Here, we investigate the Fibonacci numbers whose sum of aliquot divisors is also a Fibonacci number (the prime Fibonacci numbers have this propert

Luca, Florian

Stănică, Pantelimon

Calhoun, Institutional Archive of the Naval Postgraduate School

Calhoun: The NPS Institutional ArchiveFaculty and Researcher Publications Faculty and Researcher Publications2010Aliquots sums of Fibonacci numbersLuca, FlorianCongressus Numerantium Vol.200, Proc. International Conf. Fibonacci Numbers & Applic.,  (W.Webb, ed.) (2010), 153-160.http://hdl.handle.net/10945/38844Aliquot sums of Fibonacci numbersFlorian LucaInstituto de Matema´ticasUniversidad Nacional Auto´noma de Me´xicoC.P. 58089, Morelia, Michoaca´n, Me´xicofluca@matmor.unam.mxPantelimon Sta˘nica˘∗Naval Postgraduate SchoolApplied Mathematics DepartmentMonterey, CA 93943, USApstanica@nps.eduJuly 24, 2007AbstractHere, we investigate the Fibonacci numbers whose sum of aliquotdivisors is also a Fibonacci number (the prime Fibonacci numbers havethis property).1 IntroductionLet (Fn)n≥1 be the sequence of Fibonacci numbers. For a positive integer nwe write σ(n) for the sum of divisors function of n. Recall that a number n iscalled multiply perfect if n | σ(n). If σ(n) = 2n, then n is called perfect. In [2],it was shown that there are only finitely many multiply perfect Fibonacci∗This paper was written while at Auburn University Montgomery, Department of Math-ematics, Montgomery, AL 36124-4023, USA1numbers, and in [3], it was shown that no Fibonacci number is perfect. Fora positive integer n, the value ϕ(n) of the Euler function is defined to be thenumber of natural numbers less than or equal to n and coprime to n.Let s(n) = σ(n) − n. The number s(n) is sometimes called the sumof aliquot divisors of n. Two positive integers m and n (with m 6= n) arecalled amicable if s(m) = n and s(n) = m. It is not known if there existinfinitely many amicable pairs, but Pomerance [5] showed that the sum ofthe reciprocals of all the members of all amicable pairs is convergent.Here, we search for Fibonacci numbers Fn such that s(Fn) is a Fibonaccinumber. In particular, prime Fibonacci numbers have the above property.We putA = {n : s(Fn) = Fm for some positive integer m}.In this paper, we give an upper bound on the counting function of A.Theorem 1. There exists a positive constant c0 such that the inequality#A(x) < c0 xlog log log xholds for all x > eee.Throughout this paper, we use the Vinogradov symbols ,  and theLandau symbols O,  and o with their usual meanings. We recall thatA  B, B  A and A = O(B) are all equivalent and mean that |A| < cBholds with some constant c, while A  B means that both A  B andB  A hold. For a positive real number x we write log x for the maximumbetween 2 and the natural logarithm of x. Note that with this convention, thefunction log x is sub-multiplicative; i.e., the inequality log(xy) ≤ log x log yholds for all positive numbers x and y. For a positive real number t and asubset B of the positive integers, we write B(t) = B ∩ [1, t]. We use p, q, Pand Q with or without subscripts to denote prime numbers.Acknowledgements. During the preparation of this paper, F. L. wassupported in part by projects PAPIIT 104505, SEP-CONACyT 46755 and aGuggenheim Fellowship.2 The Proof of Theorem 1Let x be a large positive real number.22.1 Some sievingLet ω(n) and Ω(n) be the number of prime divisors of n and the number ofprime power divisors of n (> 1), respectively. LetA1(x) = {n ≤ x : ω(n) < 0.9 log log x or Ω(n) > 1.1 log log x}. (1)By the Tura´n-Kubilius inequalities (see [8])∑n≤x(f(n)− log log x)2 = O(x log log x) for both f ∈ {ω,Ω},we infer that#A1(x) xlog log x. (2)Let y = (log log x)1/3 and letA2(x) = {n ≤ x : p 6 |n for all primes