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Old 2012-01-20, 19:37   #1
Stan
 
Dec 2011

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Question 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
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Old 2012-01-20, 20:10   #2
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Quote:
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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.
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Old 2012-01-20, 20:57   #3
Stan
 
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Originally Posted by ccorn View Post
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
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Old 2012-01-20, 21:15   #4
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Quote:
Originally Posted by Stan View Post
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.
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Old 2012-01-21, 03:20   #5
Batalov
 
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Phi(4,2^7658614+1)/2

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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
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Old 2012-01-21, 12:28   #6
Stan
 
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Originally Posted by Batalov View Post
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
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Old 2012-01-21, 21:36   #7
Stan
 
Dec 2011

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Smile 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
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Old 2012-01-22, 07:50   #8
Batalov
 
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n0 = 19 (prime)
n1 = 2^n0-1 (prime)
n2 = 2^n1-1 (?composite?)
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Old 2012-01-22, 12:05   #9
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Quote:
Originally Posted by Stan View Post
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.
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Old 2012-01-22, 15:07   #10
ccorn
 
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Quote:
Originally Posted by Stan View Post
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.
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Old 2012-01-23, 19:37   #11
Stan
 
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Quote:
Originally Posted by Stan View Post
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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