# Towards a comment on A134941 Mountain numbers

Hans Havermann pxp at rogers.com
Sun Nov 25 23:24:51 CET 2007

```On Nov 24, 2007 7:31 PM, Jonathan Post <jvospost3 at gmail.com> wrote:
> n  a(n)
> 1  1
> 2  0
> 3  8
> 4  56 (7 each beginning 12, 13, 14, 15, 16, 17, 18, 19)
> 5  <5488
>            2744 =(56 ascending substrings * 49 descending substrings)
>           + 2744 (49 ascending substrings * 56 descending substrings)
>           - those which were double-counted in the above

I get 252, confirming Hans Havermann's number.

In fact I get 1 0 8 56 252 784 1792 3108 4166 4352 3544 2232 1068 376 92 14 1 0
so it looks like I confirm all of his results.  My program might be
buggy too, and who knows, maybe buggy in the same way!

The way I wrote it can easily generate the list of all 21846 actual
numbers, too.  So I formatted that as a b.txt file and sent it offlist
to the relevant folks (seems a bit too long to post here!)

> Partial sums of this will give, ultimately, the total number of
> mountain numbers of any length of which 12345678987654321 is the
> largest.

Yup, 21846 seems to be it.

> Is it mere coincidence that the first 4 values coincide with A093134 A
> Jacobsthal trisection?

Not entirely: I think that sequence is an upper bound, if I understand
its definition correctly, something like "n digit numbers beginning
with 1 and ending with 1 and no two consecutive digits equal" should
be an equivalent description?

> All this, of course, is base 10.  What are the sequences and counts
> with other bases?

I think that doing a "base" sequence in base 10 is already
uninteresting enough. If you expect to uncover anything else
interesting here please let me know -- it's a trivial modification of
my program to replace the "9" that's currently in there with any other
value for the counting (displaying the actual numbers, as opposed to
just counting them, would take a bit of hacking in any base larger
than 10).

--Joshua Zucker

```