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On the Piatetski–Shapiro Prime Number Theorem II

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14 April 2025

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15 April 2025

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Abstract
The author prove that there are infinitely many primes of the form [nc] for $1<c< \frac{919}{775}$. Using the theory of exponent pairs, the author also show that there are infinitely many almost primes of the form [nc] with some larger c.
Keywords: 
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Contents
1. Introduction 1
2. SieveAsymptoticFormulas 3
3. TheFinalDecomposition 4
4. ExponentPairs 7
5. References 8

1. Introduction

The Euler’s conjecture, which states that there are infinitely many primes of the form n 2 + 1 , is one of Landau’s problems on prime numbers. There are several ways to attack this conjecture. One way is to consider the degree of the polynomial. In 1953, Piatetski–Shapiro [1] has proposed to investigate the prime numbers of the form [ n c ] , where c > 1 and [ n c ] denotes the integer part of n c . Clearly [ n c ] can be regarded as "polynomials of degree c". Define
π c ( x ) : = n x : [ n c ] is a prime number ,
then he has shown that π c ( x ) x ( c log x ) 1 holds for any 1 < c < 12 11 1.0909 as x . This range has been improved by many authors, and the best record now is due to Rivat and Sargos [2], where they proved the above asymptotic formula holds for any 1 < c < 2817 2426 1.1612 .
In 1992, Rivat [3] first introduced a sieve method into this problem. He established a lower bound with correct order (instead of an asymptotic formula) with 1 < c < 7 6 1.1616 . After this, many improvements were made and the range of c was enlarged successively to
1 < c < 20 17 1.1765 , 1 < c < 13 11 1.1818 , 1 < c < 45 38 1.1842 ,
1 < c < 243 205 1.18536 and 1 < c < 211 178 1.18539
by Jia [4] (and Baker, Harman and Rivat [5]), Jia [6], Kumchev [7], Rivat and Wu [8] and Li [9] respectively. In this paper, we obtain the following result.
Theorem 1.1.
For sufficiently large x and 1 < c < 919 775 1.1858 , we have π c ( x ) x ( log x ) 1 .
In 1992, Balog and Friedlander [10] considered a hybrid of the Three Primes Theorem and the Piatetski–Shapiro prime number theorem. They proved that every sufficiently large odd integer can be written as the sum of three primes of the form [ n c 0 ] for any fixed 1 < c 0 < 21 20 , and every sufficiently large odd integer can be written as the sum of two normal primes and another prime of the form [ n c 1 ] for any fixed 1 < c 1 < 9 8 . Their result has been improved by many authors. Using the same method as in [11] but with our Theorem 1.1, we can easily deduce the following.
Theorem 1.2.
Every sufficiently large odd integer can be written as the sum of two normal primes and another prime of the form [ n c 1 ] for any fixed 1 < c 1 < 919 775 .
However, if we consider the almost primes instead of primes, the results will be much better. Let P r denotes an integer with at most r prime factors counted with multiplicity. In 2021, Guo [12] proved that there are infinitely many almost primes P r of the form [ n c ] with
1 < c < 889 741 1.1997 , r = 3 , 25882 16071 1.6104 , r = 4 , 3 128 3 ( 8 r 1 ) , r 5 .
In this paper, we shall use the exponent pair processes of Sargos [13,14] together with the traditional processes to produce more efficient exponent pairs and improve the above result when r = 3 , 4 .
Theorem 1.3.
Let
π c , r ( x ) : = n x : [ n c ] = P r .
Then for sufficiently large x and
1 < c < 281563 234507 1.2006 , r = 3 , 51409 31655 1.624 , r = 4 ,
we have π c , r ( x ) x ( log x ) 1 .
Throughout this paper, we always suppose that x is a sufficiently large integer, γ and θ 0 θ 6 are positive numbers which will be fixed later. Let 37 44 < γ < 28 33 and c = 1 γ . The letter p, with or without subscript, is reserved for prime numbers. We define the sets A and B as
A = { m : m = [ n c ] , x n c < 2 x } , B = { n : x n < 2 x } ,
and we put
A d = { a : a d A } , B d = { b : b d A } , P ( z ) = p < z p , S ( A , z ) = a A ( a , P ( z ) ) = 1 1 , S ( B , z ) = b B ( b , P ( z ) ) = 1 1 .
Then we only need to show that S ( A , ( 2 x ) 1 2 ) > 0 . Our aim is to show that the sparser set A contains the expected proportion of primes compared to the bigger set B , which requires us to decompose S ( A , ( 2 x ) 1 2 ) and prove asymptotic formulas of the form
S A , z = ( 1 + o ( 1 ) ) x γ 1 2 γ 1 S B , z
for some parts of it, and drop the other positive parts. The asymptotic formulas will be given in the next section. We define the boolean function as
Boole [ X ] = 1 if X is true , 0 if X is false .

