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 2012-01-20, 19:37 #1 Stan   Dec 2011 1001002 Posts Mersenne Prime Sequence Could someone please check the attached theorem for errors and post a reply for the location of any errors?Mersenne Primes.pdf
2012-01-20, 20:10   #2
ccorn

Apr 2010

151 Posts

Quote:
 Originally Posted by Stan Could someone please check the attached theorem for errors and post a reply for the location of any errors?Attachment 7579
See here and here.

2012-01-20, 20:57   #3
Stan

Dec 2011

22×32 Posts

Quote:
 Originally Posted by ccorn See here and here.
ccorn quotes "5*phi(5) | phi(33) but 25 does not divide 33", but my statement includes that the prime 5 has to divide (33 - 1) and it does not.

Last fiddled with by Stan on 2012-01-20 at 20:59

2012-01-20, 21:15   #4
ccorn

Apr 2010

15110 Posts

Quote:
 Originally Posted by Stan ccorn quotes "5*phi(5) | phi(33) but 25 does not divide 33", but my statement includes that the prime 5 has to divide (33 - 1) and it does not.
The second "$\Rightarrow$" in the fifth line from the bottom and the following "$\Rightarrow$" aren't. (non sequitur, so to say.)
I expect that you will notice the gaps (if you have not done so already) when you try to improve the written reasoning at those points.

 2012-01-21, 03:20 #5 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 100101100011102 Posts Stan, Could you please explain how your proof path would differ for a similar sequence: n0 = 2^5-1 (prime) n1 = 2^n0-1 (prime) n2 = 2^n1-1 (composite, has known factors) What is the specific reason that we wouldn't be able to plug it in the same proof and demonstrate that n2 is actually prime? Last fiddled with by Batalov on 2012-01-21 at 03:27 Reason: a better sequence
2012-01-21, 12:28   #6
Stan

Dec 2011

3610 Posts

Quote:
 Originally Posted by Batalov Stan, Could you please explain how your proof path would differ for a similar sequence: n0 = 2^5-1 (prime) n1 = 2^n0-1 (prime) n2 = 2^n1-1 (composite, has known factors) What is the specific reason that we wouldn't be able to plug it in the same proof and demonstrate that n2 is actually prime?
2^5-1 = 1+2.3.5 and phi(2^5-1) = 2.3.5
phi(2^5-1) does not divide phi(2^n0-1), therefore no sequence.
My proof relies on the chain:
phi(2^n0-1) | phi(2^n1-1) | phi(2^n2-1) etc.

Last fiddled with by Stan on 2012-01-21 at 12:35

 2012-01-21, 21:36 #7 Stan   Dec 2011 22×32 Posts Mersenne Primes I believe the proof of my theorem to be now complete but I still need it checking. Any comments would be appreciated.Mersenne Primes.pdf The attached PDF file has been updated. Last fiddled with by Stan on 2012-01-21 at 21:38 Reason: Update of PDF file
 2012-01-22, 07:50 #8 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2×11×19×23 Posts n0 = 19 (prime) n1 = 2^n0-1 (prime) n2 = 2^n1-1 (?composite?)
2012-01-22, 12:05   #9
ccorn

Apr 2010

2278 Posts

Quote:
 Originally Posted by Stan I believe the proof of my theorem to be now complete but I still need it checking. Any comments would be appreciated.Attachment 7581 The attached PDF file has been updated.
There is still no attempt to thoroughly derive the two alleged non sequiturs. Try to actually prove those points. This will turn out to be insightful, whatever the outcome.

2012-01-22, 15:07   #10
ccorn

Apr 2010

151 Posts

Quote:
 Originally Posted by Stan ccorn quotes "5*phi(5) | phi(33) but 25 does not divide 33", but my statement includes that the prime 5 has to divide (33 - 1) and it does not.
Well then: 5*phi(5) | phi(66) and 5 | (66-1), but 25 does not divide 66.

2012-01-23, 19:37   #11
Stan

Dec 2011

22·32 Posts

Quote:
 Originally Posted by Stan 2^5-1 = 1+2.3.5 and phi(2^5-1) = 2.3.5 phi(2^5-1) does not divide phi(2^n0-1), therefore no sequence. My proof relies on the chain: phi(2^n0-1) | phi(2^n1-1) | phi(2^n2-1) etc.
.

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