Good. Now that we are on the topic of perfect numbers, I have proven that
any odd perfect number must have at least 9 factors. Another boy at my
school is trying to tell me that all perfect numbers are in the from
(a^(p-1))(a^p-1), thus showing that there are no odd perfect numbers. Can
someone please explain this to me? I am quite sure that he is wrong.
2^6972593 - 1 is prime.
e^(i*pi) + 1 = 0.
This is the e-mail address of Simon Rubinstein-Salzedo.
When you read this e-mail, Simon will probably be at a math contest.
Don't forget to check Simon's website at http://www.albanyconsort.com/simon
Thanks
SJRS
----- Original Message -----
From: <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Friday, April 28, 2000 11:41 PM
Subject: Mersenne: A conjecture on Perfect numbers
> You may have heard this one before, so if you have respond A.S.A.P., OK?
All
> right. I have conjectured the following statement:
>
> The sum of the following series, ((N)!) / ((k!)((N-k)!)) starting at k=1
to
> k=(N), is equal to (2^N). If an odd counterexample is found, then an odd
> perfect number exists and has factors that are all distinct, and the
number
> of factors is N. The only thing I found is N = 0 (0! is 1 here), which is
1,
> so the number should be prime. So, the only odd, perfect number is 1,
since
> the only factors of a prime are itself and one. 2(1) = 1 + 1, so 1 is a
> perfect number, if you take this minor exception into consideration.
>
> If anyone has proof of this conjecture, please let me know.
> _________________________________________________________________
> Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
> Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
>
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
