For all c in the grid, we can easily calculate d, e and f.

We can also easily calculate their BigN cell (where a=1 and b=c) in positive and negative space.

In row 1 (where n=1) in positive space (so not the (f,n) cells), f=(x+n)(x+n)

The values of a,b and d each increase by ONE 2n cells to the right. They decrease by one 2n cells to the left.

For every c.

We can use algebra to rearrange (x+n)(x+n) as nn + 2d(n-1) + f - 1. This means nn + 2d(n-1) + f - 1 can also be represented as either 8T(u) or 4(T(u)+T(u-1)). If we take odd (x+n)(x+n) squares as an example, which can be represented as eight times a triangle number plus one, this makes nn + 2d(n-1) + f - 2 eight times a triangle number. That makes (nn + 2d(n-1) + f - 2)/8 or nn/8 + (2d(n-1))/8 + (f-2)/8 a triangle number.

There are four different kinds of (x+n)(x+n) square, based on parity. There are odd (x+n)(x+n) squares where either x is odd and n is even or where x is even and n is odd. There are also even (x+n)(x+n) squares where both x and n are odd or where both x and n are even.

Let's apply this to the idea of representing nn + 2d(n-1) + f - 1 as eight triangles plus one. nn is also a square. This means it can be represented based on a configuration of triangle numbers. If nn is odd, it will also be eight triangles plus one. If we take nn-1, it will be eight triangles. Since nn-1 + 2d(n-1) + f - 1 is eight triangles (the -2 distributed so that nn-1, eight triangles, is also formed), we can represent (x+n)(x+n) as eight smaller triangles within eight larger triangles, with an extra unit in the middle. This picture is an example, where n=5, nn=25, nn-1=24, and (nn-1)/8=3, which is the second triangle number (1+2). The larger triangles made up of (nn+2d(n-1)+f-2)/8 are each 21, the 6th triangle number (1+2+3+4+5+6). (x+n)(x+n) in this case is 169, and x+n is 13. 169 is equal to 21*8 +1. n is equal to 5, so x is equal to 8. If you expand (x+n)(x+n) you get xx+nn+2xn, so this picture shows nn in red (including the black square) and xx+2xn (or 2d(n-1)+f-1) around the outside. For x+n=13, x=8 and n=5 is not the only valid pair of values. You could have x anywhere from 1 to 12 and n anywhere from 1 to 12 (as long as they add to 13). Speaking of which, to clarify, x is only ever 0 in (0,0), where n is also 0, so there will never be an x+n square here (at least one we can use, i.e. that isn't 0), and n is only ever 0 here and in the negative space, and we use the positive space to represent the squares and triangles, so this is irrelevant. So that's where 1-(x+n-1) comes from in this sense.

The remainder when you take nn/8 away from (x+n)(x+n)/8 in this picture is called a polite number. This is a number produced as the sum of consecutive integers. In this case, each of these is equal to 3+4+5+6, which is 18. So odd (x+n)(x+n) with an odd nn can be represented as eight times a triangle number plus eight times a polite number plus one.