If you have $N$ 1ohm resistors, how many distinct equivalent resistances can you create? Assume that only parallel and series and mixture of them is allowed and no bridging between two parallel connections is allowed.

  • $\begingroup$ What do you mean by "bridging between two parallel connections?" $\endgroup$ – George G Mar 28 '14 at 11:39
  • $\begingroup$ @George: my assumption was that you couldn't create a ladder or mesh. $\endgroup$ – Carl Witthoft Mar 28 '14 at 11:50
  • $\begingroup$ Oh, that makes sense now. $\endgroup$ – George G Mar 28 '14 at 11:51
  • $\begingroup$ This is a combinatorics problem. Since all your N resistors are indistinguishable, you can start w/ all combinations (not permutations) of spanning sets. Really, this is purely a math problem, with a little work at the end to see how many spanning collections produce the same net resistance (if any). $\endgroup$ – Carl Witthoft Mar 28 '14 at 11:52

There are basically two kinds of addition here (ordinary addition and inverse-sum-inverse), representing series and parallel arrangements.

You can represent the thing as a tree with alternating nodes of addition and ISI layers. The thing resolves pretty much down to a tree with N leaves.

The magic is dealt with here http://oeis.org/A000669 . It talks about leaves on trees, but the various branch-points represent a set of resistors in series or parallel, depending on the parity of steps between the level and the ground. Because this presuposes say, a series layer, you have to double the number.


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