# SICP Exercise 1.20

For applicative order the answer \(k=4\) is easy to find. For normal order evaluation we have \(k'=m_{k-1}+\sum_{1}^k m_i\), where \(m_k=m_{k-1}+m_{k-2}+1\), \(m_0=m_{-1}=0\). We calculate \(m_1=1\), \(m_2=2\), \(m_3=4\), \(m_4=7\), and get the final result \(k'=18\).