aliquot and mersenne

firejuggler · 2010-06-01, 23:05

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

Mini-Geek · 2010-06-01, 23:47

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

2010-06-01, 23:05	#1
firejuggler "Vincent" Apr 2010 Over the rainbow 101100011001₂ Posts	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

2010-06-01, 23:47	#2
Mini-Geek Account Deleted "Tim Sorbera" Aug 2006 San Antonio, TX USA 1000010110101₂ Posts	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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