Vtyjkyuk

I have already read this:

Number partition - prove recursive formula

But the formula from the above link requires a parameter k which is the required number of partitions, but I would like to partition it as far as it could. What I am finding is the partition number of a positive integer n, where partition number means the number of different "partitions" of n. A partition of n is an unordered list of positive integers less than or equals to n, which add to up n. Here are some examples:

1 = 1 # So T(1) = 1
2 = 2
 = 1+1 # So T(2) = 2
3 = 3
 = 2+1
 = 1+1+1 # So T(3) = 3
4 = 4
 = 3+1
 = 2+2
 = 2+1+1
 = 1+1+1+1+1 # So T(4) = 5
5 = 5
 = 4+1
 = 3+2
 = 3+1+1
 = 2+2+1
 = 2+1+1+1
 = 1+1+1+1+1 # So T(5) = 7

I would like to find T(n) for n >= 1 in terms of T(i) where 1 <= i < n (a recurrence relation), or better, in terms of n (a closed formula). I would also like to get rid of $sum$s and $prod$s, if possible. (I guess it is not avoidable for a closed formula, so I ask for a recurrence one hoping to get rid of them)

Thanks in advance!

The recurrence relation you provided above is a solution to what you require. The extra $k$ parameter will need to be evaluated $n$ times.

Using this: $p(n, k) = p(n-1, k-1) + p(n-k, k)$

You need to solve : $p(n, 1) + p(n, 2) + ... + p(n, n)$

This page has a good explanation of the algorithm and also shows a table of the calculations.
http://www.mathpages.com/home/kmath091.htm

$T(0)=1$ and, for $ngt0,$
$$T(n)=T(n-1)+T(n-2)-T(n-2-3)-T(n-3-4)+T(n-3-4-5)+T(n-4-5-6)-T(n-4-5-6-7)-T(n-5-6-7-8)+cdots$$
$$=T(n-1)+T(n-2)-T(n-5)-T(n-7)+T(n-12)+T(n-15)-T(n-22)-T(n-26)+T(n-35)+T(n-40)-cdots$$
where it is understood that $T(n)=0$ when $nlt0.$

For example:
$$T(10)=T(9)+T(8)-T(5)-T(3)=30+22-7-3=42$$
$$T(11)=T(10)+T(9)-T(6)-T(4)=42+30-11-5=56$$
$$T(12)=T(11)+T(10)-T(7)-T(5)+T(0)=56+42-15-7+1=77$$

This is Euler's pentagonal numbers theorem. The sequence
$$1, 2, 5, 7, 12, 15, 22, 26, 35, 40, dots$$
is OEIS sequence A001318: Generalized pentagonal numbers; the $(2n-1)^textst$ term is
$$n+(n+1)+cdots+(2n-1)=fracn(3n-1)2$$
and the $(2n)^textth$ term is
$$(n+1)+(n+2)+cdots+(2n)=fracn(3n+1)2.$$