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i knwo that the sum

$$ \sum_{n=0}^{\infty} |\mu (n) | f(n) $$

with $ \mu(n) $ is the mobius function

will be a sum over all integer except the squares

my question is if the sum

$$ \sum_{k=1}^{\infty}\sum_{n=0}^{\infty} |\mu (n) | f(n^{2k}) $$

is the sum over ALL the squares i mean $ f(1)+f(4)+f(9)+F(16)+f(25)+f(36) +..$

Jose Garcia
  • 8,506

1 Answers1

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Consider $144=12^2=2^4\cdot 3^2$. In order for $\mu(n)$ to be non-zero with those prime factors, $n$ would need to be $6$ , but $144$ isn't a power of $6$ so can't appear in $f(n^{2k})$, i.e. $144$ can't be written as an even power of a square-free number.

sharding4
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