Sign of Ramanujan $\tau$ function

Question

The Ramanujan $\tau(n)$ seemed to have random positive/negative signs:

    n            1     2     3     4     5      6       7        8      ...
    tau(n)       1    -24    252   -1472  4830  −6048   −16744   84480  ...

Is there a closed-form formula for the sign of $\tau(n)$?

By closed-form formula for $sign(\tau(n))$, I meant something like (1) if $n=2 mod(263)$, then $sign(\tau(n))>0$; (2) if $n=3 mod(p)$ and $p$ is any prime, then $sign(\tau(n))<0$; If $n=12398888$, an arbitrary number, then $sign(\tau(n))>0$ etc.

Or simply put: For a given number $n$, what is the sign of $\tau(n)$, 1,-1, or 0?

Thanks- mike

EDIT: $\tau(mn)=\tau(m)\tau(n)$ whenever $(m,n)=1$ (Thanks to Peter Humphries!), If $n=p_1^{r_1}\cdots p_m^{r_m}$ where $p_1,...,p_m$ are prime factors, then $sign(\tau(n))=(sign(\tau(p_1^{r_1}))\cdots (sign(\tau(p_m^{r_m}))$.

So the problem is reduced to find the formula for $sign(\tau(p^r)$ where $p$ is any prime number and $r$ is any positive number.

Answer 1

I'm quite certain there's no closed form for the sign of $\tau(n)$, but the behaviour of the sign of $\tau(n)$ is covered in some detail in this paper of Barry Mazur, and explained in some more detail by Peter Sarnak.

EDIT: Your statement about the multiplicativity of $\tau(n)$ is incorrect. The function $\tau(n)$ is multiplicative but not completely multiplicative, which is to say that $\tau(mn) = \tau(m) \tau(n)$ holds whenever $(m,n) = 1$ (that is, whenever $m$ and $n$ are coprime), but not in general.