[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.
Tony
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