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

[Hans L]

William,
Thanks for those results.  I don't have Mathematica but I was able to open
the file in the free CDF Player 11.  I don't suppose its possible to have
adjustable parameters and be able to evaluate other ranges from within the
CDF player, or is it only for viewing precomputed results?

I think the "heuristic observation" you talk about is basically the same
idea I was trying to put into a formula in my original post. I'm not sure
how clearly I came across in explaining there, but I guess my assumption
was that the ideal minimized divisor: n1!n2!n3!... would come in some form
where part counts are split between m or (m+1), or possibly 1 for the final
n'k'.  So the smaller parts 1,2,3 would be repeated (m+1) times:
n1=n2=n3=...=(m+1),  and the higher parts would be repeated m times:
...=n'k-3'=n'k-2'=n'k-1'=m
and the last part count n'k' might be equal to either 1 or m depending on
if it fits.
Its difficult to know how to even express some of these ideas in
discussion, much less prove/disprove that such a form of partition would
always results in the lowest divisor, but intuitively this seems like the
way it would be.

I guess what I was mainly trying to get at is whether it is possible to
compute without enumerating all possible partitions.

Hans


On Fri, Apr 5, 2019 at 8:57 PM William Keith <william.keith at gmail.com>
wrote:

> 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
>
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