partitions

I started thinking about the partition problem. First I thought p(n+1) would just be p(n) + 1, because to each solution of p(n) you just add 1 to make n+1, and then there is (n+1).
That is say p(3)

p(2): 1 + 1, 2
p(3): (1 + 1) + 1, (2) + 1, 1 + 1 + 1

but then it occurred to me there was more. Say p(6) you can break down the numbers > 1

p(6) = p(1) * p(5) + p(2) * p(4) + p(3) * p(3) etc. To give a partial example.

p(6) = p(4) * p (2) but then that p(4) is also p(1) * p(3) or p(2) * p(2), etc.
But then you have to think of duplicates. p(9)

p(9) could be (1 + 1 + 1 + 1 + 1 + 4) of (4 + 1 + 1 + 1 + 1 + 1) which are duplicates.

After that, my brain hurt.