Conjugate m-dimensional partitions
franktaw at netscape.net
franktaw at netscape.net
Sun May 14 17:45:57 CEST 2006
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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