>>5100

I made that. That is not completely related to the grid but its neat. So missing to the left in that pic is a column of 1's. Then each cell is determined by the sum of the cell above and the cell to the left for the rest of the grid. If we index this by (x,y) where x is the column and y is the row, it would be a triangle with side length y in the xth dimension. Maybe we can use this to help us out. 1st column is integers, 2nd is triangular numbers (which we can use to construct the squares), 3rd is pyramidal numbers etc.

Now if you change the column that is missing on the left to all 2's, this results in odd numbers in the next column. Then, the next column would be successive odd numbers. Then the next column would be a sum of successive odds which is also the square numbers, which can also be used in our square calculation stuff. The next column is cube numbers.

Essentially, if you have the left column as x and the top column as 1, then an entry in row r column c would be a (x+2)-gon of side length r in the cth dimension. Idk maybe we can use some geometry for this. Also, maybe since everything is derived from the 2 initial seeds, (x and 1), we could reduce everything to multiples of 2 and multiples of 1 and use that. I'm not sure.

The more interesting thing I discovered recently was a pattern in the continued fractions. Last week my Professor taught us continued fractions and mentioned that there were patterns of the continued fraction for sqrt(2) or something carved onto Babylonian tablets or something. He also mentioned that nobody knew anything about continued fractions. So I discovered a pattern for continued fractions that oddly relates to this whole dd+e = c thing.

So for pic related that continued fraction would be notated as [a0, a1, a2…]. Also for this, I will write this [a0, (a1, a2)] to say that the thing starts with a0, then the pattern inside the parenthesis repeats forever.

So if you have a continued fraction calculated (also if you do the calculation of these by hand, everything always conveniently cancels out, which might be good for computing) for this number sqrt(dd+e), then if (2*d) is divisible by e, then the continued fraction would be [d, (2d/e, 2d)].

(look in pic related for some good examples). If we extrapolate from there, and say ab=c, then we would know that there exists a fraction [c, (a, 2c)] which is equivalent to the square root of (c^2 + 2b). He was telling us to look at c^2 so maybe this is it?? idk. Also you could do root(c^2 + b) = [c,(2a,2c)]. These numbers would all be like the parent number but idk I've been cranking at it algebraically to solve for a and I can't poke any holes in it. Also if you do the continued fraction of [(1)] it is equivalent to the golden ratio. Also, you can calculate it out to a certain amount of terms, and you will notice that the numerator and the denominator are always fibonacci numbers. These numbers are where I was talking about the 'cancellation' because it happens with most repeated continued fractions.

>>5100

Also I didn't post it in that thread, somebody else did

>>5103

The theorem for calculating the terms out is this:

let N(t) be the numerator of the tth fraction. Guess what D(t) is. This is for [a0,a1,a2,…]

N(0) = a0

N(1) = a0*a1 + 1

D(0) = 1

D(1) = a1

Then, N(t) = atN(t-1) + N(t-2), and D(t) = atD(t-1) + D(t-2).

These formulae make it easy to see how [1,1,1,1,..] is fibonacci. This could be easier to program cuz you can cache the stuff

>>5104

Also basically the higher your t value, the more precise a fraction you will get to approximate the true value of the continued fraction. N(0)/D(0) will be a shit approximation, but N(100)/D(100) will be pretty damn close to the number you want.

>>5129

Yeah those would be sums of successive squares, which would appear in the last pic on the second graph in the third column. I think my hypothesis about the n-gon of x-dimensions may be false but it is true for the 1,1 grid on the top left.

>>5142

Recently in the qresearch Q posted USE LOGIC. Could this mean we are supposed to utilize logical functions? Such as AND, OR, NAND, or XOR. I don't know if there are any proofs about these functions and the integers but we might get something

>>5325

41/184 C(c)/EDX(x,e,d)/AB(a,b)/EDX(d,x,e)/

41/184 C(c)/EDX(d,x,e)/EDX(d,x,e)/AB(a,b)/EDX(d,x,e)/

23/184 C(c)/EDX(x,d,e)/AB(a,b)/EDX(x,e,a)/EDX(e,a,d)/

26/184 C(c)/EDX(x,e,d)/AB(a,b)/EDX(x,e,d)/EDX(x,e,d)/

28/184 C(c)/EDX(x,d,e)/AB(a,b)/EDX(x,d,e)/

These are the algorithms and the relative frequencies of success for the first however many coprimes. Nothing rock solid yet.

>>5330

Marked a lot of the semiprimes. Didn't feel like finishing it.

This highlights all the cells with same a values. Stumbled on this website but its eerily similar

http://www.divisorplot.com/5.html