2 added 75 characters in body
source | link

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
source | link

As there is a loop over $F(n-8)$, so we will have 256 times $F(n-8)$. Hence, time complexity would be: $$T(n) = 256\times T(n-8) = 256\times(256\times T(n-16)) = \Theta(256^\frac{n}{8})$$.