An elementary formula to know the number of primes in the interval ( x, 2 x) close to the exact figure for a fixed x is given here. A new elementary equation is derived (a relation between prime numbers and composite numbers distributed in the interval [1, 2 x]). An elementary method to know the number of primes in a given magnitude is suitably placed in the form of a general formula, and we have proved it. The general formula is applied to the terms of the equation, and a tactical simplification of the terms gives rise to an expressio n whose verification envisages scope for its further studies.
There are many questions concerning primes which are sophisticated. One of them is knowing how many primes are there in the interval ( x ,2 x ) for a fixed x. Joseph Bertrand, in 1845, conjectured that there is at least one prime p in the interval ( x ,2 x ) after checking this numerically up to x = 300000 in ref. ( [
Generally, to know π ( 2 x ) − π ( x ) , it is done by computing each term by means of a sieve method which involves more labor. The other option is to have a general formula for evaluation & simplify it to get an expression to provide an estimated value.
Earnest Meissel, in 1870-85, published a series of articles where he was able to count exactly the number of primes up to a given magnitude x in ref. [
The ultimate goal of this research is to show that there exists an expression that even shows that there are more than 60,000 primes in the interval [ 10 6 ,2 × 10 6 ] . We can also call it an elementary formula of primes in the interval ( x ,2 x ) for a fixed x. The idea behind this investigation is the identified relationship between composite numbers and prime numbers distributed in the interval [ 1,2 x ] .
On the basis of the computational technique of Legendre, E. Meissel, and D H Lehmer for composite numbers in ref. [
The number of primes in the interval ( x ,2 x ) for a fixed x is
π ( x , 2 x ) > π ( x ) − μ ( μ − 1 ) 2 , where μ = π ( x ) ,
a still better approximation,
π ( x , 2 x ) > π ( x ) − ( μ ′ ( μ ′ − 1 ) 2 − μ ( μ − 1 ) 2 ) where μ ′ = π [ 2 x ] .
We call these results as an expression for π ( 2 x ) − π ( x ) .
We make readers familiarize themselves with the notations and symbols used here.
● We define a new symbol “C” where ℂ ( x ) stands for the number of composite numbers not exceeding a given positive integer x, e.g. ℂ ( 10 ) = 5 , they are 4, 6, 8, 9, 10.
● π ( x ) denote the number of primes not exceeding a fixed integer x, e.g. π ( 10 ) = 4 , they are 2, 3, 5, 7.
● π ( x ,2 x ) denote the number of primes in the interval x to 2x, i.e. π ( 10,20 ) = 4 , they are 11, 13, 17, 19; π ( 11,22 ) = 3 , they are 13, 17, 19, here “11” is not included.
● Integer means positive integer, c runs for composite number, and p runs for prime number throughout.
We obtain a relation between prime numbers and composite numbers distributed in the interval [ 1,2 x ] . This identified relation was the basis for obtaining an expression for π ( 2 x ) − π ( x ) . We have by computational technique of Legendre in ref. [
π ( x ) = ( x − 1 ) − ℂ ( x ) (1)
Substitute x = 2 x , we get
π ( 2 x ) = ( 2 x − 1 ) − ℂ ( 2 x ) (2)
Multiply Equation (1) by 2, we get
2 π ( x ) = 2 x − 2 − 2 ℂ ( x ) (3)
(2)-(3) gives
π ( 2 x ) − 2 π ( x ) = ( 2 x − 1 ) − ℂ ( 2 x ) − 2 x + 2 + 2 ℂ ( x ) = − ℂ ( 2 x ) + 2 ℂ ( x ) + 1 (4)
On rearraging the terms of (4), we get
π ( 2 x ) − π ( x ) = π ( x ) − ( ℂ ( 2 x ) − 2 ℂ ( x ) ) + 1 (5)
π ( 2 x ) − π ( x ) = π ( x ) − a ( x ) + 1
π ( x ,2 x ) = π ( x ) − a ( x ) + 1 (6)
where a ( x ) = ℂ ( 2 x ) − 2 ℂ ( x ) , we call Equation (5) as our basic tool.
