TITLE: Given $n$ identical resistors $R$, find combinations of series, parallel, and series-parallel arrangements
QUESTION [8 upvotes]: Is there an algorithm to find out all possible resistance values of series, parallel, and series-parallel arrangements given $n$ identical resistors, $R$? All of them must be used.
This might even extend to differently-valued resistors, but I'll just focus on identical resistors.

REPLY [4 votes]: There is a newer (2012) pair of academic papers on this topic by by
Sameen Ahmed Khan

"How many equivalent resistances?" An older but longer preprint/version of it can be found on arxiv.
"Farey sequences and resistor networks"

He proves a theoretical upper bound for the following four sequences:

A048211, which he denotes by $A(n)$, i.e. for the "number of distinct resistances that can be produced from a circuit of n equal resistors using only series and parallel combinations",
A153588, "the total number of equivalent resistances obtained using one or more of the $n$ equal resistors [for series-parallel arrangement]", which he denotes by $C(n)$, and also 
a slightly larger set than $A(n)$, denoted by $B(n)$ and "containing bridge circuits (in addition to the configurations produced by series and/or parallel)"; no OEIS number is given for this one in the paper, but I see it's at http://oeis.org/A174283 now.
$D(n)$ is defined similarly to $C(n)$ but with $B(n)$ substituting for $A(n)$. This is now http://oeis.org/A174284

The upper bound proved is the same for all four above sequences and is denoted by 
$$ G(n) = 2\cdot(1-\frac{1}{n})\cdot \text{Farey} (F_{n+1}) -1 $$
where $\text{Farey}(n)$ is the number of fractions in the Farey sequence of order $n$ [A005728] and $F_{n}$ is the usual Fibonacci number [A000045].
Also this bit is perhaps worth noting:

A set $A(n)$ of higher order does not necessarily contain the complete sets of lower orders. For example, $2/3$ is present in the set $A(3)$, but
  it is not present in the sets $A(4)$ and $A(5)$.

and

Farey sequence is the most exhaustive set of fractions, so it is
  sure to contain some terms absent in the actual circuit configurations.

$G(n)$ itself is asymptotically $2.618^n$, which is consistent with experimental/asymptotic result from the paper by Amengual mentioned in the other answer. Also $G(n)$ got its own http://oeis.org/A176502 now.
He also proved a lower bound $ \frac{1}{4}(1+\sqrt{2})^n < A(n)$, which makes the enumeration of solutions guaranteed exponential.