From the first two threads:

Columns contain all products that have the same remainder e. The number of factors a number has determines how many times it shows up in a given column (if it’s prime, once; if it’s a semiprime, twice; if it has three prime factors, three times, and so on).

Row one contains cells which contain all factors for each column (na and nb are n apart (at a[t] and a[t+n])

a[t+n]-a[t]%n==0. These values represent na and nb of any c. If you pick any two cells in (e,1) and subtract the lower from the higher, the result will be divisible by the gap in t values. Sometimes this will produce a result that means b is less than a, so it is not always 100% valid.

The cells that have elements have a finite number of "seed" elements that are the lowest values of c (a multiplied by b) in that cell, from which all the rest of the elements in that cell can be constructed (since values of a increase by a+2x+2n from a seed element). Other seed values can be constructed based on the value of -x+2n and depending on the number of factors within n for any given cell. There is only one seed element in row one (e,1).

As a subset of this thread, let's list each individual number or cell which, if found, would solve the grid:

-a

-b

-n

-x+n

-x

-t

-xx/2 (because na is (xx+e)/2 and na appears in (e,1) — xx/2 +1 for odd e)

-an, bn, a(n-1) or b(n-1)

-the x or t values of the an/bn/a(n-1)/b(n-1) cells

-some number which is always greater than n but less than x+n (for finding n0)

-the first cell in which a[t]=c in (e,n) (n not necessarily being the n we're looking for)

We can find two cells in (e,1) in which a[t] is a multiple of c where x=c-d and where x=c+d+1. In the first of these two cells, a[t]=cBigN. In the second of these cells, a[t]=c(c-BigN+1). There are also equivalent cells in (-f,1), where a[t]=c(BigN-1) and a[t]=c(c-BigN+2). In both positive and negative space, these cells are 2d+1 apart (the d in this gap being the d of the c we're trying to factor).

>>7354

-any of the three cells where c'=cc and where n>0

For example with 1343=559, there will be four cells: where a and b are 559 (which has an n of 0, so we aren't looking for this one), 13 and (134343), 43 and (131343), and (1313) and (43*43). If we find any of these, we'll know that either a or sqrt(a) is a factor of c.