Sure, here you go:
3, 11, 17, 19, 41, 43, 59, 67, 73, 83, 89, 97, 107, 113, 131, 137, 139, 163, 179, 193, 211, 227, 233, 241, 251, 257, 281, 283, 307, 313, 331, 337, 347, 353, 379, 401, 409, 419, 433, 443, 449, 457, 467, 491, 499, 521, 523, 547, 563, 569, 571, 577, 587, 593, 601, 617, 619, 641, 643, 659, 673, 683, 691, 739, 761, 769,
47 doesn't exist in (1, 1) or (2, 1) because it belongs in the "other" category ref: >>8788
So either you're talking about treating a[t] in (0, 1) / (1, 1) as d, or using it as d.
Take 259 as an example, (1, 1), a[2] = 5. Then we either have 259 - 5 = 254 or 259 - 25 = 234, then we divide by 2 giving us 254/2, 234/2 = 127, 117
We add back to 1 giving us 128, 118. In (118, 1) we have 259 (I checked, it occurs at a[11] = 259), but not in 128. Meaning, we treat a[t] in (0, 1) / (1, 1) as d-values (giving us c - a[t]^2). This also fits with the previous discussed method for generating e's where c occurs at t=1, t=2, t=3 .. etc.
I never spent much time in (-1, 1), but I see now that I should have.
For a number k which consists of n prime factors. Find the first 2^n - 1 occurrences in (-1, 1) where a[t] % k == 0. Then compute the gcd(t, k).
Example for 259 (2 prime factors =2^2 - 1 = 3):
t's = [112, 148, 259]
gcd(112, 259) = 7
gcd(148, 259) = 37
gcd(259, 259) = 259
Two prime factors are a bit bland, so let's do the same thing, but for k = 7x37x61x101 = 1595699, 2^4 - 1 =15.
t's = [58682, 156954, 215636, 526918, 683872, 696193, 742553, 853147, 899507, 911828, 1068782, 1380064, 1438746, 1537018, 1595699] (length = 15)
gcds = [2257, 26159, 37, 427, 7, 6161, 15799, 101, 259, 227957, 3737, 43127, 61, 707, 1595699]
Note, after the last t, the pattern (periode?) starts to repeat it self.
>Where a prime factor appears in a[t] we can immediately determine where it's second appearance is, since the first two appearances of primes in a cell at [e,1] have the property where their values of their position t, summed is the value of the prime plus one. Five will appear twice within the first six elements, seven twice within eight, etc.
Ah yes. Prime numbers OCCURS twice. For composite numbers it depends on the number of factors.
Tthe number I'm using is 5 x 7 x 13 x 37 = 16835 giving us multiple combinations: { 5, 7, 13, 37, 5 x 7, 5 x 13, 5 x 37, 5 x 7 x 13, 5 x 7 x 37, 5 x 13 x 37, 7 x 13, 7 x 37, 7 x 13 x 37, 13 x 37, 5 x 7 x 13 x 37 }. All of these will exist in (e, 1) at different points of t as records (or simply factors of a). This holds true for ALL numbers (including primes, I'll show an example of those). Note the images are all capped at width = 1000 (ie 1000 generated records from (e, 1, 1) to (e, 1, 1001)). Color legend: black = 1, white = c the rest of the colors are combinations of either primes in c or combinations of them.
Imagine our (e, 1) as an infinite list of cells, spanning in a horizontal pattern. For simplicity the first pixel represents (e, 1, 1), but this ALSO works for negative x-values. In fact the a[t] = p =a[p + 1 - t] pattern is related to this.
If we draw a long line where each pixel (x-axis) represents the a-value in (e, 1, t) and colorize it based on the gcd-value of gcd(a, c) we will have what is seen in the first image. For example gcd(16835, 105) =(194, 1, 2) = 35 (7 x 5). We see how there are multiple different combinations within (e, 1) as each colored pixel represents a combination or prime.
Let's extend this to include -f as well. This is the second image, same generation and gcd computation is done, but now we have an image with e and -f as two rows. We can see what VQC has talked about, how there is a difference of 1 between several pixels, but some of these match at x - 1, others at x + 1. Ie. they diverge and converge (related to n and shadow n).
We define f = 2d + 1 - e, this gives us another perspective, we essentially INCREASE our d-value, pretending our greatest square is ONE unit bigger than it appears. We can also do this the other way around, pretending d is one unit LESS than what it appears. That is, moving in the opposite direction. In this case we will have g = 2d + e - 1 (d will shrink by one instead of growing by one).
We can think of this as …, -f, e, g, …
This is image 3, it has 3 rows with the first row representing the (g, 1), the second row (center) representing (e, 1) and the third representing (-f, 1).
We can keep growing on either side as much as we want. If we think of f_0 = 2d + 1 - e, then f_1 = 2d' + 1 - f_0 (Note f_0 is negative giving us two negatives and resulting in 2d' + 1 + f_0.).
For g we do the same, g_0 = 2d + e - 1, g_1 = 2d' + g_0 - 1.
Let's keep (e, 1) (origin e) in the center of the image and expand by 5 in both directions. This is the fourth image. To summarize:
Image 4 has the following rows:
g_4
g_3
g_2
g_1
g_0
e
f_0
f_1
f_2
f_3
f_4
The fifth image is the similar to the one above, but extended for 50 rows (g_0 .. g_49, f_0 .. f_49) with e in the center.
What we see is multiple square patterns, each factor and combination of factors are contained in squares across the perspectives of d. This is also true for prime numbers, I'll post an image of a prime number to show you. Again, this also works with negative values. I'll try and generate an example where (e, 1, 1) is at the center of the image with negative t expanding to the left of the center and positive t expands to the right.