[seqfan] Re: A093893 Subsequence

T. D. Noe noe at sspectra.com
Sat May 30 01:02:49 CEST 2009

Here is a short proof that a(n) exists for all n.  All we have to do is
show that there is a prime p such that the sums of two or more divisors of
p^(n-1) are all composite.  Note that we really only have to consider sums
of 3, 5, 7,... divisors because the sums of an even number of divisors will
be even.  Let Q be the product of the primes less than or equal to n (see
note below).  Let p be a prime of the form Qk+1.  Every power of p -- which
are the divisors of p^(n-1) -- has the same form (but with different k, of
course).  Hence, a sum of these r of powers will have the form Qk+r (for
some k).  But, because of the way we constructed Q, r and Q have a common
factor, making Qk+r composite.  QED.

Note: in practice, when n is prime, we can usually get away with using the
primes less than n because the sum of the n powers 1 to p^(n-1) is usually
composite.  For instance, a(11)=211^10.


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