Vol. 19 (2018), No. 1, pp. 641–648 DOI: 10.18514/MMN.2018.1750
ON THE ORDER OF APPEARANCE OF THE DIFFERENCE OF TWO LUCAS NUMBERS
PAVEL TROJOVSK ´Y Received 22 September, 2015
Abstract. LetFnbe thenth Fibonacci number and letLnbe thenth Lucas number. The order of appearance´.n/of a natural numbernis defined as the smallest natural numberksuch thatn dividesFk. For instance,´.Ln/D2n, for alln > 2. In this paper, among other things, we prove that
´.Lm Ln/D5Fp
p m2 n2 4 ,
for all distinct positive integersmn .mod4/, with gcd.m; n/Dp > 2prime.
2010Mathematics Subject Classification: 11B39; 11A07 Keywords: Lucas number, Fibonacci number, order of appearance
1. INTRODUCTION
Let.Fn/n0 be the Fibonacci sequence given byFnC2DFnC1CFn, forn0, where F0D0 and F1D1. These numbers are well-known for possessing amaz- ing properties (consult [4] together with its very extensive annotated bibliography for additional references and history). We cannot go very far in the lore of Fibon- acci numbers without encountering its companion Lucas sequence .Ln/n0 which follows the same recursive pattern as the Fibonacci numbers, but with initial values L0D2andL1D1.
The study of the divisibility properties of Fibonacci numbers has always been a popular area of research. Letnbe a positive integer number, theorder (or rank) of appearanceofnin the Fibonacci sequence, denoted by´.n/, is defined as the smal- lest positive integerk, such thatnjFk (some authors also call itorder of apparition, orFibonacci entry point). There are several results about´.n/in the literature. For instance, in 1975, J. Sall´e [13] proved that´.n/2n, for all positive integersn. This is the sharpest upper bound for´.n/, since for example,´.6/D12and´.30/D60 (see [10] for related results). In the case of a prime numberp, one has the better upper bound´.p/pC1, which is a consequence of the known congruenceFp .p
5/0
c 2018 Miskolc University Press
.mod p/, forp¤2, where.aq/denotes the Legendre symbol ofawith respect to a primeq > 2.
We remark that there is no a closed formula for´.n/, but by using computational methods and several calculations, one can see patterns of ´.n/ for some positive integers n. For example, with this computational approach, Marques [7–9] found explicit formulas for the order of appearance of integers related to Fibonacci and Lucas number, such as Cmk=Ck; Cm˙1; CnCnC1CnC2 and Cnk, where .Cn/nD .Fn/nor.Ln/n.
In this paper, we study the order of appearance of Lm Ln. Again, we used Mathematicato search for patterns for´.Lm Ln/. We were surprised to find out that in some cases, these values are related to the sequences
1; 2; 5; 13; 89; 233; 1597; 4181; 28657; 514229; : : : and
3; 4; 11; 29; 199; 521; 3571; 9349; 64079; 1149851; : : : :
A search in the On-Line Encyclopedia of Integer Sequences [14] is enough to conclude that, surprisingly, these sequences are the first few Fibonacci and Lucas numbers with prime indexes, respectively. In fact, this interpretation is confirmed by exhaustive calculations. More precisely, our main results are the following.
Theorem 1. Let m and n > 1 be positive distinct integers, such that m n .mod 4/. We have
(i) Ifgcd.m; n/D1, then
´.Lm Ln/D5.m2 n2/
4 :
(ii) Ifgcd.m; n/Dpis prime, then
´.Lm Ln/D5Fp
p m2 n2 4 : 2. AUXILIARY RESULTS
Before proceeding further, we recall some facts on Fibonacci and Lucas numbers for the convenience of the reader.
Lemma 1. We have
(a) FnjFmif and only ifnjm.
(b) LnjFmif and only ifnjmandm=nis even.
(c) 5FnjF5n, for all integern.
(d) IfdDgcd.m; n/, thengcd.Fm; Fn/DFd, gcd.Lm; Ln/D
Ld; if2.m/D2.n/I 1or2; otherwise
andgcd.Fm; Ln/D
Ld; ifm=d is even andn=d is oddI 1or2; otherwise.
(e) Fp .5
p/0 .mod p/, for all primes p. In particular, gcd.Fp; p/D1, if p¤5.
(f) Lp 1 .mod p/, for all primesp. In particular,gcd.Lp; p/D1.
Here, as usual, .aq/ denotes the Legendre symbol of a with respect to a prime q > 2.
Proof of these assertions can be found in [4]. We refer the reader to [1,3,6,11] for more details and additional bibliography.
The second lemma is a consequence of the previous one Lemma 2(Cf. Lemma 2.2 of [7]). We have
(a) IfFnjm, thennj´.m/.
(b) IfLnjm, then2nj´.m/.
(c) IfnjFm, then´.n/jm.
