>>6585

>>6586

(31,1,8) = {31:1:143:15:128:160}

(-2,1,8) = {-2:1:143:16:127:161}

(-f,1)'s a is one less than (e,1)'s a, and (-f,1)'s b is one greater than (e,1)'s b, but only for this particular t value. At other t values, they're further apart. For example:

(31,1,2) = {31:1:23:3:20:28}

(-2,1,2) = {-2:1:11:4:7:17}

and

(31,1,12) = {31:1:303:23:280:328}

(-2,1,12) = {-2:1:311:24:287:337}

So at this particular t value, (e,1) and (-f,1)'s a and b values are one unit apart. Just blindly looking for unverified patterns here, but the difference is |f|. If these rules apply to all other values (or at least the first one), this t value would seem to act like a mirror point or the turning point of a parabola with a predictable gap where the gap between these a and b values from (e,1) and (-f,1) gets larger either side of this t value (and probably harder to predict). I've spent about 5 minutes on this so I don't know how that applies to factorizing c, but other than that, both infinite sets do that thing with the a and b values swapping (i.e. a[t] = b[t-1], I don't know about b). If nobody else does I might have a look at these other latest VQC clues at some point tonight.

All of that said and done, yet again, I feel like I'm completely out of the loop: VA, how did you get a t value of 8 with the na transform?

>>6588

Good stuff. So this means you can find the product of b and (n-1) of the correct (e,n) record just using (e,1) and (-f,1). That's pretty crazy. That doesn't mean this is solved, does it? I'm not sure how you'd factor one of the two out.

>>6593

Oh wait, recursively? If you treat b and n-1 as a and b, you could do the same thing until you're applying it to an (e,1) record.

>>6597

How are you getting the correct t for that record? I think I might have just missed a couple posts a while ago about that or something.

>>6597

>>6598

I should elaborate. I mean the correct t for the (e,1) and (-f,1) records from d, e and f.

>>6600

Do you mean you can find the t of the (e,1) record using the d and e from the (e,n) record?

>I haven't actually solved anything yet

If (e,n)'s b * (e,n)'s (n-1) = (e,1,t-1)'s d - (e,n)'s d, and you can find (e,1,t-1) using only (e,n)'s d, e and f, then I think you actually have. You're using this to factorize the product of two unknown numbers, and b(n-1) is two unknown numbers. If you apply the d[T-1] - d thing to new_c = b(n-1), you're factorizing that too. All you have to do is apply this recursively until you're applying it to something where n=1, in which case the (e,1) transform will be the same cell, f will be a square equal to (x+n)(x+n), allowing you to find a and b of this c=b(n-1), thus allowing you to find the b(n-1) of the previous c=b(n-1) and so on until you reach your original cell. So you are finding the t value of the transform without needing to know any more than d and e from (e,n)'s c, right? I can't imagine how confusing this paragraph would be to the uninitiated