The Average Number of Divisors of the Euler Function

Transcription

1 The Average Number of Divisors of the Euler Function Sungjin Kim Santa Monica College/Concordia University Irvine Department of Mathematics Dec 17, 2016

2 Euler Function, Carmichael Function - Definitions and Notations Let n 1 be an integer. Denote by φ(n), λ(n), the Euler Phi function and the Carmichael Lambda function, which output the order and the exponent of the group (Z/nZ) respectively. Sungjin Kim (SMC/CUI) Dec 17 2 / 11

3 Euler Function, Carmichael Function - Definitions and Notations Let n 1 be an integer. Denote by φ(n), λ(n), the Euler Phi function and the Carmichael Lambda function, which output the order and the exponent of the group (Z/nZ) respectively. Let n = p e 1 1 per r be a prime factorization of n, then we can compute φ(n) and λ(n) as follows: φ(n) = r φ(p e i i ), and λ(n) = lcm (λ(p e 1 1 ),..., λ(per r )) i=1 where φ(p e i i ) = p e i 1 i (p i 1) and λ(p e i i ) = φ(p e i i ) if p i > 2 or p i = 2 and e i = 1, 2, and λ(2 e ) = 2 e 2 if e 3. Sungjin Kim (SMC/CUI) Dec 17 2 / 11

4 Definitions and Notations We write P z = p z p. We also use the following restricted divisor functions: τ z (n) := p e n p>z τ(p e ), τ z,w (n) := p e n z<p w τ(p e ), and τ z(n) := p e n p z τ(p e ). Moreover, for n > 1, denote by p(n) the smallest prime factor of n. Sungjin Kim (SMC/CUI) Dec 17 3 / 11

5 Previous Results Sungjin Kim (SMC/CUI) Dec 17 4 / 11

6 Previous Results Luca, Pomerance (LP, 2007): As x, x exp ( 1 γ 7 e 2 log x τ(φ(n)) x exp ( 1 + O ( 2 2e γ 2 ( )) ) log τ(λ(n)) ( ( log x log 1 + O )) ). Sungjin Kim (SMC/CUI) Dec 17 4 / 11

7 Previous Results Luca, Pomerance (LP, 2007): As x, x exp ( 1 γ 7 e 2 log x τ(φ(n)) x exp ( 1 + O ( 2 2e γ 2 ( )) ) log τ(λ(n)) ( ( log x log 1 + O Luca, Pomerance (LP, 2007): As x, 1 1 τ(λ(n)) = o max τ(φ(n)). x y x y n y )) ). F. Luca, C. Pomerance, On the Average Number of Divisors of the Euler Function, Publ. Math. Debrecen, 70/1-2 (2007), pp Sungjin Kim (SMC/CUI) Dec 17 4 / 11

8 Main Results (Theorem 1.1) As x, we have τ(φ(n)) ( ) τ(λ(n)) x exp 2e γ log x 2 (1 + o(1)). Sungjin Kim (SMC/CUI) Dec 17 5 / 11

9 Main Results (Theorem 1.1) As x, we have τ(φ(n)) ( ) τ(λ(n)) x exp 2e γ log x 2 (1 + o(1)). (Theorem 1.2) As x, τ(λ(n)) = o τ(φ(n)). Sungjin Kim (SMC/CUI) Dec 17 5 / 11

10 Heuristics (Conjecture 1.1) As x, we have ( τ(λ(n)) = x exp 2 ) 2e γ log x 2 (1 + o(1)). Thus, it is expected that both sums τ(φ(n)) and τ(λ(n)) satisfy ( x exp 2 ) 2e γ log x 2 (1 + o(1)). Sungjin Kim (SMC/CUI) Dec 17 6 / 11

11 The Method - Theorem 1.1 (Lemma 5 in LP) Let A > 0 and 1 < z A log x log 4 2 x. Then S z (x) := p x τ z (p 1) p ( log x log x = c 1 log z + O log 2 z ). Sungjin Kim (SMC/CUI) Dec 17 7 / 11

12 The Method - Theorem 1.1 (Lemma 5 in LP) Let A > 0 and 1 < z A log x log 4 2 x. Then S z (x) := p x τ z (p 1) p ( log x log x = c 1 log z + O log 2 z ). (Corollary 2.1) Let A > 1 and log 1 A x z log A x. Then as x, log x S z (x) = c 1 (1 + o(1)). log z Sungjin Kim (SMC/CUI) Dec 17 7 / 11