We revert back to the notations, symbols, and definitions used here.
● We define a new symbol π ′ where π ′ ( x ) denote number of primes strictly less than x, e.g. π ′ ( 5 ) = 2 , they are 2,3, here “5” is not included.
● p 1 , p 2 , p 3 , p 4 , ⋯ denotes primes 2,3,5,7, ⋯ , x denote a fixed integer.
● m denote a positive integer < x.
● [ x p ] stands for an integral part of x when x is divided by a “p”, e.g. 10 3 = 3.33 , here [ 10 3 ] = 3 , similarly for [ x m ] .
● p r 1 is the largest prime in [ x p 1 ] means when x is divided by p 1 , p r 1 is the largest prime less than [ x p 1 ] .
Definition 1. We define an integer m = p 1 a ⋅ p 2 b ⋅ p 3 c ⋯ p r l , where a , b , c , ⋯ , l ≥ 0 , and p r is the largest prime |m.
Keeping in mind the computational techniques of ref. [
Lemma 1. The number of composite numbers not exceeding a given integer x is computed as given below.
ℂ ( x ) = ∑ 2 ≤ m < x π [ x m ] − π ′ [ p r ] > 0 ( π [ x m ] − π ′ [ p r ] ) (7)
an elaborated form of Lemma (1), expressed as in (7), is given below.
ℂ ( x ) = ∑ a ≥ 1 ( π [ x p a ] − π ′ [ p ] ) > 0 ∑ 2 ≤ p ≤ x ( π [ x p ] − π ′ [ p ] ) > 0 ( π [ x p a ] − π ′ [ p ] ) step I
+ ∑ a , b ≥ 1 ( π [ x p a ⋅ q b ] − π ′ [ q ] ) > 0 ∑ 6 ≤ p ⋅ q < x q > p ( π [ x p ⋅ q ] − π ′ [ q ] ) > 0 ( π [ x p a ⋅ q b ] − π ′ [ q ] ) step II
Similarly for m = p a ⋅ q b ⋅ r c and so on.
We write a lemma necessary in proving the lemma 1. This lemma is based on the standard pattern of the fundamental theorem of arithmetic as in ref. ( [
Lemma 2. Every c ≤ x can be expressed as
∑ 1 < c ≤ x 1 = ℂ ( x ) = { { p 1 ⋅ p 1 , p 1 ⋅ p 2 , p 1 ⋅ p 3 , p 1 ⋅ p 4 , ⋯ , p 1 ⋅ p r 1 } , { p 2 ⋅ p 2 , p 2 ⋅ p 3 , p 2 ⋅ p 4 , p 2 ⋅ p 5 , ⋯ , p 2 ⋅ p r 2 } , { p 3 ⋅ p 3 , p 3 ⋅ p 4 , p 3 ⋅ p 5 , p 3 ⋅ p 6 , ⋯ , p 3 ⋅ p r 3 } , ⋯ , { p 1 2 ⋅ p 1 , p 1 2 ⋅ p 2 , p 1 2 ⋅ p 3 , ⋯ } , { p 2 2 ⋅ p 2 , p 2 2 ⋅ p 3 , p 2 2 ⋅ p 4 , ⋯ } , { p 3 2 ⋅ p 3 , p 3 2 ⋅ p 4 , p 3 2 ⋅ p 5 , ⋯ } , { p 1 ⋅ p 2 2 , p 1 ⋅ p 2 ⋅ p 3 , p 1 ⋅ p 2 ⋅ p 4 , ⋯ } , { p 2 ⋅ p 3 2 , p 2 ⋅ p 3 ⋅ p 4 , p 2 ⋅ p 3 ⋅ p 5 , ⋯ } , ⋯ , { p 1 ⋅ p 2 ⋅ p 3 2 , p 1 ⋅ p 2 ⋅ p 3 ⋅ p 4 , p 1 ⋅ p 2 ⋅ p 3 ⋅ p 5 , ⋯ } , ⋯ , { ∀ p 1 a ⋅ p 2 b ⋅ p 3 c ⋯ ≤ x ; a , b , c , ⋯ ≥ 0 } }
