[seqfan] Re: K(n)—>a(n).

[M. F. Hasler]

On Tue, Dec 10, 2019 at 10:30 AM David Sycamore via SeqFan <
seqfan at list.seqfan.eu> wrote:

> For integer n let K(n) be the permutation of the digits of n formed by
> sequentially combining the greatest and smallest digits in adjacent pairs
> until running out of digits to play with. Examples: K(1)=1, K(10)=10,
> K(123)= 312, K(277272)=727272, K(539204)=905243.
>
> We compute a(n) as follows:
> Take the absolute difference |n-K(n)| to get a new number. Repeat the
> process with that number until reaching a number m for which K(m)=m, then
> a(n)=m.
>

Maybe a more "moderate" suggestion could be:
in general, first consider the most simple / natural variant, before going
further to what could be considered as too contrived.
In this example, rather than concatenating pairs of largest & smallest
digits, one might first think of sorting the digits in (weakly) increasing
/ decreasing order.
Do the corresponding sequences "K(n)", n-K(n), and  iterations of that
already exist?
Or is there a "drawback" making irrelevant these simpler variants and
motivating your more exotic approach? (If so, it should be explained.)

Then, prior to considering *iterations* of *n-K(n)*, the first step would
be to consider K(n) on its own.
If K(n) is not worth being in OEIS, then *iterations* of  *n-K(n)*  (two
levels of further "complication") won't be "worth", either.
The next step would be to consider/add/study  b(n) = K(n)-n  (no need for
absolute value when we know that K(n)>n;
in other cases I'd suggest to keep the sign, which is ignored in search and
easier to remove than to reconstruct),
and only then the iterations of n -> b(n).

If any of the more basic sequences indeed are in OEIS, then it would be
appreciable to include a reference to the A-numbers in the post to the
SeqFan list,
if possible (to please the readers) in the "clickable" form oeis.org/Axxx.

- Maximilian