SOME REMARKS ON SUM OF EULER’S TOTIENT FUNCTION Es-Said En-Naoui To cite this version: Es-Said En-Naoui. SOME REMARKS ON SUM OF EULER’S TOTIENT FUNCTION. 2020. hal- 03105850 HAL Id: hal-03105850 https://hal.archives-ouvertes.fr/hal-03105850 Preprint submitted on 11 Jan 2021 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Copyright SOME REMARKS ON SUM OF EULER’S TOTIENT FUNCTION A PREPRINT Es-said En-naoui [email protected] January 7, 2021 ABSTRACT Euler’s totient function counts the positive integers up to a given integer n that are relatively prime n to n. The aim of this article is to give a result about the sum : φ(k) , for every prime number p . k=1 Pp|k Keywords : Euler’s totient , Dirichlet product, additive function , multiplicative function , Dirichlet series. 1 Introduction Euler totient function φ is the function defined on the positive natural numbers N∗ in the following way: if n ∈ N∗, then we have : φ(n)= x ∈ N∗ :1 ≤ x ≤ n , gcd(x, n)=1 . Clearly φ(p) = p − 1, for any prime number p and, more generally, if α ∈ N ∗, then φ(pα) = pα − pα−1. This follows from the fact that the only numbers which are not coprime with pα are multiples of p and there are pα−1 such multiples x with 1 ≤ x ≤ pα. It is well-known that φ is multiplicative, i.e., if m and n are coprime, then φ(mn) = φ(m)φ(n). If n ≥ 3 and the prime decomposition of n is α1 αs n = p1 ...ps , then from what we have seen s s αi αi−1 1 φ(n)= (pi − p )= n (1 − ). pi Yi=1 Yi=1 . arXiv:2101.02040v1 [math.GM] 28 Dec 2020 The following property is simple but very useful in this paper. Proposition 1. If p,m ∈ N∗, with p prime, and n = pm, then φ(n) = (p − 1)φ(m), if gcd(p,m)=1, and φ(n)= pφ(m), if gcd(p,m) 6=1. Proof. If gcd(p,m)=1, then we have φ(n)= φ(p)φ(m) = (p − 1)φ(m). Now suppose that gcd(p,m) 6=1. We may write m = pαm′, with α ≥ 1 and gcd(p,m′)=1. Thus φ(n)= φ(pα+1)φ(m′)= pα(p − 1)φ(m′). However φ(m)= pα−1(p − 1)φ(m′) and so φ(n)= pφ(m) . Some remarks on sum of Euler’s totient function A PREPRINT n 2 Calcul the sum φ(k) k=1 Pp|k Let x be a real number , then ⌊x⌋ is defined to be the unique integer n satisfying n ≤ x<n +1.This integer ⌊x⌋ is called the floor of x. Then we have rhis results : Corollary 2. Let m be an integer, and let x be a real number. Then, m ≤ x holds if and only if m ≤⌊x⌋ holds. Proof. Recall that ⌊x⌋ is the unique integer n satisfying n ≤ x<n +1. Thus, ⌊x⌋ is an integer satisfying ⌊x⌋≤ x< ⌊x⌋ +1 . If m ≤ x holds, then m ≤ ⌊x⌋ holds . Conversely, if m ≤ ⌊x⌋ holds, then m ≤ x (because m ≤ ⌊x⌋ ≤ x). Combining these two implications, we conclude that m ≤ x holds if and only if m ≤ ⌊x⌋ holds. Corollary 2 is thus proven. R N∗ ⌊x⌋ x Proposition 3. Let x ∈ and m ∈ . Then , m = m . x x x Proof. By the definition of ⌊x⌋, that m is the unique integer n satisfying n ≤ m <n +1.Thus, m is an integer x x x satisfying m ≤ m < m +1 . N∗ x x x But m ∈ and thus m≥ 1> 0 . Hence, we can multiply the inequality m ≤ m by m. We thus obtain m m ≤ x. x x x But the Corollary 2 (applied to m m instead of m) shows that m m ≤ x holds if and only m m ≤ ⌊x⌋holds x x x (since m m is an integer). Thus,m m ≤ ⌊x⌋ holds (since m m ≤ x holds ).Dividing this inequality by m, we x ⌊x⌋ ⌊x⌋ x obtain m ≤ m . We can now divide the inequality ⌊x⌋≤ x by m (since m> 0).We thus obtain m ≤ m . ⌊x⌋ x x x ⌊x⌋ x x Hence , m ≤ m < m +1.So we have m ≤ m < m +1.In other words, m is an integer n satisfying ⌊x⌋ ⌊x⌋ ⌊x⌋ ⌊x⌋ n ≤ m < n +1.But m is the unique integer n satisfying n ≤ m < n +1 (because this is how m is ⌊x⌋ ⌊x⌋ x defined).Hence, if n is any integer satisfying n ≤ m <n +1, then n = m .We can apply this to n = m (since x x ⌊x⌋ x m is an integer n satisfying n ≤ m <n +1), and thus obtain m = m .Proposition 3 is proven . n Proposition 4. Let n a positive integer and p a number prime, put f(n)= p , Then we have : n ∀k ∈ N∗ , fo ··· of(n)= (1) pk k times | {z } n Proof. we prove by induction on k that , then When k =1, it is clear that f(n)= p . Assume that fo ··· of(n)= n . Then we have : pk k times | {z } n k fo ··· of (n)= fofo ··· of(n)= f p p k+1 times k times n | {z } | {z } We let x = pk to substitute into proposition 3, then We will find : n fo ··· of (n)= pk+1 k+1 times | {z } n Now after we give some property useful in this paper ,then , let n be a positive integer ,and put :Ψ(n)= φ(k) , So k=1 for every prime number p we put : P n n Υ(n,p)= φ(k) and ∆(n,p)= φ(k) Xk=1 Xk=1 (k,p)=1 p|k 2 Some remarks on sum of Euler’s totient function A PREPRINT Then we have : Ψ(n)=Υ(n,p) + ∆(n,p) (2) Theorem 5. For every positive integer n ≥ 2, and a prime number p , we have: logp(n) n ∆(n,p) = (p − 1) Ψ (3) pα αX=1 Proof. we have : n n φ(k)= φ(p)+ φ(2p)+ ··· + φ(jp) , with j ≤ p Xk=1 p|k n n n p p p = φ(jp)= φ(jp)+ φ(jp) Xj=1 Xj=1 Xj=1 p|j (p,j)=1 From the proposition 1 , we have φ(jp) = (p − 1)φ(j) if (j, p)=1 and φ(jp)= pφ(j) if (j, p) > 1 then : n n n p p φ(k)= p φ(j) + (p − 1) φ(j) kX=1 Xj=1 Xj=1 p|k p|j (p,j)=1 n n = p∆ ,p + (p − 1)Υ ,p p p n n n = p∆ ,p + pΥ ,p − Υ ,p p p p n n n = p ∆ ,p +Υ ,p − Υ( ,p p p p =Ψ n by 2 | p{z } n n = pΨ − Υ ,p p p n n n Since Υ p ,p =Ψ p − ∆ p ,p by equality 2 , Then : n n n ∆(n,p)= pΨ − Ψ − ∆ ,p p p p so that : n n ∆(n,p) = (p − 1)Ψ + ∆ ,p (4) p p n In the equality 4, Substituting n by p , and by using the proposition 4 we have : n n n ∆ ,p = (p − 1)Ψ + ∆ ,p (5) p p2 p2 n Now , Substituting this ∆ p ,p into 4 we find that: n n n ∆(n,p) = (p − 1)Ψ + (p − 1)Ψ + ∆ ,p (6) p p2 p2 n In the same way if we repeat this operation up to an integer α such that pα ≤ 1 finds this equality : n n n n ∆(n) = (p − 1)Ψ + (p − 1)Ψ + (p − 1)Ψ + ··· +Ψ p p2 p3 pα logp(n) n = (p − 1) Ψ pα αX=1 3 Some remarks on sum of Euler’s totient function A PREPRINT Remark 6. For every prime number p we have : ∆(pn,p) = Ψ(n) + ∆(n,p). p − 1 So for p =2 , we have: ∆(2n, 2) = Ψ(n) + ∆(n, 2). References [1] Lehmer D.H., On Euler’s totient function, Bull. Amer. Math. Soc., 38, 745-751(1932). [2] Childs L.N., A Concrete Introduction to Higher Algebra 3rd edition, Springer 2008. [3] Tom M. Apostol, Introduction to Analytic Number Theory, SpringerVerlag, New York, Inc, USA, 1976 . [4] L. Dickson, History of the theory of numbers. Vol. I: Divisibility and primality, Chelsea Publishing Co., New York, 1966 . 4.