Evening all.

So as you may have noticed, I haven't posted much lately, because I haven't made any progress.

But I think I just had a mini breakthrough.

I can't explain it better than this at the moment:

We have the grid organized by (e,n). Any number that is e bigger than a perfect square will be listed in column e.

What about f though? For each entry in the grid, what is the f value and how does it change in the column e?

Our semiprime value c has a constant e & f. There are many many entries in column e, but only 2 entries will share the same f.

So lets look at f in (e, 1).

If e is odd, the series of f is:

f = 4, 16, 36, 64

f = 2^2, 4^2, 6^2, 8^2

If e is even, the series of f is:

f = 1, 9, 25, 49, 81

f = 1^2, 3^2, 5^2, 7^2, 9^2

This is super interesting, since we should be able to find our axb record by simply searching for a matching f.

I'm trying to figure out a number of things as next steps:

• How does this pattern change as n grows?

• This seems to be related to x+1, more analysis needed for n>1

• How do we use the na cell in (e,1) to locate our axb for n>1?

A little more info. This came out of analyzing pythagorean triplets. They can only be 3 even numbers, or an even and 2 odds. We only care about the second case.

In the c=145 example, the na records in (e,1) both have the even side of the triangle = f (24), and the hypotenuse = a value in (e, 1).

>>2664

Correct, this is in the case that you have f = a perfect square, also n = 1.

>>2666

Sorry, maybe i said it wrong. It wont match in (e,1). It will just match elsewhere in e. In the (e,1), all f's are perfect squares.

There are patterns in f too for higher n. I think I'll have an equation soon. Once we have that we can calculate n from f.

>>2668

Take our example c=145.

d=12

e=1

f=-24 (or 24 if you use absolute value)

If we look at (1,1), we get the following elements:

{1:1:2:1:1:5} (f=-4)

{1:1:8:3:5:13} (f=-16)

{1:1:18:5:13:25} (f=-36)

{1:1:32:7:25:41} (f=-64)

{1:1:50:9:41:61} (f=-100)

{1:1:72:11:61:85} (f=-144)

{1:1:98:13:85:113} (f=-196)

{1:1:128:15:113:145} (f=-256)

{1:1:162:17:145:181} (f=-324)

In each cell -f = (x+1)^2. This equation works for any (e,1).

I don't have the full process yet for jumping to the correct n, thats the piece thats missing.

But what I'm testing out right now, is to jump to the n = a record for the largest square under our f.

Then traverse that list.

Its obvious that thats not all there is to it, because c=85 doesn't work.

But if we figure out how f grows with n, we should be able to find the correct n given f.

If you have code to generate the grid, add f to the output and you'll see the patterns i'm talking about.