I was working on my software so it could draw the exclusion lines for me... for a couple of reasons:
 So I could update this site with pixel perfect images of the exclusion lines (which still hasn't happened yet... I got a bit side tracked along the way)

To be able to analyse at the process of degredation of the exclusion lines as the iterations evolve... when layering the Ulam iterations the exlcusion lines appear strongly in the first iterations, then slowly degrade as the iterations continue, the degradation always moving out from the centre (further out they are always perfect... not a single prime lies on them). I suspected that this is because the first value of the quadratics may or may not be prime... the ones that are prime are destroying the perfection of the exclusion lines.
Because I don't at this point know how to locate the x  y coordinates of an integer in the Ulam spiral I decided the best way to create the lines was to know how far they were offset from the center of the spiral.... it's not hard to create a vertical line that is always, for example, 10 pixels to the left of the center of the Ulam spiral.
I started by looking at the vertical exclusion lines in the top left quadrant. Below is an image borrowed from another page in this site... the context is not quite right because this one is looking at the exclusion lines in an Ulam spiral centred at 1and moving the x and Y axis through the iterations... I will update this image later, for now I will use this one because it shows the spacing of the vertical exclusion lines (red arrows)... the offset from the center is also 1more than it is when starting the Ulam spiral at 0.... the analysis is based on the Ulam spiral starting at 0... but at least it shows the spacing between the exclusion lines as they are in the Ulam spiral when starting at 0 (by the time I've explained all this I could have updated the image... oh well).
If I know the offset of those vertical lines from the centre I can easily draw them with my software, it's not a simple matter of plotting every point that is, say, 10 pixels to the left of the center because the lines start at an offset in the vertical axis (y axis) as well... but that y axis offset is the same value as the x offset so it's also not that hard.
Anyway, going out from the center of the Ulam spiral (the right hand side of this picture... remember, this is a zoom into the top left quadrant) I find that the offsets are 1,7,10,22,27,45,52,76....
I noticed that there are actually two quadratic series of numbers in that sequence:
 1,10,27,52.... etc.
 7,22,45,76.... etc.
The quadratic equations that correspond to those series of numbers are:
 4x^{2}  3x + 0
 4x^{2} + 3x + 0
So here we have two quadratic equations that tell us the hortizontal (x axis) offsets of the vertical (y axis) exclusion lines in the top left quadrant.
I updated my software to draw these lines... and to catch the first three integers that were drawn on those points... this way I can determine the quadratic equations that define the vertical exclusion lines... using the good old method of common differences.
The method of common differences works by seperating a quadratic equation into three simultaneous equations. I won't go into the details right now (note to self, provide seperate page with details) but the end result is that it will give you a quadratic equation that starts at the first value you put into it. Let me explain.
When looking at the vertical exclusion line at offset 7 I pick up the integers 210,273,344,423,510,605.... etc.
If I plug those numbers into the method of common differences process I get the quadratic equation 4x ^{2} + 51x + 155.
When you put x = 1 into that equation you get 210.... when you put x = 2 into that equation you get 273... etc. great, it does the job... but, it doesn't exactly satisfy our initial premise that the 'c' part of the quadratic equation y(x) = a * x ^{2} + b * x + c must be zero for us the say that the equation will always produce composites (except possibly for the first non zero value remember).
There is, what I call, a base quadratic that generates the same series of numbers and when you put x = 0 into it you get zero out. This doesn't apply to all quadratics... only those that describe parabolas that cross the point (0,0).
If I put x = 0 into the equation 4x ^{2} + 51x + 155 I get 155. To find the base quadratic I run the value of x into the negative values until I get a zero value, I then take the first three non zero values and use those values in the method of common differences to find the 'base quadratic'. Following on with the example will help explain.
So we've got y(x) = 4x ^{2} + 51x + 155, if I use x = 0 I get 155, if I use x = 1 I get 108, if I keep going with that I eventually find that x = 5 will give 0.
Then we take the values x = 4, x = 3 and x = 2 to get the first three values that we need... 15,38 and 69
Putting those numbers back into the method of common differences we get the quadratic equation 4x ^{2} + 11x + 0
I call 4x ^{2} + 11x + 0 the base quadratic equation of 4x ^{2} + 51x + 155.... they both generate the same series of numbers, they simply differ in position that they start at for the value of x.
My software takes care of all of this... picking up the first integers that fall on the offset line and then extracting the base quadratic. Here are the results for the first few vertical exclusion lines in the top left quadrant:
Offset amount 
Base Quadratic equation 
1 
4x^{2}  5x + 0 
7 
4x^{2} +11x + 0 
10 
4x^{2}  13x + 0 
22 
4x^{2} + 19x + 0 
27 
4x^{2}  21x + 0 
45 
4x^{2} + 27x + 0 
52 
4x^{2}  29x + 0 
76 
4x^{2} + 35x + 0 
Remember that these offset lines are being generated by two series of quadratic equations... so we should seperate the above table into....
Offsets generated by 4x^{2}  3x + 0 
Base Quadratic equation 
1 
4x^{2}  5x + 0 
10 
4x^{2}  13x + 0 
27 
4x^{2}  21x + 0 
52 
4x^{2}  29x + 0 
Offsets generated by 4x^{2} + 3x + 0 
Base Quadratic equation 
7 
4x^{2} + 11x + 0 
22 
4x^{2} + 19x + 0 
45 
4x^{2} + 27x + 0 
76 
4x^{2} + 35x + 0 
Note that the 'b' value differs by 8 between each succesive exclusion line.
So far we are looking only at the top left quadrant and the first Ulam iteration.
Let's follow the evolution of the first offset line, 4x^{2}  5x + 0, as we go through the Ulam iterations...
Ulam Iteration 
Base Quadratic equation
for first offset exclusion line in top left quadrant 
1 
4x^{2}  5x + 0 
2 
8x^{2}  5x + 0 
3 
???

