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Old 2010-06-01, 23:05   #1
firejuggler
 
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Apr 2010
Over the rainbow

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Default aliquot and mersenne

is it verified that for the 48 mersenne know prime,
Mx*(M(x-1)+1) end in a cycle of a 1 period?

M2*(M1+1) =6 know cycle
M3*(M2+1) = 28
M5*(M4+1) = 496
M7*(M6+1) = 8128
...
....
M43112609*(M43112608+1) =?? will be a cycle too?
and it seem that they are perfect drivers too

Last fiddled with by firejuggler on 2010-06-01 at 23:10
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Old 2010-06-01, 23:47   #2
Mini-Geek
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"Tim Sorbera"
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Yes, it's proven.
A number whose aliquot sequence is a cycle with a period of one is just a roundabout way to refer to a perfect number.
A link has been proven (quite a long time ago too!) between Mersenne primes and even perfect numbers:

If and only if 2^p-1 is prime, then (2^{p-1})(2^p-1) is a perfect number.
All even perfect numbers are of the form (2^{p-1})(2^p-1).

(Note that (2^{p-1})(2^p-1)=(M(p-1)+1)*Mp, so this is the same as your forms like M7*(M6+1), just using the more common form)
And perfect drivers are, by definition, perfect numbers. So yes, 2^{43112608}*(2^{43112609}-1) is a perfect number, and so it's also a perfect driver, and the aliquot sequence 2^{43112608}*(2^{43112609}-1) will cycle back to itself immediately.

Further reading:
http://en.wikipedia.org/wiki/Perfect_number
http://en.wikipedia.org/wiki/Mersenne_number
http://en.wikipedia.org/wiki/Aliquot_sequence
http://mersennewiki.org/index.php/Aliquot_Sequences

Last fiddled with by Mini-Geek on 2010-06-01 at 23:52
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