On Landau's Fourth Problem

On Landau's Fourth Problem

Are there infinitely many primes of the form, $(2n)^{2}+1$, where n is a positive integer?

Guest

Re: On Landau's Fourth Problem

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 $n \ge i$ is

$\prod_{n=i}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{(\pi(\sqrt{(2n)^2+1})+1)}= 0.$

Guest

Re: On Landau's Fourth Problem

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 $n \ge i$ is

$\prod_{n=i}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{(\pi(\sqrt{(2n)^2+1})+1)}= 0.$

"Please observe $\pi(x)$ is the prime-counting function." -- Dave the antworker.

Guest

Re: On Landau's Fourth Problem

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 $n \ge i$ is

$\prod_{n=i}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{\pi(\sqrt{(2n)^2+1})+1}= 0.$

"Please observe $\pi(x)$ is the prime-counting function." -- Dave the antworker.

We also observe that

$k_{n }p_{n}$ = $(2n)^{2}$ + 1

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

Note: $k_{n }$ = 1 is a violation.

Dave the antworker

Guest

Re: On Landau's Fourth Problem

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 $n \ge i$ is

$\prod_{n=i}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{\pi(\sqrt{(2n)^2+1})+1}= 0.$

"Please observe $\pi(x)$ is the prime-counting function." -- Dave the antworker.

We also observe that

$k_{n }p_{n}$ = $(2n)^{2}$ + 1

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

Note: $k_{n }$ = 1 is a violation when $n \ge i.$

Dave the antworker

Guest

Re: On Landau's Fourth Problem

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 $n \ge i$ is

$\prod_{n=i}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{\pi(\sqrt{(2n)^2+1})+1}= 0.$

"Please observe $\pi(x)$ is the prime-counting function." -- Dave the antworker.

We also observe that

$k_{n }p_{n}$ = $(2n)^{2}$ + 1

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

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

Dave the antworker

Guest

Re: On Landau's Fourth Problem

Final Observation: $\pi(x)$ is the odd prime-counting function since we do not count the even prime two.

Guest

Re: On Landau's Fourth Problem

$\prod_{n=1000}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})}{\pi(\sqrt{(2n)^2+1})+1}$

Guest

Re: On Landau's Fourth Problem

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

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;

...

[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.

Guest

Re: On Landau's Fourth Problem

Prob($10^{10^{10^{10000}}}$: $\infty$)

= $\prod_{n=10^{10^{10^{10000}}}}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})-1}{\pi(\sqrt{(2n)^2+1})}$

to account for prime two.

Dave the antworker.

Guest

Re: On Landau's Fourth Problem

Prob($10^{10^{10^{10000}}}$: $\infty$)

=

$\prod_{n=10^{10^{10^{10000}}}}^{\infty}\frac{\pi(\sqrt{(2n)^2+1})-1}{\pi(\sqrt{(2n)^2+1})}$

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

Guest

Re: On Landau's Fourth Problem

$\prod_{m=1}^{\infty}(\sum_{n=1}^{10^{100000} }9×10^{-n})$

Guest

Re: On Landau's Fourth Problem

Guest wrote:

$\prod_{m=1}^{\infty}(\sum_{n=1}^{10^{100000} }9×10^{-n})$

Guest

Re: On Landau's Fourth Problem

Guest wrote:
Guest wrote:

$\prod_{m=1}^{\infty}(\sum_{n=1}^{10^{10^{10^{10^{100000}}}} }9×10^{-n})$

Guest

Re: On Landau's Fourth Problem

[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

$.9999999999^{10^{13}}$ = 5.07550973604988 × $10^{-435}$

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

Guest