On Landau's Fourth Problem

On Landau's Fourth Problem

Postby Guest » Sat Feb 24, 2018 2:20 am

Are there infinitely many primes of the form, [tex](2n)^{2}+1[/tex], where n is a positive integer?
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Re: On Landau's Fourth Problem

Postby Guest » Sat Feb 24, 2018 3:42 am

The answer for Landau's Fourth Problem is affirmative!

Here are two important hints for our proof:

1. The gap between two consecutive primes can be arbitrarily long.

2. The probability there is a negative answer for Landau's Fourth Problem when [tex]n \ge i[/tex] is

[tex]\prod_{n=i}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{(\pi(\sqrt{(2n)^2+1})+1)}= 0.[/tex]
Guest
 

Re: On Landau's Fourth Problem

Postby Guest » Sat Feb 24, 2018 4:16 am

Guest wrote:The answer for Landau's Fourth Problem is affirmative!

Here are two important hints for our proof:

1. The gap between two consecutive primes can be arbitrarily long.

2. The probability there is a negative answer for Landau's Fourth Problem when [tex]n \ge i[/tex] is

[tex]\prod_{n=i}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{(\pi(\sqrt{(2n)^2+1})+1)}= 0.[/tex]


"Please observe [tex]\pi(x)[/tex] is the prime-counting function." -- Dave the antworker.
Guest
 

Re: On Landau's Fourth Problem

Postby Guest » Mon Feb 26, 2018 11:53 am

Guest wrote:
Guest wrote:The answer for Landau's Fourth Problem is affirmative!

Here are two important hints for our proof:

1. The gap between two consecutive primes can be arbitrarily long.

2. The probability there is a negative answer for Landau's Fourth Problem when [tex]n \ge i[/tex] is

[tex]\prod_{n=i}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{\pi(\sqrt{(2n)^2+1})+1}= 0.[/tex]


"Please observe [tex]\pi(x)[/tex] is the prime-counting function." -- Dave the antworker.


We also observe that

[tex]k_{n }p_{n}[/tex] = [tex](2n)^{2}[/tex] + 1

where prime, [tex]p_{n}[/tex] is:
[tex]3 \le p_{n}\le\sqrt{k_{n}p_{n}} =\sqrt{(2n)^2+1}\le k_{n }.[/tex]

Note: [tex]k_{n }[/tex] = 1 is a violation.

Dave the antworker
Guest
 

Re: On Landau's Fourth Problem

Postby Guest » Mon Feb 26, 2018 11:59 am

Guest wrote:
Guest wrote:
Guest wrote:The answer for Landau's Fourth Problem is affirmative!

Here are two important hints for our proof:

1. The gap between two consecutive primes can be arbitrarily long.

2. The probability there is a negative answer for Landau's Fourth Problem when [tex]n \ge i[/tex] is

[tex]\prod_{n=i}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{\pi(\sqrt{(2n)^2+1})+1}= 0.[/tex]


"Please observe [tex]\pi(x)[/tex] is the prime-counting function." -- Dave the antworker.


We also observe that

[tex]k_{n }p_{n}[/tex] = [tex](2n)^{2}[/tex] + 1

where prime, [tex]p_{n}[/tex] is:
[tex]3 \le p_{n}\le\sqrt{k_{n}p_{n}} =\sqrt{(2n)^2+1}\le k_{n }.[/tex]

Note: [tex]k_{n }[/tex] = 1 is a violation when [tex]n \ge i.[/tex]

Dave the antworker
Guest
 

Re: On Landau's Fourth Problem

Postby Guest » Tue Feb 27, 2018 10:30 am

Guest wrote:The answer for Landau's Fourth Problem is affirmative!

Here are two important hints for our proof:

1. The gap between two consecutive primes can be arbitrarily long.

2. The probability there is a negative answer for Landau's Fourth Problem when [tex]n \ge i[/tex] is

[tex]\prod_{n=i}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{\pi(\sqrt{(2n)^2+1})+1}= 0.[/tex]


"Please observe [tex]\pi(x)[/tex] is the prime-counting function." -- Dave the antworker.


