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Proof of a stronger statement of the Cramér Conjecture

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Expand view Topic review: Proof of a stronger statement of the Cramér Conjecture

Re: Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Thu Sep 26, 2019 5:01 pm

Yes! G(X) [tex]\le log^{2} X[/tex] for large X where [tex]G(X) \rightarrow log^{2} X[/tex] as [tex]X \rightarrow \infty[/tex].


Dave.

Re: Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Tue Sep 24, 2019 8:07 pm

FYI: 'On the Cramér-Granville Conjecture and finding prime pairs whose difference is 666',

https://math.stackexchange.com/questions/1972996/on-the-cram%C3%A9r-granville-conjecture-and-finding-prime-pairs-whose-difference-is-6.

Re: Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Fri Apr 05, 2019 9:42 pm

Guest wrote:Oops! Proof is still wrong!!

Dave. :-(



Note Change:

5. X = s * log X + b * [tex]log^{2} X[/tex] which does not contradict equation one! Equation five confirms equation one!

Thus, G(X) [tex]\le log^{2} X[/tex] for large X.

Hmm. I am not happy with this result! There could be more mistakes... I'll review my work again.

Dave.

P.S. I apologise for the sloppy work.

Re: Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Fri Apr 05, 2019 9:14 pm

Oops! Proof is still wrong!!

Dave. :-(

Re: Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Fri Apr 05, 2019 8:49 pm

Guest wrote:
Guest wrote:For large X, we have [tex]\pi(X) \approx X/log X[/tex] according to PNT where logX is the average gap size between consecutive primes less than or equal to X.

Now we let,

1. X = s * log X + b * [tex]log^{2}(X)[/tex], approximately.

We seek to show that equation one leads to a contradiction because of G(X) for some positive integers, s >> b.

Note: The integer constants, s and b, are the number of primes associated with the gap sizes, average and G(X), respectively.

Therefore, equation one implies,

2. [tex]X / log X \approx \pi(X) \approx s + b*log X = s + b * X / \pi(X)[/tex].

Moreover, equation two implies approximately,

3. [tex]\pi^{2}(X) - s* \pi(X) - b* X = 0[/tex].

In turn, equations, three and two, imply with the help of quadratic formula,

4. [tex]\pi(X) = \frac{s + \sqrt{s^{2} + 4bX}}{2} = s + b * log X[/tex].

However, equation four implies approximately,

5. X = (s / 2) * log X+ b * [tex]log^{2} X[/tex] which contradicts equation one!

Thus, G(X) < [tex]log^{2} X[/tex] for large X.

Dave,

https://www.researchgate.net/profile/David_Cole29/amp


Note Change:

"In turn, equations, three and two, imply with the help of quadratic formula,"


Note Change:

"5. X = (s / 2) * log X + b * [tex]log^{2} X[/tex] which contradicts equation one!"

Re: Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Fri Apr 05, 2019 8:00 pm

Guest wrote:For large X, we have [tex]\pi(X) \approx X/log X[/tex] according to PNT where logX is the average gap size between consecutive primes less than or equal to X.

Now we let,

1. X = s * log X + b * [tex]log^{2}(X)[/tex], approximately.

We seek to show that equation one leads to a contradiction because of G(X) for some positive integers, s >> b.

Note: The integer constants, s and b, are the number of primes associated with the gap sizes, average and G(X), respectively.

Therefore, equation one implies,

2. [tex]X / log X \approx \pi(X) \approx s + b*log X = s + b * X / \pi(X)[/tex].

Moreover, equation two implies approximately,

3. [tex]\pi^{2}(X) - s* \pi(X) - b* X = 0[/tex].

In turn, equations, three and one, imply with the help of quadratic formula,

4. [tex]\pi(X) = \frac{s + \sqrt{s^{2} + 4bX}}{2} = s + b * log X[/tex].

However, equation four implies approximately,

5. X = s / 2 + b * [tex]log^{2} X[/tex] which contradicts equation one!

Thus, G(X) < [tex]log^{2} X[/tex] for large X.

Dave,

https://www.researchgate.net/profile/David_Cole29/amp


Note Change:

"In turn, equations, three and two, imply with the help of quadratic formula,"

Re: Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Fri Apr 05, 2019 2:52 am

For large X we have,

[tex]c_{1 } * log^{2} X \le G(X) \le c_{2 } * log^{2} X[/tex]

where [tex]0 < c_{1 } < c_{2} < 1[/tex].

Dave.

Re: Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Fri Apr 05, 2019 2:24 am

For large X, we have [tex]\pi(X) \approx X/log X[/tex] according to PNT where logX is the average gap size between consecutive primes less than or equal to X.

