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

[M. F. Hasler]

On Mon, May 27, 2019 at 9:29 AM Neil Sloane <njasloane at gmail.com> wrote:

> 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


I think we agreed that terms with trailing 0 should be excluded, since
these are trivial solutions.


> 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's indeed not difficult to prove that 99 is in the sequence:
after the first initial terms (of 99k), the terms share in general the
leading digit, except when they cross a limit of the form m*10^p.
So one has to look at the largest multiple just below, say 99k* with k* =
floor(m*10^p), and the next term 99(k*+1) which is the smallest multiple
juste above this limit.
The former always has a digit 9 after the leading digit.
In case 10^p has an even number of digits, 99k* ends in a digit 0, so the
next term, 99(k* + 1) will end in a digit 9, same as the second digit of
the preceding term.
In case 10^p has an odd number of digits, 99k* ends in 99, 98, 97, ..., 91
(for m=1,...,9), so the next term, 99(k* + 1) will have a digit 9 in the
second place from the right ("10s digit").

A nearly identical reasoning works for 999, 9999, etc.
The principle of the proof (only consider the terms preceding and following
m*10^p, m=1...9) extends to numbers of different form, mostly requiring
just some more cases to be distinguished, depending on the possibilities
concerning the last few digits.

>    - 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.
>

I disagree with the last one, since for ANY integer n > 0, 10n has the
required property.

- Maximilian