#A5 INTEGERS 17 (2017) THE 2-ADIC ORDER OF SOME GENERALIZED FIBONACCI NUMBERS

Transcription

1 #A5 INTEGERS 7 (207) THE 2-ADIC ORDER OF SOME GENERALIZED FIBONACCI NUMBERS Tamás Lengyel Mathematics Department, Occidental College, Los Angeles, California lengyel@oxy.edu Diego Marques Departamento de Matemática, Universidade de Brasília, Brasília, Brazil diego@mat.unb.br Received: 2/2/5, Accepted: 2/4/7, Published: 2//7 Abstract Let T n = T n (k) be the generalized Fibonacci sequence of order k defined by the recurrence T n = T n T n 2 T n k, n k, with T 0 = 0 and T = T 2 = = T k =. In this paper, we fully and partially characterize the 2-adic valuations of T n (4) and T n (5), respectively. Moreover, we provide new addition formulas and congruences for the sequences {T n (k)} n 0.. Introduction Let {F n } n 0 be the Fibonacci sequence given by F n2 = F n F n, for n 0, where F 0 = 0 and F =. The p-adic order, p (r), of r is the exponent of the highest power of a prime p which divides r. The p-adic order of a Fibonacci number was completely characterized, see [4]. Much less is known about the generalized Fibonacci sequences. Let T n = T n (k), n 0, denote the generalized Fibonacci sequence of order k defined by the recurrence relation T n = T n T n 2 T n k, n k, () and the initial conditions T 0 = 0, T = T 2 = = T k the initial conditions are given by =. Note that sometimes B 0 = B = = B k 2 = 0, B k = (2) with B n = B n (k) while the recurrence B n = B n B n 2 B n k, n k, is preserved. By convention, we also set B (k) = 0. Clearly, F n = T n (2) = B n (2). corresponding author

2 INTEGERS: 7 (207) 2 Our goal is to present a systematic approach that helps establish the 2-adic order of T n (k), at least for some specialized sequences of the index n (we point out that the 2-adic valuation of T n () was fully determined in [6]). Here, we focus on T n (k) for k = 4 and 5. The first few terms of the sequence {T n (4)} n 0 are 0,,,,, 6,, 2, 4, 79, 52, 29, 565, 089, 2099, 40, 7799,... while those of {T n (5)} n 0 are 0,,,,, 4, 8, 5, 29, 57,, 222, 46, 857, 685,, 65,.... Our main results are Theorems, 2, and Lemmas 2, 5, and 6. We also suggest several conjectures, cf. Conjectures and 2. Throughout the paper, we emphasize the experimental aspects of finding and discovering relations, e.g., recurrence relations and congruences. Theorem. For n, we have 8 >< 0, if n 6 0 (mod 5), 2 (T n (4)) =, if n 5 (mod 0), >: 2 (n) 2, if n 0 (mod 0). With n and s odd, this yields that () 2 (T 5 s 2 n(4)) = n 2. (4) We refer the reader to [2] for the 2-adic valuation of B n (4). We make the following conjecture for the case of k = 5. Conjecture. For n, we have 8 0, if n 6 0 or 5 (mod 6), 2, if n 5 (mod 2), ><, if n (mod 2), 2 (T n (5)) = 2 (n 2), if n 6 (mod 2) and 2 (n 2) < 8, 2 (n 4266), if n 6 (mod 2) and 2 (n 2) 8, >: 2 (n), if n 0 (mod 2). Here, we prove it in the following weaker form. Theorem 2. For n, we have 8 0, if n 6 0 or 5 (mod 6), >< 2, if n 5 (mod 2), 2 (T n (5)) =, if n (mod 2), 2 (n 2), if n 6 (mod 2) and 2 (n 6) 6=, >: 2 (n), if n 0 (mod 2). (5) (6)

