Combinatorics: Find a recursive formula for partitions of integers into three partitions. I need to find a recursive formula for $p_n$ the number of ways to partition $n$ into three partitions.  For example if we look for the partitions of $6$ then they are $1+1+4$, $1+2+3$, and $2+2+2$.  Intuitively I look for the number of ways of partitioning $5$, which are $1+1+3$ and $1+2+2$, and try to relate this to the number of partitions of $6$.  Of course, for every partition of $n-1$ there is a corresponding partition of $n$ in which the (or a) maximal element is increased by $1$.  However, I'm having trouble finding a good characterization for the remaining partitions which are not formed in this way.  I could think about also increasing a minimal element, but then I would have to find a way to count the number of double-countings, which would mean counting the number of ways of partitioning $n=a+b+c$ such that $a\leq b\leq c$ and $a+1=b$ or $a+1=c+1$.  And I'm not even fully confidence that this description captures all of the ways that elements could get double-counted.
In the case of partitioning $5$ and $6$, increasing the maximal element takes the partitions $1+1+3$ and $1+2+2$ and yields $1+1+4$ and $1+2+3$.  Increasing the minimal element yields $1+2+3$ and $2+2+2$, so the double-counted element is $1+2+3$.    

You would better think of the problem as a star&bar setting, then you just need to divide by number of permutations that varie between $3!$,$3!/2$,$1$, but since this is more of complicated and overkill for this task, and not requirements-fit, I could sentence the recursive function of all available partitions $A+B+C$ conditioned with a pre-defined symbolic ordering, i choose it to be descendent: $A>=B>=C$
$S_A(a,n)$ is considered to be the number of descendent partitions $x_1x_2...x_n$ where $x_1<=A+1$ and $x_1+x_2+...+x_n=n+a$
In your case the partitions of 6 is calculated following $s_{3}(3,3)$
We can remove the units $1$'s to simplify calculations, which take $x_1$ decremented to be $<=A$ and $x_1+x_2+...+x_n=a$
The recursive function for this matter is reduced to:
$S_{3}(3,3)=\sum_i S_{3-i}(i,2) = S_{3}(0,2)+S_{2}(1,2)+S_{1}(2,2)$ where:
$S_{A}(a,n)=\begin{cases} 0\ \text{if  } \frac{a}{n}>A \\ 1\ \text{if  } n=1 \\ \sum_{i=0\rightarrow a-1} S_{Min(A,a-i)}(i,n-1)\ \text{otherwise} \\ \end{cases}$
I will make this formula practicable soon, I'm off my way to work now.
