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August 21st, 2017, 09:19 PM   #51
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Originally Posted by romsek View Post
No one cared when you posted this last time and no one cares now.
This statement is FALSE

PROOF:

I care

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August 22nd, 2017, 04:40 AM   #52
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Originally Posted by complicatemodulus View Post
Are numbers I invented as a Bijection with all $P\in \mathbb{N^+}$.

Are made as a pure power, I call Integer n-th Root of $P$, plus a $Rest$:

For example taking $n=2$:

$1 = 1^2+0 = 1M_2+0 $
$2= 1^2+1 = 1M_2+1 $
$3= 1^2+2 = 1M_2+2 $
$4= 2^2+0 = 2M_2+0 $
$5= 2^2+0 = 2M_2+1 $

Where $M_n=(x^n-(x-1)^n)$ since you can find the Integer Root as a recoursive difference using this function as Modulo:

For example Square root of 5 is equal to make it modulo $M_2$:

1 . 2x-1 = 1 . 5-1 = 4
2 . 2x-1 = 3 . 4-3 = 1 so 2 is the Integer Square Root of 5, so 5= 2^2+1

in fact if you can't go over because:

3 . 2x-1 = 5 . 1-5 = -4

And the same using a special trick to go in $Q$

etc...

I'm writing a book on, but you find that in several of my post here.


And lot must be written on
I'm not finding this explanation as satisfactory. For starters, what is a "$Rest$"?
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August 22nd, 2017, 05:09 AM   #53
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I've decided to reopen the thread to allow discussion of the OP by all concerned members. Aside from this, what I posted above still stands.

Thank you for your patience and good evening.
The original post was ambiguous. The ensuing discussion did lead to the interesting question below, which I am not sure got finally answered.

Counting the primes in order of magnitude, is there a procedure or formula for finding the nth prime more efficient than than just testing the odd numbers above the (n-1)th prime in succession?
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August 22nd, 2017, 05:51 AM   #54
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I believe that one can use the Riemann hypothesis for primes up to $10^{13}$. It gives a good approximation with accuracy increasing as numbers get larger. Maybe that's as good as it gets.
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August 23rd, 2017, 04:22 AM   #55
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I'm not finding this explanation as satisfactory. For starters, what is a "$Rest$"?
Yes, sorry, the good definition is written in 5 pages on the e-book I'm preparing: Two-Hand-Clock Vol.1.

On this topic you can find some more detail:

Supersymmetric Complicate Numbers

Sorry to be incomplete, but all is under development and was the 3-th very hard (unpaid) job for me...

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September 11th, 2017, 05:54 AM   #56
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The formula that generates whole prime numbers :


y=(2^(x-1)-1)/x ( formula-1)

If y is an integer , then x must be absolutely a prime number .
the set of x for any value of integer y ; x = { 3,5,7,11,13,....} and it generates all the prime numbers .

The question is that for the set of prime numbers ( x1 , x2) does the formula generates all the even numbers or not ?
y1 = (2^(x1-1) -1 ) /x1 + (2^(x2-1) -1 ) /x2
for ( x1 , x2) = ( 3,3) then y1 = 2 ;
for ( x1 , x2) = ( 3,5) then y2 = 4 ;
for ( x1 , x2) = ( 5,5) then y3 = 6 ;
for ( x1 , x2) = (5,7) then y4 = 8 ;
....................

The result for whole prime sets of ( x1 , x2) then you can generate all the even number's set .
P.S.: For the proof of formula-1 and to learn more about it please contact me . For example formula-1 must be always divided by 3 .

METE UZUN
TEL: +905315540733
e-mail: meteuzun@hotmail.com
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