[seqfan] Re: Self-stuffable numbers
M. F. Hasler
oeis at hasler.fr
Thu Dec 20 05:00:47 CET 2018
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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