Before anyone gets excited this was "". (3,2)'s d values are the triangle numbers multiplied by 2. Not sure where you were going with this.
Something interesting I found: the d and f values in (4,4) and (-2,1) are identical.
>Understanding closure under multiplication of the sum of two squares.
>This means that multiplying the sum of two squares by the sum of two squares will always give the sum of two squares.
>This explains the numbers that are in the columns with a square remainder.
I explained this here >>7551 if anyone wants more detail about it. Odd numbers that are the sum of two squares are only divisible by other smaller odd numbers that are the sum of two squares, and the a values in (1,1) are the sum of two squares (but not every odd sum of two squares shows up in (1,1) as an a value).
I wonder why he isn't coming to this board lately. He's pretty much only talking to us anyway. Why wouldn't he just come here to do it?
>I was going to
Dude come on, nobody's judging you here. It isn't egotistical at all, at least in my opinion. The only way any of this could be egotistical would be if it was tied to your identity. I'm not just saying this because of how much I wanted March 3rd to happen I wanted March 3rd to happen immensely, I have to admit. You've said yourself many times that you want to give this to the world through us, right? The way in which you do it (if you reveal it all on the 29th, or partially on the 3rd, or something else) is completely selfless if you can still walk down the street afterwards and not get mobbed by people who think you're some kind of wizard. You could even end all of this right now if you wanted to, regardless of where the moon is. It wouldn't be egotistical in the slightest. I dare say, given your conduct and intentions, it would be the exact opposite.
Now that I've had a day to think about this, I've come to the conclusion that the only way you could truly not be egotistical in this context is to release everything immediately. I don't say this based on any of my own impatience, and I'm not sure if it's even the best idea, but think about it logically. You've been leading on a bunch of people for over a year and keeping this magical thing an inch from our noses the whole time, causing us to invest a lot of time and effort into studying it and occasionally causing a few of us to have negative emotional outbursts, and you've repeatedly given us dates on which you're going to reveal significantly more details only to cancel at the last minute every single time, making us more and more pessimistic to some degree (some of us don't think March 29th is going to happen either, for example, based on all the other times). All of this is because of your personal decision making and your personal judgement. Keep in mind, I'm not saying any of this because of my own personal wants. This is just the logical conclusion I've come to in regards to your moral dilemma. You want to eliminate the influence of your own ego and personal judgement in this situation? Release everything about The End right now, regardless of what day it is.
Happy birthday from my part of the world, by the way. I really hope you come back today, read our replies and take them all into consideration.
For the sake of keeping track of the current understanding of those of us who have been here for a long time:
The information in these posts is already in the Grid Patterns thread. It looks like everything so far is known, at least to me (and I would think everyone else). If you're still awake, what's the next part?
You haven't explained X and Y yet but I've figured something out from the use of (1,n) already. Since the only factors of a[t] values in (1,1) are odd sums of squares, and since the factors of any number in (e,1) appear as n values in that column, every valid n value in (1,n) is an odd sum of squares.
With a static odd b value and increasing odd a by 8 each time, the distance between the t values of the higher of the two cells in (e,1) where the d values surround our c's d value and the (e,1) a[t]=na cell looks like this. It increases by 8 because there appear to be four separate groupings of the numbers. These graphs show the x axis as each a value from 3 to 769 (769 was the b value I chose) and the y axis as the distance from the upper d cell to the na cell. Hopefully that all makes sense. It's related to >>8663
So what's the schedule today?
"Tomorrow" he says
>I think it's a bit about how you talk to him
Yeah, I really doubt that. He has repeatedly flaked out after saying he's going to tell us something on a specific date. I don't know who you are so I don't know if you've been around as long as the rest of us and seen it happen, but he's done it a bunch of times. Him doing it again is irrelevant to me pointing it out (in a non-spiteful way, might I add; I only intend to make him aware of the fact that he's doing it if he isn't already aware, so that maybe he can stop doing it).
I'll post two questions now just in case I'm not awake when you're doing that. A few of us on Discord were trying to figure out what one of the lines from your post here >>8666 is meant to mean.
>All values in the cells at n=1 are products where you add a small square to e to make a square with c.
What does it mean to "make a square with c" (do you mean adding a square to e equals a square which is also c plus something, or is it something to do with c squared, or what)? Which values in particular are you referring to as products (c[t]? a[t]? something else maybe)? If you're referring to something other than each cell's c, what does adding a small square to e to make a square with c have to do with these products? It's a particularly confusing line.
I have one other question about what a non-trivial factorization is meant to be. Here >>8650 you said that LookupN(t, e, f) returns n. Since it takes t as an argument, that implies that you can't find n with just c, d, e, f and t (you'd have c and d since a particular pair of e and f only has one d). But you can find x with 2(t-1) or 2t-1 depending on e's parity, and a=d-x. So does that method take an incorrect t as an argument or something? What's the deal with that?
>Before this evening, have a look at large products, and the difference between n and BigN. The difference is the same as for n-1 and BigN-1, which you would find in [-f,1]
Here's a few examples. The last one is RSA100. Since we're dealing with smooth numbers, I included the prime factorization for BigN-n for each one. They're all semiprime cs, if that matters.
Oh you're the person who posted the dude with the head gear on the bunker board. That was weird, man.
>c'=abq=qc
Are we going to end up multiplying qc by another variable called v?
>how do we know what factors will also make up the two values of t where a is a factor of [e',1,t]?
Do you mean the factors of the two values of t or the factors of the two values of a'[t]? I checked the factors of a and there doesn't seem to be any obvious pattern anywhere, even if you keep a static a and change b or you keep b static and change a. Pic related. If you were actually referring to t and that wasn't a typo, that also doesn't seem to have a pattern, especially since a lot of the ones I've seen are prime.
If you're trying to teach this to us before the 29th rather than just dropping it all, you're going to have to post a bit more frequently I would think.