2. Sieve Asymptotic Formulas

In this section we provide some asymptotic formulas for sieve functions. Let ω ( u ) denote the Buchstab function determined by the following differential–difference equation
ω ( u ) = 1 u , 1 u 2 , ( u ω ( u ) ) = ω ( u 1 ) , u 2 .
Following [8] directly, we set γ = 775 919 , θ 0 = 6 γ 5 , θ 1 = 1 γ , θ 2 = 61 γ 49 11 , θ 3 = 3 3 γ , θ 4 = 3 γ 2 , θ 5 = 60 61 γ 11 , θ 6 = γ and let p j = x t j . We define the asymptotic region I as
I ( m , n ) : = θ 1 m < θ 2 or θ 3 m < θ 4 or θ 5 m < θ 6 or θ 1 m + n < θ 2 or θ 3 m + n < θ 4 or θ 5 m + n < θ 6 .
We also define a new region I 2 as
I 2 ( m , n ) : = { m + n < θ 4 , θ 1 m + θ 0 < m + n < θ 2 or θ 3 m + θ 0 < m + n < θ 4 or θ 5 m + θ 0 < m + n < θ 6 or θ 1 n + θ 0 < n + n < θ 2 or θ 3 n + θ 0 < n + n < θ 4 or θ 5 n + θ 0 < n + n < θ 6 or θ 1 m + n + θ 0 < m + n + n < θ 2 or θ 3 m + n + θ 0 < m + n + n < θ 4 or θ 5 m + n + θ 0 < m + n + n < θ 6 } .
Lemma 2.1.
We can give an asymptotic formula for
t 1 t n S A p 1 p n , x θ 0
if we have t 1 + + t n < θ 4 .
Lemma 2.2.
We can give an asymptotic formula for
t 1 t n S A p 1 p n , p n
if we can group ( t 1 , , t n ) into ( m , n ) I .
Lemma 2.3.
We can give an asymptotic formula for
t 1 , t 2 S A p 1 p 2 , p 2
if we have ( t 1 , t 2 ) I 2 .
Proof. 
By Buchstab’s identity, we have
t 1 , t 2 S A p 1 p 2 , p 2 = t 1 , t 2 S A p 1 p 2 , x θ 0 t 1 , t 2 θ 0 t 3 < min t 2 , 1 t 1 t 2 2 S A p 1 p 2 p 3 , p 3 .
We can give asymptotic formulas for the first sum on the right hand side by Lemma 2.1 and the second sum on the right hand side by Lemma 2.2. Thus, Lemma 2.3 is proved. Note that this technique was also used in [15] and [16]. □

