You lost me at "invoking some number theory you can get down to 575 resistors."

If you want to get 1.337 milliOhms from as few 1% 1 Ohm resistors as possible, you can't do better than putting them all in parallel.

1/0.001337=747.94

With 3 significant digits, 748 resistors in parallel would be within the tolerance of the resistors.

Do you know of some magical way of wiring resistors that is not a series-parallel circuit?

Edit

proof:

Premises- A resistor network is made up of series-parallel circuits
- Series: R
_{total}=R_{1}+R_{2}+R_{3}+...+R_{n} - Parallel: R
_{total}=(1)/((1/R_{1})+(1/R_{2})+(1/R_{3})+...+(1/R_{n})) - Resistor networks can be broken into sub-networks

Our goal is to lower the resistance an arbitrary amount with as few resistors as possible.

Claim: The resistor network of the chosen resistance with the fewest resistors will have no resistors in series iff extra accuracy is not needed.

- The smallest possible resistor network is a single resistor. (by 1 and 4)
- Resistor networks in series always raise the resistance. (by 2 and A)
- Resistor networks in parallel always lower the resistance. (by 3 and A)
- The simplest resistor network is formed with a single resistor. (by A)
- To lower the resistance, extra resistors must be added in parallel. (by 1, B and C)
- Recursively add a single resistor in parallel until desired resistance is overshot or met. (until resistance <= target resistance)
- At the end of recursion we have the simplest possible resistor network with the desired resistance. (By D, E and F)
- The simplest possible resistor network with the desired resistance has no resistors in series. (by F and G)

Proof by restatement of Claim from Premises.Edit3: Well, not quite: I introduced the

Goal at step E.