Proof. Let m < x as per definition in (1). Let p r 1 is the largest prime in [ x p 1 ] , p r 2 is the largest prime in [ x p 2 ] , ⋯ . Let us obtain the pattern as in the lemma
(2) ∀ c ≤ x . We start with m = p 1 , ⋯ , and continue for m = p 1 2 , ⋯ , and so on ∀ m ≤ x , conditionally. We write
π [ x p 1 ] = { p 1 ⋅ p 1 , p 1 ⋅ p 2 , p 1 ⋅ p 3 , p 1 ⋅ p 4 , ⋯ , p 1 ⋅ p r 1 } (as required in the lemma 2)
π [ x p 2 ] = { p 2 ⋅ p 1 , p 2 ⋅ p 2 , p 2 ⋅ p 3 , p 2 ⋅ p 4 , p 2 ⋅ p 5 , ⋯ , p 2 ⋅ p r 2 } (not as required in lemma 2)
∴ { p 2 ⋅ p 2 , p 2 ⋅ p 3 , p 2 ⋅ p 4 , p 2 ⋅ p 5 , ⋯ , p 2 . p r 2 }
= { p 2 ⋅ p 1 , p 2 ⋅ p 2 , p 2 ⋅ p 3 , p 2 ⋅ p 4 , p 2 ⋅ p 5 , ⋯ , p 2 . p r 2 }
− { p 2 ⋅ p 1 } (to make it as required in lemma 2)
= π [ x p 2 ] − ∑ p 2 ⋅ p 1 < x 1
here in the second term, instead of counting p 2 ⋅ p 1 < x , only p 1 , strictly less than p 2 is counted to reflect the counting of p 2 ⋅ p 1 , similarly in further steps, we write
= π [ x p 2 ] − ∑ p 1 < p 2 1
= π [ x p 2 ] − π ′ [ p 2 ]
… m = p j ≤ x ; such that ( π [ x p j ] − π ′ [ p j ] ) > 0 and ( π [ x p j ] − π ′ [ p j + 1 ] ) ≤ 0 because, here p j ⋅ p j ≤ x but p j + 1 ⋅ p j > x ; similarly in further steps.
Next π [ x p 1 2 ] = { p 1 2 ⋅ p 1 , p 1 2 ⋅ p 2 , p 1 2 ⋅ p 3 , p 1 2 ⋅ p 4 , ⋯ } (as required in lemma 2).
Similarly { p 2 2 ⋅ p 2 , p 2 2 ⋅ p 3 , p 2 2 ⋅ p 4 , ⋯ } = { p 2 2 ⋅ p 1 , p 2 2 ⋅ p 2 , p 2 2 ⋅ p 4 , ⋯ } − { p 2 2 ⋅ p 1 } = π [ x p 2 2 ] − ∑ p 1 ⋅ p 2 2 < x 1 = π [ x p 2 2 ] − ∑ p 1 < p 2 1 = π [ x p 2 2 ] − π ′ [ p 2 ]
… ∀ m = p a , a > 1 , such that ( π [ x m ] − π ′ [ p ] ) > 0 where p is the largest prime |m.
Similarly π [ x p 1 ⋅ p 2 ] = { p 1 2 ⋅ p 2 , p 1 ⋅ p 2 2 , p 1 ⋅ p 2 ⋅ p 3 , p 1 ⋅ p 2 ⋅ p 4 , p 1 ⋅ p 2 ⋅ p 5 , ⋯ }
& { p 1 ⋅ p 2 2 , p 1 ⋅ p 2 ⋅ p 3 , p 1 ⋅ p 2 ⋅ p 4 , ⋯ } = { p 1 2 ⋅ p 2 , p 1 ⋅ p 2 2 , p 1 ⋅ p 2 ⋅ p 3 , p 1 ⋅ p 2 ⋅ p 4 , p 1 ⋅ p 2 ⋅ p 5 , ⋯ } − { p 1 2 ⋅ p 2 } = π [ x p 1 ⋅ p 2 ] − ∑ p 1 2 ⋅ p 2 < x 1 = π [ x p 1 ⋅ p 2 ] − ∑ p 1 < p 2 1 = π [ x p 1 ⋅ p 2 ] − π ′ [ p 2 ] ; p 2 , largest p | p 1 ⋅ p 2
… ∀ m = p a ⋅ q b ; such that ( π [ x m ] − π ′ [ q ] ) > 0 where q is the largest prime |m.