3. The Final Decomposition

Before decomposing, we define non–overlapping regions U 1 U 3 as
U 1 ( m , n ) : = ( m , n ) I I 2 , m + 2 n < θ 4 U 2 ( m , n ) : = ( m , n ) I I 2 , m + 2 n θ 4 , 1 m n n < 2 , U 3 ( m , n ) : = ( m , n ) I I 2 , m + 2 n θ 4 , 1 m n n 2 .
We shall apply different techniques to the different regions above. By Buchstab’s identity, we have
S ( A , ( 2 x ) 1 2 ) = S A , x θ 0 θ 0 t 1 < 1 2 S A p 1 , x θ 0 + θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) S A p 1 p 2 , p 2 = S A , x θ 0 θ 0 t 1 < 1 2 S A p 1 , x θ 0 + θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) I I 2 S A p 1 p 2 , p 2 + θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 1 S A p 1 p 2 , p 2 + θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 2 S A p 1 p 2 , p 2 + θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 3 S A p 1 p 2 , p 2 = S 1 S 2 + S I + S U 1 + S U 2 + S U 3 .
By Lemma 2.1 and Lemma 2.2, we can give asymptotic formulas for S 1 , S 2 and S I . For S U 1 , we can use Buchstab’s identity twice more to get
S U 1 = θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 1 S A p 1 p 2 , p 2 = θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 1 S A p 1 p 2 , x θ 0 θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 1 θ 0 t 3 < min t 2 , 1 2 ( 1 t 1 t 2 ) ( t 1 , t 2 , t 3 ) can be partitioned into ( m , n ) I S A p 1 p 2 p 3 , p 3 θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 1 θ 0 t 3 < min t 2 , 1 2 ( 1 t 1 t 2 ) ( t 1 , t 2 , t 3 ) cannot be partitioned into ( m , n ) I S A p 1 p 2 p 3 , x θ 0 + θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 1 θ 0 t 3 < min t 2 , 1 2 ( 1 t 1 t 2 ) ( t 1 , t 2 , t 3 ) cannot be partitioned into ( m , n ) I θ 0 t 4 < min t 3 , 1 2 ( 1 t 1 t 2 t 3 ) ( t 1 , t 2 , t 3 , t 4 ) can be partitioned into ( m , n ) I S A p 1 p 2 p 3 p 4 , p 4 + θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 1 θ 0 t 3 < min t 2 , 1 2 ( 1 t 1 t 2 ) ( t 1 , t 2 , t 3 ) cannot be partitioned into ( m , n ) I θ 0 t 4 < min t 3 , 1 2 ( 1 t 1 t 2 t 3 ) ( t 1 , t 2 , t 3 , t 4 ) cannot be partitioned into ( m , n ) I S A p 1 p 2 p 3 p 4 , p 4 = S U 11 S U 12 S U 13 + S U 14 + S U 15 .
We can give asymptotic formulas for S U 11 S U 14 . For S U 15 we can perform Buchstab’s identity more times to make savings, but we choose to discard all of it for the sake of simplicity. Combining the above cases, we get a loss from S U 1 of
θ 0 1 2 θ 0 min t 1 , 1 t 1 2 θ 0 min t 2 , 1 t 1 t 2 2 θ 0 min t 3 , 1 t 1 t 2 t 3 2 Boole [ ( t 1 , t 2 , t 3 , t 4 ) U 15 ] ω 1 t 1 t 2 t 3 t 4 t 4 t 1 t 2 t 3 t 4 2 d t 4 d t 3 d t 2 d t 1 < 0.001993 ,
where
U 15 ( t 1 , t 2 , t 3 , t 4 ) : = ( t 1 , t 2 ) U 1 , θ 0 t 3 < min t 2 , 1 2 ( 1 t 1 t 2 ) , ( t 1 , t 2 , t 3 ) cannot be partitioned into ( m , n ) I , θ 0 t 4 < min t 3 , 1 2 ( 1 t 1 t 2 t 3 ) , ( t 1 , t 2 , t 3 , t 4 ) cannot be partitioned into ( m , n ) I .
For S U 2 , we cannot decompose further but have to discard the whole region giving the loss
θ 0 1 2 θ 0 min t 1 , 1 t 1 2 Boole [ ( t 1 , t 2 ) U 2 ] ω 1 t 1 t 2 t 2 t 1 t 2 2 d t 2 d t 1 < 0.421388 .
For S U 3 we cannot use Buchstab’s identity in a straightforward manner, but we can use Buchstab’s identity in reverse to make almost–primes visible. The details of using Buchstab’s identity in reverse are similar to those in [16] and [17]. By using Buchstab’s identity in reverse twice, we have
S U 3 = θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 3 S A p 1 p 2 , p 2 = θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 3 S A p 1 p 2 , 2 x p 1 p 2 1 2 + θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 3 t 2 < t 3 < 1 2 ( 1 t 1 t 2 ) S A p 1 p 2 p 3 , p 3 = θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 3 S A p 1 p 2 , 2 x p 1 p 2 1 2 + θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 3 t 2 < t 3 < 1 2 ( 1 t 1 t 2 ) ( t 1 , t 2 , t 3 ) can be partitioned into ( m , n ) I S A p 1 p 2 p 3 , p 3 + θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 3 t 2 < t 3 < 1 2 ( 1 t 1 t 2 ) ( t 1 , t 2 , t 3 ) cannot be partitioned into ( m , n ) I S A p 1 p 2 p 3 , 2 x p 1 p 2 p 3 1 2 + θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 3 t 2 < t 3 < 1 2 ( 1 t 1 t 2 ) ( t 1 , t 2 , t 3 ) cannot be partitioned into ( m , n ) I t 3 < t 4 < 1 2 ( 1 t 1 t 2 t 3 ) ( t 1 , t 2 , t 3 , t 4 ) can be partitioned into ( m , n ) I S A p 1 p 2 p 3 p 4 , p 4 + θ 0 t 1 < 1 2 θ 0 t 2 < min t 1 , 1 2 ( 1 t 1 ) ( t 1 , t 2 ) U 3 t 2 < t 3 < 1 2 ( 1 t 1 t 2 ) ( t 1 , t 2 , t 3 ) cannot be partitioned into ( m , n ) I t 3 < t 4 < 1 2 ( 1 t 1 t 2 t 3 ) ( t 1 , t 2 , t 3 , t 4 ) cannot be partitioned into ( m , n ) I S A p 1 p 2 p 3 p 4 , p 4 = S U 31 + S U 32 + S U 33 + S U 34 + S U 35 .
We can give asymptotic formulas for S U 32 and S U 34 , hence we can subtract them from the loss. In this way we obtain a loss from S U 3 of
θ 0 1 2 θ 0 min t 1 , 1 t 1 2 Boole [ ( t 1 , t 2 ) U 3 ] ω 1 t 1 t 2 t 2 t 1 t 2 2 d t 2 d t 1 θ 0 1 2 θ 0 min t 1 , 1 t 1 2 t 2 1 t 1 t 2 2 Boole [ ( t 1 , t 2 , t 3 ) U 32 ] ω 1 t 1 t 2 t 3 t 3 t 1 t 2 t 3 2 d t 3 d t 2 d t 1 θ 0 1 2 θ 0 min t 1 , 1 t 1 2 t 2 1 t 1 t 2 2 t 3 1 t 1 t 2 t 3 2 Boole [ ( t 1 , t 2 , t 3 , t 4 ) U 34 ] ω 1 t 1 t 2 t 3 t 4 t 4 t 1 t 2 t 3 t 4 2 d t 4 d t 3 d t 2 d t 1 < ( 0.954145 0.363595 0.019119 ) = 0.571431 ,
where
U 32 ( t 1 , t 2 , t 3 ) : = ( t 1 , t 2 ) U 3 , t 2 < t 3 < 1 2 ( 1 t 1 t 2 ) , ( t 1 , t 2 , t 3 ) can be partitioned into ( m , n ) I , U 34 ( t 1 , t 2 , t 3 , t 4 ) : = ( t 1 , t 2 ) U 3 , t 2 < t 3 < 1 2 ( 1 t 1 t 2 ) , ( t 1 , t 2 , t 3 ) cannot be partitioned into ( m , n ) I , t 3 < t 4 < 1 2 ( 1 t 1 t 2 t 3 ) , ( t 1 , t 2 , t 3 , t 4 ) can be partitioned into ( m , n ) I .
Finally, by (3)–(8), the total loss is less than
0.001993 + 0.421388 + 0.571431 < 0.995 < 1
and the proof of Theorem 1.1 is completed.

