[seqfan] Re: Fwd: Your A051217 Nonnegative numbers of the form 6^x-y^2.

[Vladimir Shevelev]

I agree with David: the sequence is, of course,  well-defined. But it is clear that without an upper estimate for x it cannot be computable. Since till now such an estimate is unknown, then we should use an empirical conjecture. For example, one can conjecture that, for the n-th term of the sequence, x  not exceeds n. In such case one can  write in the title "a(1)=0; for n>=2, a(n) is the least number >a(n-1) of the form 6^x-y^2 with x<=n  and a(n)=0, if such number does not exist". Then, e.g.,  23  pretending to be a(8)  is verified under condition  6^x-y^2-23=0, x=1,...,8. Now this sequence is  well-defined and well computable. In a comment one can pose a conjecture:" It is a monotone sequence containing all numbers of the form 6^x-y^2".

Regards,
Vladimir

----- Original Message -----
From: David Wilson <davidwwilson at comcast.net>
Date: Tuesday, November 15, 2011 1:53
Subject: [seqfan] Re: Fwd: Your A051217 Nonnegative numbers of the form 6^x-y^2.
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> Stop using that terminology.
> 
> The sequence is perfectly well-defined.
> n >= 0 is in the sequence iff it is of the form 6^x - y^2.
> Every n >= 0 is either of this form or it is not.
> 
> The question is, is the sequence computable?
> Specifically, is there an algorithm that determines if n is in 
> or out of 
> the sequence?
> 
> The real question is,
> 
> On 11/11/2011 9:57 AM, Charles Greathouse wrote:
> > No, scratch that.  I can show that the sequence is well-defined
> > through ten million.  No general proof yet, though.
> >
> > Charles Greathouse
> > Analyst/Programmer
> > Case Western Reserve University
> 
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 Shevelev Vladimir‎