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− . n n q q q n6x X
6 From Proposition 3, λq ⋆ 1 = gq ⋆τ. Consequently
(λq ⋆ 1)(n) = gq(d) τ(k) n6x 6 6 X Xd x kXx/d x x x x θ+ε = g (d) log + (2γ 1) + O q d d d d 6 − Xd x = x ζ(q)P (1) log x + qP (1)ζ′(q)+ P ′(1)ζ(q)+(2γ 1)ζ(q)P (1) q q q − q max(1/c ,θ)+ε + O x q , where θ is defined in (1) and where we used
1/cq θ 1 ε gq(d) x − , if cq− > θ ; x− | | dθ 6 ≪ (1, otherwise. Xd x 4.2 The case q 11 (mod 24) ≡± For σ > 1, we set
q 1 − m +1 m 1 ∞ h (n) H (s)= ζ(qs) 1+ + = ζ(qs)R (s) := q . q q q pms q ns p m=3 ! n=1 Y X X 1 Since q > 5, this Dirichlet series is absolutely convergent in the half-plane σ > 3 , so that
h (n) x1/3+ε. | q | ≪q,ε n6x X
From Proposition 3, λq ⋆ 1 = hq ⋆a2, hence
x (λ ⋆ 1)(n) = h (d) q q d n6x d6x r X X h (d) = x1/2 q + O x1/3+ε √ 6 d Xd x 1/2 1 1 /3+ε = x Hq 2 + O x . 4.3 The case q 5 (mod 24) ≡± In this case, it is necessary to rewrite L(s,λq) in the following shape.
7 K (s) Lemma 4. Assume q 5 (mod24). For any σ > 1, L(s,λ )= q with ≡± q ζ(s)ζ(2s)
ζ(5s), if q = 5; ζ(4s)Lq(s), if q 19, 29 (mod120); Kq(s) := ≡± ± (s) Lq , if q 43, 53 (mod120); ζ(4s) ≡± ± where
q 1 2(p2s + ps + 1) p2s +1 − m +1 m 1 L (s) := ζ(qs) 1+ + + q p7s p5s p2s 1 q q pms p m=6 ! Y − − X and 2p2s 1 q(s) := ζ(qs) 1 − L − 2s 3 2s p (p 1) (p + 1) Y − q 1 p8s − m +1 m 1 + + . 2s 3 2s q q pms (p 1) (p + 1) m=6 ! − X 1 The Dirichlet series Lq is absolutely convergent in the half-plane σ > 5 , and the Dirichlet series is absolutely convergent in the half-plane σ > 1 . Lq 6 Proof. From Proposition 3, we immediately get ζ(5s) L(s,λ )= . (3) 5 ζ(s)ζ(2s)
Now suppose q > 5 and q 5 (mod24). In this case, 3 + 2 = 2 and 4 + 3 =0 ≡± q q − q q so that we may write by Proposition 3
q 1 ζ(qs)ζ(2s) 2 − m +1 m 1 L (s,λ ) = 1 + + q ζ(s) − p2s q q pms p m=4 ! Y X K (s) = q ζ(s)ζ(2s) where
q 1 1 p4s − m +1 m 1 Kq(s) := ζ(qs) 1 + + . − 2s 2 2s 2 q q pms p (p 1) (p 1) m=4 ! Y − − X 8 Assume q 19, 29 (mod120). Then ≡± ± 5 4 6 5 + = + =2. q q q q Kq(s) can therefore be written as
q 1 ps +2 p4s − m +1 m 1 Kq(s) = ζ(qs) 1+ + + s 2s 2 2s 2 q q pms p p (p 1) (p 1) m=6 ! Y − − X q 1 2(p2s + ps + 1) p2s +1 − m +1 m 1 = ζ(qs)ζ(4s) 1+ + + p7s p5s p2s 1 q q pms p m=6 ! Y − − X = ζ(4s)Lq(s). Similarly, if q 43, 53 (mod120), then ≡± ± 5 4 6 5 + = + =0. q q q q Hence q 1 1 p4s − m +1 m 1 Kq(s) := ζ(qs) 1 + + − 2s 2 2s 2 q q pms p (p 1) (p 1) m=6 ! Y − − X (s) = Lq . ζ(4s) The proof is complete. We now are in a position to prove Theorem 1 in the case q 5 (mod24). ≡± Assume first that q 19, 29 (mod120) and let ℓq(n) be the n-th coefficient of the ≡ ± ± 1 Dirichlet series Lq(s). From Lemma 4, λq ⋆ 1 = ℓq ⋆a4 ⋆a2− and therefore 1 x x (λ ⋆ 1)(n)= ℓ (d) M = ℓ (d)L . q q m2 d q d n6x d6x m6(x/d)1/4 r d6x r X X X X Since L(z) zδ (z) for some c> 0 ≪ c
1/2 ℓq(d) x (λq ⋆ 1)(n) x | | δc ≪ √d d n6x d6x r X X
1/2 ℓq(d) x x + | | δc ≪ √d d d6√x √x