We also observe that

[tex]k_{n }p_{n}[/tex] = [tex](2n)^{2}[/tex] + 1

where prime, [tex]p_{n}[/tex] is:
[tex]3 \le p_{n}\le\sqrt{k_{n}p_{n}} =\sqrt{(2n)^2+1}\le k_{n }.[/tex]

Note: '[tex]p_{n }[/tex] = 1' and '[tex]k_{n }[/tex] is prime' are a violation when [tex]n \ge i.[/tex]

Dave the antworker
Guest
 

Re: On Landau's Fourth Problem

Postby Guest » Tue Feb 27, 2018 8:02 pm

Final Observation: [tex]\pi(x)[/tex] is the odd prime-counting function since we do not count the even prime two.
Guest
 

Re: On Landau's Fourth Problem

Postby Guest » Wed Feb 28, 2018 11:01 am

Please compute:

[tex]\prod_{n=1000}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{\pi(\sqrt{(2n)^2+1})+1}[/tex]
Guest
 

Re: On Landau's Fourth Problem

Postby Guest » Wed Feb 28, 2018 1:21 pm

Guest wrote:Please compute:

Prob(1000: [tex]\infty[/tex])
= [tex]\prod_{n=1000}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})-1}{\pi(\sqrt{(2n)^2+1})}[/tex]

to account for prime two.

Prob(1000: 10,000) = .000138989;

Prob(1000: 11,000) = .0000909983;

Prob(1000: 12,000) = 0000616167;

Prob(1000: 13,000) = 0000429001;

Prob(1000: 14,000) = 0000305871;

Prob(1000: 15,000) = 0000222699;

Prob(1000: 16,000) = .0000165075;

...

Computational Tool Link:
[url]http://m.wolframalpha.com/input/?i=Prod ... qrt%28%282×i%29%5E2%2B1%29%29+-1.+%29%2FPi%28sqrt%28%282×i%29%5E2%2B1%29%29+from+i%3D1000+to+i%3D15000[/url]


Dave the antworker.
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Re: On Landau's Fourth Problem

Postby Guest » Wed Feb 28, 2018 2:17 pm

Please compute:

Prob([tex]10^{10^{10^{10000}}}[/tex]: [tex]\infty[/tex])

= [tex]\prod_{n=10^{10^{10^{10000}}}}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})-1}{\pi(\sqrt{(2n)^2+1})}[/tex]

to account for prime two.



Dave the antworker.
Guest
 

Re: On Landau's Fourth Problem

Postby Guest » Thu Mar 01, 2018 9:48 pm

Guest wrote:Please compute:

Prob([tex]10^{10^{10^{10000}}}[/tex]: [tex]\infty[/tex])

=

[tex]\prod_{n=10^{10^{10^{10000}}}}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})-1}{\pi(\sqrt{(2n)^2+1})}[/tex]

to account for prime two.


Dave the antworker.


Two Hints:

1. Do not attempt to solve the problem by numerical calculation.

2. Please solve the problem by sound reasoning and by applying the great Prime Number Theorem.
Guest
 


Re: On Landau's Fourth Problem

Postby Guest » Sun Mar 04, 2018 11:28 pm

Guest wrote:Please compute:

[tex]\prod_{m=1}^{\infty}(\sum_{n=1}^{10^{100000} }9×10^{-n})[/tex]
Guest
 

Re: On Landau's Fourth Problem

Postby Guest » Sun Mar 04, 2018 11:30 pm

Guest wrote:
Guest wrote:Please compute:

[tex]\prod_{m=1}^{\infty}(\sum_{n=1}^{10^{100000} }9×10^{-n})[/tex]
Guest
 

Re: On Landau's Fourth Problem

Postby Guest » Sun Mar 04, 2018 11:36 pm

Guest wrote:
Guest wrote:
Guest wrote:Please compute:

[tex]\prod_{m=1}^{\infty}(\sum_{n=1}^{10^{10^{10^{10^{100000}}}} }9×10^{-n})[/tex]
Guest
 

Re: On Landau's Fourth Problem

Postby Guest » Mon Mar 05, 2018 12:05 am

Relevant Reference Link:

[url]http://m.wolframalpha.com/input/?i=%28Sum+9×10.%5E%28-n%29+from+n%3D1+to+n%3D10%29%5E%2810%5E13%29[/url]
Guest
 

Re: On Landau's Fourth Problem

Postby Guest » Mon Mar 05, 2018 12:19 am

Guest wrote:Relevant Reference Link:

[tex].9999999999^{10^{13}}[/tex] = 5.07550973604988 × [tex]10^{-435}[/tex]

http://m.wolframalpha.com/input/?i=%28Sum+9×10.%5E%28-n%29+from+n%3D1+to+n%3D10%29%5E%28
10%5E13%29
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