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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.

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Hmmm... what do you mean by "find out"? Do you want to construct them or count them? If you're after the number of series-parallel graphs with a given number of edges, you might want this: – Douglas S. Stones Dec 17 '10 at 8:30
@Douglas: Count and construct. – Kit Dec 17 '10 at 8:37
I am not sure what your definitions are (what a combination is, what your definition for when two combinations are the same) but if you could write down what you think the first few terms of the sequence is, you could look it up on the OEIS: – Qiaochu Yuan Dec 17 '10 at 8:39
@Qiaochu: I think I meant all resistor network arrangements, not some number series. I want to find out all resistance values possible with the resistors available to me. – Kit Dec 17 '10 at 8:43
up vote 1 down vote accepted

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

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This is exactly what I was looking for. Thank you! I guess the mathematical term for it is partition? Do you think there is a closed-form solution to get all the coefficients of $R$? – Kit Apr 10 '15 at 21:21
No there is not a closed form for determining the number of partitions, but there is a generating function that produces the numbers. If you need the various resistances, I'm not sure there is a good way of going about it right now. If you just need to know how many distinct resistances there are, this will do the trick. – Laars Helenius Apr 10 '15 at 21:25
I don't think this is right. The argument shows that every partition corresponds to a simple kind of series-parallel network, but not vice versa: there are series-parallel networks that don't correspond to partitions. For example, with 6 unit resistors, I can put 3 in series, 1 in parallel with those (to get 3/4); one in series with this (to get 7/4), and 1 in parallel with all that, giving a resistance of 7/11, which isn't in your list. There are many more series-parallel networks than correspond to partitions. – Tad Apr 13 '15 at 4:53
I see the problem. All my calculations assumed each partition number was in series with all the other partition numbers. Going to have to think about this more. I will update my answer with your information later today. – Laars Helenius Apr 13 '15 at 13:25

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 $n$...]

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The relevant OEIS links are and – Hans Lundmark Dec 17 '10 at 11:32
"was simply obtained by straight-line fitting of the logarithm of the exact values" - so much for scientific rigor... :'( – J. M. Dec 17 '10 at 14:11
I love it when physicists do math! I think my answer is a bit more rigorous. – Laars Helenius Apr 10 '15 at 21:38
@LaarsHelenius Why do I get the feeling that your use of 'do' there has the same connotation as in "Debbie 'does' Dallas"? ;) – Deepak Apr 11 '15 at 0:04
^^^Hahaha!!!! That's great!!! – Laars Helenius Apr 11 '15 at 0:26

There is a newer (2012) pair of academic papers on this topic by by Sameen Ahmed Khan

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 now.

  • $D(n)$ is defined similarly to $C(n)$ but with $B(n)$ substituting for $A(n)$. This is now

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 $A(5)$.


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 now.

He also proved a lower bound $ \frac{1}{4}(1+\sqrt{2})^n < A(n)$, which makes the enumeration of solutions guaranteed exponential.

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I think below is what you want: The Formula for the Equivalent Resistance of Complex Resistance Network here or here

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Hi, and welcome to Math.SE. Posting links that answer the question is welcome (in my opinion), but please write the relevant part inside your answer, so that people looking for the solution of the problem don't have to click in the link (but don't delete the link, so that people who want to get a deeper understanding can click the link, if they want). – fonini Apr 20 '15 at 3:25

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