I solved this one kind of by accident. Its a short program (most things in Haskell are) and I coded it in just a few minutes. Then I plugged in a starting series of 10^6 numbers. Thought that would be a good start. Got the result in a few seconds, and figured that I would need to do the full 10^9 digits. Left it to work over the weekend. Of course, it never finished.
After analysing the function a little more, I discovered that its an oscillating convergent function. Interestingly, after only a few thousand numbers, the varience is smaller than 10^-9. So by using 10^6, it was already overkill.
f :: Double -> Double f x = let a = 30.403243784-(x**2) in fromIntegral (floor (2**a)) * 10**(-9) u :: Double -> Double u 0 = -1 u n = f ( u (n-1)) main = let n = 10^6 in print $ (u n) + u(n+1)