Trying to solve the recurrence, but no clue how to deal with the (loglogn)^2 part

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    $\begingroup$ Substitute $n = 2^{2^k}$ in the entire expression. Can you solve the recurrence in terms of $k$? $\endgroup$
    – plshelp
    Sep 15, 2022 at 21:50

1 Answer 1



Assuming base $2$ logarithms, we can write


Hence, generalizing the pattern, you see the sum

$$\sum_{k=1}^{\lg(\lg(n))}{k^22^{\lg(\lg(n))-k}}=\lg(n)\sum_{k=1}^{\lg(\lg(n))}{k^22^{-k}}$$ appear, as well as a term $$\lg(n)T(2).$$


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