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 $$\left(\frac{1}{n_1}+\frac{1}{n_2}+\cdots+\frac{1}{n_k}\right)R.$$