[seqfan] Re: A283364 and a Diophantine equation
shevelev at bgu.ac.il
Wed Mar 8 13:38:49 CET 2017
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
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.
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
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.
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