I expected that the fractional part of f(n), n being an integer, would be distributed uniformly over [0,1] (for positive functions - otherwise take [-1,1]) for any run-of-the-mill function, except there is a good reason otherwise. I experimented a bit with MATHEMATICA and found to be dead wrong. Methodics: I computed f(n) for $0<n<=100000$ and the n-th central moments of that set. I multiplied them with $(2n+1)2^{2n}$ to normalize since a uniformly distributed U[0,1] after this operation has 1 for even and 0 for odd moments. Call f random if the moments of the set ${f(1),...,f(10⁶)}$ are about as close to 1 and 0 as those of the random number set. Some random :-) results:
$f(n)=n^r$ (trivial for r integer or r<0): Sqrt[n] is random (funnily, more random than a genuine random series :-) and I guess so is any f with non-integer r>0.
$f(n)=log(n)$: Very non-random, possibly because it's so flat.
$f(n)=n*log(n)$: Random.
$f(n)=sin(n)$: Non-random but the moments are $(2n+1)!/(n!)^2$. (This should be easy to prove since sin is periodic and you can approximate by the respective moment integrals.)
$f(n)=exp(n)$: Checked only to n<=100 for obvious reasons, slightly non-random.
Do you have a reference for me? At least the sqrt part was already analyzed 40 years ago. (And could someone verify my results, at least on sin(n)?)
