I am interesting in the asymptotic rate of growth of the following recursion:

$$ T(n) = 2T(n/2) + \frac{T(n − 1)}{\log n}, $$ with base case $T(1) = 1$.

I'm having trouble of solving this recurrence problem, it seems much more tricky than I initially thought, and I'm totally stuck after doing some unrolling. What is a good way to approach this question?

  • $\begingroup$ Do you have a guess on the solution of the recurrence? $\endgroup$
    – Steven
    Nov 4, 2019 at 19:28
  • $\begingroup$ I'm guessing nlogn but not sure how to verify it. $\endgroup$
    – ran wty
    Nov 4, 2019 at 19:39

1 Answer 1


By comparison to the recurrence $T(n) = 2T(n/2)$, we see that $T(n) = \Omega(n)$. By comparison to the recurrence $T(n) = 2T(n/2) + n$, we see that actually $T(n) = \Omega(n\log n)$.

Now let us prove by induction that the other direction holds as well. I will interpret $T(n/2)$ as $T(\lfloor n/2 \rfloor)$. Suppose that $T(m) \leq Cm\log m$ for all $m < n$. Then $$ T(n) \leq 2C\frac{n}{2}\log \frac{n}{2} + \frac{n\log n}{\log n} = Cn\log n + (1-C)n. $$ For $C \geq 1$, it follows that $T(n) \leq Cn\log n$.

It seems that we have proved that $T(n) = O(n\log n)$, but actually the base case of the induction doesn't hold. However, this shouldn't be too hard to fix (left to the reader).

  • $\begingroup$ Thanks, I figured it out. $\endgroup$
    – ran wty
    Nov 5, 2019 at 1:40

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