Partition function- without duplicates Is there a function, equivalent to the partition function, that does not allow duplication? Or, alternatively, for any N, how many partitions would there be- disallowing any that have the same integer appearing more than once.
Edit:
Sorry, to be more accurate, I'd like to know a function, or even just it's asymptotic complexity, which will produce the partitions of the input, excepting those which have integers that occur more than once. I've only just realized that this is the actual logical equivalent to my problem; I appreciate that I haven't exactly posted much effort into finding it myself. Unless someone comes back and posts a solution relatively quickly, you can expect more from me soon.
 A: The number of partitions of $n$ into distinct parts is OEIS A000009, which gives a pile of references.  It is sometimes called the partition function $Q$ (in contrast to $P$ for all partitions).  As Gerry Myerson has said, it generating function is $\prod_{m\ge 1} (1+x^m)$.
As for actually generating the partitions, you could use something already prepared such as the package partitions in R which for example gives for partitions of $12$
> library(partitions)
> Q(12)
[1] 15
> diffparts(12)
[1,] 12 11 10 9 9 8 8 7 7 7 6 6 6 5 5
[2,]  0  1  2 3 2 4 3 5 4 3 5 4 3 4 4
[3,]  0  0  0 0 1 0 1 0 1 2 1 2 2 3 2
[4,]  0  0  0 0 0 0 0 0 0 0 0 0 1 0 1

If you want to do this yourself, I find one way (if you can generate ordinary partitions) is to note that the partitions of $n$ into $k$ different positive parts are equivalent to the partitions of $n-\frac{k(k-1)}{2}$ into $k$ positive parts (possibly equal), just by adding $k-1, k-2,\ldots,1,0$ respectively to each part.  For example to find partitions of $12$ into $4$ different parts, look at partitions of $6$ into $4$ parts: there are two, namely $3+1+1+1$ and $2+2+1+1$; now add $3+2+1+0$ to these term by term to get $6+3+2+1$ and $5+4+2+1$.  
A: I believe that it is  quite instructive to derive the above generating
function  using species,  in  spite of  it  being very  simple, as  it
showcases some standard tricks in the manipulation of cycle indices.

Observe that the species $\mathcal{Q}$ that represents partitions with
unique constitutents is simply
$$\mathcal{Q} = \mathfrak{P}\left(\sum_{k\ge 1} \mathcal{Z}^k\right).$$

Now recall  the recurrence by Lovasz  for the cycle  index $Z(P_n)$ of
the set operator $\mathfrak{P}_{=n}$ on $n$ slots, which is
$$Z(P_n) = \frac{1}{n} \sum_{l=1}^n (-1)^{l-1} a_l Z(P_{n-l})
\quad\text{where}\quad
Z(P_0) = 1.$$

Let $$F_n(z) = Z(P_n)\left(\frac{z}{1-z}\right)$$
where the second parenthesis on the right represents cycle index substitution
and introduce the generating function
$$G(y) = \sum_{n\ge 0} F_n(z) y^n,$$
so that we are interested in $G(1).$

The recurrence yields
$$n Z(P_n) = \sum_{l=1}^n (-1)^{l-1} a_l  Z(P_{n-l}).$$

Substitute for $a_l$,  multiply by $y^{n-1}$ and sum  over $n\ge 1$ to
get
$$G'(y) = \sum_{n\ge 1} y^{n-1}
\sum_{l=1}^n (-1)^{l-1} \frac{z^l}{1-z^l}  F_{n-l}(z)
\\= \sum_{l\ge 1}  (-1)^{l-1} \frac{z^l}{1-z^l}
\sum_{n\ge l} y^{n-1} F_{n-l}(z).$$
This is
$$\sum_{l\ge 1}  (-1)^{l-1} \frac{z^l}{1-z^l} y^{l-1}
\sum_{n\ge l} y^{n-l} F_{n-l}(z)
= G(y) \sum_{l\ge 1}  (-1)^{l-1} \frac{z^l}{1-z^l} y^{l-1}.$$
Therefore
$$(\log G(y))' = 
\sum_{l\ge 1}  (-1)^{l-1} \frac{z^l}{1-z^l} y^{l-1}.$$
Integrating
we have
$$\log G(y) = C +
\sum_{l\ge 1}  (-1)^{l-1} \frac{z^l}{1-z^l} \frac{y^l}{l}
= C + \sum_{k\ge 1} 
\sum_{l\ge 1}  (-1)^{l-1} z^{kl} \frac{y^l}{l}
\\= C - \sum_{k\ge 1} \log \frac{1}{1+yz^k}.$$
The conclusion is that
$$G(y) = e^C \exp\left( - \sum_{k\ge 1} \log \frac{1}{1+yz^k} \right)
= e^C \exp\left( \sum_{k\ge 1} \log(1+yz^k) \right)
\\= e^C \prod_{k\ge 1} (1+yz^k).$$
Now $G(0)=1$ and hence the  constant obeys $e^C=1$, for a final answer
of 
$$G(y) = \prod_{k\ge 1} (1+yz^k).$$

In particular the generating function of partitions into unique parts
is $$G(1) = \prod_{k\ge 1} (1+z^k),$$
precisely as it ought to be.
