We describe how to compute the intersection of two Lucas sequences of the
forms $\{U_n(P,\pm 1) \}_{n=0}^{\infty}$ or $\{V_n(P,\pm 1) \}_{n=0}^{\infty}$
with $P\in\mathbb{Z}$ that includes sequences of Fibonacci, Pell, Lucas, and
Lucas-Pell numbers. We prove that such an intersection is finite except for the
case $U_n(1,-1)$ and $U_n(3,1)$ and the case of two $V$-sequences when the
product of their discriminants is a perfect square. Moreover, the intersection
in these cases also forms a Lucas sequence. Our approach relies on solving
homogeneous quadratic Diophantine equations and Thue equations. In particular,
we prove that 0, 1, 2, and 5 are the only numbers that are both Fibonacci and
Pell, and list similar results for many other pairs of Lucas sequences. We
further extend our results to Lucas sequences with arbitrary initial terms

Bilu

Cohen

Cremona

Luca

Max A. Alekseyev

Mignotte

Mátyás

Pinch

Posamentier

Siegel

Stein

Szalay

Thue

Tzanakis

English

arXiv

AbstractWe describe how to compute the intersection of two Lucas sequences of the forms</jats:p

Crossref

Integers

On the Intersections of Fibonacci, Pell, and Lucas Numbers

summary:Let $k\geq 2$ and define $F^{(k)}:=(F_n^{(k)})_{n\geq 0}$, the $k$-generalized Fibonacci sequence whose terms satisfy the recurrence relation $F_n^{(k)}=F_{n-1}^{(k)}+F_{n-2}^{(k)}+\cdots + F_{n-k}^{(k)}$, with initial conditions $0,0,\dots ,0,1$ ($k$ terms) and such that the first nonzero term is $F_1^{(k)}=1$. The sequences $F:=F^{(2)}$ and $T:=F^{(3)}$ are the known Fibonacci and Tribonacci sequences, respectively. In 2005, Noe and Post made a conjecture related to the possible solutions of the Diophantine equation $F_n^{(k)}=F_m^{(\ell )}$. In this note, we use transcendental tools to provide a general method for finding the intersections $F^{(k)}\cap F^{(m)}$ which gives evidence supporting the Noe-Post conjecture. In particular, we prove that $F\cap T=\{0,1,2,13\}$

Marques, Diego

Institute of Mathematics AS CR, v. v. i.

