[seqfan] Re: Given any partition of n into k parts, what's the maximum permutations of parts

[William Keith]

Mathematica with the old Combinatorica package can handle these
calculations very easily.  I have run a search for your maxima up to n=30,
which took just a few minutes.  I have uploaded the notebook to my Drive at
https://drive.google.com/file/d/1bWsbg25zrDK-P8exc-_58naHBl5SJU0L/view?usp=sharing
.  You should be able to retrieve it; let me know if you have trouble.  If
you do not have access to Mathematica then Wolfram's free reader will read
the file.

The number of permutations of a permutation with exactly k parts, of which
n1 are 1s, n2 are 2s, etc., is

k! / (n1! n2! n3! ... ).

When n is less than the k-th triangular number and you do not have the
maximum possible number of distinct parts, then you wish to minimize n1!
n2! n3!... .  Broadly speaking, the way to do so is to have the numbers of
parts be as equal as possible, or to have the lowest possible average.  You
end up with a very "flat" partition.  If you consider the ability to move
single units in the partition around, stepping through possible partitions,
you are always removing a part from an outer corner, reducing the number of
parts of some size A by 1 and increasing the number of parts of some size B
by 1, and then putting this single unit at an inner corner, reducing the
number of parts of some size C by 1 and increasing the number of parts of
some size D by 1.  Various conditions on the inequalities between A, B, C,
and D tell you whether (A-1)!(B+1)!(C-1)(D+1)! is larger or smaller than
A!B!C!D!. Perhaps an argument can be made which rigorizes the heuristic
observation above.

Best,
William Keith