[seqfan] Re: RE : Re:« King-walking » integers in a square box

[Andrew Weimholt]

On Fri, Jul 23, 2010 at 3:03 PM, Eric Angelini <Eric.Angelini at kntv.be> wrote:
>
> Thanks, Andrew, I will try
> to beat that by hand tomor-
> row!
> Regards,
> E.
>
>
> -------- Message d'origine--------
> De: seqfan-bounces at list.seqfan.eu de la part de Andrew Weimholt
>
> On 7/23/10, Eric Angelini <Eric.Angelini at kntv.be> wrote:
>>  This might constitute a seq S for Neil.
>>  [S starts 0, 3, 8, ...]
>
> The next term is at least 58
>
> 8 3 4 9
> 2 0 7 2
> 1 6 5 1
> 9 3 4 8
>

I looked a little closer at this one tonight, and I can now
also confirm that 58 is the next term.

For the 4x4 grid, in order to cover 0-59, we need the
multiset {0,1,1,2,2,3,3,4,4,5,5,6,7,8,9} which takes up
15 of the 16 spaces.

If the remaining tile is a duplicate
of one of {0,6,7,8,9}, then without loss of generalization,
we can say the duplicated tile is 0.

6,7,8,9, then must each be connected to a 1,2,3,4, and 5,
so they cannot go in the corners, and if they go on an edge,
they cannot be connected to each other or to one of the 0's.

If 6, 7, 8, 9, go on the edges, they must be on different edges, and then there
is no where to place the 0's so that they do not take away one of the five
required connections from some of the 6,7,8,9 tiles.

If one of 6, 7, 8, 9, goes in the interior, then only two can go on
the edges, but that
means we need to have two of them in the interior, which means only room fro
one on the edge is left, and three on the interior means no edges are
left, so we
are forced to put all 4 on the interior, but then we cannot place the
0's without
again taking away a required connection from at least two of the 6, 7,
8, 9 tiles.

The other possibility is to try to make the extra tile a 3rd instance
of one of {1,2,3,4,5},
but then we have the same problem of placing 5 tiles on the edges and
interior, such that
none of the tiles place on the edge is connected to one of the other 4
tiles. Again we see
that we are forced to have at least one in the interior, which leads
to having to have 4 in the
interior, and no where left to put the 5th.

Andrew