Mathematica BohemicaDiego MarquesOn the intersection of two distinct k-generalized Fibonacci sequencesMathematica Bohemica, Vol. 137 (2012), No. 4, 403–413Persistent URL: http://dml.cz/dmlcz/142996Terms of use:© Institute of Mathematics AS CR, 2012Institute of Mathematics of the Czech Academy of Sciences provides access to digitizeddocuments strictly for personal use. Each copy of any part of this document must contain theseTerms of use.This document has been digitized, optimized for electronic delivery andstamped with digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://dml.cz137 (2012) MATHEMATICA BOHEMICA No. 4, 403–413ON THE INTERSECTION OF TWO DISTINCTk-GENERALIZED FIBONACCI SEQUENCESDiego Marques, Brasilia(Received January 31, 2011)Abstract. Let k > 2 and define F (k) := (F(k)n )n>0, the k-generalized Fibonacci sequencewhose terms satisfy the recurrence relation F(k)n = F(k)n−1 + F(k)n−2 + . . .+ F(k)n−k, with initialconditions 0, 0, . . . , 0, 1 (k terms) and such that the first nonzero term is F(k)1 = 1. Thesequences F := F (2) and T := F (3) are the known Fibonacci and Tribonacci sequences,respectively. In 2005, Noe and Post made a conjecture related to the possible solutions ofthe Diophantine equation F(k)n = F(l)m . In this note, we use transcendental tools to providea general method for finding the intersections F (k) ∩F (m) which gives evidence supportingthe Noe-Post conjecture. In particular, we prove that F ∩ T = {0, 1, 2, 13}.Keywords: k-generalized Fibonacci numbers, linear forms in logarithms, reductionmethodMSC 2010 : 11B39, 11D61, 11J861. IntroductionSeveral problems in number theory are actually questions about the intersec-tion of two known sequences (or sets). Before giving examples, let us recall someterminology: let F := (Fn)n>0 be the Fibonacci sequence, P := {p : p prime},P := {yt : y, t ∈ Z, t > 1} (the perfect powers), F := {n! : n ∈ Z, n > 0},R := {a(10n − 1)/9: 1 6 a 6 9, n ∈ Z, n > 0} (the repdigits or unidigital num-bers). Below, we cite some results about the intersection of these sets:⊲ Erdös and Selfridge [8] proved that F ∩ P = {1}.⊲ In 2000, Luca [25] proved that F ∩R = {0, 1, 2, 3, 5, 8, 55}.⊲ Luca [26] also proved that F ∩ F = {1, 2}.⊲ In 2003, Bugeaud et al [4] showed that F ∩ P = {0, 1, 8, 144} (see [28] for ageneralization).403⊲ Let (an)n>1 be the tower given by a1 = 1 and an = nan−1 , for n > 2. Luca andthe author [27] proved that {a1 + . . .+ an : n > 1} ∩ P = {1}.However, some related questions are still open problems, as for instance the setsP ∩ F and P ∩R are unknown.Let k > 2 and denote F (k) := (F(k)n )n>0, the k-generalized Fibonacci sequencewhose terms satisfy the recurrence relation(1.1) F (k)n = F(k)n−1 + F(k)n−2 + . . .+ F(k)n−k,with initial conditions 0, 0, . . . , 0, 1 (k terms) and such that the first nonzero term isF(k)1 = 1.The above sequence is one among the several generalizations of Fibonacci numbers.Such a sequence is also called k-step Fibonacci numbers, the Fibonacci k-sequence, ork-bonacci numbers. Clearly, for k = 2 we obtain the well-known Fibonacci numbersand for k = 3, Tribonacci numbers.Recall that Tribonacci numbers have a long history. For the first time, they werestudied in 1914 by Agronomoff [1] and subsequently by many others. The nameTribonacci was coined in 1963 by Feinberg [9]. The basic properties of Tribonaccinumbers can be found in [18], [24], [36], [38]. For recent papers, we refer the readerto [3], [19], [20], [33] and to the collection [21], [22], [23].Recently, Alekseyev [2] described how to compute the intersection of two Lucassequences including the sequences of Fibonacci, Pell, Lucas and Lucas-Pell numbers.In general, we refer the reader to [34], [35], [37] for results on the intersection of tworecurrence sequences.In a very recent paper, Togbé and the author [29] proved that