[seqfan] Re: A283364 and a Diophantine equation

[Vladimir Shevelev]

Dear SeqFans,

Please, ignore the second part of 
my previous message. I submitted
A283455 which essentially differs
from A283364 beginning with a(15).
Below I prove that iterms>6  in A283455
are primes or squares of primes only.

As in A283364 one can prove that all a(n)>6 
are odd. Let a(n)>6. It is clear that a(n) is either 
prime or 
semiprime. Let us show that in the latter case 
it is square of prime. Indeed, let a(n)=p*q, p<q.
Then 2^a(n)-1 is divisible by 2^p-1<2^q-1. 
Thus both of them are Mersenne primes. Let us 
show that 2^(p*q)-1 differs from
 (2^p-1)^u*(2^q-1)^v, u,v>=1. Indeed the 
equality is possible only in case p*u+q*v=p*q.
Then p|v and q|u. Let u=q*a, v=p*b. Then
a+b=1 which is impossible for u,v>=1.
Hence, 2^(p*q)-1 has a third prime factor and 
p*q is not a member.
Are there other terms which are squares of 
primes except for 9 and 49?
 Note that, since for prime p, 2^(p^2)-1
differs from (2^p-1)^p, so  if p^2 is a term, 
then for a Mersenne prime 2^p-1 and some
 t>=1, the  number (2^(p^2)-1)/(2^p-1)^t 
should be prime or power of prime.

Best regards,
Vladimir
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 07 March 2017 15:28
To: seqfan at list.seqfan.eu
Subject: [seqfan] A283364 and a Diophantine equation

Dear SeqFans,

Yesterday I submitted A283364:
 Numbers n such that both numbers 2^n+-1
have <=2 distinct prime factors. I proved
that the terms >9 are prime.


Best regards,
Vladimir

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