Sparse sets that satisfy the Prime Number Theorem

For arbitrary real t > 1 we examine the set {


Introduction
Let x and n be strictly positive integers throughout.The prime number theorem states that π(x) ∼ x/ log x, where π(x) is the number of primes less than or equal to x.We can apply the prime number theorem to sets in the following way: Let S f,x := {f (n) : n ≤ x}, where we restrict S f,x to those sets where f is positive integer valued and not piecewise defined.We will count the number of primes in these sets using π (S f,x ) := |{s ∈ S f,x : s is prime}| .
Using this notation the prime number theorem can be restated as where | * | represents set cardinality and id is the identity function.We say a set S f,x satisfies the prime number theorem if For any real number m, denote by ⌊m⌋ the floor of (or more formally the largest integer not exceeding) m.The floor function appears prominently in Beatty sequences (see, for example, [1,3,6,10,13]) and Piatestski-Shapiro sequences (see, for example, [2,4,5,7,17,22]), both which we will consider in this paper.
Recently there has been interest in the floor function set S ⌊ x n ⌋,x .That is, the set x n : n ≤ x .
From [14] we have From [15] we have x .
An asymptotic formula with error term is also now known (see [20]) as follows: where c > 0 is a positive constant.In [20], and in correspondence with the authors, Wu and Ma made the astute observation that S ⌊ x n ⌋ ,x is a sparse set that satisfies the prime number theorem.That is, .
The authors are not aware of literature that focuses on sets that satisfy the prime number theorem.As prime numbers are an essential component of many cryptography protocols, it is conceivable that sets that satisfy the prime number theorem (particularly those less dense) may have applications.
To explore this further, define the density of a set S f,x , if it exists, as We say a function f (or its set ) is said to be sparser than a function g (or its set As mentioned above, the set S id,x satisfies the prime number theorem.But it is not sparse.In contrast, as implied from above, D ⌊x/n⌋,x ∼ 2x −1/2 and is an example of a sparse set that satisfies the prime number theorem.This paper investigates other sparse sets that satisfy the prime number theorem. Using [12] and [24], we briefly review 2 families of sets based on wellknown sequences involving the floor function.For fixed real numbers α, β the Beatty sequence is the sequence (⌊αn + β⌋) ∞ n=1 .It is known that there are infinitely many primes in the Beatty sequence if α ≥ 1.For our purposes we consider β to be positive.The Piatetski-Shapiro sequence is the sequence It is known that the sequence contains infinitely many primes for all c ∈ 1, 243 205 .Families of sets based on these sequences satisfy or do not satisfy the prime number theorem as follows: Corollary 1.With the constraints mentioned above, the family of sets S ⌊n c ⌋,x do not satisfy the prime number theorem.In contrast, the family of sets S ⌊αn+b⌋,x do satisfy the prime number theorem.Moreover, D ⌊αx+β⌋,x ∼ 1 α .While S ⌊x/n⌋,x and S ⌊αn+b⌋,x both satisfy the prime number theorem, the set S ⌊x/n⌋,x is sparser than any given set of the form S ⌊αn+b⌋,x .Our main result is as follows: Theorem 1.Let t > 1, x e be real numbers satisfying (6).Then, for all .
The error terms in the asymptotic formulas reduce with increasing N, as illustrated below.
When considered with the cardinality of S ⌊x/n⌋,x (See Lemma 1), Theorem 1 implies that the set S ⌊ x n t ⌋,x satisfies the prime number theorem.Noting that D ⌊ x n t ⌋,x x , we conclude that, for large enough t, the set S ⌊ x n t ⌋,x will be sparser than any set based on Beatty sequences.
Recall that throughout the function f in S f,x is never piecewise defined.We conjecture as follows: Conjecture 1.The family of sets S ⌊ x n t ⌋,x are the sparsest sets that satisfy the prime number theorem.
Throughout p indicates a prime number and N = {1, 2, , . ..}.The statements f (x) = O (g(x)) and f (x) ≪ g(x) denote that there exists a constant c > 0 such that |f (x)| ≤ c|g(x)| for large enough x.The statement f (x) ∼ g(x) is the assertion that lim x→∞ f (x)/g(x) equals 1.Also, Λ(n) denotes the von Mangoldt function and µ(n) denotes the Möbius function.We use ψ(x) to denote x − ⌊x⌋ − 1/2.We denote with e(x) the expression e 2πix .Finally, for all arithmetic functions f and g, the Dirichlet convolution f * g of f and g is given by

Organisation of this Paper
The proof of the main theorem (Theorem 1) begins in Subsection 4.2 where we give the split The sums S 1 (x, t) and S 3 (x, t) are estimated in Subsection 4.4 and Subsection 4.6 respectively.

Preparatory Lemmas
The following is a generalisation of [14,Thm 2].
Proof.The elements of S ⌊x/n t ⌋,x can be read off the graph f (n) = xn −t .Note that f is strictly decreasing.Let a be such that f ′ (a) = −1.Calculating the derivative and using basic algebra we see that a = (tx) 1/(t+1) .
First consider the small values of S ⌊x/n t ⌋,x .As the slope of f (n) is greater than -1 for all integers n > a, it follows that Therefore 0, 1, . . ., ⌊x/a t ⌋ ∈ S ⌊x/n t ⌋,x , which implies that the number of elements of S ⌊x/n t ⌋,x less than or equal to x/a t is We now consider the large elements of S ⌊x/n t ⌋,x .That is, elements of S t (x) greater than or equal to x/a t .Since the slope of f (n) is less than -1 for n < a it follows that each n ∈ {1, 2, . . ., (tx) 1/(t+1) } generates a unique element of S ⌊x/n t ⌋,x .Therefore the number of elements of S ⌊x/n t ⌋,x greater than or equal to x/a t is (tx) 1/(t+1) + O(1). (4) Adding ( 3) and (4), we have The following lemma will also be needed.The proof is rather long, so it will be postponed to Section 5. Lemma 2. Let t 1.For all x 2t t and ǫ > 0, we have 4 Proof of Theorem 1
where it is known [11] that we may take c ≈ 0.209 8. See Lemma 6.
4.2 Splitting π S ⌊ x n t ⌋ ,x into subsums The proof estimates This sum is split into three subsums as follows: As often in this situation, the first subsum yields the main term plus an acceptable error term, and the third one can be estimated trivially.The difficult part comes from the second sum, for which a new estimate on a sum over primes is needed.This latter sum is classically derived using the very useful Vaughan's identity, for which we provide here a slightly different version.See Lemma 9.

