Apologies if this is old to you guys, but I noticed something:

For all even numbers of 'e' in the n=1 row, cells are filled in a pattern that matches a slope of 1/e (or -1/e, since the coordinates are reversed in graphics). It's not always reduced, but it's regular; for instance, starting from e=1, n=1; you can go down 2, right two, and a cell will be occupied; you can continue this on through about half of the graphic, then onward from there at a slope of down 4, right 4. From e=2, n=1; you move down one, right two, pretty much through the entire graphic–every cell following that pattern has contents. For every even 'e' in a 64-iteration grid, the pattern holds to the edge of the graphic (up to e=52, n=1 in that case).

I've made a huge image with color overlay to help identify the trends. I haven't specifically identified trends because I'm not sure which ones you're already aware of yet…but there's a lot of "b = a" at t+2, etc.:

https://anonfile.com/Xcpeg4f5b8/outputRSA0_20480x10240_i_64_xMin_0_xMax_128_yMin_0_yMax_64_t_0_scalar160xWithTextTrendOverlay.png

You can also overlay pics related over the grid to see what I'm talking about.

I go into things in greater detail in this post:

>>6636

Again, sorry if old.

>>6707

Hi VA! I think you got snared into the wrong thread, but it's nice to hear from you :)

>>6871

thanks for the tip–I'll have to look. There can't be many of those? If a[t] = bn, if 'a' is always less than or equal to 'b', but both 'a' and 'b' are positive in a regular grid, then that would mean that it would have to be perfect squares (a = b) in the 'n = 1' row…but all of the (a = b) squares are in (0,0). Do you mean any bn? Because that would be amazing ;)

I'll look into it further…

I'll check it out–