HEY YOU GUYS PAY ATTENTION
I’ve made several posts all at once, but I want to draw very specific attention to this one. I’ve noticed while I spent the last month or two lurking far more than I was contributing that whenever VQC has brought something that was seemingly a change of subject from whatever we were focusing on up until that point, we’ve started focusing on the new thing mostly by itself. This is what this anon >>6407 was talking about. Having begun work on an explanation sticky, I’ve been bringing as many different possible calculations that we’ve figured out over the last 7 months together. I’ve noticed that this is a bit of a running theme: we work on one thing, VQC mentions something new, we start focusing on that, once or twice an anon will mention that they remembered something interesting that VQC said a while before that, VQC will mention another new thing, repeat the cycle. I found a specific crumb that sums up why it’s important for us to avoid doing this:
Once you state everything you know about the problem, the solution presents itself.
If we find every single thing we know how to find for a given c, and take into account everything we know how to find if we also know a and b, we will most likely have enough information to find the answer. I’ve been up for the last 5 hours or so combing through every thread trying to get absolutely everything that we have learned to calculate that we’ve been told is useful and turn it into the following list (although some of that time was spent talking to Topol on /qresearch/). I’m tired enough that I can’t entirely pay attention anymore, so it’s probably about time I stopped until “tomorrow” (today). I got up to >>2990 so I know for sure there’s a lot missing (such as parity of d and e being used to calculate parity of n and x, as well as triangle-related calculations). If anyone would like to pick up where I left off feel free to but either way I'll try to finish this off soon.
What I’m proposing is that we should pick any semiprime c and apply absolutely everything we know to it, rather than just focusing on triangles or just focusing on na transforms by themselves. VQC has mentioned this several times. This list is definitely not complete so if you notice anything that needs to be added please say so.
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)
All the things we can find if we know a and b:
>(e, n)
>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
All the things we can find that have been suggested to be important but not looked into in very much detail as far as I can tell:
>negative values of x for cell (1, 1) and then cells 2 to the right
>how many times numbers that are the product of three, four, five etc prime numbers turn up for a given e (it’s not linear, apparently)
>All products (integers) c that are the sum of two squares appear (only) in columns where e=0,1,4,9,16,25,..
>The value of x at (e,1) for na (a[t]) is the value of x at (e,n) (I only put this in the last section because I don’t completely understand the na (a[t]) part)
>think about the distribution of n and (n-1) in cell (e,1) in the a[t] and d[t] values respectively. This is the quick way to utilising f.