[seqfan] Re: Self-stuffable numbers
Neil Sloane
njasloane at gmail.com
Thu Dec 20 14:32:06 CET 2018
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
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 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> wrote:
>
> > 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!
> >
>
> Indeed a great find!
>
>
> > 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?
> >
>
> I don't claim that this is a full understanding, but the following proves
> that it works for any length >= 12 and explains some structure of this
> solution:
> Let x(k) = ((103*10^k+R(k))*1000 + 8)*10^6 with R(k)=(10^k-1)/9
> and S(x) = the result of "stuffing" x with itself (assuming it's possible),
> i.e., inserting its own digits in the spaces made after each nonzero digit.
> Then,
> S(x(k)) = ((110303*100^k+R(2k))*10^6 + 808)*10^12
> and one has (see also further below)
> S(x(k)) / x(k) = (10^(k+3) + 1) * 1 069 750 000
> which proves that x(k) is self-stuffable.
>
> 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).
>
> I did a quick scan for for other families of solutions which are multiples
> of repunits,
> but m = 928 (* 10^k, k>=6) is the only one I found, which would work for
> arbitrarily long repunits.
> (Single digit numbers m = 2, ..., 9 (multiplied by some 10^k) work only for
> repunits with a few selected lengths <= m,
> namely: 2 * {1, 11} ; 3 * {1, 111} ; 4 * {1, 11, 111, 1111} ; 5 * {1,
> 11111} ;
> 6 * R({1, 2, 3, 4, 5, 6}) ; 7* R({1, 3, 5, 7}) ; 8* R({1, 2, 4, 5, 7, 8})
> ; 9* R({1, 3, 7, 9}),
> for example, m = 6, R(2) => x = 66 * 10^10, S(x) =66600006 * 10^16 = x *
> 101 * 9991 * 10^6.)
>
> That said, the question of a different example analogous to your family
> {x(k) = R(k+3)*928*10^6, k >= 0} remains completely open.
> (So far we knew that A322323 is the union of the infinite families of terms
> associated to each of the roots A322002, but any root M will produce only
> terms whose lengths are multiples of sumdigits(M), or, for the exceptional
> roots 22 and 126, also terms of length { k*sumdigits(M) + length(M), k >= 0
> }.
> This did not guarantee that terms of length p exist for any large prime p.
> Using A322002(1..23) one gets terms with lengths being any multiple of any
> prime in {2, 3, 5, 7, 11}. This together with the roots {R(k+3)*928, k>=0}
> ensures that A322323 has terms of any given length > 1.)
>
> - Maximilian
>
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