Estimate Of Maximum Prime Gap Between Consecutive Primes...

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Estimate Of Maximum Prime Gap (MPG) Between Consecutive Primes For Primes Less Than N:

If we let the average prime gap,a(n), in the interval, [tex][0, n] =[/tex]{ [tex]x, 0 \le x\le n \ge 97[/tex] }, be

a(n) = n / [tex]\pi(n)[/tex],

then we estimate:

mpg [tex]\le (1 + a(n) - ⌊a(n)⌋) * \sum_{i=1}^{⌊a(n)⌋ - 1 }2 * i[/tex]

to the nearest positive even integer.

Please verify this result. Thank you!

Keywords: [tex]\pi(n)[/tex] is prime-counting function, Uncertainty Principle For Calculating Primes (UCP) (see another post under number theory)

primework123

[Guest]

Estimate Of Maximum Prime Gap (MPG) Between Consecutive Primes For Primes Less Than N:

If we let the average prime gap,a(n), in the interval, [tex][0, n] =[/tex]{ [tex]x, 0 \le x\le n \ge 97[/tex] }, be

a(n) = n / [tex]\pi(n)[/tex],

then we estimate:

mpg [tex]\le (1 + a(n) - ⌊a(n)⌋) * \sum_{i=1}^{⌊a(n)⌋ - 1 }2 * i[/tex]

to the nearest positive even integer.

Please verify this result. Thank you!

Keywords: [tex]\pi(n)[/tex] is prime-counting function, Uncertainty Principle For Calculating Primes (UCP) (see another post under number theory)

primework123

Update:

Let

[tex]mpg' \approx .5 * (mpg + \sum_{i=1}^{⌊a(n)⌋ - 1 }2 * i)[/tex]

to the nearest positive integer where

mpg [tex]= (1 + a(n) - ⌊a(n)⌋) * \sum_{i=1}^{⌊a(n)⌋ - 1 }2 * i[/tex].

David Cole (aka primework123)

[Guest]

Estimate Of Maximum Prime Gap (MPG) Between Consecutive Primes For Primes Less Than N:

If we let the average prime gap,a(n), in the interval, [tex][0, n] =[/tex]{ [tex]x, 0 \le x\le n \ge 97[/tex] }, be

a(n) = n / [tex]\pi(n)[/tex],

then we estimate:

mpg [tex]\le (1 + a(n) - ⌊a(n)⌋) * \sum_{i=1}^{⌊a(n)⌋ - 1 }2 * i[/tex]

to the nearest positive even integer.

Please verify this result. Thank you!

Keywords: [tex]\pi(n)[/tex] is prime-counting function, Uncertainty Principle For Calculating Primes (UCP) (see another post under number theory)

primework123

Update:

Let

[tex]mpg' \approx .5 * (mpg + \sum_{i=1}^{⌊a(n)⌋ - 1 }2 * i)[/tex]

to the nearest positive even integer where

mpg [tex]= (1 + a(n) - ⌊a(n)⌋) * \sum_{i=1}^{⌊a(n)⌋ - 1 }2 * i[/tex].

David Cole (aka primework123)

[Guest]

Keywords: Prime Number Theorem (PNT), the sound Goldbach Conjecture (GC),
[tex]\pi(*)[/tex] is the prime counting function; , Fundamental Theorem of Arithmetic, Number Theory, and Analysis/Synthesis,...

Prime Work:

(1) There are infinitely many more positive integers (even or odd) than there are prime numbers, or prime numbers have a zero density relative to the positive integers,
and
(2) prime numbers generate the positive even integers so efficiently that gaps between two consecutive prime numbers increase without bound.

(3) Prime Parity Law (PPL):

[tex]\pi(e = mg = 1 + p_{2n }) = 2 * \pi(g = 1 + p_{n }) = 2n[/tex] where

[tex]p_{n } > 2, p_{2n }[/tex] are odd prime numbers; [tex]2 < m ≤ 3[/tex]; and as [tex]g\rightarrow\infty[/tex], [tex]m \rightarrow 2[/tex].

...

Note: mpg' ( the maximum prime gap between two consecutive prime numbers in the interval, [0, n ≥ 97] ) is calculated in accordance to the above laws for the distribution of prime numbers.

[Guest]

Estimate Of Maximum Prime Gap (MPG) Between Consecutive Primes For Primes Less Than N:

If we let the average prime gap,a(n), in the interval, [tex][0, n] =[/tex]{ [tex]x, 0 \le x\le n \ge 97[/tex] }, be

a(n) = n / [tex]\pi(n)[/tex],

then we estimate mpg:

[tex]\sum_{i=1}^{⌊a(n)⌋ - ⌊log(n)⌋ }2 * i \le mpg < (1 + a(n) - ⌊a(n)⌋) * \sum_{i=1}^{⌊a(n)⌋ - ⌊log(n)⌋ }2 * i[/tex]

where mpg is rounded to the nearest positive even integer.


Please verify this latest result. Thank you!

