It’s high time we solved this bloody thing. It’s kind of sad reading through the last 7 months and remember the many times VQC said we were pretty much there and everyone saying “oh nice it’ll be done in the next week or two”. Here’s the list updated, with one thing definitely missing (explained at the bottom), and potentially other things I may have missed (which I guess you guys will have to spot and point out). I’ve gone through every thread, so it should be pretty much everything. Logic would dictate that if we apply every single one of those things to a given c, the answer will be in there somewhere.
All the things we can calculate/find/whatever with our given c:
>d, e and f (d and e being the variables necessary for calculation, “the collapse of the superposition”)
>the cell where (a, b) = (1, c)
>the cell in (e, 1) with our e and n=1
>the cell in (-f, n-1) with our f and n=n-1 (0)
>the (e, 1) and (-f, n-1) cells related to our (1, c) cell’s e and f values
>if positive f is a square and increasing d by 1 and e by 2 results in the same positive f value, we can calculate all of endxab
>the place in column 0 (e, 0) at which our c^2 is (which may be in more than just (0, 0), in which case I’m not completely sure this is calculable without ironically having to factorize)
>whether or not e is 0
>the parity of e
>the pre-factorization of the decision tree (dividing by two until odd)
>n0
>what would be missing from the triangles if we could only fill them with n0 squared and multiples of 2d
>every different case of the width of the (f-2) chunks we used to calculate n0 being doubled (which is relevant because “if doubling the width of the chunks of (f-2) didn't give us a solution, the number c would be prime, the worst case (doubling the width of (f-2) chunks very quickly determines this as the number of times it doubles is logarithmic in the length of (f-2))”
>the remainder of f when divided by 8 (which is a key to what values of d and (n-1) can be used)
>an uneven distribution of f-1 among the eight triangles
All the things we can find if we know a and b:
>(e, n)
>(d+n)(d+n) and (x+n)(x+n), as well as (d+n) and (x+n)
>(xx+e) and 2na
>the cell in (-f, n) with our f and the actual n from our c
>big_N, which is n at the cell where (1, a) or (1, b) = (1, c) (i.e. every prime number can only be calculated with 1 and itself as the a and b values, and so they’ll only ever have one n value)
>the (e, 1) and (-f, n-1) cells related to our (1, a) and (1, b) cells’ e and f values
>the cells in (e, 1) at which na and nb are n cells apart (n being the correct n for our c)
>t
>with the cells in n=1 that we can find related to our c (and all others), the cell at n+=2 has the same n and x values but a, b and d have increased by 1
>for our given c’s (e, n), there will be values in a cell at (e+2n,n) (which is meant to be an important pattern too)
>the cell at which a = a^2 and b = b^2, the increasing powers, and the other binomial-related combinations of a and b
>the x value at (e,n) is equal to the x value where d[t] = na+x
>a pattern of (n-1) as a factor of each d[t] value at (e,1) which is different (increasingly) from the pattern of factors of n in a[t] which gives us the offset that is used to solve the problem and thus get the cell at (e,1) to do all the work for you
>big_N’s D minus the real d to our semiprime c contains (n-1) as a factor
>however these trees >>3654 were made (I can’t figure it out)
>the factor tree is meant to find x or x+n, so if we can generate a factor tree and we know x and x+n we’d be able to find where they fit in
>eight triangle numbers that are equal to (x+n)(x+n)-1
>the base of each triangle for odd (x+n)(x+n) (which is ((x+n)-1)/2)
>(n-1)(n-1) is another odd square so you can use the same 8Tu + 1 rules on it, meaning you would recursively find the base, the triangle numbers etc for (x+n)(x+n) = (n-1)(n-1)
>the maximum and minimum amount to add and subtract from (f-2)/40 (another thing I didn’t completely understand, >>4344 here)
>a square made up of dd + e + f + 2d(n-1) + (nn – 1)
>another square made up of nn + 2d(n-1) + f – 1 (this and the above being a rearrangement of c being the difference between two squares)
>the permutations of possible values of f,d and (n-1) for each odd (x+n)(x+n)
>the range of possible triangle base values which are larger than (n-1)/8 but smaller than (x+n)/8, and the relationship between this range of values and (f-1)/8
>the range of possible x and n values for a given (x+n)
>the range of possible e, d, a, b, c, f and t values for a given (x+n)
>the sequence of variables in relation to (x+n) squares when you move around the grid by moving across with a constant x and steps of 2n and when you move up and down keeping d constant
>(-f,n-1),(-f+2(n-1),(n-1)),(-f+4(n-1),(n-1)) etc and (e,n),(e-2n,n),(e-4n,n) etc