Another way of thinking about the question is, "how close do you need to be for a dim star (our Sun) to outshine a brighter one?"

Consider our Sun, and another star a distance $d$ away that is $m$ times brighter than the Sun (in terms of luminosity or absolute magnitude).  Let's look a location in space that is a distance $x$ from the Sun in the direction of the second star, and a distance $y$ in the perpendicular direction.

![enter image description here][1]

The amount of light seen at the location is proportional to the inverse square of the distance.  The relative amount of light from each star is therefore just:

$$
\frac{1}{x^2+y^2} \qquad\qquad \frac{m}{(d-x)^2+y^2}.
$$

We want to know when the first is greater than the second:

$$
\frac{1}{x^2+y^2} > \frac{m}{(d-x)^2+y^2} \\
(d-x)^2+y^2 > m x^2 + m y^2 \\
d^2 - 2dx > (m-1)x^2 + (m-1)y^2 \\
\left(1+\frac{1}{m-1}\right)d^2 > \left(\frac{1}{m-1}\right)d^2 + 2dx + (m-1)\left(x^2 + y^2\right) \\
\left(\frac{m-1+1}{m-1}\right)d^2 > (m-1)\left(\frac{d^2}{(m-1)^2} + 2x\frac{d}{m-1} + x^2 + y^2\right) \\
m\left(\frac{d}{m-1}\right)^2 > \left(\frac{d}{m-1} + x\right)^2 + y^2 \\
$$

This inequality describes a circle of radius $r=d \sqrt{m}/(m-1)$, centered around the point $x=-d/(m-1),\ y=0$.

---

Let's take the specific case of Sirius, for which $m=25.4$ and $d=8.60~\text{ly}$.

Applying the above equations, we get $r=1.77~\text{ly}$ and $x=-0.35~\text{ly}$.  Therefore, even at half the distance of the next-closest star Sirius is brighter than the Sun.  This means that the Sun is the brightest star in the sky for exactly eight planets.

  [1]: https://i.sstatic.net/knbdP.png