20120120, 19:37  #1 
Dec 2011
2^{2}×3^{2} 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

20120120, 20:10  #2  
Apr 2010
151 Posts 
Quote:


20120120, 21:15  #4  
Apr 2010
227_{8} Posts 
Quote:
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. 

20120121, 03:20  #5 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{5}·3·101 Posts 
Stan,
Could you please explain how your proof path would differ for a similar sequence: n_{0} = 2^51 (prime) n_{1} = 2^n_{0}1 (prime) n_{2} = 2^n_{1}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 n_{2} is actually prime? Last fiddled with by Batalov on 20120121 at 03:27 Reason: a better sequence 
20120121, 12:28  #6  
Dec 2011
100100_{2} Posts 
Quote:
phi(2^51) does not divide phi(2^n_{0}1), therefore no sequence. My proof relies on the chain: phi(2^n_{0}1)  phi(2^n_{1}1)  phi(2^n_{2}1) etc. Last fiddled with by Stan on 20120121 at 12:35 

20120121, 21:36  #7 
Dec 2011
2^{2}·3^{2} 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 20120121 at 21:38 Reason: Update of PDF file 
20120122, 07:50  #8 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{5}·3·101 Posts 
n_{0} = 19 (prime)
n_{1} = 2^n_{0}1 (prime) n_{2} = 2^n_{1}1 (?composite?) 
20120122, 12:05  #9  
Apr 2010
151 Posts 
Quote:


20120122, 15:07  #10 
Apr 2010
151 Posts 

20120123, 19:37  #11 
Dec 2011
2^{2}×3^{2} Posts 

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