3 INTEGERS: 7 (207) With n and s odd, this yields that We also propose the following conjecture. 2 (T 6 s 2 n(5)) = n. (7) Conjecture 2. For n and k 2 integers and s odd integer, we have 2 (T s (k) 2 n(k)) = n c(k) where c(2) = 2 and otherwise, 8 2, if k 0 (mod 4);, if k (mod 4); >< 2 (k 2), if k 2 (mod 8); c(k) =, if k (mod 8);, if k 6 (mod 8); >:, if k 7 (mod 8). Remark. Conjecture 2 can be easily verified for k = 2 and. In fact, we proved for n that 2 (T n (2)) = 2 (n) 2 if n 0, 6 (mod 2) in [4] and 2 (T n ()) = 2 (n) if n 0, 8 (mod 6) in [6]. In this paper, we prove the conjecture for k = 4 and 5. We outline a plan that can be followed in order to prove Conjecture 2. In fact, we will apply the plan in the cases of k = 4 and 5 in Sections 4 and 5, respectively. Step. First we establish an addition formula for T qr (k) in terms of T q 0(k) and T r 0(k) with q 0 and r 0 close to q and r, respectively; more precisely, with q k 2 apple q 0 apple q k and r apple r 0 apple r k. Step 2. The second step is to come up with a set of induction hypotheses for T s (k) 2n i(k) (mod 2 nc(k) ) for all i : 0 apple i apple k and n n 0 (k) with some functions c(k) and n 0 (k), e.g., T s (k) 2 n s 2 nc(k) (mod 2 nc(k) ), n, in Lemmas 5 and 6 and prove it simultaneously by using the recurrence relation for T qr (k) from the first step. Note that the congruence T s (k) 2n i(k) (mod 2 nc(k) ) will follow for any i apple and i k by the recurrence (). Step. In the induction proof, first we deal with the case s = and we prove this case by induction on n. The same procedure will work for other values of s. In conclusion, this process yields that if m = s (k ) 2 n and s 2 (T m (k)) = n c(k) for n n 0 (k). is odd then

4 INTEGERS: 7 (207) 4 We illustrate the actual steps in Sections 2 and. Section 2 is devoted to the process of obtaining recurrence relations while Section contains the congruences that are the essential tools in proving Theorems and 2. The actual calculations and proofs in the cases of k = 4 and 5 are presented in Sections 4 and 5. They lead to identities () and (2) that are crucial in proving the congruences (4), (5), and (6). 2. Obtaining a Recurrence by an Addition Formula As a reminder, we note the addition formula, given in Lemma 4 of [6], which yields a recurrence for T qr (). For all integers q and r with q and r 0, we have that T qr = T q 2 T r (T q T q 2 )T r T q T r2. Note that T q = T q 4 T q T q 2. It is determined in Theorem 2. of [7] that with T n = T n (4) and B n = B n (4), we have T q = B q 2 T (B q 2 B q )T 2 (B q 2 B q B q 4 )T B q T 4, (8) for q 5 where B q = B q 2 B q B q 4 B q 5. The formula (8) can be easily generalized to Lemma. For q 5 and r 0 with T n = T n (4) and B n = B n (4), we have that T qr = B q 2 T r (B q 2 B q )T r2 (B q 2 B q B q 4 )T r B q T r4. To obtain similar identities for a general k, we use the fact that one can relate the sequences {T n (k)} n 0 and {B n (k)} n 0. In fact, we have the following general result Lemma 2. Let k 2 be an integer and set T n = T n (k) and B n = B n (k). For integers q > k and r 0, we have that 0 0 kx Xi kx Xi T q A T i and T qr A T ri. (9) i= j=2 B q j Remark 2. We also use identity (9) in its equivalent form 0 kx Xi T qr A T ri. (0) i=0 j= with q k 2 and r 0, cf. () and (2). B q j i= j=2 B q j

5 INTEGERS: 7 (207) 5 We omit the proof which can be easily done by mathematical induction on q > k for every fixed r 0. Remark. Identity (9) also works for sequences T n (k) of real numbers satisfying () with arbitrary initial conditions. Our next step is to determine B q 0(k) in (9) in terms of the sequence {T n (k)} n 0. We note that although B n () = T n (), usually there is a non-trivial linear relationship between the two sequences. We use the approach outlined in []. The result is derived in (8) and (9) as well in (22) and (2), and used by () and (2) in Lemmas and 4, respectively. Lemma. For T qr (4) with q 2 and r 0, we have the recurrence 5 T qr = T q T 4 q 2T q2 T q T r 5 T q 2T q 7 T q 2 T 4 q2 T q T r 5 T q 2 2T q 4T q T q 2 T 4 q2 T q 5 T q T 4 q2 2T q T q4 T r. T r2 () Lemma 4. For T qr (5) with q and r 0, we have the recurrence 5Tq T qr = T q 5T q2 8T q 27T q4 T r 2 2 5Tq 57T q 7T q 5T q2 9T q 27T q4 T r 5Tq 2 9T q 57T q 27T q4 T r2 6T q 2 7T q 2T q2 2 (2) 5Tq 2T q2 2 5Tq 57T q 2 9T q T q2 2 6T q 2 27T q4 5T q T r 54T q 2 T q 8T q4 2 27T q5 T r4.