Basic tools
In this subsection we give the Lemmas used for the estimation of t,1 (x), used in turn to estimate S 2 (x, t).
We will make frequent use of the following simple bound.
Lemma 3. Let j ∈ Z 0 , t > 1 and x e such that x t 2t .Then Proof.The left-hand side is equal to , and the result follows using the inequality (1 − X) −1 2 whenever X 1 2 .A more accurate estimate can be proved if we used the series expansion of (1 − X) −K .Lemma 4. Let j ∈ Z 0 , t > 1 and x e such that x t 2t .Then, for all N, j ∈ Z 0 Proof.As above, the lhs is written as , and we conclude the proof using the series expansion (1 Proof.Follows at once from [23, Lemma 3.2].
Lemma 6.Let 1 < β β 0 be fixed.Then, for all N ∈ Z 0 and all x > max e The error term depends only on N.
Proof.By partial summation and the prime number theorem, there exits by successive integrations by part.

The sum S 1 (x, t)
By the prime number theorem, there exists c > 0 such that so that, using the asymptotic expansion of Li, we derive for all N ∈ Z 0 , Now using Lemma 3 for the error term and Lemma 4 for the main term yields provided that x t 2t .

The sum S 2 (x, t)
The sum S 2 (x, t) is split into 2 subsums.The subsum Σ t,1 (x) is estimated in subsubsection 4.5.1.The estimation of subsum Σ t,2 (x) is the result of Lemma 2 and is proven in Section 5.
We first write We have For x max (t 2t , 5(5t) t ), using Lemma 5 with β = 1 + 1 t in the 1st sum of the error term and β = 2 + 1 t in the 2nd sum of the error term, and also using Lemma 3, the error term does not exceed For x max t 2t 2 , t t e 818(t+1)(N +2) 2 , using Lemma 6 with β 0 = 2 and β = 1 + 1 t , the main term is where we used Lemma 4 and Lemma 3 in the last line.Note that the condition (6) implies that the error term in ( 9) is absorbed by the error term in (10).Hence, for all N ∈ Z 0 and all x satisfying (6), we obtain Using Lemma 2 we derive at once Σ t,2 (x) ≪ x for all x 2t t and all ε > 0 small.Note that if x satisfies (6) and 0 < ε < 1 500(t+1) , then the bound ( 12) is absorbed by the error term in (11) above.
By (11) and (12) we immediately derive the next estimate.Assuming x, t > 1 satisfying (6), we have 4.6 The sum S 3 (x, t) which is absorbed in the error term of (13) if x, t satisfy (6).

Proof of Lemma 2
This section estimates the sum t,2 (x).In subsection 5.1 we combine the Dirichlet hyperbola principle with Vaughan's identity for the Möbius function to equate a von Mangoldt function twisted exponential sum with 4 triple exponential sums and a double exponential sum.These five sums are estimated in Subsection 5.2.Finally, in subsection 5.3, we use the previous lemma with specified exponential to estimate t,2 (x).

Technical lemmas
The first tool is a direct application of the celebrated Dirichlet hyperbola principle.See [8,Corollary 3.4].
with a m ≪ m ǫ and b m ≪ m ǫ .
Proof.Lemma 7 with f = µ and g = log yields R<n R 1 and we treat the inner sum of the 2nd sum with Lemma 8.The assumption Finally, we will make use of the following estimate which is currently one of the best for estimating triple exponential sums of type II.See [25].

Exponential sums over primes
The aim of this section is the proof of the next estimate.
and the exponent pair (k, ℓ) yields Finally, using 1 V U 1/2 , the full sum S does not exceed where we also used the fact that the condition R c 0 x 6 6α+1 implies that . This inequality implying that (R 3 V 2 ) 1/4 RV −1/4 , and therefore and we conclude the proof noticing that the term x −1/2 R 1+α/2 is absorbed by R 11/12 since R c 0 x 6 6α+1 .

Proof of Lemma 2
We will use Lemma 11 with van der Corput's exponent pair (k, ℓ) = 1 2 , 1 2 , which yields provided that ( 15) is fullfiled with R replaced by N and where T := xN −α .Set S(x) := t,2 (x).Then and recall that t 1.First, an usual splitting argument yields Next, for all integers (t −t x) Since δ 0, we see that the function ϕ : u → t x u − t x u + δ is decreasing, and satisfies |ϕ(u)| t −1 δx 1/t N −1−1/t for all u ∈ [N, 2N], so that we derive by Abel summation Now using ( 16) with x replaced by hx 1/t and α = 1 t we derive  and the result follows by substituting the estimate (20) into (18) with δ = 0 and δ = 1 respectively, and then in (17), yielding (5) as required.

Proof of Corollary 1
For the sets based on the Piatestski-Shapiro sequences we infer from [12] that π(S ⌊n c ⌋,x ) ∼ (x c ) 1/c log x c = x c log x .