johnson counter sine wave resistor network
The output voltage is fairly easy to calculate as:

$$
V_{output} = \sum_i\frac{V_iR_p}{R_i}
$$

Where \$V_i\$ is the voltage at \$Q_i\$, and \$R_i\$ is the resistance connected to \$Q_i\$, and \$R_p\$ is the parallel connection of all the resistors.

Explanation: 

Using superposition the output voltage can be calculated for one source at a time and summed using resistive division. In the following \$R_{pxi}\$ represents "all resistors in parallel except \$R_i\$" :
$$
V_{output} = \sum_i\frac{V_iR_{pxi}}{R_i + R_{pxi}}
$$

Another way to express all resistors in parallel (\$R_p\$) except \$R_i\$ is:

$$
\frac{1}{(1/R_p-1/R_i)}
$$

So we have:
$$
V_{output} = \sum_i\frac{V_i}{R_i + 1/(1/R_p-1/R_i)}\frac{1}{(1/R_p-1/R_i)}
$$

Luckily, the above expression reduces to:
$$
V_{output} = \sum_i\frac{V_iR_p}{R_i}
$$

I don't think it gets easier than that.