[seqfan] Re: R: Re: Self-stuffable numbers
M. F. Hasler
seqfan at hasler.fr
Fri Dec 21 15:26:22 CET 2018
On Fri, Dec 21, 2018 at 7:22 AM Mason John <john.mason at lispa.it> wrote:
> I had decided to stop calculating new values until some optimisation had
> been found.
However as there has been renewed interest, I have calculated values
> through 10^18 and I have posted a new b-file.
> Here are the values post a(84) of the previous file. It took my PC about 8
Al of these are given in terms of the roots in the b-file
and the larger roots [1115008, 112621104, 1031111008, 10311111008,
(not in the b-file because it stops at 10^6) ;
among these only the first two come in addition to the infinite sequence
The main challenge is to find larger (primitive) roots, i.e.,
terms of A322323 (without repetitions and) with trailing zeros removed.
Arbitrarily many larger terms of A322323 are then computed
> -----Messaggio originale-----
> Da: SeqFan <seqfan-bounces at list.seqfan.eu> Per conto di Neil Sloane
> Inviato: giovedì 20 dicembre 2018 14:32
> A: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Oggetto: [seqfan] Re: Self-stuffable numbers
> I'm following this discussion with great interest.
> Looks like a pot of bouillabaisse !
> When the soup is cooked, I hope you (Maximilian) will write up a report,
> maybe put it on the arXiv, or at least add it to the main sequence A322323.
> Best regards
> On Wed, Dec 19, 2018 at 11:01 PM M. F. Hasler <oeis at hasler.fr> wrote:
> > On Wed, Dec 19, 2018 at 9:23 AM Hans Havermann <gladhobo at bell.net>
> > > https://oeis.org/A322323
> > > The appearance of 103008000000 at digit-size 12 was an unexpected
> > delight.
> > > At digit-size 13 we have 1031008000000; at 14, 10311008000000; at
> > > 15, 103111008000000; and yes, the pattern continues!
> > >
> > > Would understanding why this particular infinite family of
> > > self-stuffable numbers exists allow one to find (or rule out the
> > > existence of) other, larger examples?
> > We can understand a little more why & how this works by noticing that
> > x(k) = 928 * R(k+3) * 10^6, i.e., x(0) = 111*(...), x(1) = 1111*(...)
> > etc and S(x(k)) = 4279 * 928/4 * R(2(k+3)) * 10^12.
> > From here we see that
> > S(x(k)) / x(k) = R(2(k+3))/R(k+3) * 10^6 * 11*389/4 (and of course
> > R(2(k+3))/R(k+3) = 10^(k+3) + 1).
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