4. Exponent Pairs

In this section we shall give a proof of Theorem 1.3. Using the same arguments as in [12], we only need to find an exponent pair ( k , l ) to give an upper bound for c. The corresponding upper bounds when r = 3 , 4 are
c < 70 117 l 117 k + 1 when t = 3 and c < 362 487 l 487 k + 1 when t = 4 .
For the definition of exponent pairs, one can see [18], Definition 11]. We know that if ( k , l ) is an exponent pair, then both
A ( k , l ) = k 2 k + 2 , l 2 k + 2 + 1 2
and
B ( k , l ) = l 1 2 , k + 1 2
are exponent pairs.
Sargos [13,14] mentioned two different transformation processes. If ( k , l ) is an exponent pair, then both
C ( k , l ) = k 12 ( 1 + 4 k ) , 11 ( 1 + 4 k ) + l 12 ( 1 + 4 k )
and
D ( k , l ) = 5 k + l + 2 8 ( 5 k + 3 l + 2 ) , 29 k + 21 l + 10 8 ( 5 k + 3 l + 2 )
are also exponent pairs. These four processes can also be seen in [18]. Note that in [12] only processes A and B are used.
Now we shall complete our proof of Theorem 1.3. For r = 3 we take the exponent pair
B C D 13 84 , 55 84 = 6013 13008 , 137 271 ,
and for r = 4 we take the exponent pair
B D 13 84 , 55 84 = 195 796 , 235 398 .
By (9) and the arguments in [12], the proof of Theorem 1.3 is complete.

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