Conjugate m-dimensional partitions

[franktaw at netscape.net]

I should, first of all, have stated that more clearly.  The actual 
condition is that n <= m+2+floor((m+1)/2)

If T(n,m) is the number of m-dimensional partitions of n, up to 
conjugacy, then T(n,m) - T(n,m-1) is the number which actually require 
all m+1 dimensions for the diagram.  In other words, each of the m+1 
points with a single 1 coordinate and all others 0 must be present; the 
origin must also be present.  This is m+2 points.

Now as we increase n above m+2, the first points we can add are of two 
types: those with a single 2 coordinate, and those with two 1 
coordinates.  This means that at most 2(n-(m+2)) dimensions can be used 
by these additional points.  Any other points don't add any new 
dimensions.  (It doesn't matter which ones, because we are only 
concerned up to conjugacy.)  Since m+1 dimensions are available, the 
value is fixed when m+1 >= 2(n-(m+2)); this gives the specified 
condition.

Franklin T. Adams-Watters

-----Original Message-----
From: Max <maxale at gmail.com>
  > Your three "guessed" values are correct, assuming the accuracy of 
your
 > other numbers. The final increments in the columns become fixed (as
 > you conjecture) once you get to twice the dimension;

 Do you have a proof for that?


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