# Can the Ruler function be expressed in a closed-form expression?

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? –  Ｊ. Ｍ. Sep 1 '11 at 7:34
$f(n)=k$ where k is such that $n=2^k q$ with $q$ odd –  Heike Sep 1 '11 at 7:43
@Heike: $f(n)=\underline{k+1}$ where $k$ is such that $n=2^kq$ with $q$ odd. –  robjohn Sep 1 '11 at 9:19
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 $1+\nu_2(n)$. –  Gerry Myerson Sep 1 '11 at 12:47
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 $f(k)$. –  Ｊ. Ｍ. Sep 2 '11 at 9:02