[seqfan] Re: Arithmetic progressions such that adjacent terms have a common non-zero digit

[Neil Sloane]

Interesting sequence that:  99, 990, 999, 1998, 2997, 3996, 3999, 4992,
4995, 5994, 6875, 6993, 6996,
7992, 8125, 8704, 8991, 9856, 9900, 9984, 9990, 9999

If you can find a proof that the first five (say) terms are correct then
you could (and should) add it to the OEIS.

It is certainly in the same spirit as many other sequences we have.

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 Mon, May 27, 2019 at 2:08 AM David Radcliffe <dradcliffe at gmail.com>
wrote:

> Daniel Griller asked the following question:
>
>     Does there exist a positive integer n such that every term in the
> sequence n, 2n, 3n, 4n, 5n, ... has a non-zero digit in common with the
> next term?
>
>     (Source: https://twitter.com/puzzlecritic/status/1125035277557354497)
>
> If we allowed 0 as a common digit then every multiple of 10 would be a
> solution.
>
> I was able to prove the following:
>
>    - 99 is the smallest positive integer with this property.
>    - 10^k - 1 has this property for every k > 1.
>    - A positive integer n has this property if and only if 10*n has this
>    property.
>
> Note that the adjacent terms k*n and (k+1)*n usually have the same leading
> digit, and it suffices to look at the boundaries where the leading digit
> changes.
>
> The sequence of numbers having this property might begin as follows, but I
> have not proved this.
>
> 99, 990, 999, 1998, 2997, 3996, 3999, 4992, 4995, 5994, 6875, 6993, 6996,
> 7992, 8125, 8704, 8991, 9856, 9900, 9984, 9990, 9999
>
> Is there an efficient algorithm to generate the terms of this sequence?
>
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