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I(2^k) &\le 2I(2^{k-1}) + 6M(2^{k-1})+ 4 A(2^{k-1})\\
&\le 2^2I(2^{k-2}) + 6(M(2^{k-1})+2M(2^{k-2})) + 4(A(2^{k-1}) + 2A(2^{k-2}))\\
&\
\end{align}</math>
If <math>M(n)\le cn^\omega,</math> and <math>\alpha=2^\omega\ge 4,</math> one gets eventually
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