Lemma 3. For all primesp¤5, we have thatgcd.´.p/; p/D1.
Proof. By combining Lemma1(e) together with Lemma2(c), we conclude that
´.p/jp .p5/. Thus, when p ¤5, one has that .p5/D ˙1 and so ´.p/ divides p 1 or pC1. This yields that ´.p/DpC1 or ´.p/p 1 and in any case
gcd.´.p/; p/D1.
The next lemma will be very useful in the proof of our theorems. First, we recall the identities
FaLbDFaCbC. 1/bFa b;
5FaFbDLaCb . 1/bLa b; (2.1) LaLbDLaCbC. 1/bLa b
for all integersaandb. These identities can be easily deduced from Binet’s formulas:
FnD˛n ˇn
˛ ˇ andLnD˛nCˇn, forn0, where˛D.1Cp
5/=2andˇD.1 p 5/=2.
Lemma 4. Letmandnbe positive integers, such thatmn .mod 4/. Then (a) F.m n/=2L.mCn/=2DFm Fn:
(b) F.mCn/=2L.m n/=2DFmCFn: (c) L.m n/=2L.mCn/=2DLmCLn: (d) 5F.mCn/=2F.m n/=2DLm Ln:
Proof. It suffices to use suitable choices of a andb in the identities (2.1). For example, ifaD.m n/=2andbD.mCn/=2, thenaCbDmanda bD n. By (2.1), we have
F.m n/=2L.mCn/=2DFmC. 1/mC2nCnC1Fn;
where we used that F nD. 1/nC1Fn. The proof of (a) is complete, since.mC n/=2CnC1is odd. In fact, this follows from
mn .mod 4/)mCn
2 n .mod 2/)mCn
2 CnC12nC1 .mod2/:
The other items can be proved in the same way.
Thep-adic order (or valuation) ofr,p.r/, is the exponent of the highest power of a primep which dividesr. Throughout the paper, we shall use the known facts that p.ab/Dp.a/Cp.b/and thatajbif and only ifp.a/p.b/, for all primesp.
We remark that thep-adic order of Fibonacci and Lucas numbers was completely characterized, see [2,5,12]. For instance, from the main results of Lengyel [5], we extract the following two results.
Lemma 5. Forn1, we have
2.Fn/D 8 ˆˆ
<
ˆˆ :
0; ifn1; 2 .mod3/I 1; ifn3 .mod 6/I 3; ifn6 .mod 12/I 2.n/C2; ifn0 .mod 12/;
5.Fn/D5.n/, and ifpis prime¤2or5, then p.Fn/D
p.n/Cp.F´.p//; ifn0 .mod´.p//I 0; otherwise.
Lemma 6. Letk.p/be the period modulop of the Fibonacci sequence. For all primesp¤5, we have
2.Ln/D 8
<
:
0; ifn1; 2 .mod3/I 2; ifn3 .mod 6/I 1; ifn0 .mod 6/
p.Ln/D
p.n/Cp.F´.p//; ifk.p/¤4´.p/andn´.p/2 .mod ´.p//I 0; otherwise.
With all of the above tools in hand, we now move to the proof of the theorem.
3. THE PROOF OFTHEOREM1 3.1. Item (i)
By Lemma4(d), we have
Lm LnD5FmCn 2
Fm n
2 :
SinceF.m˙n/=2dividesLm Ln, then.m˙n/=2j´.Lm Ln/. Note that.m n/=2 and.mCn/=2are coprime which yields
m2 n2
4 j´.Lm Ln/: (3.1)
Also,F.m˙n/=2jF.m2 n2/=4and since
gcd.F.mCn/=2; F.m n/=2/DFgcd..mCn/=2;.m n/=2/DF1D1 we conclude thatF.mCn/=2F.m n/=2jF.m2 n2/=4. Thus
Lm LnD5FmCn 2 Fm n
2 j5Fm2 n2 4 jF
5
m2 n2 4
; where we used Lemma1(c). Therefore
´.Lm Ln/j5
m2 n2 4
: (3.2)
The combination between (3.1) and (3.2) yields
´.Lm Ln/2
m2 n2 4 ; 5
m2 n2 4
:
Now, it suffices to prove that5F.mCn/=2F.m n/=2−F.m2 n2/=4. In fact, this follows because
5.5F.mCn/=2F.m n/=2/D1C5..m2 n2/=4/
> 5..m2 n2/=4/D5.F.m2 n2/=4/:
This completes the proof.
3.2. Item (ii)
Proceeding as before, we can easily deduce that m2 n2
4p j´.Lm Ln/: (3.3)
Here we used that gcd..m n/=2; .mCn/=2/Dpfor all primepDgcd.m; n/.
