[seqfan] Re: Monotonic ordering of nonnegative differences

[Neil Sloane]

Max,  Thanks very much for the explanation.  (I will have to think about
how the reductions mod 2 and mod 3 work.)

You did not include your calculations in A173671, presumably because it was
not
a matter of simply running a program, but required
a series of steps, each one needing a separate program.  Would it be
possible for you to
document what you did?  For example, in a text file attached to that
sequence?
Giving enough details so that someone could check it, and carry out similar
calculations for the other 50 sequences that Clark looked at.

If we had that, then we could delete the dubious programs in A173671.

Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com



On Fri, Oct 11, 2019 at 1:31 PM Max Alekseyev <maxale at gmail.com> wrote:

> Correction: I meant to say - there was NO "search limit" in my submission.
>
> On Fri, Oct 11, 2019 at 12:33 PM Max Alekseyev <maxale at gmail.com> wrote:
>
> > Hi Neil,
> >
> > I'd like to comment on A173671 -- there was "search limit" in my
> > submission.
> > I now see such a limit in a code submitted by someone else, but this
> > simply means that this code may produce incorrect results.
> >
> > I know two methods of proving that 3^m-2^n=k for a given k is insoluble
> in
> > m,k.
> > First is to find a suitable M (if it exists) such that the
> > congruence 3^m-2^n == k (mod M) is insoluble (which is easy to verify).
> >
> > Second is to find the integral points on the following 6 elliptic curves
> > corresponding to residues of m and n modulo 2 and 3, respectively:
> > y^2 = x^3 + k
> > y^2 = 2x^3 + k
> > y^2 = 4x^3 + k
> > 3y^2 = x^3 + k
> > 3y^2 = 2x^3 + k
> > 3y^2 = 4x^3 + k
> > If in none of the integral points y is a power of 3 and x is a power of
> 2,
> > then 3^m-2^n=k does not have integer solutions in m,n.
> > Computing integral points in many cases can done routinely in
> > Sage/Magma/etc.
> >
> > So, I did prove the numbers in my submission A173671, but I cannot say
> > much about the later-on additions (e.g., b-file) though.
> >
> > Regards,
> > Max
> >
> >
> > On Fri, Oct 11, 2019 at 11:37 AM Neil Sloane <njasloane at gmail.com>
> wrote:
> >
> >> Robert, thank you for catching those errors.  Yes, we will need to add a
> >> comment.
> >> Sadly, there are b-files too.  Should they be deleted, do you think?
> >> Another thing: the complementary sequences are also in the OEIS, e.g.
> >> A173671 ,
> >> which is the complement of A192111, and was submitted by Max Alekseyev.
> >> With a different search limit.  I will handle this, once we decide what
> to
> >> do.  Any comments, anyone?
> >>
> >> We have a rule that programs and b-files should not be based on
> >> conjectures, so should the
> >> programs be deleted too?
> >>
> >> I really hope we can keep the sequences, and obviously if we keep the
> >> sequences then we need to keep the programs, to show how they were
> >> calculated.  But the b-files?
> >>
> >> Best regards
> >> Neil
> >>
> >> Neil J. A. Sloane, President, OEIS Foundation.
> >> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> >> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway,
> NJ.
> >> Phone: 732 828 6098; home page: http://NeilSloane.com
> >> Email: njasloane at gmail.com
> >>
> >>
> >>
> >> On Thu, Oct 10, 2019 at 5:31 PM <israel at math.ubc.ca> wrote:
> >>
> >> > There are 52 sequences from A192110 to A192202, contributed by Clark
> >> > Kimberling, with Name of the form "Monotonic ordering of nonnegative
> >> > differences a^i-b^j, for i>=0, j>=0" for various values of a and b.
> >> >
> >> > From the Mathematica code, it seems they are all computed by assuming
> i
> >> <=
> >> > 40. I'm not aware of any theoretical justification for the assumption
> >> that
> >> > any term in the range of the Data (which might go up to several
> million)
> >> > will arise from i <= 40, although I have no counterexample and it may
> be
> >> > unlikely that there is one. These are related to Catalan's conjecture
> >> > (proved by Mihailescu), according to which 1 is not a member of any of
> >> > these sequences unless i=1 or j<=1 works. There are also modular
> reasons
> >> > for excluding some values (e.g. if prime p divides b but not a, then
> all
> >> > terms divisible by p are of the form a^i-1). But for many values > 1,
> I
> >> > don't think much is known rigorously.
> >> >
> >> > Should these sequences all get a Comment that the Data are
> conjectured?
> >> >
> >> > Cheers,
> >> > Robert
> >> >
> >> >
> >> > --
> >> > Seqfan Mailing list - http://list.seqfan.eu/
> >> >
> >>
> >> --
> >> Seqfan Mailing list - http://list.seqfan.eu/
> >>
> >
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>