User blog:Nayuta Ito/Calculating values of SIR function, by inventor himself.

SIR(1,n,d)
SIR(1,n,d) is exactly d and you cannot go higher. Here's the proof:

There is a theorem that if a polynomial has only integer coefficients, all the coefficients of the factors of it are integers.

If an equation whose root goes higher than d (let it e), the absolute value of the of the constant term of the original equation must be e or greater.

The exception to the statement above is when one of the roots is zero but in that case, you can divide the whole equation by x and start it over.

Anyway, the absolute value of the constant term (or the lowest non-zero term) must be e or greater, which means it is bigger than d.

It's a contradiction because we can only use the integer between -d and d for the original coefficients.

Therefore, the absolute of the root cannot go higher than d, which means SIR(1,n,d) is d.

SIR(2,n,d)
This is getting supercalifragilisticexpialidociously difficult. Actually, finding one computable function that grows higher than SIR(2,n,d), or proving SIR(2,n,d) is as strong as Σ(n) will be one of the most problems in algebra (not just in Googology!).

SIR(2,1,d)
This is still comprehensible. All the equations are the form of ax+by+c=0. Since the coefficients cannot go higher than d, there must be a solution in 0<=x<d and 0<=y<d. So, SIR(2,1,d) cannot grow faster than 2d.