Let f(1) = a
Given:
f(n) = f(n/2) if n is even
f(n) = f(n + 5) if n is odd
Now, f (2) = f (2/2) = f(1) = a [Because 2 is even]
For n = 3; 3 is odd
f(3) = f (3 + 5) = f (8) = f(8/2) = f(4) = f(4/2) = f (2) = f(2/2) = f(1) = a
For n = 4; 4 is even
f(4) = f (4/2) = f(2) = f(2/2) = f(1) = a;
For n = 5, 5 is odd
f(5) = f(5 + 5) = f(10) = f(10/2) =f(5) = f(5 + 5) = f(10) = f(10/2) = f(5) = b [As f(5) is repeating ,let f(5) = b]
for n= 6
f(6) = f(6/2) = f(3) = x
So, there are two values only a and b. All multiples of 5 are b and others have value a.
So, the maximum possible size of R is 2.