Hello,
What is the formula for the following recursive definition...
$\displaystyle f(0)=1, f(1)=0, f(2)=2, f(n)=2f(n-3), n>=3 $
The sequence is 2, 0, 4, 4, 0, 8, 8, 0 ...
Actually its a tricky way of combining three separate sequences.
Let $\displaystyle n = 3k$ and $\displaystyle f(3m) = a_m$ then:
$\displaystyle f(n)=2f(n-3), n \geq 3 \Rightarrow a_k = f(3k)=2f(3k-3) = 2f(3(k-1)) = 2a_{k-1}, k \geq 1$ with $\displaystyle a_0 = 1$
Now, for $\displaystyle n = 3k+1 $ and $\displaystyle f(3m + 1) = b_m$ then:
$\displaystyle b_k = f(3k + 1)=2f(3k+1 - 3) = 2f(3(k-1) + 1) = 2b_{k-1}, k \geq 1$ with $\displaystyle b_0 = 0$
Finally $\displaystyle n = 3k+2 $ and $\displaystyle f(3m + 2) = c_m$ then:
$\displaystyle c_k = f(3k + 2)=2f(3k+2 - 3) = 2f(3(k-1) + 2) = 2c_{k-1}, k \geq 1$with $\displaystyle c_0 = 2$
So $\displaystyle a_k = 2^k , b_k = 0, c_k = 2^{k+1}$