a possible period of 124 for the sign of Ramanujan $\tau(3^n)$ function 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  ...

$\tau(mn)=\tau(m)\tau(n)$ whenever $(m,n)=1$ (Thanks to Peter Humphries!), If $n=p_1^{k_1}\cdots p_m^{k_m}$ where $p_1,...,p_m$ are prime factors, then $\mathrm{sign}(\tau(n))=(\mathrm{sign}(\tau(p_1^{k_1}))\cdots (\mathrm{sign}(\tau(p_m^{k_m}))$.
So the problem is reduced to find the formula for $\mathrm{sign}\left(\tau(p^k)\right)$ where $p$ is any prime number and $k$ is any positive number.
According to answer to this related question, there is no closed-form expression for $\mathrm{sign}\left(\tau(p^k)\right)$.
I did some numerical calculation using Mathematica 9.0 and found out that $\mathrm{sign}\left(\tau(3^k)\right)$ seemed to have a period of $124=2^2 31$:
$$\mathrm{sign}\left(\tau(3^{k+1})\right)=\mathrm{sign}\left(\tau(3^{k+1+124})\right)\tag{1}$$
I verified (1) for $k\in[1,10000]$.
I am looking for reference papers on similar results.
Thanks-
mike
EDIT:
Just found out exceptions (counter examples) at $k=13023,13085$ for (1) for $k\in[1,20000]$.
So the claim does not hold anymore.
 A: According to Wikipedia's page on the tau function, the following identity has been proven:
$$\tau(p^{r+1})=\tau(p)\tau(p^{r})-p^{11}\tau(p^{r-1})$$
the relevance of this is pretty clear, since, computing the constants there and letting $a_r=\tau(3^r)$, we end up with:
$$a_{r+1}=252a_r - 177147 a_{r-1}.$$
and the initial conditions
$$a_0=1$$
$$a_1=252$$
which using the standard techniques for manipulating linear homogenous recurrence relations (i.e. Mathematica), we end up with
$$a_n=\left(\frac{1}2+\frac{7i}{\sqrt{1991}}\right)(126-9i\sqrt{1991})^n+\left(\frac{1}2-\frac{7i}{\sqrt{1991}}\right)(126+9i\sqrt{1991})^n$$
and since we see a conjugate pair here, we end up with
$$a_n=\Re\left(2\left(\frac{1}2+\frac{7i}{\sqrt{1991}}\right)(126-9i\sqrt{1991})^n\right)$$
and, if all we want to know is whether $a$ is positive or not, it suffices to compute the argument of the inner value, which is
$$z_n=\arg\left(\frac{1}2+\frac{7i}{\sqrt{1991}}\right)+n\arg(126-9i\sqrt{1991})$$
which is a nice linear equation - and $a$ is positive exactly when the argument is (up to a multiple of $2\pi$) in the interval $(-\pi/2,\pi/2)$. Thus, we expect it to be roughly periodic, since, for any $k$ there is a $c$ so that $z_{n+k}=z_n + c$ and if $c$ is close to $2\pi$, then $\text{sgn}(\Re(z_{n+k}))$ will usually equal $\text{sgn}(\Re(z_n))$. However, it is only truly periodic if $\arg(126-9i\sqrt{1991})$ is a rational multiple of $\pi$.
