I miss you all, triangular anons!

Sorry - I became very busy with work.

I'm trying to catch up a bit, and still trying to understand the triangular number formulas - if anyone has any specific posts they recommend I read first, I'm all ears.

I've been thinking a bit about the relationship between triangles and squares, so here's a list of things I've been thinking about regarding that topic (not sure if my whitespace formatting is going to work as intended):

1. We know that the series of all odd squares are described as 8 triangles + 1. (8T(u) + 1).

series of odd squares:

1, 9, 25, 49, 81, 121, 169, 225, 289, 361

minus 1, divide 8:

0, 1, 3, 6, 10, 15, 21, 28, 36, 45

Interesting to note the parity of the numbers here. Similar to the columns of e.

1. We know that the series of all even squares can be divided by 4.

series of even squares:

0, 4, 16, 36, 64, 100, 144, 196, 256, 324, 400

divide by 4:

0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100

So we have a way to break a square into either a smaller square or a triangular number. I've been thinking about how a triangular number can be expressed as a square, so my own 3 observations (although I think others have already expressed maybe some of these).

1. Any triangular number can be expressed as a square minus the triangle of number -1.

T(u) = u^2 - T(u-1)

I'm not sure if I like this expression, because in algorithmic terms the movement of u is linear.

Conversely, a square can be expressed as the sum of 2 consecutive triangular numbers.

u^2 = T(u) + T(u-1)

1. An even triangle number can be expressed as the largest square that fits under it (u/2) plus 2 smaller triangles of size (u/2).

T(u) = (u/2)^2 + 2T(u/2)

1. An odd triangle number can be expressed as the largest square that fits under it ((u+1)/2) plus 2 smaller triangles of size (u-1)/2.

T(u) = ((u+1)/2) + 2T((u-1)/2)

What I like about these last two is the u/2 in the recursive part of the algorithmic interpretation.

Ok, so another couple of known relationships:

1. The series of differences of 2 consecutive squares is the series of odd numbers:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144

• 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121

=1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23

1. The series of differences of 2 consecutive triangle numbers is the series of whole numbers:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91

• 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78

=1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13

1. The relationship between the two series above is 2n - 1.

So I'm not quite sure what to conclude of all this. I can imagine growth in several aspects of the relationship. I'm going to continue to get caught up. I hope this helps someone!

>>5167

Hey PMA.

I too am confused with the description of the last steps in the process.

I'm confused about the triangle stuff.

There are many cases when n-1 is larger than (x+n-1)/2, and therefore 1+8T(n-1) is larger than (x+n)^2.

Also, there many cases where (f-2)/8 doesn't lie between n-1 and (x+n-1)/2.

I'm very confused indeed.

I've been focused on understanding this post:

>>4678

In particular how the multiples of 2d can be turned into a square, or parts of the triangles.

This last series of posts in RSA 10, were super messy, and contained many errors.

Very unlike Chris.

And now his twitter is gone.

>>5171

I'm trying to understand now.

What about this image that VQC posted:

>>4678

Does your solution explain this post? I can't make the connection.

Based on that image, I was assuming that the 2 halves of (f-2) would be a band of the triangle, or maybe a band of the square, as in the image attached.

I like what you've done with the nn-1 on the outside, but originally I was thinking that the nn-1 + f is basically a recursive version of the problem.

>>5173

Ah, ok, I'll take another look at your images with that in mind.

You're saying that 2d(n-1) is a square? In which cases? In the solution and the prime? That could really help me.