In how many ways can I sum integers from $1$ to $N-1$ to obtain $N$? I'm looking for the exact formula for $f(N) =$ number of ways to sum $1, 2, ..., N-1$ to obtain $N$. 
$N$ is an integer $> 0$. Integers $1, 2, ..., N-1$ can be used $0$ or 1 time as an element in the sum.
For example:
$f(6) = 3$. There are 3 ways to obtain 6 from integers less than $6$:
$1+2+3$
$1+5$
$2+4$
 A: A summation of $n$ from integer summands is called a partition.
The number of distinct partitions defines the partition function $p(n)$.
It includes the case $n = n$ and summands can show up more than once.
So your $f(n) < p(n)$. One of the OEIS comments calls partitions with summands not repeating "restricted partitions", see $A_9$ below.
Searching OEIS, gives: A111133.

A111133 Number of partitions of $n$ into at least two distinct parts.
Old name: Number of sets of natural numbers less than $n$ which sum to
  $n$.

So we have
$$
f(n) = A_{111133
}(n)
$$

EXAMPLE
  $a(6) = 3$ because $1+5$, $2+4$ and $1+2+3$ each sum to $6$. That is,
  the three sets are $\{1,5\}$,$\{2,4\}$,$\{1,2,3\}$. 

Formula:
It seems there is only a recursive function available, A000009 with
$$
f(n) = A_9(n) - 1
$$ 
features the comment:

For $n>1$, 
  $$
a(n)=\frac{1}{n} \sum_{k=1}^n b(k) \, a(n-k)
$$
  with $a(0)=1$, $b(n)= A_{000593}(n)$ = sum of odd divisors of $n$; cf. A000700. - Vladeta Jovovic, Jan 21 2002

Here is a link to A000593. That is already enough informatio to come up with a piece of code. Otherwise continue to follow the references. 
Or you go down the road of the generating functions, where the coefficients of a power series of $G$ give the sequence. 
A: The power series for
$$
\prod_{k=1}^\infty\left(1+x^k\right)
$$
which is equal to
$$
\prod_{k=1}^\infty\frac1{1-x^{2k-1}}
$$
is the generating function for your number plus $1$. The beginning of the power series is
$$
\small1+x+x^2+2x^3+2x^4+3x^5+4x^6+5x^7+6x^8+8x^9+10x^{10}+12x^{11}+15x^{12}+\dots
$$
