[seqfan] R: Re: Self-stuffable numbers
Mason John
john.mason at lispa.it
Fri Dec 21 12:22:41 CET 2018
Hi
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 hours.
If only someone could prove that all values were even, or some other criterion ...
85 1031111008000000
86 1115008000000000
87 2072500000000000
88 2455000000000000
89 4444000000000000
90 5425000000000000
91 8000800000000000
92 8080000000000000
93 8800000000000000
94 10311111008000000
95 12176000000000000
96 103111111008000000
97 112621104000000000
98 131328000000000000
99 140625000000000000
100 143136000000000000
101 148500000000000000
102 150000150000150000
103 161280000000000000
104 172800000000000000
105 190440000000000000
106 202020202020202020
107 208800000000000000
108 217800000000000000
109 220022002200220022
110 231000231000231000
111 235008000000000000
112 240000240000240000
113 262800000000000000
114 281250000000000000
115 288000000000000000
116 290700000000000000
117 297000000000000000
118 460800000000000000
119 501372000000000000
120 506250000000000000
121 526320000000000000
122 549000000000000000
123 562500000000000000
124 576000000000000000
125 580500000000000000
126 600000600000600000
127 606060000000000000
128 622080000000000000
129 666000000000000000
130 671130000000000000
131 734400000000000000
132 765000000000000000
133 853200000000000000
134 854001000000000000
john
-----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
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
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>
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