Fibonacci and Lucas Numbers which have Exactly

Fibonacci and Lucas Numbers which have Exactly Three Prime Factors and Some Unique Properties of F18 and L18 Prapanpong Pongsriiam Silpakorn University, Thailand The 18th Fibonacci Conference 1-7 July 2018

Fibonacci and Lucas Numbers which have Exactly Three Prime F Prapanpong Pongsriiam

Acknowledgment The organizers, Dalhousie University, and AARMS



1. Graduate and start following The Fibonacci Quarterly in 2012, which is 6 years ago. 6 | 18


1. Graduate and start following The Fibonacci Quarterly in 2012, which is 6 years ago. 6 | 18 2. Publish 2 articles in The Fibonacci Quarterly. 2 | 18


1. Graduate and start following The Fibonacci Quarterly in 2012, which is 6 years ago. 6 | 18 2. Publish 2 articles in The Fibonacci Quarterly. 2 | 18 What else? I should find more.


Arithmetic Progression


Theorem (Green-Tao Theorem 2008) The sequence of prime numbers contains arbitrarily long arithmetic progressions.

B. Green and T. Tao, The primes contain arbitrarily long arithmetic progression, Annals of Mathematics, 167 (2008), 481–547.


Arithmetic progressions in the least positive reduced residue systems. A(n) = {a ∈ N | 1 ≤ a ≤ n and (a, n) = 1} ℓ(n) = the length of longest arithmetic progressions in A(n)


Example A(1) = {1}, A(2) = {1}, A(3) = {1, 2},

ℓ(1) = 1 ℓ(2) = 1 ℓ(3) = 2


Example A(1) = {1}, A(2) = {1}, A(3) = {1, 2},

ℓ(1) = 1 ℓ(2) = 1 ℓ(3) = 2

A(5) = {1, 2, 3, 4}, A(6) = {1, 5}, A(7) = {1, 2, 3, 4, 5, 6},

ℓ(5) = 4 ℓ(6) = 2 ℓ(7) = 6


Example A(1) = {1}, A(2) = {1}, A(3) = {1, 2},

ℓ(1) = 1 ℓ(2) = 1 ℓ(3) = 2

A(5) = {1, 2, 3, 4}, A(6) = {1, 5}, A(7) = {1, 2, 3, 4, 5, 6},

ℓ(5) = 4 ℓ(6) = 2 ℓ(7) = 6

A(18) = {1, 5, 7, 11, 13, 17}, |A(F18 )| = φ(F18 ) = 1152,

ℓ(18) = 3

(1, 7, 13) or (5, 11, 17)

ℓ(F18 ) =??


Problem: (Chapter B40 in Guy’s book) Bernado Recamán asks if ℓ(n) → ∞ as n → ∞.

R. K. Guy, Unsolved Problems in Number Theory, Springer, 2004


Theorem (Stumpf (2017)) For all n > 1, { } { } P(n) − 1 n n max , ≤ ℓ(n) ≤ max P(n) − 1, , 2 γ(n) γ(n) ∏ where P(n) is the largest prime factor of n and γ(n) = p|n p.

P. Stumpf, A short note on reduced residues, Integers 17 (2017), A4


ℓ(F18 ) = ??


Theorem (PP 2018) Suppose that n > 1 is squarefree and p is the largest prime factor of n. Then   if n is a prime;  p − 1, p+1 if n = 2p, p ≥ 3, and p ≡ 3 (mod 4); ℓ(n) = 2 ,⌊ ⌋   p − p2 − 1, otherwise. n

Theorem (PP (2018)) Suppose that n > 1 is not squarefree and p is the largest prime factor of n. Then { } n ℓ(n) = max ,p − 1 . γ(n) P. Pongsriiam, Longest arithmetic progressions in reduced residue systems, Journal of Number Theory, 183 (2018), 309–325.


The Third Point: |A(F18 )| = 1152 ℓ(F18 ) = 18


Find all n such that ℓ(n) = 18.


