There exists a cell in (e,1) where a[t]=na (the n we want * the a we want). There also exists a cell in (-f,1) where a[t]=a(n-1), and this f is 2d+1 less than e. This is also true for b (so a[t]=nb in (e,1) and a[t]=b(n-1) in (-f,1)), and each of these values also exists in b[t] since a[t]=b[t-1] in the n=1 row. The an and bn values (and a(n-1) and b(n-1)) are n elements apart (the n we're looking for).

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All whole numbers (integers) c, that give a remainder of 0,1,3 when divided by 4, are the difference of two squares.

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Any number that leaves 2 when divided by four can either be divided OR multiplied by 2 to make it the difference of two squares.

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The values of a in the first cell where e=1 contains ALL factors for odd numbers that are the sum of two squares.

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All products (integers) c that are the sum of two squares appear (only) in columns where e=0,1,4,9,16,25…

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If a number at position t has a factor s, then s is a factor at (t+s), (t+2s) and so on for a at (e,1).

Also, if a number at position t has a factor s at (e+1), then s is a factor at (s+1-t), (2s+1-t), etc for a at (e,1).

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For (e,n): B(e,2,t) = A(e,2,t+e)

For (e,n) with a = 1, we have:

B(e,n) = B(e+2,n+1) - 2

A(e,n) = A(e+2,n+1) = 1

X(e,n) = X(e+2,n+1)

D(e,n) = D(e+2,n+1)

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For all numbers in (e,1) we have:

A(e,1,t) = B(e,1,t-1)

A(e,1,t) + 1 = A(e+2,1,t)

B(e,1,t+1) = 2 * B(e,1,t) - A(e,1,t) + 4

A(e,1,t) = D(e,1,t+1)

A(e,1,t+1) = D(e,1,t) + X(e,1,t+1) = D(e,1,t) + X(e,1,t) + 2

D(e,1,t) = A(e,1,t+1) - 1

D(e,1,t) = D(e-2,1,t) + 1

In addition, for a=1, we have:

For (e,n,d,x,1,b), (e+2, n+1, d, x, 1, b-2)

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Choose any cell (e,n).

Take a look at the corresponding cell (e, a).

The a value of the 2nd cell = the n value of the 1st cell.

The 2nd cell is the "parent" of the tree.

Choose any cell p0 = (e, n, 1).

For that cell, there are 2 corresponding cells p1 = (e, a, 1) and p2 = (e, b, 1).

For p1 at some t1, you'll find a = n of p0. (I believe you'll also find b=n in that list too).

For p2 at some t2, you'll find a = n of p0. (I believe you'll also find b=n in that list too).

Example 1:

p0 = (6,11,1) ={6:11:5:4:1:31}

p1 = (6,1,3) ={6:1:15:4:11:21}

p2 = (6,31,1) ={6:31:37:26:11:125}

This pattern exists everywhere I believe for t0 = 1.

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a[t+1]=a[t]+2x[t]+2n

x[t+1]=x[t]+2n

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Generate the (e,1) for any c.

x = e % 2

a = (e/2)+x

n = 1

b = a + 2x + 2n

c = a*b

d = floor_sqrt(c);

t = 1