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Old 2015-01-08, 17:56   #1
Owl
 
Aug 2014

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Default Is There A Breakthrough For The Mersenne Primes?

As the consequence of the Riemann Hypothesis see the complete generalization at the Theorem 11 (30), (31), i would submit to you the following.

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The questions are how far the relation holds true and is it very helpful for the Mersenne Primes Project. Thank you.

ps: The mentioned article will be updated soon. I apologize for any confusion.

Last fiddled with by Owl on 2015-01-08 at 17:56
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Old 2015-01-08, 18:25   #2
Batalov
 
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Blech...

inb4miscmath
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Old 2015-01-08, 18:36   #3
Batalov
 
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In case you were wondering, here's a professional opinion that sums it all up nicely
http://randomprocessed.blogspot.com/...ndividual.html

Dozens of boards around the net are swamped with that nonsense. We have been honored to be added to the list.
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Old 2015-01-09, 00:18   #4
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Old 2015-01-09, 05:54   #5
LaurV
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grrr....
+1 for ban
(edit: wait to see if he insists with is, maybe it is just a guy asking?)

Last fiddled with by LaurV on 2015-01-09 at 05:54
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Old 2015-01-09, 17:03   #6
R.D. Silverman
 
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Quote:
Originally Posted by LaurV View Post
grrr....
+1 for ban
(edit: wait to see if he insists with is, maybe it is just a guy asking?)
BTW,

This nonsense is not new. This crank posted the same crap a number of years ago.
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Old 2015-01-09, 17:11   #7
petrw1
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Default With only High School math

Seems to me that function holds true for any p; Mersenne or not; Prime or not; Even or odd.
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Old 2015-01-09, 17:52   #8
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Quote:
Originally Posted by petrw1 View Post
Seems to me that function holds true for any p; Mersenne or not; Prime or not; Even or odd.
False.
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Old 2015-01-09, 20:24   #9
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OK, so for what p does \frac{2^{(p+4)} + 2^3}{2^4} - \frac{3}{2} \ \ne\  2^p-1 ?
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Old 2015-01-09, 21:02   #10
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Quote:
Originally Posted by danaj View Post
OK, so for what p does \frac{2^{(p+4)} + 2^3}{2^4} - \frac{3}{2} \ \ne\  2^p-1 ?
That part is correct, of course. But the equation also says that 2^p-1 = \zeta(p+4), which is obviously untrue.

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Old 2015-01-09, 21:25   #11
danaj
 
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Quote:
Originally Posted by Mini-Geek View Post
That part is correct, of course. But the equation also says that 2^p-1 = \zeta(p+4), which is obviously untrue.
Of course that isn't right. I assumed he was making up another symbol like many of his texts, not that he meant the actual Riemann Zeta function. We're left wondering if it's amazingly wrong or amazingly trivial and non-useful.
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