Greatest power of two dividing an integer

See Sloane's entry for the sequence, which is A007814. which includes a recursive by-cases formula, and the generating function $A(x) = \sum_{k=1}^{\infty}(x^{2^k})/(1-x^{2^k})$,

I suspect this answer will not be found satisfactory, but here goes. Write $[x]$ for the integer part of $x$. Then $[n/2]-[(n-1)/2]$ is 1 if $n$ is a multiple of 2, 0 otherwise. $[n/4]-[(n-1)/4]$ is 1 if $n$ is a multiple of 4, 0 otherwise. Etc. So the function you want is $$[n/2]-[(n-1)/2]+[n/4]-[(n-1)/4]+[n/8]-[(n-1)/8]+\dots$$ where the sum isn't really infinite, it has $r$ terms, where $r$ is something like $\log_2n$.

Note that this $f$ is computable, and thus corresponds to a (finite size) lambda term. So, in that sense, it has a closed-form. Of course, if you use a different base language for what you mean by closed-form, this may disappear. On the other hand, by using Goedel numbering, we can encode this lambda term as an arithmetic function -- which will most surely be absolutely hideous.

Now, if the real question is, is there an easy way to compute this, the answer is yes, very easy: represent your $n$ in binary, and count the number of trailing 0s -- that number is the $f(n)$ you seek. This is a representation-dependent answer though. But that's because the representation-independent answers to this question will tend to be hopelessly inefficient.