Some results on divisor problems related to cusp forms
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Some results on divisor problems related to cusp forms Wei Zhang1 Received: 4 September 2018 / Accepted: 27 July 2019 © Springer Science+Business Media, LLC, part of Springer Nature 2020
Abstract Let λ f (n) be the normalized Fourier coefficients of a holomorphic Hecke cusp form of full level. We study a generalized divisor problem with λ f (n) over some special sequences. More precisely, for any fixed integer k ≥ 2 and j ∈ {1, 2, 3, 4}, we are interested in the following sums Sk (x, j) :=
λk, f (n j ) =
n≤x n=n 1 n 2 ···n k
n≤x
j
j
j
λ f (n 1 )λ f (n 2 ) · · · λ f (n k ).
Keywords Fourier coefficients · Cusp forms · Automorphic L-function · Divisor problem Mathematics Subject Classification 11N37 · 11F70
1 Introduction Let λ f (n) denote the normalized Fourier coefficients of a holomorphic cusp form f (z) of even positive integral weight κ ≥ 2 for the full modular group S L 2 (Z), and f (z) is also an eigenform of all the Hecke operators. Then we have Tn f = λ f (n) f for n ∈ N, and λ f (1) = 1. The Fourier series expansion of f (z) at the cusp ∞ is (κ−1)/2 e(nz). Associated to f , there is an L-function L(s, f ), f (z) = ∞ n=1 λ f (n)n which is defined as L(s, f ) =
∞ λ f (n) n=1
ns
=
−1 −1 1 − α p p −s 1 − β p p −s , s > 1, p
v+1 where α p +β p = λ f ( p), α p β p = |α p | = 1, and λ f ( p v ) = (α v+1 p −β p )/(α p −β p ) for ν ∈ N. It is well known that the multiplicative function λ f (n) ∈ R (see [5]) and
B 1
Wei Zhang [email protected] School of Mathematics, Shandong University, Jinan 250100, Shandong, China
123
W. Zhang
|λ f (n)| n for the holomorphic cusp forms. For any fixed integer k ≥ 1 and 1 ≤ j ≤ 4, we are interested in the estimates for following sums Sk (x, j) := n≤x λk, f (n j ). j j j Here, for s > 1, λk, f (n j ) = f (n 1 )λ f (n 2 ) · · · λ f (n k ) is defined n=n 1 n 2 ···n k λ ∞ ∞ j s k j s by L j (s) = n=1 λ f (n )/n and L j (s) = n=1 λk, f (n )/n . In fact, when k = 2, j = 0, this is the classical divisor problem, which is the conjecture that S2 (x, 0) x 1/4+ε . On the other hand, when k = 1, j = 1, this is a classical problem, Hecke [6] and continued with Walfisz [20], Harfer and Ivic [5] which began with 1/3 , Rankin [17], and Wu [21]. The up-to-date result of Wu [21] λ (n) x n≤x f is S1 (x, 1) = n≤x λ f (n) x 1/3 logδ x, where δ = −0.118 . . .. For k ≥ 2, j = 1, we have L(s, f )k =
∞ λk, f (n) n=1
ns
, s > 1.
Here λk, f (n) = n=n 1 n 2 ···n k λ f (n 1 )λ f (n 2 ) · · · λ f (n k ). The sum Sk (x, 1) = λ (n) has been investigated by some experts. For example, Fomenko [2] n≤x k, f and Kanemitsu et al. [11] got some non-trivial results for Sk (x, 1), where Sk (x, 1) x 1−3/(2+2k)+ with k ≥ 2. Moreover, Landau’s classical results imply that Sk (x, 1) x 1−2/(1+2k)+ with k ≥ 1. Recently, for k ≥ 3, these results were improved by Fomenko [3] and Lü [14] by showing that Sk (x, 1)
x k/(2+k)+ for 3 ≤ k ≤ 6, x 1−3/2k+ for k ≥ 6
(1.1)
and Sk (x, 1)
x 3/5+ for k = 3, 1−3/2k+ for k ≥ 4, x
(1.2)
respec
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