<s>[INST] Find sum $ \sum\limits_{k=2}^{2^{2^n}} \frac{1}{2^{\lfloor \log_2k \rfloor} \cdot 4^{\lfloor \log_2(\log_2k )\rfloor}} $ Calculate sum
$$ \sum_{k=2}^{2^{2^n}} \frac{1}{2^{\lfloor \log_2k \rfloor} \cdot 4^{\lfloor \log_2(\log_2k )\rfloor}}  $$
I hope to solve this in use of Iverson notation:
my try
$$ \sum_{k=2}^{2^{2^n}} \frac{1}{2^{\lfloor \log_2k \rfloor} \cdot 4^{\lfloor \log_2(\log_2k )\rfloor}} = \sum_{k,l,m}2^{-l}4^{-m} [2^l \le k < 2^{l+1}][2^{2^m} \le k < 2^{2^m+1}]  $$
and now:
$$ [2^l \le k < 2^{l+1}][2^{2^m} \le k < 2^{2^m+1}] \neq 0 $$ if and only if $$2^l \le k < 2^{l+1} \wedge 2^{2^m} \le k < 2^{2^m+1} $$
I can assume that $l$ is const (we know value of $l$) and treat $m$ as variable depence from $l$. Ok so: 
$$2^l \le 2^{2^m} \wedge 2^{2^m+1} \le 2^{l+1} $$
but it gives me that $l=2^m$
I think that it is not true (but also I don't see mistake). Even if it is true, how can be it finished? 
 [/INST] Let's write
$$\sum_{k=2}^{2^{2^n}} \frac{1}{2^{\lfloor \log_2(k) \rfloor}4^{\lfloor \log_2(\log_2(k))\rfloor}} = \sum_{i=0}^{n-1} \sum_{k=2^{2^i}}^{2^{2^{i+1}}-1} \frac{1}{2^{\lfloor \log_2(k) \rfloor}4^{\lfloor \log_2(\log_2(k))\rfloor}} + \frac{1}{2^{2^n}4^{n}}$$ 
$$= \sum_{i=0}^{n-1} \sum_{k=2^{2^i}}^{2^{2^{i+1}}-1} \frac{1}{2^{\lfloor \log_2(k) \rfloor}4^{i}} + \frac{1}{2^{2^n}4^{n}}$$
Moreover for all $i=0, ..., n-1$, 
$$\sum_{k=2^{2^i}}^{2^{2^{i+1}}-1} \frac{1}{2^{\lfloor \log_2(k) \rfloor}} = \sum_{j=2^i}^{2^{i+1}-1} \sum_{k=2^j}^{2^{j+1}-1} \frac{1}{2^{\lfloor \log_2(k) \rfloor}} = \sum_{j=2^i}^{2^{i+1}-1} \sum_{k=2^j}^{2^{j+1}-1} \frac{1}{2^j}  =  \sum_{j=2^i}^{2^{i+1}-1}  \frac{2^j}{2^j} = 2^i $$
You deduce that 
$$S = \sum_{i=0}^{n-1} \frac{1}{2^i} + \frac{1}{2^{2^n}4^{n}} = 2 - \frac{1}{2^{n-1}} + \frac{1}{2^{2^n}4^{n}}$$
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