I have a question related with number of additive partition or method similar like this: $$p(5)=1+4=2+3=1+1+1+1+1=1+1+1+2=1+2+2=1+1+3$$

For a given number $n$,if we are trying to calculate number of additive partition of $p(n\cdot k)$ where $k$ is some integer,or $p(n+a+c)$ , $a,c \in \mathbb Z$, is there commutative rules or something similar to calculate number of partition of sum of some numbers?or number of partition of some number multiplied by another number? For example like this $$p(n+k)\overset{?}{=}p(n)+p(k)?$$

The partition function grows very rapidly. (Please see Wikipedia for some details.) There certainly is no simple relationship between $p(n+k)$ or $p(n\cdot k)$ and $p(n)$, $p(k)$. – André Nicolas Jan 26 '12 at 15:50
Perhaps the closest thing to what you want is Euler's pentagonal number formula, which is in that wikipedia article: $p(k)=p(k−1)+p(k−2)−p(k−5)−p(k−7)+p(k−12)+p(k−15)−p(k−22)−\dots$, where the numbers $1,2,5,7,12,15\dots$ are the pentagonal numbers $m(3m-1)/2$, $m$ running through the (positive and negative) integers. – Gerry Myerson Jan 26 '12 at 23:28