Abstract
In this article, the error term of the mean value theorem for binary Egyptian fractions is studied. An error term of prime number theorem type is obtained unconditionally. Under Riemann hypothesis, a power saving can be obtained. The mean value in short interval is also considered.
1 Introduction
Let
with positive integers
is always soluble. This conjecture is still open, although much work has been carried out. See [2,3,4], for example, for more details about the ternary conjecture. Refer to [5] for more information on Egyptian fractions.
Considering the binary Egyptian fractions when
The Diophantine equation is not always necessary to have a solution. For example, for n with all its prime factor p of the form
Huang and Vaughan [6] proved that
with
and
where
and
Here
In [6], Huang and Vaughan employed for the first time in this area, of complex analytic technique from multiplicative number theory, and gave an innovative counting function with a different criterion with Croot et al. in [11]. Their estimate (3) holds uniformly for X and
Theorem 1
Let
with
and some absolute constant
Remark 1
Theorem 1 is better than (3) when a is small. Obviously, Theorem 1 is a PNT (prime number theorem) type result, which depends on the zero-free region of the Riemann zeta-function. So in some sense, it is the best possible result under the present methods in the analytic number theory. It is impossible to improve the exponent 1/2 in Theorem 1 without better zero-free region of the Riemann zeta-function.
Under Riemann hypothesis (RH), we can prove the following power saving result.
Theorem 2
Let
with
Here the implied constant depends only on
Remark 2
One main tool in [6] is the well-known Perron formula. Our proof relies on the convolution method, the results concerning divisor problems with Dirichlet characters (see [12,13]) and moment results of the Dirichlet L-functions.
In the authors’ another work [14], the mean square of the error term under RH was studied. An asymptotic formula can be obtained, which suggests that the average size of the error term is
Both Theorems 1 and 2 depend on zeros of
Then we can get the following estimate.
Theorem 3
Let
with same
Remark 3
For the proof of Theorem 3, we use the technique of Zhai [15] who considered the short interval distribution of a class of integers, and his lemma (see Lemma 8) can help us to get some saving in short interval. The constraint for
In what follows,
2 Preliminary
In this section, we list some lemmas which will be needed in the proof.
We cite a theorem of Friedlander and Iwaniec [12].
Lemma 1
Let
with
and the implied constant depends only on
For primitive character
whence we get
and
Also for
The following lemma is due to Nowak [13], which has more precise estimation than the theorem given by Friedlander and Iwaniec for the special case given below.
Lemma 2
Let
with
and
Here the main term is given by
and the implied constant depends only on
Lemma 3
For any
with the implied constant depending only on
where
For
Then we have
Lemma 4
Assuming RH, then we have
Proof
Well-known.□
The estimation for the mean value of the Möbius function depends on the zero-free region of Riemann zeta function (see for example Theorem 12.7 [16]).
Lemma 5
Let
The following results about moments of zeta function and Dirichlet L-functions are well-known (see [17] and Theorem 10.1 of [18]).
Lemma 6
Let
and
where
Lemma 7
Let
where
Proof
Let
Then according to Lemma 6, we get
□
The following lemma is due to [15], which gives an upper bound for some summation in short intervals.
Lemma 8
Let
3 Proof of Theorem 1
First of all, we assume that
Otherwise, the theorem follows from the upper bound (3).
We start from the identity (see (3.1) in [6])
where
For
Then
where
If
If
Noting that
converges absolutely for
Hence, we can write
where
and for
with
Proposition 1
Let
Proof
When
where
For the first sum, by using (6) and Lemma 3, we get
Similarly, we have
Since
with
Then, for
Taking
for some sufficiently small
Remark 4
If we remove the constraint
Proposition 2
Let
Proof
We can apply similar argument to the proof above and get
with
where
which implies (14) by taking z as the proof in Proposition 1.□
By an appeal to (8), (10) and (12), we have
with
and
Thanks to (7) and (9), the first term on the right is
and the second term on the right is
Therefore, we have
For the main term,
Therefore, we obtain
By an appeal to (8), (10) and (14), we have
where
with
and
It is easy to check that the main term in (20) equals to
Therefore, we obtain
Combining (18) and (21), and according to the computation of Huang and Vaughan in [6] about the residue, we get
where
4 Proof of Theorem 2
It is easy to get the upper bound from (3) when
W start from (11). Let
We write according to Lemmas 2 and 3
where
and for
Then
By (23) and Lemma 2 (here the exponent 18 in the following is chosen to make sure that
and for
It remains to compute
with
for
Move the line of integral in (27) to
with
and
Assuming RH, using convexity bound for Dirichlet L-function and Lemma 4, the contribution from the horizontal path is
Thus, we have
Combining (22), (24), (25), (29) and (30), we can get for
with
It implies
Similarly, we can get for
with
and
with the help of (26).
Insert (32) and (33) into (10) and take
and for
Let
where
and
According to (31), we have
Thus, we have
Thus, under RH, we can improve (18) by
Let
where
with
and
Then, under RH, we can improve (21) by
In order to estimate
Noting that
Then Hölder’s inequality indicates
Therefore, we obtain by Lemmas 6 and 7
Taking
5 Mean value in short interval
Appeal to (8), (10) and (11), we can write
Observing that
we have
For the cases where characters are non-principal, denote by
and
For principal character case, denote by
and
Thus, we have
Proposition 3
We have
Proof
Rewrite
Then according to Lemma 2, for
The main term above is
Then we can get easily
The error term in (39) is
□
Proposition 4
We have
Proof
We can write
Lemma 3 implies
Observing that
it is easy to check that
Removing the constraint for
which infers
□
Proposition 5
We have
and
Proof
In order to use Lemma 8, we shall cut the length of the summation over
According to Lemma 8, we get
Then it is easy to check
Similar argument can get the second assertion.□
Combining (38) and Propositions 3–5, we can deduce Theorem 3.
Acknowledgement
Xuanxuan Xiao was supported in part by National Natural Science Foundation of China (11701596) and The Science and Technology Development Fund, Macau SAR (File no. 0095/2018/A3). Wenguang Zhai was supported by National Natural Science Foundation of China (Grant no. 11971476).
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© 2020 Xuanxuan Xiao and Wenguang Zhai, published by De Gruyter
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