Assume $n=5$.
We have $p(n)=$
5
4 + 1
3 + 2
3 + 1 + 1
2 + 2 + 1
2 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1
What I want to get is the number of partitions in which the maximum integer is $m$, for each $1\leq m \leq n$.
For the previous example, #5=1, #4=1, #3=2, #2=2, #1=1.
Can anybody help for the general case where $n$ is unknown? Thanks.
On
Your problem is equate to finding factor of $x^n$ in :
$(1+x+x^2+...+x^n)^n = (\frac{1-x^{n+1}}{1-x})^n$
And subtract $n!-1 $ from that.
Because if you difine $f = (1+x+x^2+...+x^n)$ for every summation that is equal $n$ we have one term in polynomial $f^n$ which consist of $x^n$. for example for $1+1+1+...1 = n $ we have term $x.x.x.....x = x^n $ that's enough to choose $x $ from each parantes of $f^n $ or for $n$ we have $x^n .1.1.1....1 $ that's enough to choose $x^n$ from one pranteses of $f^n$ and choose $1$ from others.
But we count some of the summations several time and the number of what we count it several is $n! -1$
So your answer is : (factor of $x^n$ in $f^n$) $- (n!-1)$
You will not find a simple formula for the quantity you want. If $P_m(n)$ is the number of partitions of $n$ for which the largest part is $m$, then you have $p(n)=\sum_{m=1}^n P_m(n)$, where $p(n)$ is the number of partitions of $n$. This would then give a nice formula for $p(n)$, but we know that this does not exist. You can check Wikipedia for a generating functions though.