Originally Posted by

**emakarov** I think 14, 16 and 22(a) are correct.

For 22(b) I suggest the substitution $\displaystyle m = \log\log n$ instead of $\displaystyle m = \log n$. I.e., $\displaystyle n=2^{2^m}$ and $\displaystyle \log n=2^m$.

Then $\displaystyle f(n)=f(2^{2^m})=2f(2^{2^{m-1}})+2^m=2(2f(2^{2^{m-2}})+2^{m-1})+2^m$. Let's pause here: this is $\displaystyle 2^2f(2^{2^{m-2}})+2\cdot2^m$. This easily generalizes from two to $\displaystyle k$ iterations. Then find the limit case, when $\displaystyle f(2)=1$ has to be used; you'll get an expression for $\displaystyle f(n)$ in terms of $\displaystyle 2^m$ and $\displaystyle m$, i.e., $\displaystyle \log n$ and $\displaystyle \log\log n$.