Work in progress...

I will show that the composites of a perfect square interval are at least two fewer than the integers for any interval greater than {1,4}.

i
Consider the integers in a given interval to be a set that excludes the endpoints in the following steps and examples.

Thus, the integers we will consider for interval {81,100} are:

82 = 41 * 2

83 = 83 * 1 (P)

84 = 7 * 3 * 2 * 2

85 = 17 * 5

86 = 43 * 2

87 = 29 * 3

88 = 11 * 2 * 2 * 2

89 = 89 * 1 (P)

90 = 5 * 3 * 3 * 2

91 = 13 * 7

92 = 23 * 2 * 2

93 = 31 * 3

94 = 47 * 2

95 = 19 * 5

96 = 3 * 2 * 2 * 2 * 2 * 2

97 = 97 * 1 (P)

98 = 7 * 7 * 2

99 = 11 * 3 * 3

ii
Any odd number greater than 1 can be represented as the difference of two squares. The interval size is the sum of the interval’s endpoint square roots:

100 – 81 = 19
81 = 9 and 100 = 10

With respect to prime factors, notice that for any given interval there are one or more prime factors of every composite that must be less than the square root of the larger perfect square (that is, less than 10 in this example).

iii
We can try cancellation of the composites with prime factors up to half the interval. Just eliminate the composites with prime factor 2 (every second composite), eliminate the composites with prime factor 3 (every third composite), and so on through all the composites with a prime factor less than half the interval. This will eliminate all the composite integers, and it will leave only the primes.

82 = 41 * 2

83 = 83 * 1

84 = 7 * 3 * 2 * 2

85 = 17 * 5

86 = 43 * 2

87 = 29 * 3

88 = 11 * 2 * 2 * 2

89 = 89 * 1

90 = 5 * 3 * 3 * 2

91 = 13 * 7

92 = 23 * 2 * 2

93 = 31 * 3

94 = 47 * 2

95 = 19 * 5

96 = 3 * 2 * 2 * 2 * 2 * 2

97 = 97 * 1

98 = 7 * 7 * 2

99 = 11 * 3 * 3

iv
We can say that the occurrence of every prime less than the interval is determined by its cardinal number. Thus, the frequency of 2 is one half (every second composite), the frequency of 3 is one-third (every third composite), the frequency of 5 is one-fifth (every fifth composite).

For integers 82 through 99, the occurrence of prime factors 19 and less are as follows:

2 : 8

3 : 6

5 : 3

7 : 3

11: 2

13: 1

17: 1

19: 1

v
Without considering their individual prime factors, we have the ability to eliminate those integers that are composite. Divide the interval by each prime factor, beginning with 2, and subtract the quotient from the remainder. Precisely:

1. Take the difference of two consecutive perfect squares.

2. Divide the difference by 2.

3. Divide this result by 3 and subtract the quotient from the result.

4. Repeat Step 3 for every prime less than or equal to the difference.

Example:

19 / 2 = 9.5

9.5 / 3 = 3.17

6.33 / 5 = 1.27

5.06 / 7 = 0.72

4.34 / 11 = 0.39

3.95 / 13 = 0.3

3.65 / 17 = 0.21

3.44 / 19 = 0.18

3.08

The remainder (3.08 in this case) represents a close approximation of how many integers (3 in the case of {81,100}) are not composite. We can call this formula fractional elimination. The logic is simply this:

a. The frequency of every prime factor can be considered a fraction.
b. The frequency of 2 is ½, the frequency 3 is ⅓, the frequency of 5 is ⅕, and so on.
c. Repeat subtraction of these fractional amounts eliminates all the composites.
d. The remainder must approximate the integers that are prime.

vi
We know that the number of primes less than a given number is about /log , and therefore is greater for each successive square number. By the fractional elimination procedure in the preceding step, we can calculate the number of integers that are prime for each interval and compare it with that suggested by the asymptotic curve of prime number distribution. However, fractional elimination doen not rely on this limit; it relies solely on the size of the interval.

Can we establish a deterministic correlation between the prime count less than or equal to the interval and the prime count within the interval? No (for the reason stated in Step 7).

Can we establish a lower bound between the prime count less than or equal to the interval and the prime count within the interval? Yes.

There are two steps that can establish a proof by contradiction for setting a lower bound is feasible.

vii

[To be completed]
For every perfect square interval there is one pronic number, a composite of form (+1), comprising two consecutive integers that are coprime. For example, for interval {100,121}, the integers are 10*11 (in this case, a composite and a prime). Every prime factor less than half the interval occurs two or more times - at least once before and once after the pronic. Every prime factor greater than half the interval and less than the interval must occur once in a composite, either before or after the pronic. However, we cannot know through the method of fractional elimination whether these larger-than-half-interval prime factors will occur more than once.

viii

[To be completed]
We know that the method of fractional elimination applied to any interval greater than {25,36} produces a remainder greater than 2:

11 / 2 = 5.5
5.5 / 3 = 1.83
3.67 / 5 = 0.73
2.94 / 7 = 0.42
2.52 / 11 = 0.23
2.06

It does not matter what the magnitude of the interval is: The hypothetical elimination of a prime factor from the cardinal set of an interval does not decrease the remainder - it increases it.

For example,

a. Apply this formula to the interval {100,121}, which contains five primes. The formula produces a number that rounds to 3.

b. Suppose four or five primes of this interval were to vanish, and you were to apply again the formula to the interval {100,121}.

c. Again, the formula eliminates all the composites of the interval with prime factors less than 21.

d. Thus there would be remaining one or two (subtract 3 from 4 or 5) integers in the interval {100,121} that cannot be composite or prime.

That would be an impossible result.


This shows there is no case where the integers are not prime at least twice for every square interval greater than {0,1}.