1 2 Journal of Integer Sequences, Vol. 20 (2017), 3 Article 17.7.1 47 6 23 11 Long and Short Sums of a Twisted Divisor Function Olivier Bordell`es 2, all´ee de la Combe 43000 Aiguilhe France [email protected] Abstract τ Let q > 2 be a prime number and define λq := q where τ(n) is the number of divisors of n and q· is the Legendre symbol. When τ(n) is a quadratic residue modulo q, then the convolution (λq ⋆ 1)(n) could be close to the number of divisors of n. The aim of this work is to compare the mean value of the function λq ⋆ 1 to the well known average order of τ. A bound for short sums in the case q = 5 is also given, using profound results from the theory of integer points close to certain smooth curves. 1 Introduction and main result If Ω(n) stands for the number of total prime factors of n and λ = ( 1)Ω is the Liouville function, then − ζ(2s) L(s,λ)= (σ > 1) . ζ(s) This implies the convolution identity (λ⋆ 1)(n)= x1/2 , n6x X where, as usual, F ⋆G is the Dirichlet convolution product of the arithmetic functions F and G given by (F ⋆G)(n) := F (d)G(n/d). d n X| 1 τ Define λ3 := 3 where τ(n) is the number of divisors of n and 3· is the Legendre symbol. Then from Proposition 3 below ζ(3s) L(s,λ )= (σ > 1) , 3 ζ(s) implying the convolution identity 1/3 (λ3 ⋆ 1)(n)= x . n6x X τ Now let q > 2 be a prime number and define λq := q where q· is the Legendre symbol. Our main aim is to investigate the sum (λq ⋆ 1)(n). n6x X When τ(n) is a quadratic residue modulo q, one may wonder if (λq ⋆ 1)(n) has a high prob- ability to be equal to the number of divisors of n. Note that this function is multiplicative, 2 and, for any prime p, (λq ⋆ 1)(p)=1+ q . Consequently, when 2 is a quadratic residue modulo q, then (λq ⋆ 1)(n)= τ(n) for all squarefree numbers n. On the other hand, when 2 is a quadratic nonresidue modulo q, then (λq ⋆ 1)(n) = 0 unless n =1 or n is squarefull, so 1/2 that n6x (λq ⋆ 1)(n) x in this case. It then could be interesting to compare this sum to the average order of≪ the function τ, i.e., P τ(n)= x (log x +2γ 1) + O xθ+ε , (1) − n6x X where 1 6 6 131 4 θ 416 , (2) the left-hand side being established by Hardy [5], the right-hand side being the best estimate to date due to Huxley [6]. To state our first main result, some specific notation are needed. For any prime q > 3, let c , respectively d , be least positive integer m 1,...,q 2 for which m = m+1 , q q ∈ { − } q 6 q respectively m = m+1 . Note that c and d are well-defined, since it is known from q 6 − q q q [3, p. 75-76] that the number of m for which m = m+1 and m = m+1 are q q q − q respectively 1 (q 3) and 1 (q 1). Hence there is at least 1 (q 3) integers mfor which 2 − 2 − 2 − m = m+1 . For convenience, set d = 3. q 6 ± q 3 As usual in number theory, we adopt Riemann’s notation s = σ + it C and ζ is the Riemann zeta function, and define the Euler products ∈ q 1 − m +1 m 1 1 Pq(s) := 1+ σ > , q − q pms cq p m=c Y Xq 2 and q 1 − m +1 m 1 R (s) := 1+ + σ > 1 . q q q pms 3 p m=3 ! Y X Theorem 1. Let q > 3 be a prime number. (a) If q 1 (mod8) ≡± ζ′ P ′ (λ ⋆ 1)(n)= xζ(q)P (1) log x +2γ 1+ q (q)+ q (1) +O xmax(1/cq,θ)+ε , q q − ζ P q,ε n6x q X where θ is defined in (1) and (2). (b) If q 11 (mod24) ≡± 1/2 q 1 1/3+ε (λq ⋆ 1)(n)= x ζ 2 Rq 2 + Oq,ε x . n6x X (c) If q 5 (mod24), there exists c> 0 such that ≡± 3/5 −1/5 1/2 c log x1/4 log log x1/4 (λ ⋆ 1)(n) x e− ( ) ( ) . q ≪q n6x X Furthermore, if the Riemann hypothesis is true, then for x sufficiently large 1 2 / 5/2+ε (λ ⋆ 1)(n) x1/4e(log √x) (log log √x) . q ≪q,ε n6x X Example 2. 