\[
  \begin{cases}
    b_0 = 1 \\
    b_1 = 2 \\
    b_{n+1} = b_n + 2 \cdot b_{n-1}
  \end{cases}
\]


\begin{align*}
  b_2 &= b_1 + 2 \cdot b_{0} = 2 + 2 \cdot 1 = 4 \\
  b_3 &= b_2 + 2 \cdot b_{1} = 4 + 2 \cdot 2 = 8 \\
  b_4 &= b_3 + 2 \cdot b_{2} = 8 + 2 \cdot 4 = 16 \\
  b_5 &= b_4 + 2 \cdot b_{3} = 16 + 2 \cdot 8 = 32 \\
  b_6 &= b_5 + 2 \cdot b_{4} = 32 + 2 \cdot 16 = 64 \\
  b_7 &= b_6 + 2 \cdot b_{5} = 64 + 2 \cdot 32 = 128
\end{align*}

Jeg gjetter at $f(n) = 2^n$. 

\begin{align*}
  b_n + 2 \cdot b_{n-1} &= 2^n + 2 \cdot 2^{n-1} \\
  &= 2^n + 2^n \\
  &= 2 \cdot 2^{n} \\
  &= 2^{n+1} \\
  &= b_{n+1}
\end{align*}