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 2011-10-21, 17:58 #1 fivemack (loop (#_fork))     Feb 2006 Cambridge, England 33×239 Posts Garambois's site now has an explanation in English Jean-Luc just mailed me to say that www.aliquotes.com now links to a summary in English of the work presented on the site; he would be interested in our opinion on it.
 2011-11-13, 17:31 #2 garambois     "Garambois Jean-Luc" Oct 2011 France 2×433 Posts Hello everybody, I rode a lot of pages on mersenneforum, and I learned a lot of things about aliquot sequences, but slowly, because of my english. I didn't know the article written by Richard Guy and John Selfridge. Now, I know the difference between a driver or a guide. So, I corrected my english summary on my website using the good words. But now, I have an important question. At the beginning of the article, it is written : "It is shown that no driver can be expected to persist indefinitely." What does it mean ? Does it mean that in an aliquot sequence, there is no driver or no guide which can persist indefinitely ? For example, it is impossible that the divisor 3 persist in each terms of an aliquot sequence, isn't it ? Is this really shown ? Is it the same with the divisor 2 or 2 is an exception ? This is what I understood when I rode this article... I am really surprised by those results, but OK ! Why ? I invite you too see on db the guide of the aliquote sequence 30006066277577135239100006400=z*154345556085770649600 where z is an integer and 154345556085770649600 is the first 6-perfect number. So, I think that all k-perfect numbers are better guides than other numbers. I have other examples on my website. You can also see what LaurV says. Thank you to people who can answer to my questions ! Jean-Luc
2011-11-13, 18:01   #3
Andi47

Oct 2004
Austria

2×17×73 Posts

Quote:
 Originally Posted by garambois I invite you too see on db the guide of the aliquote sequence 30006066277577135239100006400=z*154345556085770649600 where z is an integer and 154345556085770649600 is the first 6-perfect number.
*puh* - a 2^15*3^5*5^2*7^2*11*13*17*19*31*43*257 driver?

2011-11-13, 18:46   #4
garambois

"Garambois Jean-Luc"
Oct 2011
France

86610 Posts

Quote:
 Originally Posted by Andi47 *puh* - a 2^15*3^5*5^2*7^2*11*13*17*19*31*43*257 driver?
Hello Andi47,

I think we can not call 2^15*3^5*5^2*7^2*11*13*17*19*31*43*257 a driver but a guide because the article written by Richard Guy and John Selfridge, theorem 2 : "Only 2, 24=2^3*3, 120=2^3*3*5, 672=2^5*3*7, 523776=2^9*3*11*31 and the even perfect numbers are drivers".

But I note than 120, 672 and 523776 are the first, the second and the third 3-perfect numbers, ie sigma(120)=3*120, sigma(672)=3*672 and sigma(523776)=3*523776.

Jean-Luc

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