Hi fags, I'm the guy in previous threads concerned not about cool stuff, but the fundamental efficiency of reaching the correct cells. So with that in mind, I've been looking at {0:n:d:x:1:c^2}, because in a factoring situation, we don't have a, b, or n, but we do have c and c^2.

There seems to be an efficiently discoverable pattern in {0:n:d:x:1:c^2}. Let's take a 3, b 3, c 9 simply because it appears on the downloadable spreadsheets and .png files for The End. c^2 81.

Now let's look through column 0 for "1:81}", and you'll find {0:32:9:8:1:81}. This can be quickly constructed via the general pattern {0:n:c:c - 1:1:c^2}. You can solve for n = ((a + b) / 2) - d. For this case:

n = ((81 + 1) / 2) - 9

n = (82 / 2) - 9

n = 41 - 9

n = 32

So, we can immediately construct a starting point, here {0:32:9:8:1:81}. Now search column 0 for ":81}". You will find:

{0:2:63:14:49:81}

{0:8:45:20:25:81}

{0:18:27:18:9:81}

{0:32:9:8:1:81}

[cont]

>>934

Now let's look at the following values of n in increasing order: 2, 8, 18, 32. Let's start by looking at the differences between elements:

8 - 2 = 6

18 - 8 = 10

32 - 18 = 14

Or, put differently,

6 = 2 + 4

10 = 2 + 4 + 4

14 = 2 + 4 + 4 + 4

Let's just look at the 4's:

4

44

444

This is called a Triangular number (https://en.wikipedia.org/wiki/Triangular_number). If these were all ones, the formula for summing them (where h is height of tree, here 3, and where s is the sum) is:

s = (h(h + 1)) / 2

So for:

1

11

111

… we would have:

s = (3(4)) / 2

s = 12 / 2

s = 6

Which you can easily verify.

[cont]

>>935

The formula for each element being 4 and not 1 would be simply 4 times the previous equation:

s = 4((h(h + 1)) / 2)

So for:

4

44

444

… we would have:

s = 4((3(4)) / 2)

s = 4(12 / 2)

s = 4(6)

s = 24

So let's go back to the sequence above:

6 = 2 + 4

10 = 2 + 4 + 4

14 = 2 + 4 + 4 + 4

Here, h = 3. We can sum up all of the 4's with the above equation. All we need to add in are the 2's, and there are exactly h 2's. So:

4((h(h + 1)) / 2) + 2h

Recall the original sequence 2, 8, 18, 32. Since it starts at 2 and not 0, we have to add 2 to the final result:

4((h(h + 1)) / 2) + 2h + 2

Let's now denote the first element 2 as element 0 and 32 as element 3, or h = 3.

s = 4((3(4)) / 2) + 2(3) + 2

s = 4(12 / 2) + 6 + 2

s = 4(6) + 8

s = 24 + 8

s = 32

>>936

Since the sum 's' is equal to the n field in {0:32:9:8:1:81}, n = 32, and in general, we can use n above where we used s. If we simplify:

n = 4((h(h + 1)) / 2) + 2h + 2

n = 2h^2 + 4h + 2

0 = 2h^2 + 4h + (2 - n)

We can apply the quadratic formula:

(-b [+-] sqrt(b^2 - 4ac)) / 2a

a = 2 (the constant in 2h^2)

b = 4 (the constant in 4h)

c = (2 - n) (the plain numbers)

(-4 [+-] sqrt(4^2 - 4(2)(2 - n))) / 2(2)

(-4 [+-] sqrt(16 - 8(2 - n))) / 4

To work the example above with n = 32, we can find "where we are" in the series in {0:n:d:x:1:c^2}. So for {0:32:9:8:1:81}, where n = 32:

(-4 [+-] sqrt(16 - 8(2 - 32))) / 4

(-4 [+-] sqrt(16 - 8(-30))) / 4

(-4 [+-] sqrt(16 - (-240))) / 4

(-4 [+-] sqrt(16 + 240)) / 4

(-4 [+-] sqrt(256)) / 4

(-4 [+-] 16)

(-4 + 16) / 4 [choosing +]

12 / 4

3

So at {0:32:9:8:1:81}, we know that we're at index 3 (zero based), or if you want to talk -th numbers, add 1, so we're at the 4th.

Interesting, right?

>>938

>A is a perfect square

This is part of the pattern

>N is a perfect square * 2

This is random luck and not part of a pattern

>>959

Sorry to post this here, but these are my comprehensive suggestions: >>962

>>1069

>>1070

Nobody has a problem with the occasional inspirational shitpost. What we have problems with is when it's 20% of total posts.

>>1078

Nobody wants bans. But nobody wants every other post to be four huge pictures of deep dream wankery plus stoner memes. Especially when we're crunching the numbers and formulas in each others posts. One in twenty I think we can absorb.

So it's a matter of degree. Like, for example, bumps on .5 chan.

>>1082

Forgot your name, VeritasAequitas? Why are you hopping around on multiple overlapping IDs, and posting with and without your "name"?

Nobody is saying ban the spiritual or ban even shitposting itself. The bans should be reserved for when there are high signal levels, such as when VQC is posting or when there is significant technical progress being made, and spiritposting or shitposting every other post is disruptive.

>>1099

VQC, can you please take a quick look at my post >>934 where I show how to calculate any n for any column 0 record where c^2 appears?

>>1120

Thanks. Yeah I should have been more clear. You calculate the 1 * c^2 value in column 0, then convert that n to "index", aka "height". This is zero-based, so it goes index 0, index 1, index 2, index 3…

My example had n = 32 and I calculated a index or height of 3.

The point I should have better stated is that from that you can calculate the n values of index 2, 1, 0 in constant time (which are 18, 8, and 2 respectively).

This is interesting because any 0, n cell you generate this way has perfect squares for a and b. Work through it once or twice on paper; it's fucking magic.

>>1696

>Is anyone in touch with Inner/Outer Heaven?

>>1697

>Do tell, MA. Inner/outer Heaven? What’s your intuition saying?

It appears VA has outed himself as a newfag?