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From: franka@mmintl.UUCP (Frank Adams)
Newsgroups: net.math
Subject: Re: abnormal perfect numbers
Message-ID: <865@mmintl.UUCP>
Date: Tue, 10-Dec-85 15:41:44 EST
Article-I.D.: mmintl.865
Posted: Tue Dec 10 15:41:44 1985
Date-Received: Sun, 15-Dec-85 00:22:31 EST
References: <1283@ihuxi.UUCP> <1741@utcsri.UUCP>
Reply-To: franka@mmintl.UUCP (Frank Adams)
Distribution: net
Organization: Multimate International, E. Hartford, CT
Lines: 42
Summary: 


In article <1741@utcsri.UUCP> gclark@utcsri.UUCP (Graeme Clark) writes:
>We were interested in just this question.  We called a number n such
>that s(n)=n+k a k-perfect number.  Hence you are asking about 1-perfect
>numbers.  We wrote a simple program to examine the numbers from 1 to 100,000
>looking for k-perfect numbers for k in the range -100 to 100, and found
>some curious results:
>
>    There were no 1-perfect numbers.
>
>    All (-1)-perfect numbers were powers of two (it's clear that all
>    powers of two are (-1)-perfect, but I don't know about the converse).
>    
>    For most values of k there were not very many (on the order of 10)
>    k-perfect numbers in the range 1-100,000, but for two particular
>    values of k there were (12 and 53 I think) there were lots and lots
>    of k-perfect numbers (on the order of 1000).  

I think it must have been 12 and 56.  Generally, if N is a perfect number,
and p is prime which does not divide N, then pN is a (2N)-perfect number.
(This is a matter of simple arithmetic.)

Generally, one would expect k-perfect numbers with k odd to be rather rare.
Note that n is k-perfect if sigma(n)-2n = k.  (sigma is the sum of the
divisors function.)  This can be odd only if sigma(n) is odd.  The only
prime powers with sigma(n) odd are (1) powers of two, and (2) even powers
of odd primes.  Since sigma is multiplicative (if a and b have no common
factors, sigma(ab)=sigma(a)sigma(b)), only numbers of the form (2^n)(u^2),
where u is odd, will have odd values of sigma.

For the cases k = 1 and k = -1, the constraints are rather severe.  We are
trying to solve sigma(n) = 2n + 1 or 2n - 1.  This means that n and sigma(n)
must have no common factors.  Since numbers with sigma(n) close to 2n must
have a fair number of factors, and therefore sigma(n) will also have a fair
number of factors, it is hard for this to be the case.  It is therefore a
likely conjecture that these equations have no solutions except for powers
of two.

I don't really see how to start constructing a proof, however.  I would
guess that analytic methods would be required.

Frank Adams                           ihpn4!philabs!pwa-b!mmintl!franka
Multimate International    52 Oakland Ave North    E. Hartford, CT 06108