(λ ⋆ 1)(n) x1/2δ x1/4 + x7/20+ε x1/2δ x1/4 . q ≪ c ≪ c n6x X Now suppose that the Riemann hypothesis is true. By [1], which is a refinement of [9], we 1/2 know that M(z) ε z ω(z). The method of [9, 1] may be adapted to the function L yielding ≪ L(z) z1/2 ω(z) log z. ≪ε Observe that, for any a > 2, ε> 0 and z > eee
log z exp log z (log log z)a 6 exp log z (log log z)a+ε p p so that L(z) z1/2 ω(z) and hence ≪ε ℓ (d) x ℓ (d) (λ ⋆ 1)(n) x1/4 | q | ω x1/4ω √x | q | x1/4ω √x q ≪ d1/4 d ≪ d1/4 ≪ n6x d6x r d6x X X X 1 completing the proof in that case. The case q = 5 is similar but simpler since λ5 ⋆1 = a5 ⋆a2− by (3). Finally, when q 43, 53 (mod120), we proceed as above. Let νq(n) be the n-th ≡ ± ± 1 1 coefficient of the Dirichlet series (s). Then λ ⋆ 1 = ν ⋆a− ⋆a− from Lemma 4, so that Lq q q 4 2 1 x (λ ⋆ 1)(n)= ν (d) µ(m)M q q m2 d n6x d6x m6(x/d)1/4 r X X X and estimating trivially yields
1/2 νq(d) 1 1 x (λq ⋆ 1)(n) x | | δc ≪ √d m2 m2 d n6x d6x m6(x/d)1/4 r X X X and we complete the proof as in the previous case.
10 Remark 5. Let us stress that a bound of the shape
(λ ⋆ 1)(n) x1/4+ε q ≪ n6x X for all x sufficiently large and small ε > 0, is a necessary and sufficient condition for the Riemann hypothesis. Indeed, if this estimate holds, then by partial summation the series s 1 ∞ n=1 (λq ⋆ 1)(n)n− is absolutely convergent in the half-plane σ > 4 . Consequently, the 1 function Kq(s)ζ(2s)− is analytic in this half-plane. In particular, ζ(2s) does not vanish Pin this half-plane, implying the Riemann hypothesis, proving the necessary condition, the sufficiency being established above.
5 A short interval result for the case q =5 5.1 Introduction This section deals with sums of the shape
(λ5 ⋆ 1)(n) x 3/5 −1/5 1/2 c log x1/4 log log x1/4 (λ ⋆ 1)(n) x e− ( ) ( ) 5 ≪ x 1 2 / 5/2+ε (λ ⋆ 1)(n) x1/4e(log √x) (log log √x) . 5 ≪ε x (λ ⋆ 1)(n) τ(n) y log x. 5 ≪ ≪ x 11 5.2 Tools from the theory 1 In what follows, N Z>1 is large and δ 0, 4 . The first result is [7, Theorem 5] with k = 5. See also [2, Theorem∈ 5.23 (iv)]. ∈ Lemma 6 (5th derivative test). Let f C5 [N, 2N] such that there exist λ > 0 and λ > 0 ∈ 4 5 satisfying λ = Nλ and, for any x [N, 2N] 4 5 ∈ f (4)(x) λ and f (5)(x) λ . ≍ 4 ≍ 5 Then 1/15 1/6 1 1/4 (f,N,δ) Nλ + Nδ + δλ− +1. R ≪ 5 4 Remark 7. The basic result of the theory is the following first derivative test (see [2, Theorem 1 5.6]): Let f C [N, 2N] such that there exist λ > 0 such that f ′(x) λ . Then ∈ 1 | |≍ 1 1 (f,N,δ) Nλ + Nδ + δλ− +1. (4) R ≪ 1 1 This result is essentially a consequence of the mean value theorem. The second tool is [4, Theorem 7] with k = 3. 