Keywords: [tex]\pi(n)[/tex] is prime-counting function and Uncertainty Principle For Calculating Primes (UCP) (see another post under number theory)

Reference Data Link: https://primes.utm.edu/notes/gaps.html

David Cole (aka primework123)

[Guest]

Estimate Of Maximum Prime Gap (MPG) Between Consecutive Primes For Primes Less Than N:

If we let the average prime gap,a(n), in the interval, [tex][0, n] =[/tex]{ [tex]x, 0 \le x\le n \ge 97[/tex] }, be

a(n) = n / [tex]\pi(n)[/tex],

then we estimate mpg:

[tex]\sum_{i=1}^{⌊a(n)⌋ - ⌊log(n)⌋ }2 * i \le mpg < (1 + a(n) - ⌊a(n)⌋) * \sum_{i=1}^{⌊a(n)⌋ - ⌊log(n)⌋ }2 * i[/tex]

where mpg is rounded to the nearest positive even integer;


Please verify this latest result. Thank you!

Keywords: [tex]\pi(n)[/tex] is prime-counting function and Uncertainty Principle For Calculating Primes (UCP) (see another post under number theory)

Reference Data Link: https://primes.utm.edu/notes/gaps.html

David Cole (aka primework123)

The term, [tex](1 + a(n) - ⌊a(n)⌋)[/tex], can probably be replaced by 1.25 so that we have:

[tex]\sum_{i=1}^{⌊a(n)⌋ - ⌊log(n)⌋ }2 * i \le mpg < 1.25 * \sum_{i=1}^{⌊a(n)⌋ - ⌊log(n)⌋ }2 * i[/tex]

where mpg is rounded to the nearest positive even integer;

[tex]mpg' = 1.125 * \sum_{i=1}^{⌊a(n)⌋ - ⌊log(n)⌋ }2 * i[/tex]

where mpg' is rounded to the nearest positive even integer.


Please verify these latest results. Thank you!

Reference Data Link: https://primes.utm.edu/notes/gaps.html

David Cole (aka primework123)

[Guest]

Estimate Of Maximum Prime Gap (MPG) Between Consecutive Primes For Primes Less Than N:

If we let the average prime gap, a(n), in the interval, [tex][0, n] =[/tex]{ [tex]x, 0 \le x\le n \ge 97[/tex] }, be

a(n) = n / [tex]\pi(n)[/tex],

then we estimate mpg:

[tex]\sum_{i=1}^{⌊a(n)⌋ - ⌊log(n)⌋ }2 * i \le mpg < (1 + a(n) - ⌊a(n)⌋) * \sum_{i=1}^{⌊a(n)⌋ - ⌊log(n)⌋ }2 * i[/tex]

where mpg is rounded to the nearest positive even integer;


Please verify this latest result. Thank you!

Keywords: [tex]\pi(n)[/tex] is prime-counting function and Uncertainty Principle For Calculating Primes (UCP) (see another post under number theory)

Reference Data Link: https://primes.utm.edu/notes/gaps.html

David Cole (aka primework123)

The term, [tex](1 + a(n) - ⌊a(n)⌋)[/tex], can probably be replaced by 1.25 so that we have:

[tex]\sum_{i=1}^{⌊a(n)⌋ - ⌊log (a(n))⌋ }2 * i \le mpg < 1.25 * \sum_{i=1}^{⌊a(n)⌋ - ⌊log (a(n))⌋ }2 * i[/tex]

where mpg is rounded to the nearest positive even integer;

[tex]mpg' = 1.125 * \sum_{i=1}^{⌊a(n)⌋ - ⌊log (a(n))⌋ }2 * i[/tex]

where mpg' is rounded to the nearest positive even integer.


Please verify these latest results. Thank you!

Reference Data Link: https://primes.utm.edu/notes/gaps.html

David Cole (aka primework123)

Correction!!! I have replaced log(n) with log(a(n)). See changes above.

[Guest]

Estimate Of Maximum Prime Gap (MPG) Between Consecutive Primes For Primes Less Than N:

If we let the average prime gap, a(n), in the interval, [tex][0, n] =[/tex]{ [tex]x, 0 \le x\le n \ge 97[/tex] }, be

a(n) = n / [tex]\pi(n)[/tex],

then we estimate mpg:

[tex]\sum_{i=1}^{⌊a(n)⌋ - ⌊log( a(n) )⌋ }2 * i \le mpg < (1 + a(n) - ⌊a(n)⌋) * \sum_{i=1}^{⌊a(n)⌋ - ⌊log( a(n) )⌋ }2 * i[/tex]

where mpg is rounded to the nearest positive even integer;


Please verify this latest result. Thank you!

Keywords: [tex]\pi(n)[/tex] is prime-counting function and Uncertainty Principle For Calculating Primes (UCP) (see another post under number theory)

Reference Data Link: https://primes.utm.edu/notes/gaps.html

David Cole (aka primework123)[/quote]

The term, [tex](1 + a(n) - ⌊a(n)⌋)[/tex], can probably be replaced by 1.25 so that we have:

[tex]\sum_{i=1}^{⌊a(n)⌋ - ⌊log (a(n))⌋ }2 * i \le mpg < 1.25 * \sum_{i=1}^{⌊a(n)⌋ - ⌊log (a(n))⌋ }2 * i[/tex]

where mpg is rounded to the nearest positive even integer;

[tex]mpg' = 1.125 * \sum_{i=1}^{⌊a(n)⌋ - ⌊log (a(n))⌋ }2 * i[/tex]

where mpg' is rounded to the nearest positive even integer.


Please verify these latest results. Thank you!

Reference Data Link: https://primes.utm.edu/notes/gaps.html

David Cole (aka primework123)

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