& { p 1 ⋅ p 2 ⋅ p 3 2 , p 1 ⋅ p 2 ⋅ p 3 ⋅ p 4 , p 1 ⋅ p 2 ⋅ p 3 ⋅ p 5 , ⋯ } = { p 1 2 ⋅ p 2 ⋅ p 3 , p 1 ⋅ p 2 2 ⋅ p 3 , p 1 ⋅ p 2 ⋅ p 3 2 , p 1 ⋅ p 2 ⋅ p 3 ⋅ p 4 , ⋯ } − { p 1 2 ⋅ p 2 ⋅ p 3 , p 1 ⋅ p 2 2 ⋅ p 3 } = π [ x p 1 ⋅ p 2 ⋅ p 3 ] − ∑ p 1 2 ⋅ p 2 ⋅ p 3 , p 1 ⋅ p 2 2 ⋅ p 3 < x 1 = π [ x p 1 ⋅ p 2 ⋅ p 3 ] − ∑ p 1 , p 2 < p 3 1 = π [ x p 1 ⋅ p 2 ⋅ p 3 ] − π ′ [ p 3 ] ;
where p 3 is the largest p | p 1 ⋅ p 2 ⋅ p 3 .
… ∀ m = p a ⋅ q b ⋅ r c ⋯ ; such that ( π [ x m ] − π ′ [ p r ] ) > 0 , where p r is the largest prime |m. Now, we ascertain that the entire pattern of lemma 2 is obtained. The condition ( π [ x m ] − π ′ [ p r ] ) > 0 , is by determination of the computational technique. This completes the proof of lemma 1. □
We illustrate the lemma 1 by taking a trivial example for example No “100”, in the way it is written in the elaborated form. Following are the steps how to proceed.
● The fixed number (say 100 here) is to be divided by m < x and find out the integral part.
● Choose all m = p ≤ 100 for which the condition ( π [ x p ] − π ′ [ p ] ) > 0 , satisfies.
● Choose all m = p ⋅ q , product of two primes, say 2 ⋅ 2,2 ⋅ 3,2 ⋅ 5 , ⋯ for which the condition ( π [ x p ⋅ q ] − π ′ [ q ] ) > 0 , q > p , satisfies.
● Choose all m = p ⋅ q ⋅ r , product of three primes, say 2 ⋅ 2 ⋅ 2,2 2 ⋅ 3,2 ⋅ 3 2 , ⋯ for which the condition ( π [ x p ⋅ q ⋅ r ] − π ′ [ r ] ) > 0 , r > q > p , satisfies.
All possibilities of m = p a ⋅ q b ⋅ r c ⋯ , for all combinations of p , q , r , ⋯ , a , b , c , ⋯ are to be covered and care is taken that no non-trivial m is left out, ultimately, these are m < x for which the condition ( π [ x m ] − π ′ [ p r ] ) > 0 , satisfies where p r is the largest prime |m.
