Formubla-bla-bla to calculate the sum of two Prime numbers just by knowing the product - Page 9

Godzilla · 2016-11-04, 22:43

Quote:

Originally Posted by science_man_88

I don't even know how to interpret this you don't show how to get n2 etc.

n2 = 1 to $\infty$

LaurV · 2016-11-05, 15:16

Quote:

Originally Posted by Godzilla

n2 = 1 to $\infty$

I think you should forget this thread for a while, then go to Nick's theory thread, read all what is written there for a while, try to solve all the exercises, follow with the other chapters yet unpublished, and in a year or two, you may understand why your efforts in this thread (to find a magic factorization formula) are not only futile, but also hilarious.

Godzilla · 2016-12-06, 12:44

Before i have wrong . sorry

Godzilla · 2016-12-06, 14:10

Now the new formula is :

$p1 * p2 = product$

formula :

$\frac{(p1)\pi}{product} * \frac{p2}{3\pi} = 0.33333333 = \frac{1}{3}$ always $\frac{1}{3}$

or this (it is equal)

$\frac{(p2)\pi}{product} * \frac{p1}{3\pi} = 0.33333333 = \frac{1}{3}$ always $\frac{1}{3}$

And i think that in few test is possible to know the factors $p1$ and $p2$ , because the result is always $\frac{1}{3}$ .

What do you think about it ?

thanks

science_man_88 · 2016-12-06, 14:30

Quote:

Originally Posted by Godzilla

Now the new formula is :

$p1 * p2 = product$

formula :

$\frac{(p1)\pi}{product} * \frac{p2}{3\pi} = 0.33333333 = \frac{1}{3}$ always $\frac{1}{3}$

or this (it is equal)

$\frac{(p2)\pi}{product} * \frac{p1}{3\pi} = 0.33333333 = \frac{1}{3}$ always $\frac{1}{3}$

And i think that in few test is possible to know the factors $p1$ and $p2$ , because the result is always $\frac{1}{3}$ .

What do you think about it ?

thanks

it doesn't tell you the factors then as it doesn't matter what p1 and p2 are the value is the same ...

Godzilla · 2016-12-06, 14:48

Quote:

Originally Posted by science_man_88

it doesn't tell you the factors then as it doesn't matter what p1 and p2 are the value is the same ...

Better is :

$\frac{BiggerFactor}{product} = 0.33333333 = \frac{1}{3}$ always

and in few test i think is possible to found it ?

science_man_88 · 2016-12-06, 14:57

Quote:

Originally Posted by Godzilla

Better is :

$\frac{BiggerFactor}{product} = 0.33333333 = \frac{1}{3}$ always

and in few test i think is possible to found it ?

the problem is any ratio or formula you have given either depends on knowing the factors in which case we don't need it or is so independent of them that you can't tell what they are this one only works when p1 is 3.

Godzilla · 2016-12-06, 15:10

Quote:

Originally Posted by science_man_88

the problem is any ratio or formula you have given either depends on knowing the factors in which case we don't need it or is so independent of them that you can't tell what they are this one only works when p1 is 3.

Yes but the concept is equal , the last is wrong ok , but the first i think is probable to resolve in few time with an algorithm , because the result is always $\frac{1}{3}$

science_man_88 · 2016-12-06, 15:31

Quote:

Originally Posted by Godzilla

Yes but the concept is equal , the last is wrong ok , but the first i think is probable to resolve in few time with an algorithm , because the result is always $\frac{1}{3}$

if there was a way to convert x*y to x+y you could in theory convert a geometric mean ( sqrt(x*y)) to an arithmetic mean ( 1/2*(x+y)) but here's a table to compare the product versus sum:

product:
$ \begin{tabular} & 2 & 3 & 5 & 7 \\ 2& 4 & 6 & 10 & 14 \\ 3& 6 & 9 & 15 & 21 \\ 5 & 10 & 15 & 25 & 35\\ 7 & 14 & 21 & 35 & 49 \\ \end{tabular}$

sum:

$\begin{tabular} & 2 & 3 & 5 & 7 \\ 2& 4 & 5 & 7 & 9 \\ 3& 5 & 6 & 8 & 10 \\ 5 & 7 & 8 & 10 & 12\\ 7 & 9 & 10 & 12 & 14 \\ \end{tabular}$

there are is more than one way to get to ten in the sum so you have to account for that so going from product to sum is a many to one function if it exists at all even within the primes. all the even valued ones are showing the goldbach partitions of an even number. so you have to relate the product of the numbers to the goldbach partitions of a number.

Godzilla · 2016-12-06, 15:42

I do not understand very well, but I speak of this formula, have you tried? The formula clearly says that the exact result is $\frac{1}{3}$ and should move p1 and p2 until it approaches $\frac{1}{3}$ .

Quote:

Originally Posted by Godzilla

Now the new formula is :

$p1 * p2 = product$

formula :

$\frac{(p1)\pi}{product} * \frac{p2}{3\pi} = 0.33333333 = \frac{1}{3}$ always $\frac{1}{3}$

or this (it is equal)

$\frac{(p2)\pi}{product} * \frac{p1}{3\pi} = 0.33333333 = \frac{1}{3}$ always $\frac{1}{3}$

And i think that in few test is possible to know the factors $p1$ and $p2$ , because the result is always $\frac{1}{3}$ .

What do you think about it ?

thanks

science_man_88 · 2016-12-06, 16:01

Quote:

Originally Posted by Godzilla

I do not understand very well, but I speak of this formula, have you tried? The formula clearly says that the exact result is $\frac{1}{3}$ and should move p1 and p2 until it approaches $\frac{1}{3}$ .

the pi's cancel out the p1*p2 and product cancel out you get left with 1/3 regardless of p1 and p2 assuming their product is not 0.

2016-12-06, 12:44	#91
Godzilla May 2016 242₈ Posts	Before i have wrong . sorry Last fiddled with by Godzilla on 2016-12-06 at 12:55

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2016-12-06, 14:10	#92
Godzilla May 2016 242₈ Posts	Now the new formula is : $p1 * p2 = product$ formula : $\frac{(p1)\pi}{product} * \frac{p2}{3\pi} = 0.33333333 = \frac{1}{3}$ always $\frac{1}{3}$ or this (it is equal) $\frac{(p2)\pi}{product} * \frac{p1}{3\pi} = 0.33333333 = \frac{1}{3}$ always $\frac{1}{3}$ And i think that in few test is possible to know the factors $p1$ and $p2$ , because the result is always $\frac{1}{3}$ . What do you think about it ? thanks