I understand the last post you had. It's the same as PMA's calculation for the next record but worded a bit differently. Also if you go into negative x then back into positive you can get the record in between c and c'. Could be irrelevant though
So I've been looking into the frequency of repeating A[t] values and B[t] values in (e,1), (-f,1) and other cells with same c but different d. Look at the following records images. Pay attention to the block on the right. The center dot is (e,1,t) for whatever record we have at the moment with the same x or t. Column to the left would be -f, then the next would be for d=d+2, etc. Then if you go to the right each column is d=d-1. Going down you have higher t values, and going vertical up you have lower t values. If there is a red dot that means it shares a as factor of A[t], green dot means it shares b as factor of A[t] and cyan is c as factor. Notice that for any entry in with the t=t+n shift, the location of that c row (which is origianlly the cyan one going down to the right) turns into a B line for the next record at t+n, then it is always a B line for the rest of those records. This is pretty neato I think.
Guys I looked at the frequencies with which a, b, c and their squares appear as factors in A[t] in the (e,1) cell. The gist of the pictures is each column represents a column (e,1). The column to the left is NOT (e-1,1), but it is (-f,1). The next to the left of that is the cell with the same a,b but different d, so each column represents a different e for a d value for the same c value. The row/column that are white are the row that has the same x value as the goal cell. For (1,5,12,7,5,29) this would be (1,1) with x=7 (or t=4). In the top half, red means that A[t] is divisible by a, green means A[t] is divisible by b, and cyan means its divisible by c. Then in the bottom half, this is the same but the divisibility of each square. There are very cool patterns in this square part.
Looking at this, we can see that the factors of A and B are fairly easy to predict, but the factors of AA, BB, and D*D are these weird patterns. That repeat.
Also for each shift in a column the e shifts by 2d+1 changes d by one, so this is not a linear relation but the pattern looks linear. Maybe its a derivative or something?
So I'm looking at the A[t] divisibilities for (e,1) and related e columns (namely the e's for various d's). So here is a gif with the divisibilities for (1,c) for c=1,3,5,…,149
Its symettrical. Also you'll notice those little 3x2 boxes. They switch axes. It starts they switch after 2 numbers, then they switch after 4 numbers twice, then they switch after 6 numbers twice, then after 8 numbers twice etc. Basically each row is ABD(1,c,d) for various d's. (the original D is in the center of the pic). Also this just looks dank.
So the colors are yellow if its divisible by c and black if its divisible by a (a=1 in all these cases).
black = divisible by a
red = divisible by b
yello = divisible by c
white = divisible by nothing
Here are the maps for
5*7
5*9
5*11
respectively
Here is 5*29.
Is that a sine curve I see??
This graph is centered horizontally at d=0 and vertically at t=0. Down on the graph is a higher t. To the right on the graph is a higher d. So these correspond to (e,1) values for e's that are changing depending on the d values.
This is basically the post because each d value is paired with a certain e, and they are both unique to c values. Center dot is (c,1,0) for (e,n,t), to the right is (c-1,1,0), down from that would be (c-1,1,2) etc..
The first picture is (1,c) so black means that the A[t] value is divisible by a for that (e,1) cell [note in this first one its black because a=1]. Then yellow means it is divisible by c. Then red means it is divisible by b and white means its not divisible by anything. This first picture is just for(1,c) values, so this is kind of what we're working with for the moment. You'll notice they're in little pods of 6 and then there are lines from the center of the pods to the next one diagonally. These are ways to visualze the factors of any value in (e,1) rows for any d. How it starts is with c=3. If the top left corner of one of the little 6 packs is at (d,t) then there is the top left corner at (d+2ci, t+cj) for any integers i,j. Also if I'm at (d,t) and im' on the top left corner of a little pod, I can always do (d-2i, t-i) for any i and the same goes for the other corners. You're in like a little rectangle web. These create the diamonds and the diamonds are always centered at d=0,t=0. Moreover, these little 6 packs are only on top and bottom of the center cell, or the left and right. If we have c=5 or 7, they start horizontally, on the left and right. Then as the web expands out to c=9, 11, 13, 15, the little 6 packs are above and below the origin. Then for c=17, 19, 21, 23 they are horizontal again. Then for c=25,27,29,31,33,35 they are vertical again, then for c=37,39,41,43,45,47 they are horizontal again. I think it would continue on this pattern out switching above and below. Also for these there are random little dots that I can't figure out the pattern on (I'll look into it soon enough).
Now since these are all factors of odd numbers, in the second pic we can see how they fit together, becaues we have the product of 2 odd numbers. So to visualize these we just overlay two other graphs like this. These are multiples of 5 notice how the grid mostly stays the same and the other grid shifts out. The black is the 5 grid and the red is other odd values. yellow is still c. So we can see that the c values are just the juxtaposition of these two regular graphs that we can solve for for any c (mostly).
The third pic is the factorization of all odd c up to like 150 I think. If I could factor it I did and the pictures are very cool for them. These patterns are just a superposition of the two regular graphs and the c values come from where they intersect. This one is pretty cool
Last pic is squares and seeing that these juxtapositions of these graphs makes cool almost sinusoidal movement.
It looks like if e is odd, then bn and b(n-1) have the same t value. If that is the case then t for an is one less than the t for a(n-1).
If e is even, then an and a(n-1) have the same t value and bn has t one greater than b(n-1).
Remember 2na = xx + e. If n=1 then xx + e = 2a and so a = (xx+e)/2. Since we know that we can reverse engineer this to put our c value in the a position, so we can find where c is in A[t] in n=1.
c = (xx + e)/2 - 2c = xx + e
So we do the same calculation the square root then the remainder with the first thing. So for 145 we would get 2c = 290 = xx + e = 17*17 + 1.
Also 145 = 1212 + 1. Also we know that (of course) 2c - c = c. This gives us the equation 1717 + 1 - (1212 + 1) = 145, which simplifies to 1717 - 12*12 = 145, the correct answer to our problem, because d+n = 17, x+n = 12, d-x = 5 (I think this is why VQC liked this number). This got me to look into more equations and they aren't always that simple. Here is another:
17*41 = 697
697 = 26 * 26 + 21
697*2 = 37 * 37 + 25
3737 + 25 - 2626 - 21 = 697
3737 - 2626 + 4 = 697
So if anything this may just give us a close approximation of a difference of two squares. But we may be able to notice that when you shift d up one by either of these e gets shifted by 2d+1 or 2d-1 depending on the direction. Maybe we could use that and algebra to match up the columns to whatever makes e values equal then when e is equal we have the solution.
Also there are always 2 columns that match up with the values
Okay so this is a snippet of the code and it isn't necessarily the beginning, the opening bracket indicates that it is the start of a function or a loop or an if statement or something, it seems like there have been a few things defined already, namely e and minus_f and they have been defined before the '{'. Again this could be after an 'if' in a loop or something where e and minus_f change in the loop. Then you take the sqrt of 2d which could mean to recursively do the function on 2d [145 = 1212 + 1, 24 = 44 + 8, 8 = 22 + 4, 4 = 22, 4 = 2*2, etc]. under some circumstance. I'm trying to read between the lines here
But also he creates the return value so its probably the start of the function