Example. For x = 100 , to work out for π ( 100 ) , from (1), we have
π ( 100 ) = 100 − 1 − ℂ ( 100 ) (8)
Computation of C(100): For Step I, m = p a : a = 1 , m = p ≤ 100 = { 2,3,5,7 } ; not for m = 11 because π [ 100 11 ] − π ′ [ 11 ] ≤ 0 ; for a = 2,3 , ⋯ , m = { 2 2 ,3 2 ,2 3 ,3 3 ,2 4 ,2 5 } ; not for m = 2 6 because ( π [ 100 2 6 ] − π ′ [ 2 ] ) ≤ 0 , similarly for m = 3 4 and so on; we proceed for doing calculations, we have
+ π [ 100 2 ] − π ′ [ 2 ] + π [ 100 3 ] − π ′ [ 3 ] + π [ 100 5 ] − π ′ [ 5 ] + π [ 100 7 ] − π ′ [ 7 ] = π [ 50 ] − 0 + π [ 33 ] − 1 + π [ 20 ] − 2 + π [ 14 ] − 3 = 15 − 0 + 11 − 1 + 8 − 2 + 6 − 3 = 15 + 10 + 6 + 3 = 34 (9)
+ π [ 100 2 2 ] − π ′ [ 2 ] + π [ 100 3 2 ] − π ′ [ 3 ] = π [ 25 ] − 0 + π [ 11 ] − 1 = 9 − 0 + 4 − 1 = 9 + 4 = 13 (10)
+ π [ 100 2 3 ] − π ′ [ 2 ] + π [ 100 3 3 ] − π ′ [ 3 ] = π [ 12 ] − 0 + π [ 3 ] − 1 = 5 − 0 + 2 − 1 = 5 + 1 = 6 (11)
+ π [ 100 2 4 ] − π ′ [ 2 ] + π [ 100 2 5 ] − π ′ [ 2 ] = π [ 6 ] − 0 + π [ 3 ] − 1 = 3 − 0 + 2 − 0 = 5 (12)
For Step II: m = p a ⋅ q b = { 2 ⋅ 3 , 2 ⋅ 5 , 2 ⋅ 7 , 2 2 ⋅ 3 , 2 2 ⋅ 5 , 2 3 ⋅ 3 , 2 ⋅ 3 2 , 3 ⋅ 5 } ; not for m = 2 ⋅ 11 because ( π [ 100 2 ⋅ 11 ] − π ′ [ 11 ] ) ≤ 0 ; similarly for m = 2 2 ⋅ 7 , 2 3 ⋅ 5 , 3 ⋅ 7 , 2 ⋅ 5 2 and so on; we proceed for doing calculations, we have
+ π [ 100 2 ⋅ 3 ] − π ′ [ 3 ] + π [ 100 2 ⋅ 5 ] − π ′ [ 5 ] + π [ 100 2 ⋅ 7 ] − π ′ [ 7 ] = π [ 16 ] − 1 + π [ 10 ] − 2 + π [ 7 ] − 3 = 6 − 1 + 4 − 2 + 4 − 3 = 5 + 2 + 1 = 8 (13)
+ π [ 100 2 2 ⋅ 3 ] − π ′ [ 3 ] + π [ 100 2 2 ⋅ 5 ] − π ′ [ 5 ] + π [ 100 2 3 ⋅ 3 ] − π ′ [ 3 ] = π [ 8 ] − 1 + π [ 5 ] − 2 + π [ 4 ] − 1 = 4 − 1 + 3 − 2 + 2 − 1 = 3 + 1 + 1 = 5 (14)
+ π [ 100 2 ⋅ 3 2 ] − π ′ [ 3 ] + π [ 100 3 ⋅ 5 ] − π ′ [ 5 ] = π [ 5 ] − 1 + π [ 6 ] − 2 = 3 − 1 + 3 − 2 = 2 + 1 = 3. (15)
No Step III: for m = 2 ⋅ 3 ⋅ 5 , because ( π [ 100 2 ⋅ 3 ⋅ 5 ] − π ′ [ 5 ] ) ≤ 0 and so on; the sum of all (9) to (15) = 34 + 13 + 6 + 5 + 8 + 5 + 3 = 74 substituting in (8), we get π ( 100 ) = 100 − 1 − 74 = 25 .
We close this section with the following assertion.
Assertion 3. Primes near 2x “thin out” compared to near x. We write π ( x , 2 x ) < π ( x ) . There are a few rare intervals like ( 2,4 ) , ( 4,8 ) , ( 10,20 ) where this assertion is not true is without loss of generality. Further, adding π ( x ) to both sides of this, we have π ( x , 2 x ) + π ( x ) < π ( x ) + π ( x ) which ⇒ π ( 2 x ) < 2 π ( x ) .
