I am in need of a way to calculate a few things about partitions. If we say there are 4 objects (n=4), then the partitions of 4 are:
4
3 + 1
2 + 2 *
2 + 1 + 1
1 + 1 + 1 + 1 *
* these partitions do not have a unique mode.
I'm looking for a function, p'(n), that will tell me how many partitions there are with unique maximum values (i.e. a unique mode). So p(4) = 5, but p'(4) = 3.
Also, I am looking for a function that can tell me what the unique maximum value is for a particular partition, i, in the set of partitions. For instance, in the partition 3+1, the unique maximum/mode is 3.
Can anyone help me with this?
Many thanks.
Partitions with a unique maximum part
Re: Partitions with a unique maximum part
The only definition of mode that I know is "the value that appears most often in a set of data" but that is clearly not what you mean here.
- PurpleBlu3s
- Posts: 73
- Joined: Mon Sep 19, 2011 5:49 pm
Re: Partitions with a unique maximum part
If you think of partitioning data based on value, then what I mean is the number of occurrences of the mode.
For example, we have a set of 8 data points:
1 1 2 3 2 2 2 4
Grouping them based on their value:
(2 2 2 2) (1 1) (3) (4)
Counting the size of the groups:
4 2 1 1
This is a partition of 8, and the number of occurrences of the mode is 4.
For example, we have a set of 8 data points:
1 1 2 3 2 2 2 4
Grouping them based on their value:
(2 2 2 2) (1 1) (3) (4)
Counting the size of the groups:
4 2 1 1
This is a partition of 8, and the number of occurrences of the mode is 4.

Re: Partitions with a unique maximum part
The classic way of calculating the partitions is recursive using the Pk(n) = Partitions of n with k as maximum:
https://en.wikipedia.org/wiki/Partition ... r_of_parts
So you can use these results and calc p*(n) = ∑i=2n Pi-1(n-i)
https://en.wikipedia.org/wiki/Partition ... r_of_parts
So you can use these results and calc p*(n) = ∑i=2n Pi-1(n-i)

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Re: Partitions with a unique maximum part
I know this is old, but I just saw this on reviewing the forums and I wanted to share my thoughts (spoilered for anyone who wants to work on it).
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