ON THE SUM OF DIVISORS FUNCTION

Transcription

1 Annales Univ. Sci. Budapest., Sect. Comp ) ON THE SUM OF DIVISORS FUNCTION N.L. Bassily Cairo, Egypt) Dedicated to Professors Zoltán Daróczy and Imre Kátai on their 75th anniversary Communicated by Bui Minh Phong Received December 15, 2012; accepted January 10, 2013) Abstract. The following assertion is proved. Let Q 1, Q 2 be odd primes, A Q1,Q 2 ) be the number of those n for which Q 1 σn), Q 2 σn + 1) simultaneously hold. Then A Q1,Q 2 ) c, if > X log ) 0. 5 c, X 0 are positive constants. 1. Introduction 1.1. Notation P = set of primes. ϕn) = Euler s totient function, σn) = sum of divisors function, τn) = number of divisors. Let 1 = log, 2 = log 1,..., k+1 = log k Formulation of the theorems In his paper [4] Kátai Theorem 4) proved the following assertion: [ λ Let λ > 2, I = 2 3, 2 ], Q 1, Q 2 P, E Q1,Q 2 ) := # {n : Q 1 ϕn), Q 2 ϕn + 1)}. Key words and phrases: Sum of divisor function Mathematics Subject Classification: 11N25, 11N60.

2 130 N.L. Bassily Then, uniformly for Q 1, Q 2 I, 1 E Q 1,Q 2 ) = 1 + O 1) ) B 2 κ 1 κ 2, where and κ j = ep ) 2, j = 1, 2), Q j 1 B = p 3 p P 1 ) 2. pp 1) Kátai notes that similar theorem can be proved for σ instead of ϕ, and that he is unable to count the asymptotic of those n for which 3 ϕn), 3 ϕn + 1) simultaneously holds. In this paper we shall investigate the function A Q1,Q 2 ) := # {n : Q 1 σn), Q 2 σn + 1)}, where Q 1, Q 2 are arbitrary odd primes, Q 1 = Q 2 is included. Theorem 1. If Q 1, Q 2 P, Q 1 2, Q 2 2, then there are constants c > 0 and X 0 such that A Q1,Q 2 ) c log ) 5, if > X 0. Theorem 2. Let Q P, Q 2, B Q ) = # {p : p P, Q σp + 1)}. Then there are constants c > 0 and X 0 such that B Q ) c log ) 5. Our theorems follow from some variants of known, deep theorems.

3 On the sum of divisors function Main auiliary results 2.1. Let Q P, Q 2, χn) be a character mod Q, such that χ 1) = 1. Let rn) = d n χd), T ) = p+1 0 mod Q) Theorem 3. We have T ) = A 0 log + O where δ and A 0 are positive constants. rqp + 1) µqp + 1). log ) 1+δ 2.2. Let Q 1 Q 2, Q 1, Q 2 be odd primes, χn) be a character mod Q 2, such that χ 1) = 1. Let rn) = d n χd). Let A = Qa 1Q b 2, a {1, 2}, b {1, 2} such that Q 1 σa). Let S) = rap + 1) µap + 1). p+1 0 mod Q 1) Theorem 4. We have S) = B 0 log + O where δ and B 0 are suitable positive constants. log ) 1+δ ), ), 3. Deduction of Theorem 1 and 2 from Theorem 3 and 4 In the case Q 1 = Q 2 = Q let A = Q, χ mod Q be a Dirichlet character such that χ 1) = 1, and rn) = d n χd). If p mod Q), and µqp + 1) rqp + 1) 0, then Qp + 1 is a squarefree number, and π Qp+1, π P implies that 1+χπ) = 2, consequently Q π + 1, thus Q σqp + 1). Furthermore Q σqp) = Q + 1)p + 1). Assume that Q 1 Q 2. Let A = Q a 1Q b 2, where a and b are such positive integers that Q Q Q b 2, Q Q Q a 1. Observe that

4 132 N.L. Bassily a, b {1, 2} is a suitable choice. Let χ be a character mod Q 2, such that χ 1) = 1, and rn) = d n χd). Let p 1 mod Q 1 ). Then Q 1 σap), and if µap + 1) rap + 1) 0, then Q 2 σap + 1). We have T 2 ) A Q1,Q 2 ) p 1 mod Q) r 2 Qp + 1) µqp + 1). Since r 2 Qp+1) τ 2 Qp+1), and n τ 2 n) c 3 1, therefore A Q,Q ). 5 1 We can obtain similarly that S 2 ) B Q ) and hence that B Q ). 5 1 p+1 0 mod Q) r 2 Ap + 1) µap + 1), Remark. We could improve these inequalities by using some sieve results. 4. Sketch of the proof of Theorem 3 and 4 The main ingredient of the proof is the inequality due to E. Bombieri and A.I. Vinogradov which is quoted now as Lemma 1. Let πz, D, l) = #{p z : p l mod D)}, li z = z 2 du log u. Lemma 1. See Elliott [1], Chapter 7.) ma z πz, D, l) li z ϕd), D 1 B where B 2A ma l mod D) l,d)=1 A 1

5 On the sum of divisors function 133 I. Kátai considered in [4] the sum T ) = r 4 p 1) µp 1), where r 4 n) = d n χ 4d), χ mod 4) is the character satisfying χ 1) = 1, and proved that ) T ) = A 0 log + O log ) 1+δ. To prove Theorem 3 we can follow his argument. Let T, k) = rqp + 1). p+1 0 mod Q) Qp+1 0 mod k) Arguing as in [4], we have T ) = ) µd)t, d 2 ) + O 1,5. 1 Let k 6 1, k, Q) = 1. Then d 3 1 T, k) = Qp+1=uv Qp+1 0k) p+1 0Q) χu). Thus T, k) = u Q+1 u,q)=1 χu) p 1Q) Qp+1 0[k,u]) 1. Since Q, u) = 1, therefore the right most sum equals Q 2 j=1 where l j are those residues for which π, Q[k, u], l j ), Ql j mod [k, u]), l j mod Q) holds. With this modification we can proceed further. The enveloping sieve of C. Hooley can be applied. The proof of Theorem 4 is similar.

6 134 N.L. Bassily References [1] Elliott, P.D.T.A., Arithmetic functions and integer proucts, Springer Verlag, [2] Hooley, C., Applications of sieve methods to the theory of numbers, Cambridge University Press, [3] Kátai, I., A note on a sieve method, Publications Math. Debrecen, ), [4] Kátai I., On the prime divisors of the Euler phi and the sum of divisors functions, Annales Univ. Sci. Budapest., Sect. Comp., ), N.L. Bassily Department of Mathematics Faculty of Sciences Ain Shams University Cairo Egypt