[seqfan] Re: A245486

[israel at math.ubc.ca]

For odd prime p, this leads to a criterion for 2*p to 
occur in the sequence, which I have entered into a Comment:

More generally, let m = A014664(i), i >= 2. If m is odd, 2*A000040(i) 
occurs in the sequence iff A000040(i) = A006530(2^m-1), in which case it is 
a(2^m-1). If m is even, 2*A000040(i) occurs in the sequence iff A000040(i) 
= A006530(2^(m/2)+1), in which case it is a(2^m).

Are there any odd members of A006881 that do not occur in the sequence?

Cheers,
Robert Israel

On Jul 24 2014, Jack Brennen wrote:

>Actually, it's impossible.  2^n+1 is never divisible by 23, and
>when 2^n-1 is divisible by 23, it's also divisible by 89.
>
>So 46 cannot occur in the sequence.
>
>
>
>On 7/23/2014 3:25 PM, Jack Brennen wrote:
>> When do you get the number 46 in the sequence?
>>
>> That would imply 2^n-1 or 2^n+1 with largest prime factor 23, which
>> seems unlikely.
>>
>>
>>
>>
>> On 7/23/2014 2:59 PM, Frank Adams-Watters wrote:
>>> I have a new sequence in editing state: https://oeis.org/draft/A245486 -
>>> Greatest prime factor of n times greatest prime factor of n+1.
>>>
>>> 1) I think it's the case that recent results show that this sequence
>>> goes to infinity; equivalently, each member of A006881(products of 2
>>> distinct primes) occurs only finitely many times. Can someone confirm
>>> this?
>>>
>>> 2) I can almost prove that every member of A006881 does occur in this
>>> sequence. Can anyone find a proof? (Or a counterexample.)
>>>
>>> Franklin T. Adams-Watters
>>>
>>> _______________________________________________
>>>
>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>
>>>
>>>
>>
>
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