only finitely manyterms of a linear recurrence sequence whose characteristic polynomial has a dominantroot can be repdigits. As an application, since the characteristic polynomial of therecurrence in (1.1), namely xk − xk−1 − . . . − x − 1, has just one root α such that|α| > 1 (see for instance [39]), hence F (k)∩R is a finite set, for all k > 2. See also thearticle [32] for some results on the set F (k) ∩ P and a conjecture on the intersectionF (k)∩F (m). We point out that this last intersection is, to the best of our knowledge,not known even in the easiest case (k,m) = (2, 3), that is, for numbers that are bothFibonacci and Tribonacci. A possible way to find this intersection is to look at theFibonacci and Tribonacci sequences modulo pt, where p is a prime number. We referthe reader to [5], [13], [16], [17] for results of this nature. However, this approachseems to be hard to work in practice. This observation prompted the author to lookfor a more interesting and constructive approach which could be useful in the generalcase.It is important to notice that Mignotte (see [31]) showed that if (un)n>0 and(vn)n>0 are two linearly recurrence sequences then, under some weak technical as-404sumptions, the equationun = vmhas only finitely many solutions in positive integersm,n. Moreover, all such solutionsare effectively computable. Therefore, it seems reasonable to think that F (k) ∩F (m)is a finite set for all k 6= m.The goal of this paper is to apply transcendental tools to provide a method forstudying the intersection F (k) ∩ F (m), for integers 2 6 k < m and determine com-pletely this set for (k,m) = (2, 3) (confirming the expectation). More precisely, ourresult is the following.Theorem 1. The only solution of the Diophantine equation(1.2) Fn = Tmin positive integer numbers m and n with n > 3, is (n,m) = (7, 6). Hence, F ∩ T ={0, 1, 2, 13}.We organize this paper as follows. In Section 2, we will recall some useful propertiessuch as a result of Matveev on linear forms in three logarithms and the reductionmethod of Baker-Davenport that we will use in the proof of Theorem 1. In Section 3,we first use Baker’s method to obtain a bound for n, then we completely proveTheorem 1 by means of the Baker-Davenport reduction method.2. Auxiliary resultsWe recall the well-known Binet’s formula:(2.1) Fn =ϕn − (−ϕ)−n√5for all n > 0,where ϕ = (1 +√5)/2. It is almost unnecessary to stress that this is a very helpfulformula which moreover allows to deduce thatϕn−2 < Fn < ϕn−1 for all n > 1.In 1982, Spickerman [36] found the following “Binet-style” formula for the Tri-bonacci sequence:(2.2) Tn =αn−α2 + 4α− 1 +βn−β2 + 4β − 1 +γn−γ2 + 4γ − 1 for all n > 0,405where α, β, γ are the roots of x3 − x2 − x− 1 = 0. Explicitly, we haveα =13+13(19 − 3√33)1/3+13(19 + 3√33)1/3,β =13− 16(1 + i√3)(19 − 3√33)1/3 − 16(1 − i√3)(19 + 3√33)1/3,γ =13− 16(1 − i√3)(19 − 3√33)1/3 − 16(1 + i√3)(19 + 3√33)1/3.Another interesting formula due to Spickermann isTn = Round[ α(α− β)(α − γ)αn],where, as usual, Round[x] is the nearest integer to x.Since α−2 < α/(α − β)(α − γ) = 0.33622 . . . < α, the above identity yields theboundsαn−3 < Tn < αn+2 for all n > 1.The Fibonacci and Tribonacci numbers can also be computed using the generatingfunctionsz1 − z − z2 = 1 + z + 2z2 + 3z3 + 5z4 + 8z5 + 13z6 + 21z7 + 34z8 + . . . ,(2.3)z1 − z − z2 − z3 = 1 + z + 2z2 + 4z3 + 7z4 + 13z5 + 24z6 + 44z7 + 81z8 + . . .