$T(n) = T(\sqrt n) + 5$

I know that each $k$-th level is equal to $n^{1/2^k}$, and intuitively, $T(n) = \theta(\log_2(\log_2 n))$. I am just not certain where the constant 5 goes in the substitution method, and how I can generate tight lower and upper bounds for this recurrence relation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.