Now we let,

1. X = s * log X + b * [tex]log^{2}(X)[/tex], approximately.

We seek to show that equation one leads to a contradiction because of G(X) for some positive integers, s >> b.

Note: The integer constants, s and b, are the number of primes associated with the gap sizes, average and G(X), respectively.

Therefore, equation one implies,

2. [tex]X / log X \approx \pi(X) \approx s + b*log X = s + b * X / \pi(X)[/tex].

Moreover, equation two implies approximately,

3. [tex]\pi^{2}(X) - s* \pi(X) - b* X = 0[/tex].

In turn, equations, three and one, imply with the help of quadratic formula,

4. [tex]\pi(X) = \frac{s + \sqrt{s^{2} + 4bX}}{2} = s + b * log X[/tex].

However, equation four implies approximately,

5. X = s / 2 + b * [tex]log^{2} X[/tex] which contradicts equation one!

Thus, G(X) < [tex]log^{2} X[/tex] for large X.

Dave,

https://www.researchgate.net/profile/David_Cole29/amp

Re: Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Thu Apr 04, 2019 5:26 pm

Guest wrote:How does [tex]G(X) \approx log^{2}X[/tex] for large X affect the Fundamental Theorem of Arithmetic (FTA)?

Are there too few primes being generated which may violate FTA when applied to composites greater than or equal to [tex]X^{2}[/tex]?

Re: Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Thu Apr 04, 2019 5:22 pm

How does [tex]G(X) \approx log^{2}X[/tex] for large X affect the Fundamental Theorem of Arithmetic (FTA)?

Are there too few primes being generated which may violate FTA when apply to composites greater than or equal to [tex]X^{2}[/tex]?

Re: Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Thu Apr 04, 2019 3:38 pm

Guest wrote:
Guest wrote:Hmm. Can we improve our 'proof'?

Hmm. Here's a minor improvement:

[tex]0 < 1 / b \le 1[/tex].


Here's a significant development in support of our proof:

If [tex]b * G(X) \approx b* log^{2}X[/tex], then

b * G(X) is not << X and therefore,

[tex]s * logX \approx X[/tex] is false

for 1 << b << [tex]\pi(X)[/tex] and for large X.

Re: Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Thu Apr 04, 2019 2:16 pm

Guest wrote:Hmm. Can we improve our 'proof'?

Hmm. Here's a minor improvement:

[tex]0 < 1 / b \le 1[/tex].

Re: Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Wed Apr 03, 2019 9:50 pm

Hmm. Can we improve our 'proof'?

Re: Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Wed Apr 03, 2019 12:46 am

Furthermore, for large X, we have:

[tex]c_{1 } * log^{2} X \le G(X) \le c_{2 } * log^{2} X[/tex]

where [tex]0 < c_{1 } < c_{2} \le 1[/tex].

Dave,

https://www.researchgate.net/profile/David_Cole29/amp

Proof of a stronger statement of the Cramér Conjecture

Post by Guest » Tue Apr 02, 2019 11:04 pm

Proof of a stronger statement of the Cramér Conjecture,

[tex]G(X) < log^{2}X[/tex],

for large X where G(X) is the largest prime gap between consecutive primes less than or equal to X:

https://en.m.wikipedia.org/wiki/Cramér%27s_conjecture

Keywords: Prime Number Theorem (PNT)

What constrains the size of G(X)?

The answer is the number, s, of smaller prime gaps between consecutive primes less than or equal to X.

But the average prime gap is size, [tex]\approx log X[/tex], since we exclude all prime gaps of size, G(X).

So, [tex]s * logX \approx X[/tex].

And b is the number of prime gaps of size, G(X).

Therefore, s * log X + b * G(X) = X. But, b << s,
generally.

So, we have,

s * log X + b * G(X) = X which implies

G(X) = (X - s * log X)/b = (X /(b * log X) - s/b) * log X

[tex]\approx[/tex] ([tex]\pi(X)/ b - s /b[/tex]) * log X.

(1). We observe that, [tex]\pi(X) - s \ge 1[/tex],

[tex]\pi(X) - s << \pi(X)[/tex],

and 0 < 1 / b < 1.

(2). Furthermore, we observe that, log X << [tex]\pi(X)[/tex].

We combine the ideas of (1) and (2) to conclude:

[tex]G(X) \approx ( \pi(X) / b - s /b ) * log X < (log X) * log X = log^{2}X[/tex]


Therefore, [tex]G(X) < log^{2}X[/tex] for large X.

Dave.

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