13 The Method - Theorem 1.1 We use p 1, p 2,..., p v to denote prime numbers. We define the following multiple sums for 2 v x: T v,z (x) := p 1 p 2 p v x and for u = (u 1,..., u v ) with 1 u i x, T u,v,z (x) := p 1 p 2 p v x i, p i 1 mod u i τ z (p 1 1)τ z (p 2 1) τ z (p v 1) p 1 p 2 p v, τ z (p 1 1)τ z (p 2 1) τ z (p v 1) p 1 p 2 p v, Define T v := {(t 1,..., t v ) : i, t i [0, 1], t t v 1}. We adopt the idea from Gauss Circle Problem. p 1 p 2 p v x log p 1 log x + log p 2 log x + + log p v log x 1. Sungjin Kim (SMC/CUI) Dec 17 8 / 11

14 The Method - Theorem 1.1 Let v = c log x log 2 x Use vol(t v ) = 1 v! T v,z (x) := for some positive constant c to be determined. to approximate p 1 p 2 p v x τ z (p 1 1)τ z (p 2 1) τ z (p v 1) p 1 p 2 p v, by 1 v! S z(x) v (1 + o(1)) v. Sungjin Kim (SMC/CUI) Dec 17 9 / 11

15 The Method - Theorem 1.1 Let v = c log x log 2 x Use vol(t v ) = 1 v! T v,z (x) := for some positive constant c to be determined. to approximate p 1 p 2 p v x τ z (p 1 1)τ z (p 2 1) τ z (p v 1) p 1 p 2 p v, by 1 v! S z(x) v (1 + o(1)) v. This approximation allows a lower bound of ( ) τ(λ(n)) log x exp n log 2 x (2c + c log c 1 2c log c + o(1)). Maximizing above by the first derivative, the optimal choice for c is e γ 2. This proves Theorem 1.1. Sungjin Kim (SMC/CUI) Dec 17 9 / 11

16 The Method - Theorem 1.2 (Lemma 4.1) For any 2 y x, we have n x y τ(φ(n)) n log5 x x τ(φ(n)). Sungjin Kim (SMC/CUI) Dec / 11

17 The Method - Theorem 1.2 (Lemma 4.1) For any 2 y x, we have n x y τ(φ(n)) n Define E 1 (x), E 2 (x) and E 3 (x): log5 x x τ(φ(n)). and E 1 (x) := {n x : 2 k n or there is a prime p n with p 1 mod 2 k }, E 2 (x) := {n x : ω(n) ω}, E 3 (x) := {n x} (E 1 (x) E 2 (x)). Use Lemma 4.1 to estimate sums τ(λ(n)) over the above three sets and obtain the estimate in Theorem 1.2. Sungjin Kim (SMC/CUI) Dec / 11

18 The Binomial Model For z = log x, let τ z,z 2(lcm(p 1 1, p 2 1,..., p v 1)) τ z,z 2(p 1 1)τ z,z 2(p 2 1) τ z,z 2(p v 1). Let the number X q of primes p 1,..., p v such that q p i 1. We model X q by a binomial distribution with parameters v (number of trials), 2 q (probability of success). Sungjin Kim (SMC/CUI) Dec / 11

19 The Binomial Model For z = log x, let τ z,z 2(lcm(p 1 1, p 2 1,..., p v 1)) τ z,z 2(p 1 1)τ z,z 2(p 2 1) τ z,z 2(p v 1). Let the number X q of primes p 1,..., p v such that q p i 1. We model X q by a binomial distribution with parameters v (number of trials), 2 q (probability of success). A Difficulty: Achieving independence of X q with various primes q. Sungjin Kim (SMC/CUI) Dec / 11

20 The Binomial Model For z = log x, let τ z,z 2(lcm(p 1 1, p 2 1,..., p v 1)) τ z,z 2(p 1 1)τ z,z 2(p 2 1) τ z,z 2(p v 1). Let the number X q of primes p 1,..., p v such that q p i 1. We model X q by a binomial distribution with parameters v (number of trials), 2 q (probability of success). A Difficulty: Achieving independence of X q with various primes q. A Direction: Assuming the EH to prove the heuristics. Sungjin Kim (SMC/CUI) Dec / 11

21 The Binomial Model For z = log x, let τ z,z 2(lcm(p 1 1, p 2 1,..., p v 1)) τ z,z 2(p 1 1)τ z,z 2(p 2 1) τ z,z 2(p v 1). Let the number X q of primes p 1,..., p v such that q p i 1. We model X q by a binomial distribution with parameters v (number of trials), 2 q (probability of success). A Difficulty: Achieving independence of X q with various primes q. A Direction: Assuming the EH to prove the heuristics. Thank you for listening! Sungjin Kim (SMC/CUI) Dec / 11