Okay, let's do some simplification:

- assume a = 2*n + 1, since a is a prime number and all primes except for 2 are odd numbers
- then floor(a/2) = n
- assume b >= a, therefore b >= 2*n + 1

Substitute values accordingly:

r = (1 - b - 2*b*n + abs(3 - 2*9^n + 2*b + 4*b*n))%(2*n+1)

Now, there are two options for abs(3 - 2*9^n + 2*b + 4*b*n)

abs <= 0: r = (-2 + 2*9^n - 3*b - 6*b*n) % (2*n + 1)
abs > 0: r = (4 - 2*9^n + b + 2*b*n) % (2*n + 1)

Rearrange that, and you get

abs <= 0: r = (2/3*3^(2*n + 1) - (2*n + 1)*b - 2) % (2*n + 1)
abs > 0: r = (4 - 2/3*3^(2*n+1) + b*(2*n+1)) % (2*n + 1)

Remember that a = 2*n + 1:

abs <= 0: r = (2/3*3^a - a*b - 2) % a
abs > 0: r = (4 - 2/3*3^a + a*b) % a

(a*b)%a is always 0, so

abs <= 0: r = ((2*3^a)/3 - 2) % a
abs > 0: r = (4 - (2*3^a)/3) % a

Let's also insert initial condition:

when b > (2*3^a - 9)/(6*a): r = (4 - (2*3^a)/3) % a
otherwise: r = ((2*3^a)/3 - 2) % a

Here's a modified paste: https://pastebin.com/42uW6kEZ

Now, if you look closely, (4 - (2*3^a)/3) % a == 0 only holds when a = 1 or a = 2, so you'll always get an error if b > (2*3^a - 9)/(6*a).

You'll also get an error whenever a = 3, because ((2*3^3)/3 - 2) % 3 == 1.

For all other cases you have ((2*3^a)/3 - 2) % a == 0, which I believe could be proven in a similar way as the RSA encryption algorithm.