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A generating function for the total number of partitions of $n$ is: $$\prod_{i=1}^n\sum_{j=0}^n x^{ij}$$ The polynomial generated by this generating function will have some term $x^n$, the coefficient of which will equal the total number of partitions of $n$.

Let $n$=6. The eleven partitions of 6 are the following: 6, 5+1, 4+2, 3+3, 4+1+1, 3+2+1, 2+2+2, 3+1+1+1, 2+2+1+1, 2+1+1+1+1, 1+1+1+1+1+1. I have set created two cases.

Case 1: where $k\le3$ and the maximum value of a term in a partition is 4 (where 4 is equal to $\frac{2n}{k}$). The partitions of 6 that follow this are: 4+2, 3+3*, 4+1+1*, 3+2+1, 2+2+2.

Case 2: where $k\le4$ and the maximum value of a term in a partition is 3 (where 3 is equal to $\frac{2n}{k}$). The partitions of 6 that follow this are: 3+3, 3+2+1, 2+2+2*, 3+1+1+1*, 2+2+1+1.

I am also investigating the specific partitions (noted with a "*") where every term in the partition subtracted from $\frac{2n}{k}$ creates another partition of $n$. For example, in case 1, where $\frac{2n}{k}=4$, subtracting the terms 4, 1, and 1 (the partition 4+1+1) from 4 gives 0, 3 and 3 (the partition 3+3). Also in case 2 where $\frac{2n}{k}=3$, subtracting the terms 3, 1, 1, and 1 (3+1+1+1) from 3 gives 0, 2, 2 and 2 (the partition 2+2+2). The other partitions create themselves when each term is subtracted from $\frac{2n}{k}$.

What generating function can be used to

1) find the number of partitions that satisfy the parameters where $k\le$ some number?

2) find the number of partitions where maximum value for a term in a partition does not exceed $\frac{2n}{k}$?

3) find the number of partitions that, when the terms are subtracted from $\frac{2n}{k}$, create a different partition of $n$?

I am very new to combinatorics and would appreciate dumbed-down answers. Thank you!

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