If e is even and f is odd, the t values in the prime solution cell, the an cell in (e,1) and the a(n-1) cell in (f,1) are all equal. If e is odd and f is even, the t values in the prime solution cell and the an cell in (e,1) are equal, and the a(n-1) cell in (f,1) has a t value one greater.

>>7647

These rules carry over into (f,1). The fact that 2d+1 is odd (and is the gap between (e,1) and (f,1) obviously) means that an will fit either the odd or the even rule and a(n-1) will fit the other (but with f/2 or (f-1)/2 instead of e). If an is equal to twice a square plus e/2, a(n-1) will be equal to two consecutive squares plus (f-1)/2, and vice versa (vice versa with the correct rules I mean).

Also, there are rules to the squares as well. (fyi, I’m using e to describe both e and f. f being added to something is akin to |f| being taken away, since f is negative) In odd e (e,1) cells, the a values are 0+1+(e-1)/2, 1+4+(e-1)/2, 4+9+(e-1)/2, etc, from t=1, t=2, t=3. In even e (e,1) cells, the a values are 0+0+(e/2), 1+1+(e/2), 4+4+(e/2), etc, from t=1, t=2, t=3. So when e is odd, the a values are equal to (t-1)(t-1)+(tt)+(e-1)/2, and when e is even, the a values are equal to 2(t-1)(t-1) + (e/2). This explains why (e,1) has t values starting from 1 while (f,1) has gradually increasing minimum t values. The a value at t=1 in e will be 0 plus something positive. After e=0, it’ll always have a value. The a value at t=1 in f will be 0 plus something negative. The negative value being added will gradually increase as f increases, and as it becomes bigger than each of the squares, it’ll increase the first t value at which there’s a valid (positive) a.

an, bn, BigN, c*BigN and c(c-BigN+1) all appear in (e,1). If e is even, each of these values minus e/2 will be twice a square. If e is odd, each of these values minus (e-1)/2 will be equal to two consecutive squares added together. In all cases, they’ll be different squares, but these squares will all be based around their t values. a(n-1), b(n-1), BigN-1, c(BigN-1) and c(c-BigN+2) are all in (f,1). Same rules.

(x+n) = (b-a)/2. You can rearrange this with algebra to show that the (x+n) from the cell where a=1 and b=b minus the (x+n) from the cell where a=1 and b=a is equal to the (x+n) from the cell we're looking for (since these (x+n) values are equal to (a-1)/2 and (b-1)/2 respectively).

Polite numbers are numbers that can be expressed as the sum of consecutive integers.

The set of impolite numbers are numbers that cannot.

The impolite numbers are the powers of 2.

The number of ways you can define a given polite number as a sum of consecutive numbers is equal to the number of odd factors (including 1 and itself).

Triangle numbers are obviously a subset of the polite numbers (whereby the sequence starts at 1), and a square is the sum of 2 consecutive triangle numbers.

An odd square is the 8 triangles + 1.

An even square can be reduced by dividing by 4.

Therefore the problem can be reduced down to differences of triangle numbers where one triangle is off by 1, and this difference of triangles can be expressed as polite numbers…

If you have a particular an in (e,1), there will be another cell in (0,1) or in (1,1) with the same t value. a from these cells and f from our semiprime have a relationship with a(n-1).

If e is odd, you find the cell in (0,1) with the same t as the (e,1) cell. That (0,1) cell’s a minus (f/2) is equal to a(n-1).

If e is even, you do the same thing, but with (1,1) and with (f+1)/2.

Apparently there's a useful values in the cell in (e,1) where x=f or f-1. VQC didn't say which value though. Also it appears that it doesn't make sense to say where x=f based on parities, but he said it, so I'm putting it here.

The VQC is a lookup table, and there's meant to be a "lookup x" for our given c that "gives you na or (n-1)". There's some x value somewhere in the grid that can applied to c in some way to give us na or n-1. Since there are two possibilities, it's probably based on e and f parities, given all of the patterns based around e and f parities. Also, since the x value in (e,1) where a[t] = na will always be equal to the x value in the prime solution record, finding that x value would bypass (e,1) and na completely, since we wouldn't need to factor na to know what a and b are (since we'd have the prime solution record). So this x value is most likely not the x value from the (e,1) record where a[t] = na.

In regards to the frequency of repeating A[t] values and B[t] values in (e,1), (-f,1) and other cells with same c but different d. Look at the following records images. Pay attention to the block on the right. The center dot is (e,1,t) for whatever record we have at the moment with the same x or t. Column to the left would be -f, then the next would be for d=d+2, etc. Then if you go to the right each column is d=d-1. Going down you have higher t values, and going vertical up you have lower t values. If there is a red dot that means it shares a as factor of A[t], green dot means it shares b as factor of A[t] and cyan is c as factor. Notice that for any entry in with the t=t+n shift, the location of that c row (which is origianlly the cyan one going down to the right) turns into a B line for the next record at t+n, then it is always a B line for the rest of those records.

For e=0 you can get to all records where a and b are a multiple of n just by starting at (0, n, 1) and jumping by t = t + n. First image is using c6107's d value of 78 to get to (0, 78, 1), and then jumping.

Somehow this works for 2d as well, where jumping by t+78 from (0, 156, 1), the a and b values are still always divisible by 78. Next level up you could use 156 * 2 = 312 and go to (0, 312, 1+312), and a and b is always divisible by 156.

This diagonal movement increases c 4 times for each step, and has some really interesting patterns in the generated records.