[seqfan] Re: Very nice new sequence A329126 [1, 6, 42, 60, 139810, 126, ...]

[Neil Sloane]

Allan, very nice conjecture, very simple, has to be correct!    Even
without a proof, just finding the conjecture is a major step forward.
Please add it to the sequence A329126.

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 Tue, Nov 12, 2019 at 10:59 PM Allan Wechsler <acwacw at gmail.com> wrote:

> There's a clear pattern here, and I wonder if it persists. Express n in the
> form 2^k * d, where d is odd. Then (conjecture) the solution is 1 (0^(d-1)
> 1)^(d-1) 0^k. This fits all the examples given, and seems plausible.
>
>
> On Tue, Nov 12, 2019 at 10:32 PM Neil Sloane <njasloane at gmail.com> wrote:
>
> > A new sequence sent in by Alon Ran, A329126,
> > which starts 1, 110, 101010, 111100, 100010001000100010, 1111110,
> >
> > a(n) is defined to be the lex. earliest string of numbers such that when
> > read in every base b >= 2, a(n) is divisible by n.
> >
> > It follows from the definition that all the digits are 0 and 1.  The
> author
> > gives a number of other remarks, but it was not completely clear to me
> > which were empirical observations and which were theorems. I do not have
> > time to work on it, but it looks like a lovely problem. He gives 21
> terms.
> >
> > The strings are huge. If they are rewritten in base 10 you get A329000
> > = 1,6,42,60,139810,126,..., which makes it a lot easier to remember the
> > first few terms.
> >
> > I hope someone will look into this and figure out what is going on.
> >
> > It has keyword "base", but it does not depend on any one base - it
> depends
> > on all bases (so maybe that keyword is not justified).
> >
> > Neil
> >
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> >
>
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