In (e,1), at the t where a[t] = nb, the d[t - 1] - d = b(n-1)

• Goto to cell (c, 1)

• Find the row where the x is equal to a

• Factor the a value at that record

• Result should be a*(d+n)

• Goto to cell (c, 1)

• Find the row where the x is equal to b

• Factor the a value at that record

• Result should be b*(d+n)

• Goto to cell (-c, 1)

• Find the row where the x is equal to a

• Factor the a value at that record

• Result should be a*(x+n) <- this is not perfect, but always a * a factor of (x+n)

• Goto to cell (-c, 1)

• Find the row where the x is equal to b

• Factor the a value at that record

• Result should be b*(x+n) <- this is not perfect, but always b * a factor of (x+n)

Moreso, it seems that these 4 records are the only records whereby the x value is a factor of the a value.

Because at any record d=x+a:

• at (c,1) where x = our original a

• the a value in that record = a*(d+n)

• the d value in that record = a*(d+n+1)

Then go back 1 record, so that x = a-2:

• the d value in that record = a*(d+n-1)

Therefore, there are 2 records in (c,1) next to each other, where d is divisible by the higher x value.

I believe this is only true where x is a divisor of c, so 4 times per (c,1) (1, a, b, c) for our semiprimes.

If a cell contains an element c, another element in the cell can be constructed from it.

Call it c'

e' = e

n' = n

x' = x + 2n

a' = b

d' = a' + x'

b' = a' + 2x' + 2n

Once c' is constructed, c' becomes c and the process is repeated ad infinitum. e and n don't change, which make sense since these are the coordinates of the cell.

All odd numbers are the difference of two squares.

The product of two primes, is the difference two sets of squares.

Every odd number is the difference of two consecutive squares.

The value of n for the product of 1 and c is defined now as BigN for odd numbers.

Since d-a=x and x+n is the smaller square in the difference of two squares, then for odd numbers, BigN is ((c-1)/2)-x

x is d-1, since a=1