coverage of plane partitions
David Wasserman
dwasserm at earthlink.com
Wed Sep 8 06:13:31 CEST 2004
The main difficulty I see here is that sequence authors have different methods of handling trivial values. I would search for this by picking out a small nontrivial piece. The 5th row is the first one that has more than 2 consecutive nonzero values, so we can lookup 3,15,3,3. This produces a small set of matches. When one defines a triangular table, often there's another natural way to define it that makes the rows read in the opposite direction. So we should also lookup 3,3,15,3. This produces another small set of matches. Since the table is not obviously triangular, we can also view it as a square table, which would be read by antidiagonals. The 9th antidiagonal is the first one that has more than 2 consecutive nonzero values, so we can lookup 3,21,3,1 and 1,3,21,3. These also produce small sets of matches. None of these match your table, so I think we can conclude that it's not in OEIS.
v/r,
David
>dear seqfanners,
>
>I came across a table that is not evidently triangular, but might hold some appeal.
>It is the classification of plane partitions of n according to the number of ways it can be
>'extended' to a plane partition of n+1 by adding 1 element to it.
>Stated differently, it counts how many partitions of n have k different partitions of n+1 just
>covering it :
>
> k:=1 k:=2 k:=3 k:=4 k:=5 k:=6 k:=7 k:=8 k:=9 k:=10 k:=11 k:=12
>
> n:=1 0 0 1 0 0 0 0 0 0 0 0 0
> n:=2 0 0 3 0 0 0 0 0 0 0 0 0
> n:=3 0 0 3 3 0 0 0 0 0 0 0 0
> n:=4 0 0 6 6 0 1 0 0 0 0 0 0
> n:=5 0 0 3 15 3 3 0 0 0 0 0 0
> n:=6 0 0 9 21 6 12 0 0 0 0 0 0
> n:=7 0 0 3 34 21 25 3 0 0 0 0 0
> n:=8 0 0 10 45 36 54 15 0 0 0 0 0
> n:=9 0 0 6 54 72 108 36 6 0 0 0 0
> n:=10 0 0 9 84 102 172 117 15 0 1 0 0
> n:=11 0 0 3 84 174 306 228 54 7 3 0 0
> n:=12 0 0 18 114 225 483 447 162 18 12 0 0
> n:=13 0 0 3 114 348 724 824 369 66 37 0 0
> n:=14 0 0 9 171 453 992 1476 774 193 96 3 0
> n:=15 0 0 9 153 558 1566 2289 1566 483 240 15 0
> n:=16 0 0 15 219 750 2029 3634 2949 1086 552 63 0
>
>
>row sums are A000219, the weighted products (dot product with the k's) is A090984.
>If I would drop columns k=1 and k=2 (easy to show that any plane partition can be extended in at
>least 3 ways), then I got the ugliest of Offset problems.
>
>If any of you were to do a search for a freak like this, what strategy would you use?
>
>W.
>
>ps. I'd never thought to get so much irregularity from such a 'clean' source.
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