In (f,1) where a=BigN-1, x is equal to our starting d value.
t is the order of members in a cell
In the first row where n=1, t gives a values of x that alternates between odd and even.
The value of x in e and -f will never both be even of odd since the gap between the column is always odd or 2d+1
a[t] is half the square of x plus e for the column
When e is larger than x by more than 2x+1, the value at a[t] will be found in a previous column in the first row.
For odd e, x=2t-1; even e, x=2(t-1)
a[t]=(((2t-1)(2t-1)+e)/2, (((2(t-1)2(t-1))+e)/2 for odd, even e
d[t]=a[t]+x[t]
In the rows below the first, where n>1, have you noticed that there is sometimes more than one set of products in the cell?
If there are two, both are growing by 2n added to x in that cell. The first set you can spot the same starting value of x from the value of x in the first cell at n=1. The second begins at -x until adding 2n makes it positive.
Since we're using floors of square roots, 2d doesn't equal d of 4c in every case. Here's an example:
c=145, d=12, e=1, 2d=24
4c=580, d=24, e=4, 2d=48
4c=2320, d=48, e=16, 2d=96
4c=9280, d=96, e=64, 2d=192
4c=37120, d=192, e=256, 2d=384
This is where it stops following the pattern.
4c=148480, d=385, e=255, 2d=770
4c=593920, d=770, e=1020, 2d=1540
It then also resets here too.
4c=2375680, d=1541, e=999, 2d=3082
4c=9502720, d=3082, e=3996, 2d=6164
And so on, every time e becomes greater than d.
Forgot to mention, the take-home from this is that when e d, the difference is 1.
By multiplying c with 4 we're creating a new record and distributing the 4 (2x2) to both a and b, giving us 2a, 2b. This also gives us an even square to work with 2(x+n)^2.
Odd (x+n)^2 = 8(Tu) + 1
Even (x+n)^2 = 4(Tu + T(u-1))
We now have two triangle problems to solve. For an even square, half of the triangles is one unit longer than the other half.
Now for some insight into our equations (This is just for insight, not sure if it can be used for solving):
8Tu = 8(u*(u+1)/2) =4u(u+1) => 2(2u*(u+1)).
2u(u+1) should be something we all recognize as it's the same method we calculate a's or d's depending on parity of e (Missing + (e + 1)/2 or e/2). This should mean (I might be wrong) that we are using values from (0, 1) as the 4T(u), as in 4T(u) should appear in (0, 1) as d. Just to re-iterate T(u) is the triangle with a base of u, has the equation u(u+1)/2. But by multiplying 4 we get 4(u(u+1)/2) =2u(u+1). This should mean that our odd (x+n)^2 - 1 = 2(d[u + 1]) from (0,1).
For even squares we have the equation:
4(Tu + T(u-1)) which is also equal to 4Tu + 4T(u-1) =2u(u+1)) + 2(u)(u-1). Here we have two d's from (0, 1) with a difference of t by 1. Essentially d[u+1] + d[u].
I think this might be why VQC has said that column (0, 1) is so special. Our x+n squares, regardless of parity, can be expressed using d's from (0, 1).
(Note from AA - this isn't necessarily what VQC meant but it could well be)
When you decrease e by 1 but increase n by 1, the increase in f is equal to 2d-1.
When you increase e and n by 1, at the same t, the increase in f is equal to 2*(the increase in d)+1.
In every single case with a semiprime, (BigN-n)/(b-1) = (a-1)/2 (where a and b are the factors we're trying to find, so not BigN's a and b).
We can directly calculate the cell in (e,1) in which cbigN first appears as a[t]. In this cell, x=c-d (the c and d we already know). Another cell in (e,1) where a[t]%c==0 is where x=c+d+1. a[t]=c(c-BigN+1) in this cell. These cells are 2d+1 apart (just like the (e,n) and (f,n) cells). There are also equivalents in (f,1). In the equivalent first relevant cell in (-f,1) where x=c-d, a[t]=c(BigN-1). In the equivalent second relevant cell in (-f,1) where x=c+d+1, a[t]=c(c-BigN+2).
The difference between the x value of the cells in (0,n) where c=1c^2 and where c=a^2b^2 is always divisible by 12.
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The u of the cell in (0,n) where a=b and b=a^2b is always divisible by the b we're looking for.
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Take a[t] from (e, 1) and subtract from a in (e, 1). You'll get smooth numbers. Take d[t] from d in (f, 1) and you'll get smooth numbers. Smooth numbers are made from triangles and squares.
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You can represent a given c with its e and f values, right? Well, this unique pair of e and f values in row 1 for the an and a(n-1) cells also represent valid cs themselves. These new c values are actually divisible by a or b (depending on which the e-f/f-e gap is divisible by) and they're divisible by BigN-n.
When a number is the sum of two squares and it's odd, it will only be divisible by other numbers that are also the sum of two squares and odd. So if you have a c value and it's in an e column where e is a square (because this means it's the sum of two squares, based on another crumb), its factors will be in columns where e is a square too.
Every value of a in (1,1) is the sum of two squares and is odd. VQC's crumb said that these values "contain ALL factors for odd numbers that are the sum of two squares", but they are also themselves odd numbers that are the sum of two squares.
While these a values are each odd sums of two squares, some odd sums of two squares are missing. Here's the first ten odd sums of squares: 1, 5, 13, 17, 25, 29, 37, 41, 45, 53. Here are the first ten a values in (1,1): 1, 5, 13, 25, 41, 61, 85, 113, 145, 181. So, as you can see, while a in (1,1) are all odd square sums, some of the odd square sums are missing, such as 17, 29, 37, 45 and 53.
The sequence of a values starts at 1 and increases by 4(x-1) where x is the place in the sequence. In other words, 1, +4=5, +8=13, +12=25, +16=41, etc. This follows the pattern of sums of squares that are odd in that they are all equal to a multiple of 4 plus 1.
These odd numbers that are the sum of two squares appear as factors of others in the series in a pattern. Where a number appears as a factor, it will appear as a factor again the number of numbers away that is equal to itself. If that was worded confusingly, take this example: where 5 appears as a factor, it will appear again 5 numbers in the sequence away, then 5 away again, and so on. It will also always appear once between these times. This seems to hold for all of them. For 5, it appears two away, then three away (this second one being 5 away from the origin), and this pattern repeats. When 13 appears as a factor, it appears 8 away, and then appears again 5 away, and so on. Where 17 appears as a factor, it appears 4 away, and then appears again 13 away, and so on. For 25, it's 18 and 7. For 29, it's 12 and 17. This seems to hold for all numbers that are odd and the sum of two squares.
In summary, the crumb "the values of a in the first cell where e=1 contains ALL factors for odd numbers that are the sum of two squares" is true because all of these a values are themselves odd numbers that are the sum of two squares, and because every number that is odd and the sum of two squares can only be divided by other numbers that are odd and the sum of two squares. I'm not sure if every c value in the columns where e is a square are the sum of two squares, but every c value that is the sum of two squares does appear in these columns. If you have a c value that is the sum of two squares, it will appear as an a value in (1,1) and its factors will appear in (1,1).
Aside from a 16 page word document I need to go through and turn into digestible stuff, I've gone through every post in every thread. I've accepted the fact that none of you are going to help me with this at this point, so once I've gone through that word document (most of it is triangles I'm pretty sure), I'll sort/enumerate all of these into groups based on the things VQC was talking about (all cells in a row, all cells in a column, etc).