Stars and Bars with an odd constraint. Stars and bars is a classic combinatorics question, but I've run into a variant I've never seen before.

I have $n$ stars. Rather than group them into piles using $k - 1$ bars, I want to group them up by any number of bars, such that no pile of stars contains less than $c$ stars. Order of piles doesn't matter. Order of stars doesn't matter.

In this case, I don't know how many bars there are in general.
I know that there can be $0$ and I know that there can be $\frac n c - 1$ bars (for nice values of $n,c$).
The issue here is that the minimum number of stars per pile is throwing a wrench into how I know to approach the problem. I could sum over $k = 0 \text{ to } \frac n c - 1$ and solve this sub-problem.

I have $n$ stars and $k$ bars. How many combinations have no pile with less than $c$ stars?

This problem, I'm also not familiar with how to solve. How do I approach either of these?
 A: I don't know about the first count in which the number of bars is not specified. However if there are $k$ bars then there are $k+1$ summands, and since each one has size at least $c$ the least $n$ one can make is $c(k+1).$ That can be made in only one way (one star in each pile). For $n>c(k+1)$ we could let $m=n-c(k+1)$ and then partition $m$ into $k+1$ or fewer parts using the usual partition function.
For example there are various partitions of $6$ into three or fewer parts $1+1+4,1+2+3,2+2+2,1+5,2+4,3+3,6$ One notation for partitions of $n$ into $k$ parts used is $p(n,k),$ and this function has been studied (and there are recursive ways to compute it. It would then be summed for an initial segment of $k$ values to apply to your situation. [Sorry about the notation clash here, using $k$ for number of bars, and also in the $p(n,k)$ function...]
So to apply this to your situation, if you had $k=2$ bars, and therefore three summands, and your required $c$ was say $4,$ one could not make any $n$ less than 12. $12$ itself could be made in only 1 way as $4+4+4,$ but after subtracting $c(k+1)=12$ one would be "partitioning" zero (which usually isn't defined). However to use the above partitions of $6$ into 3 or fewer parts, it would be applied to $n=18$ given the above $k,c.$ and give for example $4+4+10$ from the partition $6=6$ of 6, and $5+6+7$ from the partition $1+2+3$ of $6.$
So I believe in case the number of bars is known, the count can be found as outlined using the information about the partition function $p(n,k)$ [Maybe look up "partitions of n into k parts".]
