,...,1. This completes the proof. 5. Acknowledgement The author is grateful to Pavel Trojovský for nice discussions on the subject. References [1] A. Benjamin and J. Quinn, The Fibonacci numbers exposed more discretely, Math. Mag., 7.3 (003), [] J. H. Halton, On the divisibility properties of Fibonacci numbers, The Fibonacci Quarterly, 4.3 (19), [3] D. Kalman and R. Mena, The Fibonacci numbers exposed, Math. Mag., 7.3 (003), [4] T. Koshy, Fibonacci and Lucas Numbers with Applications, Wiley, New York, 001. [5] A. M. Legendre, Theorie des Nombres, Firmin Didot Freres, Paris, [] D. Marques, On integer numbers with locally smallest order of appearance in the Fibonacci sequence, Internat. J. Math. Math. Sci., Article ID (011), 4 pages. 138 VOLUME 50, NUMBER

8 ORDER OF APPEARANCE OF PRODUCTS OF FIBONACCI NUMBERS [7] D. Marques, On the order of appearance of integers at most one away from Fibonacci numbers, (to appear in The Fibonacci Quarterly). [8] D. Marques, On the order of appearance of powers of Fibonacci and Lucas numbers, Preprint. [9] T. Lengyel, The order of the Fibonacci and Lucas numbers, The Fibonacci Quarterly, 33.3 (1995), [10] E. Lucas, Théorie des fonctions numériques simplement périodiques, Amer. J. Math., 1 (1878), , [11] P. Ribenboim, My Numbers, My Friends: Popular Lectures on Number Theory, Springer-Verlag, New York, 000. [1] D. W. Robinson, The Fibonacci matrix modulo m, The Fibonacci Quarterly, 1. (193), 9 3. [13] N. J. A. Sloane (Ed.), The On-Line Encyclopedia of Integer Sequences, njas/sequences/. [14] J. Vinson, The relation of the period modulo m to the rank of apparition of m in the Fibonacci sequence, The Fibonacci Quarterly, 1. (193), [15] N. N. Vorobiev, Fibonacci Numbers, Birkhäuser, Basel, 003. MSC010: 11B39 Departamento de Matemática, Universidade de Brasília, Brasília, DF, , Brazil address: diego@mat.unb.br MAY