2 added 75 characters in body edited Feb 5 '18 at 8:24 OmG 1,67055 silver badges1515 bronze badges As there is a loop over $$F(n-8)$$, so we will have 256 times $$F(n-8)$$. Hence, time complexity would be: $$T(n) = 256\times T(n-8) = 256\times(256\times T(n-16)) = \Theta(256^\frac{n}{8})$$. As mentioned in comment $$256 = 2^8$$, we will have $$T(n) = \Theta(2^n)$$. As there is a loop over $$F(n-8)$$, so we will have 256 times $$F(n-8)$$. Hence, time complexity would be: $$T(n) = 256\times T(n-8) = 256\times(256\times T(n-16)) = \Theta(256^\frac{n}{8})$$. As there is a loop over $$F(n-8)$$, so we will have 256 times $$F(n-8)$$. Hence, time complexity would be: $$T(n) = 256\times T(n-8) = 256\times(256\times T(n-16)) = \Theta(256^\frac{n}{8})$$. As mentioned in comment $$256 = 2^8$$, we will have $$T(n) = \Theta(2^n)$$. 1 answered Feb 4 '18 at 21:47 OmG 1,67055 silver badges1515 bronze badges As there is a loop over $$F(n-8)$$, so we will have 256 times $$F(n-8)$$. Hence, time complexity would be: $$T(n) = 256\times T(n-8) = 256\times(256\times T(n-16)) = \Theta(256^\frac{n}{8})$$.