The values of a[t] at (0,1) are twice the square numbers.

The values of d[t] at (0,1) are 4 multiples by the triangular numbers. All other cells at row 1 can be constructed from these values by adding to them or subtracting from them.

In (0,n), there is a pattern of a values in relation to squares, dependent on parity. This pattern applies to most cells. When n is even, the a values are all (n/2)each square in ascending order (e.g. in (0,6) it's 3, 12, 27, 48, 75 etc, which are 31, 34, 39, 316, 325 etc, and this n/2 applies to all even ns in (0,n)). When n is odd, the a values are all 2neach square in ascending order (so same thing as before; e.g. in (0,5) it's 10, 40, 90, 160, 250 etc, which is 101, 104, 109, 1016, 1025).

This pattern does not apply to every cell. At (0,8), the a values are one times the squares again (1, 4, 9, 16, 25 etc). At (0,9), the a values are all twice the squares (2, 8, 32, 50 etc). From (0,10) the pattern from the previous paragraph continues. At (0,16), the a values become twice the squares again. (0,17) is like in the previous paragraph. (0,18) is one times the squares. It's normal then again until (0,24), which is three times the squares (3, 12, 27, 48, 75, etc, like (0,6)). There seem to be further patterns to these special cases, but since I'm not aware of there being anything specific to use this for yet, I'll leave it for now.

• in (0, 1) we know a[t] = tt2.

• we also know how to "move" up and down (0,1) whereby t=p+t (I think thats the equation… please correct me if i'm off-by-one)

• This means that if we try to find all the a[t] that are divisible by 3, we can simply list:

t = 3, 6, 9, 12, etc.

• This means that at (0,9) for example, we'll see the x[t] values are equal to the x[t] values in (0,1) at t = 3, 6, 9, etc.

d[t] at (0,n) and its relationship to triangle numbers…

so, lets say we're at n = 15, what are the set of d[t]?

which rows are in (0,1) have 15 as a factor?

start with t = 16:

t=16: a = 450 = 21515 x = 30 = 215 d = ( 15 +1)( 15)2

t=31: a = 1800 = 421515 x = 60 = 2215 d = (215 +1)(215)2

t=46: a = 4050 = 921515 x = 90 = 3215 d = (315 +1)(315)2

t=51: a = 7200 = 1621515 x = 120 = 4215 d = (415 +1)(415)2

move to (0,15):

t= 1: a = 30 = 215 x = 30 = 215 d = 2215 *1

t= 2: a = 120 = 4215 x = 60 = 2215 d = 2215 *3

t= 3: a = 270 = 9215 x = 90 = 3215 d = 2215 *6

t= 4: a = 480 = 16215 x = 120 = 4215 d = 2215 *10

What this is all saying is, for odd n, d[t] = n4T(t)