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

PROOF:

I care

August 22nd, 2017, 04:40 AM   #52
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Quote:
 Originally Posted by complicatemodulus 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 "$Rest$"?

August 22nd, 2017, 05:09 AM   #53
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Quote:
 Originally Posted by greg1313 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?

 August 22nd, 2017, 05:51 AM #54 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond 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. Thanks from JeffM1
August 23rd, 2017, 04:22 AM   #55
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Quote:
 Originally Posted by greg1313 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...

Thanks
ciao
Stefano

 September 11th, 2017, 05:54 AM #56 Newbie   Joined: Jan 2010 Posts: 9 Thanks: 0 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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