1/2+ε (λ7 ⋆ 1)(n) =0.454 x (log x +2γ +0.784) + Oε x . n6x X . (λ ⋆ 1)(n) =0.899 x (log x +2γ 0.678) + O x131/416+ε . 23 − ε n6x X . 1/2 1/3+ε (λ13 ⋆ 1)(n) =1.969 x + Oε x . n6x X 3/5 −1/5 1/2 c log x1/4 log log x1/4 (λ ⋆ 1)(n) x e− ( ) ( ) . 5 ≪ n6x X 3 2 Notation In what follows, x > e4 is a large real number, ε 0, 1 is a small real number which does ∈ 4 not need to be the same at each occurrence, q always denotes an odd prime number, q· is the Legendre symbol and τ λ := , q q where τ(n) := d n 1. Also, 1 is the constant arithmetic function equal to 1. | 1 For any arithmeticP functions F and G, L(s,F ) is the Dirichlet series of F , and F − is the Dirichlet convolution inverse of F . If r Z> , then ∈ 2 1, if n = mr; ar(n) := (0, otherwise. For some c> 0, set c(log x)3/5(log log x)−1/5 (log x)1/2(log log x)5/2+ε δc(x) := e− and ω(x) := e . Finally, let M(x) and L(x) be respectively the Mertens function and the summatory function of the Liouville function, i.e. M(x) := µ(n) and L(x) := λ(n). n6x n6x X X 3 The Dirichlet series of λq Proposition 3. Let q > 3 be a prime number. For any s C such that σ > 1 ∈ ⊲ If q 1 (mod8) ≡± q 1 − m +1 m 1 L (s,λ )= ζ(qs)ζ(s) 1+ . q q − q pms p m=c Y Xq ⊲ If q 3 (mod8) ≡± q 1 ζ(qs)ζ(2s) − m +1 m 1 L (s,λ )= 1+ + . q ζ(s) q q pms p Y mX=dq 4 Proof. Set χq := q· for convenience. From [8, Lemma 2.1], we have ∞ χ (α + 1) ∞ χ (α) L (s,λ ) = 1+ q = 1+ ps q q psα psα p α=1 ! p α=2 ! Y X Y X 1 q 1 − s 1 − m 1 s = 1+ p 1 p− − pqs q pms − p m=1 !! Y X 1 q 1 1 − − m 1 = 1 − pqs q p(m 1)s p m=1 − ! Y X q 1 − m 1 = ζ(qs) 1+ . q p(m 1)s p m=2 − ! Y X If q 1 (mod8), then 2 = 1 and ≡± q q 1 1 1 − m 1 L (s,λ )= ζ(qs)ζ(s) 1 + 1 , q − ps − ps q p(m 1)s p m=2 − ! Y X where q 1 q 1 1 − m 1 − m 1 1 1 = − ps q p(m 1)s q p(m 1)s − pms m=2 − m=2 − X Xq 2 q 1 − m +1 1 − m 1 = q pms − q pms m=1 m=2 X q 1 X 2 1 − m +1 m 1 q 1 = + q ps q − q pms − q p(q 1)s m=2 − q 1 X − m +1 m 1 1 = + . q − q pms ps m=2 X Similarly, if q 3 (mod8), then 2 = 1 and ≡± q − q 1 ζ(qs)ζ(2s) 1 1 − m 1 L (s,λ )= 1+ + 1+ , q ζ(s) ps ps q p(m 1)s p m=2 − ! Y X 5 where q 1 q 1 1 − m 1 − m 1 1 1+ = + ps q p(m 1)s q p(m 1)s pms m=2 − m=2 − X Xq 2 q 1 − m +1 1 − m 1 = + q pms q pms m=1 m=2 X q 1 X 2 1 − m +1 m 1 q 1 = + + q ps q q pms − q p(q 1)s m=2 − q 1 X − m +1 m 1 1 = + . q q pms − ps m=2 X We achieve the proof noting that, if q 1 (mod24), then 3 2 = 4 3 = 0 ≡ ± q − q q − q and, similarly, if q 11 (mod24), then 3 + 2 = 0 whereas 4 + 3 = 2. ≡± q q q q 4 Proof of Theorem 1 4.1 The case q 1 (mod 8) ≡± For σ > 1, we set q 1 − m +1 m 1 ∞ g (n) G (s)= ζ(qs) 1+ = ζ(qs)P (s) := q . q q − q pms q ns p m=c n=1 Y Xq X This Dirichlet series is absolutely convergent in the half-plane σ > 1 , so that cq g (n) x1/cq+ε. | q | ≪q,ε n6x X By partial summation, we infer gq(n) 1+1/cq+ε = ζ(q)P (1) + O x− , n q n6x X gq(n) x 1+1/cq+ε log = ζ(q)P (1) log x + qP (1)ζ′(q)+ P ′(1)ζ(q)+ O x− .