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Old 2011-10-21, 17:58   #1
fivemack
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Default 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.
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Old 2011-11-13, 17:31   #2
garambois
 
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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

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Old 2011-11-13, 18:01   #3
Andi47
 
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Quote:
Originally Posted by garambois View Post
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?
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Old 2011-11-13, 18:46   #4
garambois
 
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Quote:
Originally Posted by Andi47 View Post
*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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