(2.4)In order to prove Theorem 1, we will use a lower bound for a linear form inthree logarithms à la Baker and such a bound was given by the following result ofMatveev [30].Lemma 1. Let α1, α2, α3 be real algebraic numbers and let b1, b2, b3 be nonzerorational numbers. DefineΛ = b1 logα1 + b2 logα2 + b3 logα3.Let D be the degree of the number field Q(α1, α2, α3) over Q and let A1, A2, A3 bepositive real numbers which satisfyAj > max{Dh(αj), |logαj |, 0.16} for j = 1, 2, 3.Assume thatB > max{1,max{|bj |Aj/A1; 1 6 j 6 3}}.406Define alsoC = 6750000 · e4(20.2 + log(35.5D2 log(eD))).If Λ 6= 0, thenlog |Λ| > −CD2A1A2A3 log(1.5eDB log(eD)).As usual, in the above statement, the logarithmic height of an s-degree algebraicnumber α is defined ash(α) =1s(log |a| +s∑j=1log max{1, |α(j)|}),where a is the leading coefficient of the minimal polynomial of α (over Z), (α(j))16j6sare the conjugates of α and, as usual, the absolute value of the complex numberz = a+ bi is |z| =√a2 + b2.After finding an upper bound on n which is generally too large, the next step is toreduce it. For that, we need a variant of the famous Baker-Davenport lemma, whichis due to Dujella and Pethö [6]. For a real number x, we use ‖x‖ = min{|x−n| : n ∈N} = |x−Round[x]| for the distance from x to the nearest integer.Lemma 2. Suppose that M is a positive integer. Let p/q be a convergent of thecontinued fraction expansion of γ such that q > 6M and let ε = ‖µq‖ −M‖γq‖,where µ is a real number. If ε > 0, then there is no solution to the inequality0 < mγ − n+ µ < AB−min positive integers m,n withlog(Aq/ε)logB6 m < M.See Lemma 5, a) in [6]. Now, we are ready to deal with the proofs of our results.4073. The proof of Theorem 13.1. Finding a bound on n. By Binet’s formulae (2.1) and (2.2) we getϕn − (−ϕ)−n√5=αmα′+βmβ′+γmγ′.Let us denote by α′, β′, γ′ the values of Q(x) = −x2 + 4x − 1 at x = α, β, γ,respectively. By (2.2) and equation (1.2), we haveϕn√5− αmα′=(−1)nϕ−n√5+βmβ′+γmγ′, m, n > 1.More precisely,(3.1)∣∣∣ϕn√5− αmα′∣∣∣ 6∣∣∣ϕ−1√5∣∣∣ + 2∣∣∣ββ′∣∣∣ < 0.67 for any m,n > 1where in the last inequality we have used |β| = |γ| = 0.73735 . . . and |β′| = |γ′| =3.84631 . . ..DefineΛ = Λ(m,n) = m logα− n logϕ+ log(√5α′).ThenΛ = log(αmϕ−n√5α′),which yields|eΛ − 1| =∣∣∣∣αmϕ−n√5α′− 1∣∣∣∣.On the other hand, from (3.1) we get∣∣∣∣ϕn − αm√5α′∣∣∣∣ < 0.67 ·√5 < 1.5.Hence|eΛ − 1| = 1ϕn∣∣∣∣ϕn − αm√5α′∣∣∣∣ <1.5ϕn.Since ϕ = 1.61803 . . ., we have 1.5/ϕn < ϕ−n+1 and then(3.2) |eΛ − 1| < ϕ−n+1.408We claim that Λ 6= 0. In fact, towards a contradiction, suppose that Λ = 0 andthus αm√5/α′ = ϕn. Therefore α2m/α′2 is a quadratic algebraic number. Howeverα2m/α′2 ∈ Q(α) which is absurd, because α is a 3-degree algebraic number.If Λ > 0, then Λ < eΛ−1 < ϕ−n+2 (see (3.2)). If Λ < 0, then 1−e−|Λ| = |eΛ−1| <ϕ−n+2. Thus, for Λ < 0, we get|Λ| < e|Λ| − 1 < ϕ−n+11 − ϕ−n+1 < ϕ−n+2,where we have used the fact that 1 − ϕ−n+1 > 1/ϕ for all n > 3.Hence, we have |Λ| < ϕ−n+2 for any Λ 6= 0, which yields(3.3) log |Λ| < −(n− 2) logϕ.Now, we will apply Lemma 1. Takeα1 = α, α2 = ϕ, α3 =√5/α′, b1 = m, b2 = −n, b3 = 1.Then Q(α1, α2, α3) = Q(α, ϕ), D = 6 and C < 1.2 · 1010.It is easy to verify that 1/α′ is a root of 44x3 − 2x− 1 and that√5/α′ is a root of1936x6−880x4+100x2−125. Since√5/α′ is a 6-degree algebraic number, its minimalpolynomial (over Z) is 1936x6 − 880x4 + 100x2 − 125. Using direct calculation, weverify that the absolute value of every root of the minimal polynomial is less than1. Hence h(α3) < (log 1936)/6 < 1.262. Next, we have h(α1) = (logα)/3 = 0.204and h(α2) = (logϕ)/2 < 0.241. We then take A1 = 1.22, A2 = 1.45 and A3 = 7.58.Since (1.2) implies n > m, we havemax{1,max{|bj|Aj/A1; 1 6 j 6 3}}= max{m, 1.2n} = 1.2n =: B.Hence, Lemma 1 yields(3.4) log |Λ| > −6.8 · 1012 log(82n).Combining the estimates (3.3) and (3.4), we get6.8 · 1012 log(82n) > (n− 2) logϕ,and this inequality implies n < 6 · 1014 and, by the trivial estimate m < n, we havem < 6 · 1014. In order to improve the estimates, we use the bounds on Fn andTm together with Equation (1.2) to obtain αm−3 < Tm = Fn < ϕn−1, which yields409m < 0.8n+ 2.2. Hence, m < 4.8 · 1014. Similarly, ϕn−2 < Fn = Tm < αm+2 yieldsn < 1.3m+ 4.6.3.2. Reducing the bound. The next goal is to reduce the bound on m. Forthat, let us suppose, without loss of