1/s Lemma 8. Let s Q∗ 2, 1 and X > 0 such that N 6 X . Then there exists a constant c := c (s)∈ 0,\1 {±depending± } only on s such that, if 3 3 ∈ 4 2 N δ 6 c3 (5) then X 3 s 1/7 59 s 1/21 ,N,δ XN − + δ XN − . R ns ≪ Our last result relates the short sum of λ5 ⋆ 1 to a problem of counting integer points near a smooth curve. Lemma 9. Let 1 6 y 6 x. Then x y 1/2 1/5 2/5 1 − − (λ5 ⋆ )(n) max 5 ,N, log x + yx + x y . ≪ 2 −1 1/5 6 1/5R n √ 5 x x 1/5 (λ ⋆ 1)(n)= µ(d) 5 d2 n6x X dX6√x 12 so that x + y 1/5 x 1/5 (λ ⋆ 1)(n) = µ(d) + µ(d) 5 d2 − d2 x x + y x 1/5 2/5 1/2 + x− y + yx− ≪ n5 − n5 2 −1 1/5 6 1/5 $r % r ! (16y x )X 2 1 1/5 1/5 and for any integers N (16y x− ) , (2x) and n [N, 2N] ∈ ∈ i i x + y x y 1 < < n5 − n5 √ 5 4 r r N x so that the sum does not exceed x y 1/5 2/5 1/2 − − max 5 ,N, log x + x y + yx ≪ 2 −1 1/5 6 1/5R n √ 5 (16y x ) 5.3 The main result 6 11/20 5 Theorem 10. Assume y c3 x where c3 := c3 2 is given in (5). Then 1/12 4/9 (λ ⋆ 1)(n) x + yx− log x. 5 ≪ x (λ ⋆ 1)(n) x1/12 log x. 5 ≪ x 2 1 1/5 1/10 1/10 1/6 1/6 1/5 16y x− x y 1/12 1/40 1/6 3/20 1/4 − − max 5 ,N, x + x y + x y . 2 −1 1/5 6 1/10 R n √ 5 ≪ (16y x ) 5 1/2 1/2 5 − For the second range, we use Lemma 8 with X = x , s = 2 and δ = y (N x) . Notice 1/10 6 11/20 1 2 6 that the conditions N > 2x and y c3 x ensure that δ < 4 and N δ c3. We get x y 1/12 4/9 − max 5 ,N, x + yx . 2x1/10 x y 1/12 3/4 − max 5 ,N, x + yx . 1/6 6 1/5 R n √ 5 ≪ 2x 1/12 1/40 1/6 3/20 1/4 4/9 1/5 2/5 (λ ⋆ 1)(n) x + x− y + x− y + yx− log x + x− y 5 ≪ x 6 Acknowledgments The author deeply thanks Prof. Kannan Soundararajan for the help he gave him to adapt his result to the function L(x), and Benoit Cloitre for bringing this problem to his attention. The author gratefully acknowledges the anonymous referee for some corrections and remarks that have significantly improved the paper. References [1] M. Balazard and A. de Roton, Notes de lecture de l’article “Partial sums of the M¨obius function” de Kannan Soundararajan, preprint, 2008, https://arxiv.org/abs/0810.3587. [2] O. Bordell`es, Arithmetic Tales, Springer, 2012. [3] H. Davenport, The Higher Arithmetic, 5th edition, Cambridge University Press, 1982. [4] M. Filaseta and O. Trifonov, The distribution of fractional parts with applications to gap results in number theory, Proc. London Math. Soc. 73 (1996), 241–278. 14 [5] G. H. Hardy, On Dirichlet’s divisor problem, Proc. London Math. Soc. 15 (1916), 1–25. [6] M. N. Huxley, Exponential sums and lattice points III, Proc. London Math. Soc. 87 (2003), 591–609. [7] M. N. Huxley and P. Sargos, Points entiers au voisinage d’une courbe plane de classe Cn, II, Functiones et Approximatio 35 (2006), 91–115. [8] R. K. Muthumalai, Note on Legendre symbols connecting with certain infinite series, Notes on Number Theory and Discrete Mathematics 19 (2013), 77–83. [9] K. Soundararajan, Partial sums of the M¨obius function, J. Reine Angew. Math. 631 (2009), 141–152. 2010 Mathematics Subject Classification: Primary 11N37; Secondary 11A25, 11M41. Keywords: Number of divisors, Legendre symbol, mean value, Riemann hypothesis. (Concerned with sequences A000005, A008836, and A091337.) Received January 14 2017; revised versions received June 1 2017; June 26 2017. Published in Journal of Integer Sequences, July 1 2017. Return to Journal of Integer Sequences home page. 15