6 INTEGERS: 7 (207) 6. Congruences We note that for k = the congruences in (4) of Lemma 6 in [6] are equivalent to the following statement. For s, n, and T m = T m (), we have the congruences T s 2 n s 2 n (mod 2 n ), T s 2n (mod 2 n ), T s 2n 2 s 2 n (mod 2 n ). () Now we establish similar congruences for k = 4. Lemma 5. For s, n 2, and T m = T m (4), we have that T 5 s 2 n s 2 n2 (mod 2 n ), T 5 s 2n s 2 n (mod 2 n ), T 5 s 2n 2 s 2 n s 2 n2 (mod 2 n ), T 5 s 2n (mod 2 n ), (4) while for n =, we have that T 0 s 8s (mod 6), T 0 s 4s (mod 6), T 0 s2 4s (mod 6), T 0 s (mod 6), (5) which yields that 2 (T 5 s 2 n(4)) = n 2 if n and s odd. Proof of Lemma 5. We closely follow the steps of the proof of Lemma 6 of [6]. First, we deal with the basis case s =. We have to prove (4) for n 2. We use induction on n. Clearly, the congruences hold for n = 2. We suppose that they are true for n 2, and then we use () for T 5 2 n i = T (5 2n )(5 2 n i), 0 apple i apple, to obtain the required congruences for T 5 2 n i. Next, by the induction hypothesis, we suppose that the congruences (4) hold for s. Then, we use exactly the same procedure and () as before for T 5 (s) 2n i = T (5 s 2n )(5 2 n i). In a similar fashion, we use induction on s to prove the congruences (5), corresponding to the case with n =. We omit the details. Example. We illustrate the above proof in the case of k = 4, n 2, s, and i = 0. With the setting r = 5 s 2 n and q = 5 2 n, we obtain by () that T 5 2n (s) = 5 T 5 2 n T n 2T 5 2n 2 T 5 2 n T 5 2n s 2T 5 2 n 5 T 5 2 n 7 T 5 2 n T 5 2n s 4T 5 2 n 5 T 5 2 n 22T 5 2n T 5 2n 2 T 5 2 n 2 2 T 5 2 n 2 4 T 5 2 n

7 INTEGERS: 7 (207) 7 4 T n T 5 2n s2 T 5 2 n T 5 2 n 2 2T 5 2n 4 T 5 2 n 4 T 5 2n s, which results in 2n2 s 2n s 2n4 s 5 22n s 2 2n4 s 22n6 s 22n7 s 22n8 s 22n9 s 2n2 2n (mod 2 n ) by the induction hypothesis. We get 2n2 (s ) 2 n2 (s ) (mod 2 n ) by replacing any term including a factor with a high power of 2 with 0. More precisely, any term including 2 c nd with d or c > combined with d is dropped. It implies that the statement T 5 (s) 2 n (s ) 2 n2 (mod 2 n ) in (4) is also true. Note that the substitutions and simplifications above can be easily preformed by using Mathematica. In the case of k = 5 we proceed similarly. Lemma 6. For s, n, and T m = T m (5), we have that T 6 s 2 n s 2 n (mod 2 n2 ), T 6 s 2n (mod 2 n2 ), T 6 s 2n 2 s 2 n (mod 2 n2 ), T 6 s 2n (mod 2 n2 ), T 6 s 2n 4 (mod 2 n2 ), (6) which yields that 2 (T 6 s 2 n(5)) = n if n and s odd. The proof essentially duplicates the steps of the proof of Lemma 5 and we leave the details to the reader. 4. The Case of k = 4 Before we present the proof of Lemma, we explore an approach given in []. In fact, we use it with some modifications and with n 0 and m 4. We start with the matrix 0 T n T n T n2 T n T mn T n T n2 T n T n4 T mn T n2 T n T n4 T n5 T mn2 T n T n4 T n5 T n6 T mn C A. (7) After experimenting with di erent values of m and row reducing the matrix in (7), we successfully obtain the recurrence relation T mn = B m T n (B m 2 B m ) T n (B m B m 2 B m ) T n2 B m T n suggesting (9) of Lemma 2 in its equivalent form (0) for k = 4 with m 4 and n 0. In a similar fashion, we establish the