Theorem Let n be a positive integer. Then ℓ(n) = 18 if and only if n satisfies one of the following conditions: n = 19, 74, 115, n = 19A where A is a positive divisor of

∏

(1) p, ω(A) ≥ 2,

p≤17

and A ̸= 6, 10, 14, 15,

(2)

n = 19mγ(m)Bm where m = 2, 3, . . . , 17 and ∏ p≤17 p Bm is a positive divisor of , γ(m) ∏ n = 108C where C is a positive divisor of p.

(3) (4)

5≤p≤19

Note F18 satisfies (3) with m = 4 and Bm = 17.


Corollary ℓ(Fm ) = 18 if and only if m = 18.



Watch Time


6 | 18 2 | 18 ℓ(Fm ) = 18 ⇔ m = 18 What else?


The number of prime factors. The number of divisors. ω(n) = the number of distinct prime factors of n d(n) = the number of positive divisors of n


Bugeaud, Luca, Mignotte, and Siksek give a description of Fn for which ω(Fn ) ≤ 2. We extend the investigation on ω(Fn ), ω(Ln ), d(Fn ), and d(Ln )

Y. Bugeaud, F. Luca, M. Mignotte, and S. Siksek, On Fibonacci numbers with few prime divisors, Japan Academy Proceedings. Series A, 81 (2005), 17–20.


Theorem The only solutions to the equation ω(Fn ) = 3 are given by n = 16, 18, or 2p for some prime p ≥ 19, 2

3

(5)

n = p, p , p for some prime p ≥ 5,

(6)

n = pq for some distinct primes p, q ≥ 3.

(7)


Idea of proof: For m ̸= 2, Fm | Fn if and only if m | n. (Carmichael) The Fibonacci number Fn has a primitive divisor for every n ̸= 1, 2, 6, 12 and the Lucas number Ln has a primitive divisor for every n ̸= 1, 6. Let m be a positive integer and let x(m) be the number of elements in the set {1, 2, 6, 12} ∩ {d : d | m}. Then the following statements hold: (i) ω(Fm ) ≥ d(m) − x(m). (ii) If ω(m) ≥ 3, then ω(Fm ) ≥ 5. (iii) ω(F2m ) = m − 1 for m < 6 and ω(F2m ) ≥ m for m ≥ 6.

n = 2a , pa , 2a pb , pa qb


Example (i) ω(Fn ) = 3 and n = 2p: n = 2 × 19, Fn = 37 × 113 × 9349; n = 2 × 23, Fn = 139 × 461 × 28657; n = 2 × 29, Fn = 59 × 19489 × 514229. (ii) ω(Fn ) = 3 and n = p, p2 , p3 : n = 37, Fn = 73 × 149 × 2221; n = 72 , Fn = 13 × 97 × 6168709; n = 53 , Fn = 53 × 3001 × 158414167964045700001. (iii) ω(Fn ) = 3 and n = pq: n = 3 × 5, 3 × 7, 3 × 11, 5 × 7 and Fn = 2 × 5 × 61, 2 × 13 × 421, 2 × 89 × 19801, 5 × 13 × 141961, respectively.


Theorem If ω(Ln ) = 3, then n satisfies one of the following conditions: n = 2a for some a > 7, n = p, p2 , p3 for some odd prime p, n = 2a p, 2a p2 for some odd prime p and positive integer a, n = pq for some distinct odd primes p, q. In addition, if ω(Ln ) = 3 and n = 9 · 2a for some a ≥ 1, then n = 18.


Example (i) ω(Ln ) = 3 and n = 2a : n = 28 and Ln = 34303 × 73327699969 × p where p is a prime with 39 digits. (ii) ω(Ln ) = 3 and n = p, p2 , p3 : n = 59, Ln = 709 × 8969 × 336419; n = 52 , Ln = 11 × 101 × 151; n = 33 , Ln = 22 × 19 × 5779. (iii) ω(Ln ) = 3 and n = 2a p, 2a p2 : n = 2 × 11, 22 × 3, 23 × 5, 2 × 32 and Ln = 3 × 43 × 307, 2 × 7 × 23, 47 × 1601 × 3041, 2 × 33 × 107, respectively. (iv) ω(Ln ) = 3 and n = pq: n = 3 × 5, Ln = 22 × 11 × 31.


