[seqfan] Re: Pairs Occurring Only Once Among # Of Divisors
T. D. Noe
noe at sspectra.com
Fri Jun 26 00:44:31 CEST 2009
At 10:42 PM +0200 6/25/09, Hagen von EItzen wrote:
>Richard Mathar schrieb:
>> On behalf of the Quetau pairs A161640 invented in
>>
>> http://list.seqfan.eu/pipermail/seqfan/2009-June/001652.html
>>
>> I started some sort of explicit proofs of these in
>> http://www.strw.leidenuniv.nl/~mathar/progs/a161460.pdf .
>> The interesting part starts at page 116. This is an unfunded
>> project and will require some Jovian years to complete.
>>
>>
It is also easy to prove 1023:
d(1023)=8
d(1024)=11
if n has 8 divisors, then it is p^7, pq^3, or pqr for distinct primes p, q, r.
if n+1 has 11 divisors, then s^10 is the only possible form with s prime.
Suppose n is even. Remember n is p^7, pq^3, or pqr. For n to be even, one
of the primes is 2. So n must be either 2^7, 2q^3, 8p, or 2qr with q and r
odd primes. The case n=2^7 is not possible because then n+1=129, which is
not a 10th power of a prime. Note that n+1 = s^10 means s must be odd and
n = s^10-1 = (s-1)(s+1)(s^4+s^3+s^2+s+1)(s^4-s^3+s^2-s+1). For odd s>1, 8
divides (s-1)(s+1). Hence 8 divides s^10-1, implying that the forms 2q^3
and 2qr are not possible. So the remaining case is n=8p. The
factorization of s^10-1 shows that after dividing by 8, we are always left
with a composite number. Hence the form 8p is also not possible.
Suppose n is odd. Then n+1 is even, which implies s^10 is even, which
implies s=2. This gives us the only solution, n=1023.
I'm guessing that a similar proof can be worked out for n = 2^(p-1)-1 where
p is prime. This proof was for p=11.
Tony
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