[seqfan] Re: nXk 1..Q matrices with each value adjacent to each other value

[Ron Hardin]

----- Original Message ----
> From: Ron Hardin <rhhardin at att.net>
> To: seqfan at seqfan.eu
> Sent: Sun, October 24, 2010 6:25:55 PM
> Subject: [seqfan] nXk 1..Q matrices with each value adjacent to each other 
>value
> 
> So far it seems like a difficult computational problem - T(n,k) = maximal 
>number 
>
> Q in a 1..Q nXk array such that every value appears at least once adjacent  to 

> every other value.

T(n,k) table somewhat expanded, with ?'s for missing values

.1..2..2..3..3..3..3..4..4..4..5..5..5.5.5..5..5..6.6.6
.2..3..4..4..5..6..6..7..7..7..?..8..?.9.9.10.10.10.?..
.2..4..5..6..6..7..8..?..9.10.10.11.11.?.?.12..?..?....
.3..4..6..7..8..?..?.10.11..?.12..?..?.?.?..?..?.......
.3..5..6..8..9.10.11.12.12.13..?..?..?.?.?..?..........
.3..6..7..?.10.11.12..?..?..?..?..?..?.?.?.............
.3..6..8..?.11.12.13..?..?..?..?..?..?.?...............
.4..7..?.10.12..?..?..?..?..?..?..?..?.................
.4..7..9.11.12..?..?..?..?..?..?..?....................
.4..7.10..?.13..?..?..?..?..?..?.......................
.5..?.10.12..?..?..?..?..?..?..........................
.5..8.11..?..?..?..?..?..?.............................
.5..?.11..?..?..?..?..?................................
.5..9..?..?..?..?..?...................................
.5..9..?..?..?..?......................................
.5.10.12..?..?.........................................
.5.10..?..?............................................
.6.10..?...............................................
.6..?..................................................
.6.....................................................

The diagonal is 2*n-1.  I wonder if there's a construction rule.

Some square solutions (each value adjacent at least once to every
other value)

A solution for 1X1  
..1
A solution for 2X2  
..1..2
..3..2
A solution for 3X3  
..1..2..3
..4..5..1
..2..3..4
A solution for 4X4  
..1..2..3..4
..5..6..7..1
..2..4..5..3
..7..7..1..6
A solution for 5X5  
..1..2..3..4..5
..6..7..8..9..1
..2..3..4..6..3
..5..9..2..8..5
..6..7..4..1..7
A solution for 6X6  
..1..2..3..4..5..6
..7..8..9.10.11..1
..2..3..4..6..7..3
..5..3.11..8..5.10
..9..6..2..4..1..8
..1.11..9..7.10..2
A solution for 7X7  
..1..2..3..4..5..6..7
..8..9.10.11.12.13..1
..2..3..4..5..6..7.11
..4..1..6..2.10.12..2
..9.12..3.13..8..4..7
..6..8..7.10..5.13..9
.11..3..5..1..9.11..8