Given $n$ identical resistors $R$, find combinations of series, parallel, and series-parallel arrangements
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.
Following the Wikipedia link given by Douglas, and the references from there to OEIS, leads to this paper which seems to answer your question: Antoni Amengual, The intriguing properties of the equivalent resistances of n equal resistors combined in series and in parallel, American Journal of Physics, 68(2), 175-179 (February 2000).
The author points out that asking for the possible values of the resistance is not the same as asking for the total number of networks. The answer to the former question is claimed to be approximately $2.55^n$ (for large $n$, I presume).
[EDIT: A look in the paper reveals that the formula $2.55^n$ was simply obtained by straight-line fitting of the logarithm of the exact values for $6 \le n \le 16$, so (being a mathematician) I'm not quite convinced about its validity for large $n$...]
Suppose we have $n$ identical resistors, then the number of different series, parallel, and series-parallel combinations is given by the number of partitions $n$. The number of partitions of $n$ is given by the sequence A000041 in OEIS.
We can interpret each partition of $n$ as a distinct way of arranging the $n$ resistors in series, parallel, or series-parallel. For example, when $n=6$ we have 11 distinct partitions:
$$ 1,1,1,1,1,1\Rightarrow 6 \text{ resistors in series}\Rightarrow \text{resistance}=6R $$ $$ 2,1,1,1,1\Rightarrow 2 \text{ resistors in parallel}, 4 \text{ resistors in series}\Rightarrow \text{resistance}=4\frac{1}{2}R $$ $$ 2,2,1,1\Rightarrow 2 \text{ resistors in parallel}, 2 \text{ resistors in parallel}, 2\text{ resistors in series}\Rightarrow \text{resistance}=3R $$ $$ 2,2,2\Rightarrow 2 \text{ resistors in parallel}, 2 \text{ resistors in parallel}, 2\text{ resistors in parallel}\Rightarrow \text{resistance}=1\frac{1}{2}R $$ $$ 3,1,1,1\Rightarrow 3 \text{ resistors in parallel}, 3 \text{ resistors in series}\Rightarrow \text{resistance}=3\frac{1}{3}R $$ $$ 3,2,1\Rightarrow 3 \text{ resistors in parallel}, 2 \text{ resistors in parallel}, 1 \text{ resistors in series}\Rightarrow \text{resistance}=1\frac{5}{6}R $$ $$ 3,3\Rightarrow 3 \text{ resistors in parallel},3 \text{ in parallel}\Rightarrow \text{resistance}=\frac{2}{3}R $$ $$ 4,1,1\Rightarrow 4 \text{ resistors in parallel}, 2 \text{ resistors in series}\Rightarrow \text{resistance}=2\frac{1}{4}R $$ $$ 4,2\Rightarrow 4 \text{ resistors in parallel}, 2 \text{ resistors in parallel}\Rightarrow \text{resistance}=\frac{3}{4}R $$ $$ 5,1\Rightarrow 5 \text{ resistors in parallel}, 1 \text{ resistors in series}\Rightarrow \text{resistance}=1\frac{1}{5}R $$ $$ 6\Rightarrow 6 \text{ resistors in parallel}\Rightarrow \text{resistance}=\frac{1}{6}R $$
Actually, determining the resistances is simply the sum of the reciprocals of each partition. But this still requires being able to write out the partition rather than just counting the partitions. So if $n_1,n_2,\ldots,n_k$ are integers such that $n_1+n_2+\cdots +n_k=n$ then the total resistance given by this partition is $$\left(\frac{1}{n_1}+\frac{1}{n_2}+\cdots+\frac{1}{n_k}\right)R.$$