trough@ihuxi.UUCP (Chris Scussel) (12/05/85)
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
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