Now, we show, how we arrive to an expression for π ( x ,2 x ) using our basic tool at (5), and the general formula for ℂ ( x ) in the lemma 1 in its elaborated form, on evaluating all the terms of ℂ ( 2 x ) − 2 ℂ ( x ) & substituting in (5), the exact value of π ( x ,2 x ) can be obtained, but due to huge labor, it is kept aside at this stage. On observation, evaluation gives three types of terms:
● Negative values & zero, say B j ( x ) ;
● Redundant;
● Positive values, say A k ( x ) .
We proceed as per methodology of lemma 1 in an elaborated form, considering Step I terms for a = 1 , for p ≤ 2 x and a = 1 , for p ≤ x for each term of ℂ ( 2 x ) − 2 ℂ ( x ) respectively, we have a term for a “p” as
( π [ 2 x p ] − π ′ [ p ] ) − 2 ( π [ x p ] − π ′ [ p ] ) (16)
= π [ 2 x p ] − π ′ [ p ] − 2 π [ x p ] + 2 π ′ [ p ]
= π [ 2 x p ] − 2 π [ x p ] is a term of negative value, by assertion 3.
− π ′ [ p ] + π ′ [ p ] is redundant.
+ π ′ [ p ] is a term of positive value.
Taking the sum of all positive values ∀ p ≤ x and balancing the term, − ∑ x < p ≤ 2 x π ′ [ p ] with other terms, we get
π ′ [ p 1 ] + π ′ [ p 2 ] + π ′ [ p 3 ] + π ′ [ p 4 ] + ⋯ + π ′ [ p r ] , where p r ≤ x and p r + 1 > x = 0 + 1 + 2 + 3 + ⋯ + r − 1 where π ′ [ p r ] = r − 1 and π [ p r ] = r , = ( r − 1 ) r 2 on substituting r = π [ p r ] = π [ x ] , we get
∑ 2 ≤ p ≤ x π ′ [ p ] = μ ( μ − 1 ) 2 where μ = π [ x ] say A 1 ( x ) ,
and is a term of positive value. that can be calculated by knowing the value of μ from the table of primes. Similarly considering all positive values and negative values of all the steps I, II, III,..., and their sum total, we have
ℂ ( 2 x ) − 2 ℂ ( x ) = ( ∑ A k ( x ) − ∑ B j ( x ) ) = a ( x ) (17)
Substituting a ( x ) in (6), we get
π ( x ,2 x ) = π ( x ) − ( A 1 ( x ) + A 2 ( x ) + ⋯ + A k ( x ) − B 1 ( x ) − B 2 ( x ) − ⋯ − B j ( x ) ) + 1 = π ( x ) − A 1 ( x ) − ( A 2 ( x ) + ⋯ + A k ( x ) − B 1 ( x ) − B 2 ( x ) − ⋯ − B j ( x ) ) + 1
Knowing the values of all the terms of a ( x ) in (17) is of huge labor. Hence, considering only A 1 ( x ) , the highest positive value, and neglecting all other values, we get the expression
π ( x ,2 x ) > π ( x ) − A 1 ( x )
π ( x , 2 x ) > π ( x ) − μ ( μ − 1 ) 2 where μ = π [ x ] (18)
