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I have this double sum

$\displaystyle\sum_{k=1}^{N}\sum_{b=2}^{L} \varphi(b)(b^{2}2^{k})$

I wish to reduce this to a single-sum function (by simplifying/getting rid of the inner summation loop so I only loop from 1 to N). Can it be done?

$\varphi(b)$ is the totient function

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If I read the formula correctly, it is the outer sum that is easiest to get rid of. – André Nicolas Feb 23 '13 at 7:08
    
@AndréNicolas L = floor(N/k) if that matters – NullOverNull Feb 23 '13 at 7:14
    
From the notation you used, I thought $L$ was unrestricted, hence my comment. – André Nicolas Feb 23 '13 at 7:29
up vote 1 down vote accepted

$$\sum_{k=1}^{N}{\sum_{b=2}^{\left\lfloor\frac{N}{k}\right\rfloor}{\varphi(b)(b^{2}2^{k})}}=\sum_{b=2}^{N}{\sum_{k=1}^{\left\lfloor\frac{N}{b}\right\rfloor}{\varphi(b)(b^{2}2^{k})}}=\sum_{b=2}^{N}{\varphi(b)b^{2}\sum_{k=1}^{\left\lfloor\frac{N}{b}\right\rfloor}{2^{k}}}=\sum_{b=2}^{N}{\varphi(b)b^{2}(2^{\left\lfloor\frac{N}{b}\right\rfloor+1}-2)}$$

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