8 INTEGERS: 7 (207) 8 Proof of Lemma. We consider the matrix 0 T n T n T n2 T n B mn T n T n2 T n T n4 B mn T n2 T n T n4 T n5 B mn2 T n T n4 T n5 T n6 B mn C A. (8) After setting m = and using di erent values of n, we observe that the row reduction always results in which yields that B n = for n, which confirms () Tn T n 4T n 2T n2 C A, (9) Note that once (20) is established, an easy induction proof justifies this identity Indeed, with n =, 2,, 4 we get that 0 = 2 = 2 6 = and = The induction step is trivial by () and (2). A natural approach to obtain the proof of Theorem is to utilize the periodicity of the underlying sequences. In some cases we can apply multisection techniques, cf. [5], to find the complete or some partial characterization of the p-adic order of the sequences. Here we combine these methods with the applications of sets of congruences for {T s (k) 2n i} k i=0 with s and n n 0(k) integers. Now we can complete the proof of Theorem. Proof of Theorem. The proof for the case n 6 0 (mod 5) is trivial by taking T n (4) (mod 2) and induction on n. In fact, the sequence {T n (4)} n 0 is periodic with period {0,,,, } modulo 2. If n 5 (mod 0) then by 5-section of the generating function P m=0 T m(4)x m (cf. [5]) we get that X T 5m (4)x 5m 2x 5 ( 2x 5 x 0 ) = 26x 5 6x 0 6x 5 x 20, m=0 which easily yields that 2 (T n (4)) =. Indeed, the denominator of the 5-sected generating function suggests the recurrence (20) T 5m0 = 26T 5m5 6T 5m 6T 5m 5 T 5m 0, m 2, (2)

9 INTEGERS: 7 (207) 9 for T r = T r (4) with r divisible by 5. We observe that 2 (T 5 ) =, 2 (T 0 ) =, 2 (T 5 ) =, and 2 (T 20 ) = 4, which yield that 2 (T 5m ) for m 0 by the initial values and (2). Now 2 (T 5m0 ) = 2 (T 5m 0 ) = with m odd also follows by recurrence (2). We note that we can extend (5) by recurrence () to obtain T 0 s4 (mod 6) and T 0 s5 6 8s (mod 6), and the latter congruence also results in 2 (T n ) = with n 5 (mod 0). In the remaining case 0 divides n, and Lemma 5 concludes the proof. 5. The Case of k = 5 Now we turn to the Proof of Lemma 4. Similarly to (8) in the case of k = 4, we now consider 0 T n T n T n2 T n T n4 B mn T n T n2 T n T n4 T n5 B mn B T n2 T n T n4 T n5 T n6 B mn2 T n T n4 T n5 T n6 T n7 B mn A. (22) T n4 T n5 T n6 T n7 T n8 B mn4 After setting m = and using di erent values of n, row reduction leads us to which results in B n = 5Tn Tn 2 5Tn2 8Tn 27T n4 2 for n, which is in agreement with (2). Its proof follows easily by induction as it was explained in the proof of Lemma for k = 4. C A We are now ready to present the proof of Theorem 2. Proof of Theorem 2. As above, the proof for the case n 6 0 and 5 (mod 6) is trivial by taking T n (5) (mod 2) and induction on n since the sequence {T n (5)} n 0 is periodic with period {0,,,,, 0} modulo 2. If n 6 (mod 2) then with n = 6 s 2 m 6, s odd and m, we get that T 6 s 2m 5 4 (mod 2 m2 ) and T 6 s 2m 6 8 s 2 m (mod 2 m2 ) by extending (6) via (). It implies that 2 (T 6 s 2m 6) = 2 (n 2) as long as either m or m =, in which cases the 2-adic order is either or 2, respectively. In a similar fashion, it follows that T 6 s 2m 222 (mod 2 m2 ). Thus, with t integer, we (2)

10 INTEGERS: 7 (207) 0 also have that T 2t5 4 (mod 8) and T 2t 222 (mod 8), which yield that 2 (T 2t5 ) = 2 and 2 (T 2t ) =. Otherwise 2 divides n, and Lemma 6 concludes the proof. Acknowledgment We thank the referee for the careful reading of the manuscript. The second author thanks CNPq for the financial support. References [] M. Bicknell and C. P. Spears. Classes of identities for the generalized Fibonacci numbers, Fibonacci Quart. 4 (996), [2] V. Facó and D. Marques. The 2-adic order of the Tetranacci numbers and the equation Q n = m!, preprint. [] J. Feng. More identities on the Tribonacci numbers, Ars Comb. 00 (20), [4] T. Lengyel. The order of the Fibonacci and Lucas numbers, Fibonacci Quart. (995), [5] T. Lengyel. Divisibility properties by multisection, Fibonacci Quart. 4 (200), [6] D. Marques and T. Lengyel. The 2-adic order of the Tribonacci numbers and the equation T n = m!, Journal of Integer Sequences 7 (204), Article 4.0., 8. [7] L. R. Natividad. On solving Fibonacci-like sequences of fourth, fifth and six order, International Journal of Mathematics and Scientific Computing (20), [8] E. M. Waddill. Some properties of a generalized Fibonacci sequence modulo m, Fibonacci Quart. 6 (978), 44 5.