(Click on graphic to have it open in a new browser window if it doesn’t display fully for you.)

Okay, after having worked on the equations for a bit more (and discovering I had left off an important set of parentheses in the previous function formulas), I have found the simplified version of each of them. These equations will once again use two variables, but since the 6n +/- 1 format is already established for primes, I’ve reworked it. For the following, n = the row you’re trying to build on the graph. This is slightly misleading because each row is actually patterned off of the 6n +/- 1 format itself. Therefore, there are two rows for each n. The 6n - 1 and the 6n + 1 value. And for purposes of the chart, a 6n - 1 number is black and a 6n + 1 number is red.

Finally, there is a controlling x value that determines how far to the right you’ll place the cell. Basically, if you were testing for a prime, you could loop x starting at 0 and running the equation, then increasing x by 1 and running the equation, repeating until x equals the n value you’re searching for. If you’re testing a 6n - 1 number, then any black values in the n column will be factors; if you’re testing a 6n + 1 number, then any red values in the n column will be factors.

Now are you ready for the massively complex new equations?

x(6n - 1) + n ; (Use to find black values on the 6n - 1 line.)
x(6n + 1) + n ; (Use to find red values on the 6n + 1 line.)
x(6n + 1) - n ; (Use to find red values on the 6n - 1 line. x must be greater than 0.)
x(6n - 1) - n ; (Use to find black values on the 6n + 1 line. x must be greater than 0.)

And just to demonstrate it, here’s the values for the first couple of rows.
n = 1
x = 0

Black finds black @ 1
Red finds red @ 1
Red finds black @ - 1*
Black find red @ - 1*

n = 1
x = 1

Black finds black @ 6
Red find red @ 8
Red find black @ 6
Black finds red @ 4

n = 2
x = 0

Black finds black @ 2
Red finds red @ 2
Red finds black @ - 2*
Black finds red @ - 2*

n = 2
x = 1

Black finds black @ 13
Red finds red @ 15
Red finds black @ 11
Black finds red @ 9

* = why X must be greater than 0 for the final two equations.

By the way, the first two equations are equivalent to the equations that I came up with before if n = S + 1. The second two would not be due to some misplaced parentheses in my original formulas :-( Oh well. This way is simpler and more “elegant.”