I have made some box joints and thus far I've found it difficult to ensure that the teeth are evenly spaced; there's usually a pin that is cut down a bit. Is there an equation or method of determining what the board's width should be? Is it as simple as "always ensure board is a multiple of 2x the kerf of the dado set?
Start by taking a multiple of the pin width, then add a little extra and cut all the slots/kerfs on all sides of the box. Once you're done, trim each side flush with the edge of a pin or slot. If you prefer, you can trim it to size after assembling and gluing it.
To guarantee that the teeth are evenly-spaced, use a jig. Popular Woodworking has a great article on how to build a box joint jig, and Fine Woodworking has a nice interactive demo illustrating how a box joint jig works.
Note that FWW calls it a finger joint jig because many people use the terms finger joint and box joint interchangeably, but personally I like to differentiate the right-angle joining version as a box joint and the straight joining version (e.g., to "lengthen" a board) as a finger joint since the straight joining version also often has tapered fingers.
Is it as simple as "always ensure board is a multiple of 2x the kerf of the dado set?
If you want half fingers at top and bottom on the same side, then yes, using an even multiple of the finger width (or dado kerf) will do it.
If you want full fingers at top and bottom of the same side, use an odd multiple of the finger width.
Given any box or drawer finger joint height, W, you must custom size the width of the fingers for it. To do this you decide you want the fingers at the top and bottom configured. As said above, you will have an odd number of fingers on any side that has both both top and bottom fingers on the same side, so if "f" equals the width of both fingers and slots on either piece, where we want a whole number integer of fingers and "slots" then for n=3 then f=5; n=4-->f=7; n=5-->f=9; n=6-->f=11; etc. Note we can say that for a given integer n, that f=n+(n-1). That reduces to f=2n-1, and f(2n-1) must equal W. Chose any f depending on how big or small you want your equal fingers to be. This needs some thinking by the questioner, but just know that you will VIRTUALLY NEVER get a whole nice number of fingers to fit into a given width with a nice neat finger width size. For example, lets say that "W" is equal to the height of drawer we want to be 8" high on the sides. Logically thinking about it, we will pick a "nice" n for the joint, say 14. So for an f of fourteen we get f=27. W/f therefore is 0.296". If you make the fingers 0.296" wide, then you can get a perfect number of them, 14 fingers on the drawer side or end that has whole outside fingers at both top and bottom and 14 matching "slots" on the drawer side or end that has a "slot" at top and bottom and ALL the fingers and slots will be the exact same width! Note that the side that has the "slots" at top and bottom has one less projecting finger than does the side that will be joined to it! There's no way around it, ya'gotta do the math. Best way to understand this is to draw a drawing and study it if you can't accept it. If you look at it long enough, you can see that the side with the "outside fingers" has to have one more projecting finger than the joining side that has the "outside slots" at top and bottom, and THAT is what makes it so that the width of the fingers plus width of the slots, whose total MUST be equal to W, must be of a common width that will result in 2n-1 being a whole integer. Being an "integer" means a whole number of fingers and slots that are not fractional. Making tops and bottoms a half finger is attacking the problem from another direction. Enough, huh?