the recursive call is :
f(N)= A1 + M1*f(M2*N/D1 - S1) Op M3*f(M4*N/D2 - S2), if N > N1
with
s1 = s2 = 0
m2 = m4 = 1
d1 = d2 > 1
we have
f(N)= A1 + M1*f(N/D1) Op M3*f(N/D1), if N > N1
The recursive call is the key point to get the asymptotic complexity, the rest is "just" constants.
So the point is to find T such as :
T(n)=2*T(n/D)
Once you find T(n) you have the number of call of your Recursion_Plus, since we are talking about asymptotic complexity it is not relevant to bother about the last calls (i.e. n<N1
).
Now it's all about mathematics, i won't describe a formal solution here but with a little intuition you can get to the result.
Each call of T induce 2 calls of T but with a # divide by D, then 4 call with a # divide by D^2 ...
the complexity is 2^(logD(n))
(with logD(n)=ln(N)/ln(D) )
particular case : with D=2, the complexity is n