[seqfan] Re: A245486

[israel at math.ubc.ca]

A candidate would be 3*89 = 267. In order for this to occur in A245486, 
there must be nonnegative integers a,b such that b == 28*a mod 44 and the 
greatest prime factor of 2^a*3^b - (-1)^(a + (b-28*a)/44) is 89. I haven't 
found any. But it's not obvious to me how to make this search finite.

Cheers,
Robert Israel


On Jul 24 2014, israel at math.ubc.ca wrote:

>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
>>>>
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>>
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