TITLE: Given N identical resistors, find the total number of series-parallel combinations (different question). please read description
QUESTION [7 upvotes]: There are multiple answers that state that the total number of equivalent resistances possible are the number of partitions of $N$.
But what I am looking at is the number of ways to combine the resistances.
For example, take $N = 4$.

All 4 resistors can be in series..
3 resistors can be in parallel, with that combination in series with
the fourth resistance.
2 resistors can be in series, in parallel with the third, which is in
parallel with the fourth.
2 resistors can be in series, and this in parallel with another 2 resistances in series.
2 resistors can be in parallel, in series with another 2 resistances in parallel.
3 resistors can be in series, with this combination in parallel with the fourth resistance.
2 resistances can be in parallel, in series with the third and fourth resistances.
All 4 resistances can be in parallel.
2 resistors can be in parallel, in series with a third, and this combination in parallel with a fourth.
2 resistors in series, in parallel with a third, this is in series with he fourth

Points 4 and 5 yield the same net resistance, but they are a different combination of resistances, hence we count them as distinct combinations.
Thus, the net number of combinations is 10.
I tried this for 5 resistors as well. 
Could someone help in generalizing this for $N$ resistors?
Consider only series and parallel combinations. do not consider bridges.
Or consider them. It's an interesting problem either way

REPLY [3 votes]: As per the comment, the sequence can be found at oeis.org/A000084 and are often called Yoke Chains.
The first 1000 terms can be found here.
One of the most readable papers on the mathematics of this question is by Finch, 2003