@leafiness0: I have another level concept ready. It's rather environmentally friendly (in terms of object conservation) and we only have to import a single enemy from the wild.

Well, you divide by 2 when a=1 (denominator's /2a for quad formula), so this gives you b/2, and a divide by 2 = a divide by 4 inside a sqrt sign, hence (b/2)^2-c. The i occurs because you extracted out one factor of sqrt(-1).

leafiness0 came up with a good idea for one button on two doors (not sure if it's compatible with your code though): stack the two buttons together.

So,
(17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1)
Start with the number n=2. Take the one leftmost that, when multiplied to n, gives an integer, and multiply that number to n and replace. Repeat process indefinitely.

2*(15/2) is the first fraction that gives me an integer when multiplied, Store 15 to n.
15*(55/1) = 825
825*(29/33) = 725
725*(77/29)=1925
1925*(13/11)=2275
2275*(17/91)=425
425*(78/85)=390
390*(11/13)=330
330*(29/33)=290
290*(77/29)=770
Repeating this process gives you
910, 170, 156, 132, 116, 308, 364, 68, 4. 4=2^2. Prime #1 found!

Formula for finding prime numbers.

(k + 2)(1 −
[wz + h + j − q]2 −
[(gk + 2g + k + 1)(h + j) + h − z]2 −
[16(k + 1)3(k + 2)(n + 1)2 + 1 − f2]2 −
[2n + p + q + z − e]2 −
[e3(e + 2)(a + 1)2 + 1 − o2]2 −
[(a2 − 1)y2 + 1 − x2]2 −
[16r2y4(a2 − 1) + 1 − u2]2 −
[n + l + v − y]2 −
[(a2 − 1)l2 + 1 − m2]2 −
[ai + k + 1 − l − i]2 −
[((a + u2(u2 − a))2 − 1)(n + 4dy)2 + 1 − (x + cu)2]2 −
[p + l(a − n − 1) + b(2an + 2a − n2 − 2n − 2) − m]2 −
[q + y(a − p − 1) + s(2ap + 2a − p2 − 2p − 2) − x]2 −
[z + pl(a − p) + t(2ap − p2 − 1) − pm]2)
> 0

Strangely enough I made a program on calc and computer that use this formula but I never got it to work.

If I remember correctly, k is prime iff there exists a Diophantine solution on the other variables. (aka: practically, this formula is "computationally useless" ) You'd need to solve for the other 25 or so integer variables such that the inequality holds; not an easy task

I think you use Ramanujan sums? Yeah, these funky summation systems (Abel, Cesaro, Euler, etc) give you results like 1+1+1+1+... = -1/2, 1-1+1-1+1-... = 1/2, and 1-2+3-4+5-6+... = 1/4.

Let's add a geometry theorem: Given a hexagon that has an incircle, its opposite diagonals meet at a single point. (Brianchon)

What algorithm are you using? ECM has breezed through a 64 bit range in less than 2 hours.