generality, that Λ > 0 (the other case can behandled in a similar way by considering 0 < Λ′ = −Λ).We know that 0 < Λ < ϕ−n+2 and therefore0 < m logα− n logϕ+ log(√5α′)< ϕ−m+2.Dividing by logϕ, we get(3.5) 0 < mγ̂ − n+ µ < 5.45 · ϕ−m,with γ̂ = logα/ logϕ and µ = log(√5/α′)/ logϕ.Surely γ̂ is an irrational number (actually, this number is transcendental by theGelfond-Schneider theorem: if α and β are algebraic numbers with α 6= 0 or 1,and β is irrational, then αβ is transcendental). So, let us denote by pn/qn the nthconvergent of its continued fraction.In order to reduce our bound on m, we will use Lemma 2. For that, takingM = 4.8 · 1014, we have thatp33q33=5373914931798006742436582738078750,and then q33 > 6M . Moreover, we get‖µq33‖ −M‖γ̂q33‖ > 0.028 =: ε.Thus all the hypotheses of Lemma 2 are satisfied and we take A = 5.45 and B = ϕ.It follows from Lemma 2 that there is no solution of the inequality in (3.5) (and thenfor the Diophantine equation (1.2)) in the range[⌊ log(Aq33/ε)logB⌋+ 1,M]= [91, 4.8 · 1014].Therefore m 6 90 and then n 6 120. To conclude, we use the formulas in (2.3) and(2.4) together with the Mathematica commandIntersection[CoefficientList[Series[x/(1− x− x2), x, 0, 120], x],CoefficientList[Series[x/(1− x− x2 − x3), x, 0, 90], x]]to find the possible solutions. Fastly, Mathematica returns us the set {0, 1, 2, 13} asits answer. This completes the proof.4104. Final remarks and a conjectureWe point out that the method in proof of Theorem 1 is quite general and thatit can be used to work on the intersection of two arbitrary k-generalized Fibonaccisequences. In fact, in a similar fashion, we found the set F (k) ∩ F (m) for 4 6 k <m 6 10. These cases suggest that the following statement (which is Conjecture 1 in[32]) should be true.Conjecture 1. Let k < m be positive integer numbers. ThenF (k) ∩ F (m) ={0, 1, 2, 13}, if (k,m) = (2, 3),{0, 1, 2, 4, 504}, if (k,m) = (3, 7),{0, 1, 2, 8}, if k = 2 and m > 3,{0, 1, 2, . . . , 2k−1}, otherwise.When working on these cases it may be helpful that the polynomials ψk(x) :=xk − xk−1 − . . .− x− 1 are irreducible over Q[x] with just one zero outside the unitcircle. That single zero is located between 2(1 − 2−k) and 2 (as seen in [39]). Also,in a recent paper, G.Dresden [7, Theorem 1] gave a simplified “Binet-like” formulafor F(k)n :F (k)n =k∑i=1αi − 12 + (k + 1)(αi − 2)αn−1ifor α1, . . . , αk being the roots of ψk(x) = 0. There are many other ways of repre-senting these k-generalized Fibonacci numbers, as can be seen in [10], [11], [12], [14],[15]. Also, it was proved in [7, Theorem 2] thatF (k)n = Round[ α− 12 + (k + 1)(α− 2) αn−1],where α is the dominant root of ψk(x).A c k n ow l e d g em e n t. The author would like to express his gratitude to theanonymous referee for carefully examining this paper and providing a number ofimportant comments and suggestions. He is also grateful to Gregory Dresden for nicetalks on the subject. The author is financially supported by CNPq and FEMAT.411References[1] M.Agronomof: Sur une suite récurrente. Mathesis 4 (1914), 125–126. (In French.)[2] M.A.Alekseyev: On the intersections of Fibonacci, Pell, and Lucas numbers. Integers11 (2011), 239–259.[3] G.Back, M.Caragiu: The greatest prime factor and recurrent sequences. Fibonacci Q.48 (2010), 358–362.[4] Y.Bugeaud, M.Mignotte, S. Siksek: Classical and modular approaches to exponentialDiophantine equations I: Fibonacci and Lucas perfect powers. Ann. Math. 163 (2006),969–1018.[5] D.Carroll, E. T. Jacobson, L. 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On the intersection of two distinct $k$-generalized Fibonacci sequences

On the intersections of Fibonacci, Pell, and Lucas numbers

On the intersections of Fibonacci, Pell, and Lucas numbers

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Institute of Mathematics AS CR, v. v. i.