4 
16x^{2}  5x + 0 
5 
20x^{2}  5x + 0 
6 
???

7 
??? 
8 
32x^{2}  5x + 0 
9 
??? 
10 
40x^{2}  5x + 0 
11 
???

12

???

13

???

14

???

15

???

16

64x^{2}  5x + 0 
17 
???

18

???

19

???

20

80x^{2}  5x + 0 
21

???

22

???

23

???

24

???

25

4x^{2}  1x + 0 
26

???

27

???

28

???

29

???

30

???

31

???

32

128x^{2}  5x + 0 
33

???

I have put ??? where my software could not work out the quadratic equation.... it double checks the quadratic equation based on the series of numbers by using the 1st,2nd and 3rd integers that land on the line, and then the 2nd, 3rd and 4th... if they don't give the same result then it decides it was not a quadratic series of integers.
Also interesting that iteration 25 results in 4x^{2}  1x + 0 instead of the expected 100x^{2}  5x + 0.... since all others that gave a quadratic as a result are of the form (4 * Iteration number)x^{2}  5x + 0.... iteration 1 is (4 * 1)x^{2}  5x + 0, iteration 2 is (4 * 2)x^{2}  5x + 0, iteration 8 is (4 * 8)x^{2}  5x + 0
Also obious that iterations that are a power of two result in quadratics:
Ulam Iteration 
Base Quadratic equation
for first offset exclusion line in top left quadrant 
1 (2^{0}) 
4x^{2}  5x + 0 
2 (2^{1}) 
8x^{2}  5x + 0 
4 (2^{2}) 
16x^{2}  5x + 0 
8 (2^{3}) 
32x^{2}  5x + 0 
16 (2^{4})

64x^{2}  5x + 0 
32 (2^{5})

128x^{2}  5x + 0 
Lets look at the first dozen or so actual integer points landing on the exclusion line in the third iteration (the first iteration that I couldn't determine a quadratic for)
2,7,25,38,72,93,143,172,238,275,357,402,500,553,667,728,858,927,1073,1150,1312,1397,1575,1668,1862.....
The first two are prime... but this doesn't seem to be the cause of the reason we don't get a quadratic out of the series of numbers because that happens with the second iteration.... the number 3 lands on the exclusion line in the second iteration... but we still can extract the quadratic 8x^{2}  5x + 0 (remember... the first value of the base quadratic may or may not be prime... hence why the exclusion lines dissolve as we layer iterations on top on one another)
So, all the numbers bar the first two are composites... but they are not defined by a quadratic.
So we've looked a little bit at the vertical exclusion lines in the top left quadrant... what about the vertical exlcusion lines in the bottom right quadrant?
Well it's all much of the same really... the offsets from the center, this time heading right, are 0,3,5,14,18,33,39,60,68....
Again there are two quadratic series in there:

3,14,33,60... etc.

5,18,39,68... etc.
The quadratic equations that correspond to those series of numbers are:
 4x^{2}  1x + 0
 4x^{2} + 1x + 0
Offset amount 
Base Quadratic equation 
0 
4x^{2}  1x + 0 
3 
4x^{2} + 7x + 0 
5 
4x^{2}  9x + 0 
14 
4x^{2} + 15x + 0 
18 
4x^{2}  17x + 0 
33 
4x^{2} + 23x + 0 
39 
4x^{2}  25x + 0 
60 
4x^{2} + 31x + 0 
^{}
And then the horizontal lines... well they are the same quadratics as the vertical exclusion lines but with negated 'b' values... so instead of the quadratic 4x^{2}  5x + 0 which is the first vertical offset in the top left quadrant, we have 4x^{2} + 5x + 0 as the first horizontal quadratic in the top right quadrant.