Value of the term μ can be known from the table of primes. On careful observation of terms, the expression (18) can further be improved as given below to get a still better approximation. On reconsidering a term for a “p” at (16) as in the proof of (18) and further simplify. Taking a term for a “p”, for ℂ ( 2 x ) − 2 ℂ ( x ) , we write
( π [ 2 x p ] − π ′ [ p ] ) − 2 ( π [ x p ] − π ′ [ p ] ) = π [ 2 x p ] − π ′ [ p ] − 2 π [ x p ] + 2 π ′ [ p ] = π [ 2 x p ] − π ′ [ p ] − 2 π [ x p ] + π ′ [ p ] + π ′ [ p ] = ( π [ 2 x p ] − π ′ [ p ] ) − { ( 2 π [ x p ] − π ′ [ p ] ) − π ′ [ p ] }
a term for a “p” (16) is >0 by the computational technique of lemma 1. Hence, π [ 2 x p ] > π ′ ( p ) and ( 2 π [ x p ] − π ′ [ p ] ) > π ′ [ p ] , taking the sum of all positive values ∀ p ≤ 2 x for the first term, ∀ p ≤ x for the second term, and proceed as in the proof of (18), we write
∑ 2 ≤ p ≤ 2 x π [ 2 x p ] > ∑ 2 ≤ p ≤ 2 x π ′ [ p ] = μ ′ ( μ ′ − 1 ) 2 where μ ′ = π [ 2 x ]
denote this value by A ″ 1 ( x ) , and
∑ 2 ≤ p ≤ x ( 2 π [ x p ] − π ′ [ p ] ) > ∑ 2 ≤ p ≤ x π ′ [ p ] = μ ( μ − 1 ) 2 where μ = π [ x ]
replacing the larger value with smaller one, and keeping the middle minus sign, the resulting term ( μ ′ ( μ ′ − 1 ) 2 − μ ( μ − 1 ) 2 ) may be taken as a better positive value, denote it by A ′ 1 ( x ) , then we write (17) as
ℂ ( 2 x ) − 2 ℂ ( x ) = ( ∑ A ′ k ( x ) − ∑ B ′ j ( x ) ) = a ( x )
where A ′ 2 ( x ) , ⋯ and B ′ 1 ( x ) , ⋯ are resulting positive & negative values here, and we write (6) as
π ( x ,2 x ) = π ( x ) − ( A ′ 1 ( x ) + A ′ 2 ( x ) + ⋯ + A ′ k ( x ) − B ′ 1 ( x ) − B ′ 2 ( x ) − ⋯ − B ′ j ( x ) ) + 1 = π ( x ) − A ′ 1 ( x ) − ( A ′ 2 ( x ) + ⋯ + A ′ k ( x ) − B ′ 1 ( x ) − B ′ 2 ( x ) − ⋯ − B ′ j ( x ) ) + 1
proceeding as before at (18), we get a better approximation
π ( x , 2 x ) > π ( x ) − ( μ ′ ( μ ′ − 1 ) 2 − μ ( μ − 1 ) 2 ) (19)
Denote the value of expression (18) as K 1 ( x ) and its error with actual figure of π ( x ,2 x ) as K ′ 1 ( x ) , of (19) as K 2 ( x ) and its error as K ′ 2 ( x ) . The value of π ( x ,2 x ) calculated from (18) & (19) and their error obtained for x = 10 n up to a trivial n = 6 , verified from the website in [
Remark. The following are calculations for other large non-trivial intervals from (18).
