On a sum involving the Mangoldt function
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On a sum involving the Mangoldt function Jing Ma1 · Jie Wu2
© Akadémiai Kiadó, Budapest, Hungary 2020
Abstract Let (n) be the von Mangoldt function, and let [t] be the integral part of real number t. In this note we prove that the asymptotic formula (d) x =x + Oε x 35/71+ε n d(d + 1) n≤x d≥1
holds as x → ∞ for any ε > 0. Keywords von Mangoldt function · Asymptotic formula Mathematics Subject Classification 11N37 · 11A25
1 Introduction As usual, denote by [t] the integral part of real number t and by (n) the von Mangoldt function, i.e., log p if n = p ν , (n) = 0 otherwise, where and in the sequel, the letter p means always primes. Motivated by the works [1,4–6], we are interested in the asymptotic behaviour of the quantity x S (x) := n n≤x
B
Jing Ma [email protected] Jie Wu [email protected]
1
School of Mathematics, Jilin University, Changchun 130012, People’s Republic of China
2
CNRS LAMA 8050, Laboratoire d’Analyse et de Mathématiques Appliquées, Université Paris-Est Créteil, 94010 Créteil Cedex, France
123
J. Ma, J. Wu
as x → ∞. In [1], Bordellès, Dai, Heyman, Pan and Shparlinski studied a more general summative function x S f (x) := f n n≤x under some reasonable hypotheses on f . More precisely, let τk (n) :=
n 1 ···n k =n
1 be the
generalised divisor function, let A > 0 and 0 < θ < 2 be two constants, and let φ = 1.618 be the Golden Ratio. Under three different types of assumptions on f :
≈
| f (n)| τk (n)
√ 1+ 5 2
(n ∈ N),
φ−1
(1.1)
−A
(log(en)) (n ∈ N), | f (n)| n 2 θ | f (n)| x (x ≥ 1),
(1.2) (1.3)
n≤x
they proved (see [1, Theorems 2.3, 2.5 and 2.6]) the asymptotic formula S f (x) = x
n≥1
f (n) + O(E(x)) n(n + 1)
for x → ∞, where ⎧ 1/2 x ⎪ ⎪ ⎪ ⎨x 1/2 (log x)k−1/2+εk+1 (x)/2 E(x) := ⎪ x(log x)−A(φ−1) ⎪ ⎪ ⎩ (θ +1)/3 (log x)(1+θ )(2+ε2 (x))/6 x and
εk (x) :=
k log3 x log2 x
under (1.1) with k = 1, under (1.1) with k ≥ 2, under (1.2), under (1.3),
1/2 k−1+
30 . log3 x
(1.4)
(1.5)
r -fold iterated logarithm. With the help of the prime number theorem, Here logr denotes the
it is easy to see that n≤x (n)2 x log x for all x ≥ 1. This shows that (n) verifies the condition (1.3) with θ = 1 + ε for any ε > 0. Applying (1.4) of [1] by Bordellès–Dai– Heyman–Pan–Shparlinski, we get S (x) = x
d≥1
(d) + Oε x 2/3+ε d(d + 1)
(1.6)
for x ≥ 2. On the other hand, by refining arguments of [1], Wu [5] improved (1.4) by establishing the following better results: (a) Assume that there is a constant ϑ ∈ [0, 1) such that | f (n)| n ϑ for n ≥ 1. Then S f (x) = x
n≥1
f (n) + O x (ϑ+1)/2 n(n + 1)
(x → ∞).
(b) Let k ≥ 2. For x → ∞, we have S f (x) = x
n≥1
123
f (n) + O(E 0 (x)), n(n + 1)
On a sum involving the Mangoldt function
with
E 0 (x) =
x 1/2 (log x)k−1+εk+1 (x)/2 under (1.1), (θ +1)/3 (1+θ )ε (x)/6 2 (log x) under (1.3). x
Since 0 ≤ (n) ≤ log n ε n ε for all n ≥ 1 and any ε > 0, then Wu’s (a) with ϑ = ε implies that (d) S (x) = x + Oε x 1/2+ε (1.7) d(d + 1) d≥1
for x ≥ 2, which sharpe
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