p < y}. (3)By Brun’s sieve,#A2(x) x∏p<y(1− 1p) xlog y xlog log log x. (4)We now writeσ(Fn) = Fn + Fm,and we look at bounds for m in terms of n, where n ≤ x does not belong toA1(x) ∪ A2(x).2.2 Bounds for m in terms of nWe start with a lower bound form. Let γ = (1+√5)/2 be the golden section.Let n ≤ x not in A1(x) ∪ A2(x). Then, there exists p < y such that p | n.Thus, Fp | Fn, thereforeγm > Fm = s(Fn) ≥ FnFp γn−p ≥ γn−y,where we used the fact that Fn  γn. Hence,m ≥ n− y +O(1),3thereforem ≥ n− 2y,once x is sufficiently large. We now look at an upper bound for m. Note thatγm−n  FmFn≤ σ(Fn)Fn≤ Fnϕ(Fn)≤∏p |Fn(1 +1p− 1). (5)For every prime number p let z(p) be its order of apparition in the Fibonaccisequence, and for a positive integer d let Pd = {p : z(p) = d}. It is knownthat p ≡ ±1 (mod z(p)) holds for all primes p > 5 and it is clear thatFd ≥∏p∈Pdp (d− 1)#Pd ,therefore#Pd  dlog d. (6)Furthermore, z(p) log p. We now get by taking logarithms in (5) thatm− n ≤∑p |Fn1p− 1 +O(1) ≤∑d |n∑p∈Pd1p− 1 +O(1).Obviously, ∑p∈Pd1p− 1 ≤∑p≡±1 (mod d)p<d21p− 1 +#Pdd2 − 2 log log dϕ(d),where in the above inequality we have used estimate (6) as well as the knownfact that the inequality∑p≡a (mod b)p<t1p− 1 ≤1p1(a, b)− 1 +O(log log tϕ(b)), (7)holds uniformly in coprime positive integers a < b and positive real numberst, where p1(a, b) is the first prime in the arithmetic progression a (mod b)4(see, for example, [4]). Since the function log log d is sub-multiplicative, weget thatm− n ≤∏pµ | |n(1 +O(µ∑ν=1log log(pν)pν))≤ expO∑p |nlog log pp+∑p≥2∑ν≥2log log(pν)pν= expO∑p |nlog log pp+ 1 .Since n 6∈ A1(x), it follows that ω(n) < 1.1 log log x. Thus, if we writep1 < p2 < . . . for the increasing sequence of all the prime numbers, then∑p |nlog log pp≤ω(n)∑i=1log log pp≤∫ pω(n)2log log ttdpi(t) (log log pω(n))2  (log log log log x)2.Hence,m− n ≤ exp (O((log log log log x)2)) < 2y,where the last inequality holds if x is large. In conclusion, if n ≤ x is not inA1(x) ∪ A2(x), then m ∈ [n− 2y, n+ 2y].2.3 More sievingLet Q = {q : z(q) < q1/3}. Note that uniformly in t > 1,2#Q(t) ≤∏q∈Qq<tq ≤∏n<t1/3Fn < γ∑n<t1/3n < γt2/3,therefore#Q(t) t2/3,5which shows that ∑q∈Qq>s1q≤∫ ∞s1td#Q(t)≤ #Q(t)t∣∣∣t=∞t=s+∫ ∞s#Q(t)t2 1s1/3+∫ ∞sdtt4/3 1s1/3. (8)We now put u = (log x)3 and letA3(x) = {n ≤ x, z(p)p | n for some p > u}. (9)For every fixed prime p > u, the number of n ≤ x which are multiples ofpz(p) is bx/pz(p)c ≤ x/pz(p). So,#A3(x) ≤∑p>uxpz(p)≤∑p>up6∈Qxpz(p)+∑p>up∈Qxz(p)p∑u1/3<d≤x∑p≡±1 (mod d)p<d3xdp+xu1/3 x∑u1/3<d≤xlog log ddϕ(d)+xu1/3 x∑u1/3<d≤x(log log d)2d2+xu1/3 x(log log x)2∑d>u1/31d2+xu1/3 x(log log x)2(log x)1/3, (10)where in the above estimates we used (8) with s = u1/3, the fact that φ(d)d/ log log d for all d, as well as estimate (7) with b = d and a = 1 and d− 1,respectively.We finally put ωu(n) for the number of prime factors p ≤ u of n, v =2 log log log x and letA4(x) = {n ≤ x : ωu(n) > v}. (11)6Again by Tura´n-Kubilius inequality,∑n<x(ωu(n)− log log u)2 = O(x log log u),and since log log u = (1 + o(1)) log log log x, we get easily that#A4(x) xlog log log x. (12)From now on, we deal only with numbers n ≤ x which are not in A1(x)∪A2(x) ∪ A3(x) ∪ A4(x).2.4 The 2-adic order of σ(Fn)Let K = b0.8 log log xc. Since n 6∈ ∪4i=1Ai(x), we get that n has ω(n) −ωu(n) > 0.9 log log x − 2 log log log x > K prime factors P > u, once x issufficiently large. Let P1 > P2 > . . . > PK be the first (largest) prime factorsof n. Then PK > u. Note thatFn =(K−1∏i=0Fn/P1...PiFn/P1...Pi+1)Fn/P1...PK ,where by convention we take P0 = 1. LetLi =Fn/P1...PiFn/P1...Pi+1for i = 0, . . . , K − 1 and LK = Fn/P1...PK .We next observe that Li and Lj are coprime for all 0 ≤ i < j ≤ K. Indeed,assume that i < j ≤ K and Q are such that Q | gcd(Li, Lj). ThenQ∣∣∣ gcd(Fn/P1...Pi+1 , Fn/P1...PiFn/P1...Pi+1).However, it is well-known that the greatest common divisor appearing abovedivides Pi+1. Hence, Q = Pi+1, and Q | Fn, therefore z(Q) | n. SinceQ > u > 5 for large x, we get that Qz(Q) | n contradicting the fact thatn 6∈ A3(x). Thus, Li and Lj are indeed coprime for all i < j.In [6], Ribenboim and McDaniel studied square-classes of Fibonacci num-bers. Given two integers m and n, they are in the same square-class if FmFn7is a square. It follows from their results that if m > 12n and n is sufficientlylarge, then m and n are not in the same square-class. In particular, if x islarge, then none of the numbers Li is a perfect square. Thus, there exists aprime Qi | Li, such that the order at which Qi appears in Li (hence, in Fn) isodd. It is also clear that Qi is odd if x is large enough (say if u > 3). Thus,∏Ki=1(Qi + 1) is a divisor of σ(Fn), which proves that σ(Fn) is a multipleof 2K .2.5 The conclusionLet A5(x) be the set of all positive integers n ∈ A(x) which are not in∪4i=1Ai(x). Let n1 < n2 < . . . < n` be all the elements in A5(x). Then thereexists ki ∈ [−2y, 2y] such that mi = ni+ ki for all i = 1, . . . , `. Furthermore,2K | σ(Fni) = Fni + Fni+ki . Let M = d4y + 1e. We show that if ` > M ,then ni+M − ni is large whenever i ≤ ` −M . Indeed, let ni < ni+1 < . . . <ni+M . Then kj ∈ [−2y, 2y] for all j = i, . . . , i +M , and since there are atmost 2b2yc + 1 < M + 1 possible values of kj and M + 1 possibilities forthe index j, it follows that there exist j1 < j2 in {i, . . . , i +M} such thatkj1 = kj2 . Let k denote the common value of kj1 and kj2 . Using the formulaFn = (γn− δn)/(γ − δ), where δ = (1−√5)/2 is the conjugate of γ, we notethat the relation 2K | Fnj1 + Fnj1+k givesγnj1 (1 + γk)− δnj1 (1 + δk) ≡ 0 (mod 2K), (13)and similarly for nj2 . Here and in what follows, we say that an algebraicinteger α is a multiple of an integer m if α/m is an algebraic integer. Writeλ = nj2 − nj1 . Then the above relation for nj2 givesγnj1γλ(1 + γk)− δnj1δλ(1 + δk) ≡ 0 (mod 2K). (14)Multiplying the congruence (13) by γλ and subtracting it from congruence(14), we get thatδnj1 (1 + δk)(γλ − δλ) ≡ 0 (mod 2K). (15)Conjugating the above relation (15) and multiplying the resulting congru-ences, we get|1 + δk||1 + γk||γλ − δλ|28is an integer which is a multiple of 22K . Noting that the above integer isnonzero, by taking logarithms we getλ log γ +O(M) ≥ 2K,therefore λ K. Thus, we just proved that ni+M − ni  K, therefore#A5(x) xMK+M  x(log log x)2/3, (16)which together with the upper bounds (2), (4), (10) and (12) completes theproof of Theorem 1.References[1] N. L. Bassily, I. Ka´tai, and M. Wijsmuller, ‘On the prime power divisorsof the iterates of the Euler-ϕ function’, Publ. Math. (Debrecen) 55 (1999),17–32.[2] F. Luca, ‘Multiply perfect numbers in Lucas sequences with odd param-eters’, Publ. Math. (Debrecen) 58 (2001), 121–155.[3] F. Luca, ‘Perfect Fibonacci and Lucas numbers’, Rend. Circ. Mat.Palermo (2) 49 (2000), 313–318.[4] C. Pomerance, ‘On the distribution of amicable numbers’, J. ReineAngew. Math. 293/294 (1977), 217–222.[5] C. Pomerance, ‘On the distribution of amicable numbers. II’, J. ReineAngew. Math. 325 (1981), 183–188.[6] P. Ribenboim and W. L. McDaniel, ‘Square classes in Lucas sequenceshaving odd parameters’, J. Number Theory 73 (1998), 14–27.[7] G. Tenenbaum, Introduction to analytic and probabilistic number theory ,Cambridge U. Press, 1995.[8] P. Tura´n, ‘On a theorem of Hardy and Ramanujan’, J. London Math.Soc. 9 (1934), 274–276.9

Aliquots sums of Fibonacci numbers

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Aliquots sums of Fibonacci numbers

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