Theorem Assume that ω(Fn ) = 3 and n = p1 p2 where p1 < p2 are odd primes. Then Fp1 = q1 , Fp2 = q2 , and Fn = qa11 q2 qa33 where q1 , q2 , q3 are distinct primes, q3 is a primitive divisor of Fn , a3 ≥ 1 and a1 ∈ {1, 2}. Furthermore a1 = 2 if and only if q1 = p2 . Idea of proof. Previous theorem and the following famous result. (Bugeaud, Mignotte, and Siksek) The only solutions to the equation Fn = ym

in integers m ≥ 2, n ≥ 0, y ≥ 1

are given by n = 0, 1, 2, 6, and 12 which correspond respectively to Fn = 0, 1, 1, 8, and 144. Moreover, the only solutions to the equation Ln = ym with m ≥ 2, n ≥ 0, y ≥ 1 are given by n = 1 and 3 which correspond respectively to Ln = 1 and 4. Y. Bugeaud, M. Mignotte, and S. Siksek, Classical and modular approaches to exponential Diophantine equations I, Fibonacci and Lucas perfect powers, Annals of Mathematics. Second Series, 163 (2006), 969–1018.



Theorem F18 is the only even Fibonacci number which has exactly three prime factors where two of the prime factors are twin.


From the table, we see that n = 18 is the only positive integer n ≤ 150 satisfying ω(Fn ) = ω(Ln ) = 3 and d(Fn ) = d(Ln ) = 16. The range n ≤ 150 can be extended further by using computer. In fact, this problem is connected to the existence or nonexistence of the prime p such that vp (Fz(p) ) > 1. Brother A. Brousseau, Fibonacci and Related Number Theoretic Tables, The Fibonacci Association, 1972.


Wall observes that vp (Fz(p) ) = 1 for all p < 104 . Mcintosh and Roettger, and Dorais and Klyve extend the range p < 104 to p < 2 × 1014 and to p < 9.7 × 1014 , respectively. For the most update information on the range of such primes p, see PrimeGrid Project. Z. H. Sun and Z. W. Sun also show that if p is odd and vp (Fz(p) ) = 1, then the first case of Fermat’s last theorem holds for the exponent p. For a survey on the conjecture that vp (Fz(p) ) = 1 for all p and other related problems, we refer the reader to Klaška. D. D. Wall, Fibonacci series modulo m, The American Mathematical Monthly, 67.6 (1960), 525–532. R. J. Mcintosh and E. L. Roettger, A search for Fibonacci-Wieferich and Wolstenholme primes, Mathematics of Computation, 76.260 (2007), 2087–2094. F. G. Dorais and D. Klyve, A Wieferich prime search up to 6.7 × 1015 , Journal of Integer Sequences, 14.9 (2011), Article 11.9.2. PrimeGrid, Wall–Sun–Sun prime search, http://www.primegrid.com. Z. H. Sun and Z. W. Sun, Fibonacci numbers and Fermat’s last theorem, Acta Arithmetica, 60.4 (1992), 371–388. J. Klaška, Donald Dines Wall’s conjecture, The Fibonacci Quarterly, 56.1 (2018), 43–51.


Theorem Suppose vp (Fz(p) ) = 1 for all p. Then ω(Fn ) = 3 implies d(Fn ) = 8, 12, 16. Moreover, ω(Fn ) = 3 and d(Fn ) = 16 if and only if n = 18 or 125.


Corollary Suppose that vp (Fz(p) ) = 1 for all p. Then ω(Fn ) = ω(Ln ) = 3 and d(Fn ) = d(Ln ) = 16 if and only if n = 18.


Small notes 18, 65, 61, 37, 58, 89, 145, 42, 20, 4, 16, 37


Small notes 18, 65, 61, 37, 58, 89, 145, 42, 20, 4, 16, 37 1818, 130, 10, 1, 1


Small notes 18, 65, 61, 37, 58, 89, 145, 42, 20, 4, 16, 37 1818, 130, 10, 1, 1 F18 = 2584, 109, 82, 68, 100, 1, 1


Small notes 18, 65, 61, 37, 58, 89, 145, 42, 20, 4, 16, 37 1818, 130, 10, 1, 1 F18 = 2584, 109, 82, 68, 100, 1, 1 1818 and F18 are happy


THANK YOU