π ( 10 12 , 2 ⋅ 10 12 ) > 34526983265
π ( 10 24 , 2 ⋅ 10 24 ) > 17728422244191190407713
here the value of the terms required in R.H.S of (18) for calculation are from ref. [
On observing the data of results obtained from the expression at (18) & (19) in
| x | π ( x ) | π ( 2 x ) | A 1 ( x ) | A ″ 1 ( x ) | A ′ 1 ( x ) | K 1 ( x ) | K 2 ( x ) | π [ x ,2 x ] | K ′ 1 ( x ) | K ′ 2 ( x ) |
|---|---|---|---|---|---|---|---|---|---|---|
| μ | μ ′ | μ ( μ − 1 ) 2 | μ ′ ( μ ′ − 1 ) 2 | A ″ 1 ( x ) − A 1 ( x ) | π ( x ) − A 1 ( x ) | π ( x ) − A ′ 1 ( x ) | (%error) | (%error) | ||
| 10 | 2 | 2 | 1 | 1 | 0 | 3 | 4 | 4 | 1 | 0 |
| (25) | (0) | |||||||||
| 102 | 4 | 6 | 6 | 15 | 9 | 19 | 16 | 21 | 2 | 5 |
| (9.5) | (23.8) | |||||||||
| 103 | 11 | 14 | 55 | 91 | 36 | 113 | 132 | 135 | 22 | 3 |
| (16.3) | (2.2) | |||||||||
| 104 | 25 | 34 | 300 | 561 | 261 | 929 | 968 | 1033 | 104 | 65 |
| (10.1) | (6.3) | |||||||||
| 105 | 65 | 86 | 2080 | 3655 | 1575 | 7512 | 8017 | 8392 | 880 | 375 |
| (10.4) | (4.6) | |||||||||
| 106 | 168 | 223 | 14,028 | 24,753 | 10,725 | 64,470 | 67,773 | 70,435 | 5965 | 2662 |
| (8.4) | (3.8) |
expression (18), (19) are inconsistent with π ( x ,2 x ) (though not showing in the initial stage), and K 2 ( x ) runs close to π ( x ,2 x ) than K 1 ( x ) is running. Knowing π ( x ) , π ( x ) , and π ( 2 x ) from the table of primes can be avoided by applying the prime number theorem as in ref. ( [
π ( x , 2 x ) > x log x − 1 2 ( x log x ( x log x − 1 ) ) = x log x − 1 2 ( x 1 2 log x ( x 1 2 log x − 1 ) ) = x log x − 1 2 ( 2 x log x ( 2 x log x − 1 ) ) = x log x − x log x ( 2 x log x − 1 ) (20)
Similarly, for (19) we write
π ( x , 2 x ) > x log x − { 2 x log 2 x ( 2 2 x log 2 x − 1 ) − x log x ( 2 x log x − 1 ) } (21)
The workouts for an estimated value of π ( x ,2 x ) from (20), (21) are shown in the
Remark. Calculation for other large intervals from (20), we have
π ( 10 12 , 2 ⋅ 10 12 ) > 33571660417
π ( 10 24 , 2 ⋅ 10 24 ) > 17440714514802363566469
π ( 10 25 , 2 ⋅ 10 25 ) > 167695213406436062322306
from (21), we have
| x | x log x | Calculations from(20) | % Error | Calculations from (21) | % Error |
|---|---|---|---|---|---|
| 103 | 144 | 107 | 20 | 119 | 11.85 |
| 104 | 1085 | 850 | 17.71 | 927 | 10.26 |
| 105 | 8685 | 7207 | 14.12 | 7521 | 10.37 |
| 106 | 72,382 | 61,997 | 11.97 | 63,899 | 9.27 |
π ( 10 12 , 2 ⋅ 10 12 ) > 33823600357
π ( 10 24 , 2 ⋅ 10 24 ) > 17472729013238329469058
π ( 10 25 , 2 ⋅ 10 25 ) > 167979542770512164505980
On observing the data of results obtained from the expressions at (20), (21) in
However, the estimated value of π ( x ,2 x ) where actual values are not yet known for a large fixed x can be known from (20), (21) to the extent machine can be used to calculate the value of the terms. Expressions (20), (21) are a suitable answer to the query raised in Section 22.9, on “Primes in an interval”, in ref. ( [
● The expressions (18), (19) are subject to some mathematical operations in x, the results of which may help in answering questions concerning primes.
● To prove the statement that the number of primes in the interval ( x ,2 x ) is asymptotic to x log x − x log x ( 2 x log x − 1 ) , and to x log x − { 2 x log 2 x ( 2 2 x log 2 x − 1 ) − x log x ( 2 x log x − 1 ) } .
The discoveries and study conducted in this article shows that the “Assertion 3”, is a truth.
Sincere thanks to the members of JAMP for their professional performance, and special thanks to the referee editor for suggesting valuable suggestions.
The author declares no conflicts of interest regarding the publication of this paper.
Kittur, S.B. (2024) On the Number of Primes in the Interval (x, 2x) by an Elementary Method. Advances in Pure Mathematics, 14, 16-29. https://doi.org/10.4236/apm.2024.141002