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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 Tags primes, proof, random Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Bromster Algebra 0 December 24th, 2015 10:09 PM skynet Probability and Statistics 1 June 18th, 2014 12:26 PM SedaKhold Calculus 0 February 13th, 2012 11:45 AM brunojo Number Theory 70 June 15th, 2009 04:37 PM

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