How many resistors are needed? I was told about the following problem. Suppose you have infinite number of resistors with only value 1$\Omega$. Question is 

What minimal number of 1$\Omega$-resistors is needed to construct given fraction resistance $R$, i.e. $R\in \mathbb{Q}_+$.

Note that you can connect two resistors ($R_1$ and $R_2$) in two ways: parallel and series. The resulting resistance ($R_p$ and $R_s$ respectively) in those cases are:
$$R_s = R_1 + R_2$$
$$R_p = R_1 \oplus R_2 = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}},$$
where we introduced $\oplus$-operation in order to simplify notes. NOTE: only schemes which are able to be written in the form:
$$(... 1 + (1\oplus 1) ... )$$
are allowed. For example, this is not allowed:

If you know how to formulate this rule better, please, say.

For example, you can use the Euclidean algorithm. 
$$\frac56 = \frac{1}{\frac65}=\frac{1}{1+\frac15} = 1 \oplus 5,$$
so you needed 6 resistors because
$$5 = \sum_{k=1}^{5}1$$
But it was not the minimum number of resistors because, for example,
$$\frac56 = \frac12 + \frac13 = (1\oplus 1)+(1\oplus 1\oplus 1),$$
where it is enough to use only 5 resistors.

I think that Euclidean algorithm often solves the problem. More over, it is needed to consider only one "half of $\mathbb{Q}_+$", for the other half it is enough to replace +'s by $\oplus$'s and vice-versa.

The one who told me about this problem adhered to the following notations.
$$[1,1]\quad \text{for}\quad 1\oplus 1.$$
$$(1,1)\quad \text{for}\quad 1 + 1,$$
so our previous example looks like
$$[1,(1,1,1,1,1)]\quad \text{and}\quad ([1,1,1],[1,1]).$$
 A: I was able to find a bound for the number of resistors required.
Given a resistance $\frac{a}{b}$, at least $n$ resistors are required where $\phi_{n+2}$ is the first Fibonacci number greater than or equal to $a+b$.
Here's my thinking process and proof:
Naturally I started out by listing things to see if I could find a pattern.
\begin{align*}
n&=1 &&1/1\\
n&=2 &&1+1=2/1\\
& &&1\oplus1 = 1/2\\
n&=3 &&1+(1+1)=3/1\\
& &&1+(1\oplus1) = 3/2\\
& &&1\oplus(1+1) = 2/3\\
& &&1\oplus(1\oplus1) = 1/3\\
n&=4 &&1+(1+(1+1)) = 4/1\\
& &&1+(1+(1\oplus1)) = 5/2\\
& &&1+(1\oplus(1+1)) = 5/3\\
\end{align*}
At this point I got bored, but I noticed a curious pattern:
Let $R_n$ be the set of possible resistances for $n$ resistors. Let $M_n = \max\{a+b: a/b \in R_n\}$
It appears to be the case that the $n$th Fibonacci number, $\phi_n$, is equal to $M_{n-2}$. Define Max($R_n$) to be the subset of $R_n$ such that $a/b \in \mathrm{Max}(R_n)$ if $a+b = M_n$. Then the maximal resistances for $n$ resistors are any resistances in Max($R_n$). It can also be observed that Max($R_n$) always seems to have two elements, and these are of the form $\frac{\phi_{n+1}}{\phi_{n}}$ and $\frac{\phi_{n}}{\phi_{n+1}}$. (Note that the first "pattern" is a direct consequence of this.)
In other words, (if this is true), given a resistance $a/b$, at least $n$ resistors are required where $\phi_{n+2}$ is the first Fibonacci number greater than or equal to $a+b$.
Claim: Given $n$ resistors, the only two maximal resistances are of the form $\frac{\phi_{n+1}}{\phi_{n}}$ and $\frac{\phi_{n}}{\phi_{n+1}}$.
Proof
Note that from here on I will use $v(R) = a+b$, where $R = a/b$.
Base case: trivial.
Inductive step: Let $k\in \mathbb{N}$. Suppose the only maximal resistances for 1 through $k$ resistors have been of the form $\frac{\phi_{k+1}}{\phi_{k}}$ and $\frac{\phi_{k}}{\phi_{k+1}}$. It follows that for $k+1$ resistors, $R = \frac{\phi_{k+2}}{\phi_{k+1}}$ can be obtained, as $\frac{\phi_{k+2}}{\phi_{k+1}} = 1\oplus\frac{\phi_{k}}{\phi_{k+1}}$. Similarly the reciprocal can be obtained. Now it must be shown that $\frac{\phi_{k+2}}{\phi_{k+1}}$ and $\frac{\phi_{k+1}}{\phi_{k+2}}$ are maximal.
Suppose $R_k = a/b$ is a resistance such that $a+b+n = \phi_k+\phi_{k+1}, n > 0$. If $b\leq \phi_{k+1}$ and $a\leq \phi_{k+1}$ then $1+a/b = (b+a)/b = (\phi_{k+2}-n)/b$, so $v(1+a/b) < v(\frac{\phi_{k+2}}{\phi_{k+1}})$. Similarly, $1\oplus a/b = a/(b+a) = a/(\phi_{k+2}-n)$ and we get $v(1\oplus a/b) < v(\frac{\phi_{k+2}}{\phi_{k+1}})$. Now suppose $a > \phi_{k+1}$ or $b > \phi_{k+1}$. This is actually impossible, as $b = \alpha + \beta$ and $a = \eta + \mu$ for some $\alpha/\beta$ and $\eta/\mu$ resistances for $k-1$ resistors. However, by the inductive hypothesis, this sum is capped at $\phi_{k+1}$. There is a second way that a resistance $R_{k+1}$ can form: Rather than adding a resistance of 1 to $R_k$, this is adding two resistances $R_{m}$ and $R_{n}$ so that $m+n=k+1$. Using similar reasoning to the special case above and the fact that $\phi_s\phi_t < \phi_{s+t}$, one can prove that $v(R_{m}+R_{n}) < v(\frac{\phi_{k+2}}{\phi_{k+1}})$. It follows that for $k+1$ resistors, there are exactly two maximal resistances, given by $\frac{\phi_{k+2}}{\phi_{k+1}}$ and $\frac{\phi_{k+1}}{\phi_{k+2}}$.
A: Since the other day I've played with resistors, and made a program similar to Martin's to enumerate all possibilities that ensure the minimum number of resistors.
You can download the result file there $\to$ All rationnals for $n\le 12$
Also from this list, I've drawn a picture of the rationnals that share the same $n$. It is quite impressive the see how the bubble expands, but the border of this bubble seems to have an asymptotic behaviour already.
As you can see it is hollow, because many rationnals are not reached for such a low $n$, eventually all the white is to be filled with a superior rank color.


