# What's the n-th term of this sequence (partition numbers)?

For the sequence of unique partitions of integer $n$: $n, \\ (n-1)+1,\\ (n-2)+2,\ (n-2)+1+1,\\ (n-3)+3,\ (n-3)+2+1,(n-3)+1+1+1,\\ (n-4)+4,\ (n-4)+3+1,\ (n-4)+2+2,\ (n-4)+2+1+1,\ (n-4)+1+1+1+1,\\ ...etc.$

the sequence of the number of items in each partition is: $1,2,2,3,2,3,4,2,3,3,4,5,2,3,3,4,4,5,6,2,3,3,4,3,4,....etc.$

What's the $nth$ term of this sequence? Does a closed form expression for it exist? My guess is not, but I can't prove it. The need for one arises from the generating function for a permutation of a set of objects in a set of containers.

• Really struggling to make sense of this question. Partitions are usually written with addition, not subtraction, e.g. partitions of $4$ are $4$, $3+1$, $2+2$, $2+1+1$, $1+1+1+1$. I don't understand how $n-1$ is a partition of $n$, nor $n-2$, $n-1-1$ etc. Perhaps you could clarify? Commented Apr 22, 2017 at 0:34
• @N Shales Apologies, don't know what I was thinking. I had meant to write these in the form (n-2)+2, etc. It's fixed now. Commented Apr 22, 2017 at 1:11
• Why do we need consider $n, n-1, n-2$ etc. instead of just looking at the partitions $1, 2, 1+1, 3, 2+1, ...$? Note that some of these will not be partitions for small values of $n$ - e.g., $(n-3, 3) = (1,3)$ is not a partition for $n = 4$ since partitions are usually defined as decreasing tuples. Commented Apr 22, 2017 at 1:30
• @Jair Taylor my reason for writing the sequence of partitions this way is that it directly relates to a problem I'm working on. I'm certain there are many other ways to write this, this way is clearest for what I'm trying to do (which is count the number of values used in each partition of $n$, starting from $n$ and ending at $n \ = \ 1 + 1 + 1 + ..$). Commented Apr 22, 2017 at 1:43

$0, 1, 1, 2, 1, 2, 3, 1, 2, 2, 3, 4, \ldots$
• Enumeration of partitions is generally hard (there's no simple formula even for the number of partitions of $n$) Commented Apr 22, 2017 at 2:55