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The Ruler function has the following property:

$$\forall n \in \mathbb{N}: f(2n) = f(n)+1,\, f(2n+1) = 1.$$

Is there any other function with this property?

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    Why do you not consider the ruler function as a closed form?2011-09-01
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    ...anyway, did you have a look at the Formula section of the [OEIS entry](http://oeis.org/A001511)?2011-09-01
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    $f(n)=k$ where k is such that $n=2^k q$ with $q$ odd2011-09-01
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    @J.M. Most of those formulas involving recursive equation and binary digits of n. What I wanted is a equation not involving those, using only +-*/^, elementary functions, and 'well known' functions like zeta function. ( I don't know what `Dirichlet g.f.: zeta(s)*2^s/(2^s-1).` means... :( )2011-09-01
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    @Heike: $f(n)=\underline{k+1}$ where $k$ is such that $n=2^kq$ with $q$ odd.2011-09-01
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    @robjohn: you're right, I miscounted.2011-09-01
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    One man's "well-known function" is another man's "what the heck is that?" The function $\nu_2(n)$ is well-known, in some circles, as the 2-adic order of $n$, that is, it's the greatest $k$ such that $n$ is a multiple of $2^k$. So in terms of that function, it's $1+\nu_2(n)$.2011-09-01
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    You could write it as $\sum _{m=0}^{n } \sum _{k=0}^{{2}^{m}-1}\cos \left( {2^{1-m} \pi \,nk} \right) {2}^{-m} $2011-09-01
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    I somehow missed your question to me; anyway, what that line in the OEIS listing means is that, letting $f(n)$ be the ruler function, $$\frac{\zeta(s)}{1-2^{-s}}=\sum_{k=1}^\infty \frac{f(k)}{k^s}$$. That is, $\frac{\zeta(s)}{1-2^{-s}}$ is the *Dirichlet generating function* for $f(k)$.2011-09-02

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