For a better rendering you can download directly the XPM file.
A: Since this method was suggested in the comments, I'm going to present it, but this is a restriction on the OP problem, in that we consider only adding a resistor either in serie, either in parallel to an existing circuit built in the same way. Thus it is far from representing all possibilities.

If there $n$ resistors then there are $2^n$ arrangements of resistors, basically they are given by the binary developpement where for instance $+$ is associated to $1$ and $\oplus$ to $0$. 
It behaves as an arithmetic stack.
Example let's look at shayne2020 listing for $n=3$.
$\begin{array}{c|c|c|c|}
\mathbb Q & \text{circuit} & \text{arithm stack} & \text{binary} \\
\hline
1/3 & (1\oplus 1)\oplus 1 & +\oplus\oplus 111 & 100\\
3/2 & (1\oplus 1)+1 & +\oplus+111 & 101\\
2/3 & (1+1)\oplus 1 & ++\oplus111 & 110\\
3/1 & (1+1)+1 & +++111 & 111\\
\end{array}$

This correspond to the  well known bijection from $\mathbb N\to\mathbb Q^+$ which has a representation as the Stern-Brocot tree : Stern-Brocot Tree

The tree also has this interesting property : Fractions on a Binary Tree II

It is not too difficult to prove that our resistors verify these properties.
Remember on the left branch of a subtree we process a $0$ in the binary developpement so this correspond to $\oplus$ operation. While on the right branch of the tree, we process a $1$, this is the $+$ operation.
Let's start from a fraction $a/b$ :


*

*on left branch of the tree we have $\frac ab\oplus 1=\frac{1}{\frac ba+1}=\frac{1}{\frac{a+b}{a}}=\frac{a}{a+b}$

*on right branch of the tree we have $\frac ab+1=\frac{a+b}{b}$


Note also that the bijection can be explicited :
$\begin{cases}
\displaystyle g(x)=\frac{1}{1+2\lfloor x\rfloor-x} \\ \\
f(n)=\underbrace{g \circ g \circ g\ \circ...\circ\;g\;}_{\text{n times}}(0)=g^{(\circ^n)}(0)
\end{cases}$
The case of more complex circuits (i.e. other parenthesizations, and there are Catalan numbers many of them) still need to be adressed. And from Martin's results it seems they lead to shorter circuits.
So even if this idea seems nice because it offers a constructive way to reach every rational resistor, it does not realizes the minimum in term of needed resistors.
A: I like the use of the "bracket - parenthesis" notation, but I will add following feature, namely to write $[3]$ for $[1,1,1]$ and $(5)$ for  $(1,1,1,1,1)$ likewise. I'd like to draw the attention of how continued fractions can be of use. Suppose we were to build a resistance of $\frac{5}{7} \Omega$. Brute force shows you that you can build this resistance as $([7][7][7][7][7])$, using $35$ resistors. If we write $\frac{5}{7}$ as a continued fraction we obtain: 
$$
5/7 = 0+\frac{1}{1+\frac{1}{2+\frac{1}{2}}}
$$ Which shows that the resistance can be built as $[1,((2),[2])]$ using only $5$ resistances. It is not sure this method is really optimal, and probably poses the same logistic problem as the travelling salesman problem.
