>>8974

Last photo for now. So we've seen how the patterns look when you look at -f, e and g's. The lines are all a bit skewed as you can see, there is an angle involved here.

But we know that our c-value occurs in multiple columns as well. Here is a 1000x1000 image with the same structure above, except instead of -f, e and g values it's the first 1000 e's where there is an a[t] = c.

So with the exception of which columns is being used, the rest is the same. Here you can clearly see the squares, the lines are sharper and less angled.

>>9014

>>7706

In case it's forgotten or people needs this to be refreshed. The BigN - n is equal to (a - 1)(b - 1)/2. This is also why the smooth numbers in (15, 6) are equal to the d in (3, 1), since the d in (3, 1) are half of c, and (15, 6) is equal to (3, 6) + 1 in all the a-values, meaning BigN - n (15, 6) will then turn into the a * b/2 from (3, 6).

Just to give an example

a=7, b=37 ={3:6:16:9:7:37}

a=8, b=38 ={15:6:17:9:8:38}

The smooth value for a=8, b=38 is equal to 7*37/2 (129). This also implies that at (e, 1, d/2) (even) and (e, 1, (d+1)/2) (odd) will be the BigN * c transformation.

>>9019

Excerpt from https://pdfs.semanticscholar.org/d51d/6f92b7bfacef49152d0961ebf1b26f5d495f.pdf, page 14.

>>9021

Excerpt from: https://vdocuments.mx/self-witnessing-polynomial-time-complexity-and-prime-factorization.html

Keep 'em coming, and I'll read and try to understand 'em.

>>9023

Haven't had time to read and digest, but I did think some more and I feel stupid again.

We know BigN - n = smooth number = (a-1)(b-1)/2. That means BigN = (a-1)(b-1)/2 + n (or shadow n of (a-1)(b-1) from the perspective of n as d).

To illustrate, take a=7, b=37 and c=259. This exists in:

{3:6:16:9:7:37}

{3:114:16:15:1:259}

The BigN - n = 114 - 6 = 108 =6*36/2.

We flip it and say 114 = 636/2 + 6 (treating 6 as d and computing the shadow n). We can then compute 216 - 66 = 180 (ie in e=180 we should have 6 and 36).

(6+36)/2 - 6 = 15, (6+36)/2 + 6 = 27

{180:15:6:0:6:36}, {180:27:-6:-12:6:36}

It should also be noted though that {180:114:-6:-7:1:216} doesn't appear to exist.

>>9024

Imagine this process:

You have a, b. Create the record for this and the record for c. Compute the BigN - n. This is half of (a - 1)(b - 1).

Do the same, but this time for a-1, b-1 and c=(a-1)(b-1). Repeat the process until you hit the "bottom" (I a=1?).

Then compute the differences between the smooth numbers, or rather the differences between ab/2, (a-1)(b-1)/2, (a - ..)(b - ..)/2, (a - a + 1)(b - a + 1)/2.

>>9026

So what the hell happened here and in /templeos?

>>9028

Who is this Defango guy?

>>9030

And MAGA Coalition == Anti Q?

>>9032

Oh okay. I see then.

Btw I made a post about something I found in the whole templeos thing, don't remember if this was known, but in (0, dd/2) in the negative space you will find a=aa, b=xx. This is the shadow cell for that record.

If only we could compute / predict x am i right

>>9034

Maybe? I don't even remember our terminology for everything.

I think record and cell overlap a lot, but element I'm not familiar with.

>>9036

I was referring to the x in (0, dd/2) or pretty much any x that achieves our goal.

So:

(e, n) = cell

(e, n, t) = element

In which case in the cell (0, dd/2) there exists an element in the negative space where a=aa, b=xx where a, x are from the original element in (e, n).

>>9039

Yeah, you need to know t, which we don't. However we know the t in many other places can be predictable, for example in (e, 1).

>>9047

For an odd d it still holds, but also not. It exists in the dd/2 which won't be an integer, since the n will exist between two integers. I haven't look that close at it, since I don't have faith in going down this path.

>>9049

If we think about the fact that the smooth numbers are equal to (a - 1)(b - 1)/2, then we know something about the numbers as we multiply prime numbers.

Say we have c=5x7x37. Then BigN - (5 - 1)(7x37 - 1)/2 = n for a=5, b=7x37.

This is also be 2BigN - (5 - 1)(7x37 - 1) = 2n for a=5, b=7x37.

What I'm trying to say / think is that we know 2n is congruent to 4 for mod 5. If we add more primes, would we be able to shorten the BigN - ? calculation, that is calculating 2BigN - 5xi and then testing the resulting numbers?

Given that we know BigN - (a-1)(b-1)/2 = n, and we know that n has to be a factor in a[t] in (e, 1), does this give us enough information to do a more informed search? Maybe not what VQC has in mind, but if it gives us a faster search it's still a step in the proper direction.

Also is it Chris who's doing the posts in templeos as VQC or someone else fucking around?

I was thinking about n and shadow n and something hit me. Like usual, I am clueless as to if it is useful, but it is an interesting though.

We have the following equations:

n = (a + b)/2 - d

shadow n = (a + b)/2 + d

We know at (e, n) and (e, shadow n) we have the records for a, b. Same principal applies for c.

What occured to me is that this also gives us another perspective on d.

d = (a + b)/2 - n

d' = (a + b)/2 + n

d = (a + b)/2 - shadow n

d' = (a + b)/2 + shadow n

Could this be of any use? Could it illuminate our problem or perspective of the problem in any significant way?

I was thinking about n and shadow n and something hit me. Like usual, I am clueless as to if it is useful, but it is an interesting though.

We have the following equations:

n = (a + b)/2 - d

shadow n = (a + b)/2 + d

We know at (e, n) and (e, shadow n) we have the records for a, b. Same principal applies for c.

What occured to me is that this also gives us another perspective on d.

d = (a + b)/2 - n

d' = (a + b)/2 + n

d = (a + b)/2 - shadow n

d' = (a + b)/2 + shadow n

Could this be of any use? Could it illuminate our problem or perspective of the problem in any significant way?

>>9147

Disregard dup, 8ch is crapping out tonight. I guess something big is going down.