# Patterns in the factorization

After some more thought I looked at how the two factors of the quadratic were changing for each consecutive exclusion line. The first line being (2x-1)(2x+3) the second line being (2x-3)(2x+5) it looked like it would be reasonable for the next line to be (2x-5)(2x+7). Some quick testing showed that it did work. Continuing the pattern we get (2x-7)(2x+9) and then (2x-9)(2x+11). I also saw that I could calculate the location along the (p=7 x=1 c=6) line based on the those factors. If we take 9*11 and add 1 we get 100. Divide that by 2 and we get 50 which is the position along the line of that exclusion line. So if we but 50 into (2x-9)(2x+11) we should get the starting number.

$(2*50-9)(2*50+11)$

$(91)(111)$

$10101$

That works and I compared it to my earlier result where I was brute force finding the line and it does match up. They also line up with the difference between the factors note earlier.

 (2x-1)(2x+3) 3–1 4 (2x-3)(2x+5) 5–3 8 (2x-5)(2x+7) 7–5 12

In looking at this I figured out the quadratic for the locations of the exclusion lines. The series 2,8,18 results in the formula

$2x^2$

That brought me back to the question of why I saw so many primes in one less than the location. So the formula for one less of the location is

$2x^2-1$

In searching I found only one site that relates primes to that equation. So I would say that equation is known in regards to primes but not that it is so deeply embedded in the Ulam Spiral.