coverage of plane partitions
wouter meeussen
wouter.meeussen at pandora.be
Tue Sep 7 22:50:22 CEST 2004
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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