Now, we claim thatLm LndividesF5Fp.m2 n2/=4p, or equivalently, that q.5F.mCn/=2F.m n/=2/q.F5Fp.m2 n2/=4p/;
holds for all primesq. Let us split the proof into four cases:
Case 1.qD5. In this case, it follows directly that
5.5F.mCn/=2F.m n/=2/D1C5..m2 n2/=4/
while
5.F5Fp.m2 n2/=4p/D1C5..m2 n2/=4/C5.Fp/ 5.p/
D1C5..m2 n2/=4/;
where we used that5.Fp/D5.p/.
Case 2.qDp¤2; 5. We have
p.5F.mCn/=2F.m n/=2/Dp..mCın/=2/Cp.F´.p//;
whereı2 f 1; 1gand we used thatp cannot divide bothF.mCn/=2andF.m n/=2, otherwise´.p/ would divide p which is an absurdity by Lemma3. On the other hand, we have
p.F5Fp.m2 n2/=4p/Dp..mCın/=2/Cp..m ın/=2/ 1Cp.F´.p//:
The result follows becausep..m ın/=2/1, sincepDgcd.m; n/ > 2.
Case 3.q¤2; 5; p. In this case, we have
q.5F.mCn/=2F.m n/=2/Dq..m2 n2/=4/Cq.F´.q//;
whereis1or2according to the value ofq..m2 n2/=4/is1or not, respectively.
For the other valuation, we obtain
q.F5Fp.m2 n2/=4p/Dq.Fp/Cq..m2 n2/=4/Cq.F´.q//:
If D1, the conclusion is immediate. For D2, then ´.q/jp and so´.q/Dp.
Thereforeq.Fp/Dq.F´.q//and the result follows.
Case 4.qD2. Since2.5F.mCn/=2F.m n/=2/D2.F.mCn/=2/C2.F.m n/=2/, we need to consider two subcases:
Subcase 4.1p¤3. In this case,3divides only one among.m n/=2and.mC n/=2. So, we have2.5F.mCn/=2F.m n/=2/D1, when3j.mCn/=2(because.mC n/=23 .mod 6/) and when3j.m n/=2one obtains2.5F.mCn/=2F.m n/=2/D3 or2.m n/C1depending on.m n/=4is odd or even, respectively. However, it is easy to infer that
2.F5Fp.m2 n2/=4p/D 8
<
:
1; if3j.mCn/=2I 3; if24−.m n/=2I 2.m n/C1; if24j.m n/=2;
yielding the desired inequality.
Subcase 4.2pD3. In this case, we get 2.5F.mCn/=2F.m n/=2/D
4; if.m n/=4is oddI
2.m n/C2; if.m n/=4is even:
Now, note that3jm˙nand8j.m2 n2/and then12j5.m2 n2/=3. Hence, by Lemma5,
2.F5.m2 n2/=3/D2..m2 n2/=3/C2D2.m n/C3 > 2.m n/C24:
The result follows.
Summarizing, we proved thatLm LndividesF5Fp.m2 n2/=4pand so, by Lemma 2(c), we obtain
´.Lm Lm/j5Fp.m2 n2/=4p: (3.4) By (3.3) and (3.4), one concludes that
´.Lm Lm/Dt .m2 n2/=4p;
wheret divides5Fp, that is,5FpDst, for some positive integers. Thus, in order to complete the proof of theorem, it suffices to prove thatsD1. Suppose, to derive a contradiction, thats > 1and letq be a prime factor of s. Note thatq¤p unless pD5.
pD5(and thenqD5). In this case,25Dt sand sotD1; 5or25. Note that 5.5F.mCn/=2F.m n/=2/D1C5.m2 n2/ > 5.m2 n2/D5.F.m2 n2/=4/:
Thus5F.mCn/=2F.m n/=2−F.m2 n2/=4and the result follows.
p¤5(and soq¤5). Since5FpDt s, thenqjFp yielding´.q/Dp. Therefore q.5F.mCn/=2F.m n/=2/Dq..mCın/=2/C2q.F´.q//;
where we used thatq divides bothF.mCn/=2andF.m n/=2, hereı2 f0; 1g. On the other hand,
q.5F.mCn/=2F.m n/=2/Dq.Fp/Cq..mCın/=2/ q.s/Cq.F´.q// Dq..mCın/=2/ q.s/C2q.F´.q//
< q.5F.mCn/=2F.m n/=2/;
sinceqjsand we used thatq.Fp/Dq.F´.q//. The proof of theorem is complete.
4. ACKNOWLEDGEMENTS
The author thanks for support to Specific Research Project of Faculty of Science, University of Hradec Kralove, No. 2101, 2017.
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Author’s address
Pavel Trojovsk´y
Faculty of Science University of Hradec Kr´alov´e, Department of Mathematics, Rokitansk´eho 62, Hradec Kr´alov´e 50003, Czech Republic
E-mail address:pavel.trojovsky@uhk.cz
