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From: trough@ihuxi.UUCP (Chris Scussel)
Newsgroups: net.math
Subject: Unboundedly multiply perfect numbers
Message-ID: <1284@ihuxi.UUCP>
Date: Thu, 5-Dec-85 09:22:34 EST
Article-I.D.: ihuxi.1284
Posted: Thu Dec  5 09:22:34 1985
Date-Received: Fri, 6-Dec-85 07:11:56 EST
Distribution: net
Organization: AT&T Bell Laboratories
Lines: 23

It was recently asked if for each k there existed n such that s(n) = k*n.
I can't answer that, but I can show that s(n)/n is unbounded.

Consider (n!). Among its divisors are 2,3,4,5,...,[sqrt(n-1)]. Each of these
has a corresponding divisor: n!/2, n!/3, n!/4,...,n!/[sqrt(n-1)]. All of
these divisors are distinct, so they each contribute to s(n!). Consider
only the second set of factors:

	n!/2 + n!/3 + n!/4 + ... + n!/[sqrt(n-1)]

which can be rewritten as

	n!*(1/2 + 1/3 + 1/4 + ... + 1/[sqrt(n-1)])

This doesn't include all of the divisors of n!, so s(n!)/(n!) is at least

        (1/2 + 1/3 + 1/4 + ... + 1/[sqrt(n-1)])

This series grows without bound as n increases, and thus s(n!)/(n!) does too.

				Chris Scussel
				AT&T Bell Labs
				ihnp4!ihuxi!trough