Find the number of ways of writing a number as the sum of positive integers?

Above is Also called the Partition Function P of n, it is denoted by P(n).

For example,5 can be written as.

5
4+1
3+2
3+1+1
2+2+1
2+1+1+1
1+1+1+1+1


The partitions of 4 are listed below:

1. 4
2. 3 + 1
3. 2 + 2
4. 2 + 1 + 1
5. 1 + 1 + 1 + 1

The partitions of 8 are listed below:

1. 8
2. 7 + 1
3. 6 + 2
4. 6 + 1 + 1
5. 5 + 3
6. 5 + 2 + 1
7. 5 + 1 + 1 + 1
8. 4 + 4
9. 4 + 3 + 1
10. 4 + 2 + 2
11. 4 + 2 + 1 + 1
12. 4 + 1 + 1 + 1 + 1
13. 3 + 3 + 2
14. 3 + 3 + 1 + 1
15. 3 + 2 + 2 + 1
16. 3 + 2 + 1 + 1 + 1
17. 3 + 1 + 1 + 1 + 1 + 1
18. 2 + 2 + 2 + 2
19. 2 + 2 + 2 + 1 + 1
20. 2 + 2 + 1 + 1 + 1 + 1
21. 2 + 1 + 1 + 1 + 1 + 1 + 1
22. 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

Approach

One way of getting a handle on the partition function involves an intermediate function p(k, n), which represents the number of partitions of n using only natural numbers at least as large as k. For any given value of k, partitions counted by p(k, n) fit into exactly one of the following categories:

1. smallest addend is k
2. smallest addend is strictly greater than k.

The number of partitions meeting the first condition is p(k, n − k). To see this, imagine a list of all the partitions of the number n − k into numbers of size at least k, then imagine appending "+ k" to each partition in the list. Now what is it a list of? As a side note, one can use this to define a sort of recursion relation for the partition function in term of the intermediate function, namely

1+ sum(P(k,n-k)=P(n)
for i=1 to floor n/2
where |_n_| is the floor function.

The number of partitions meeting the second condition is p(k + 1, n) since a
partition into parts of at least k that contains no parts of exactly k must have all parts at least k + 1.

Since the two conditions are mutually exclusive, the number of partitions meeting either condition is p(k + 1, n) + p(k, n − k). The recursively defined function is thus:

* p(k, n) = 0 if k > n

* p(k, n) = 1 if k = n

* p(k, n) = p(k+1, n) + p(k, n − k) otherwise.

Got The Hint of DP using matrix of P[n,n] :)

Euler invented a generating function which gives rise to a recurrence equation in P(n),