## A small aside concerning $\tau$ of prime powers

In case you've come here without having read the main page, here's a short summary of the definitions used: let $\sigma$ be the traditional sum of divisors function. Let $n$ be a positive integer, and factor it as $2^x \cdot m$, $m$ odd. Then define the function $f$ as $f(n) = 2^x$, and note that it is completely multiplicative ($f(xy) = f(x) f(y)$ $\forall x, y \in \mathbb{Z}$); define the function $\beta$ as $\beta = \log_2(f(n))$, and note that it is "logarithmically completely multiplicative" in the sense that $\beta(xy) = \beta (x) + \beta(y)$; and finally define the function $\tau$ as $\tau(n) = \beta(\sigma(n))$. Since $\sigma$ is multiplicative and $f$ completely so, we have that $\tau(mn) = \tau(m) + \tau(n)$ whenever $m$ and $n$ are relatively prime.

Next note that according to the definition of $\beta$, we have the criterion that $\beta(n) = x \iff n \equiv 2^x \pmod{2^{x+1}}$, and correspondingly if $p$ is a prime, then $\tau(p) = x \iff p \equiv 2^x - 1 \pmod{2^{x+1}}$. Next we consider powers of $p$: $\tau(p^a) = \beta(\sigma(p^a))$. If $a$ is even, then $\sigma(p^a)$ is odd, and so $\tau(p^a)$ is 0 (to the benefit of aliquot sequence workers trying to break the worst drivers).

Otherwise $a$ is even, and we then define $a=2k+1$ and write $\sigma(p^a) = \sum_{j=0}^a{p^j} = \sum_{i=0}^{k}({p^{2i}+p^{2i+1})}$. Consider the powers of $p \pmod {2^{x+1}}$. By the previous criterion, we have that $p \equiv 2^x - 1 \pmod {2^{x+1}}$. By simple computation, $p^2 \equiv 2^{2x} - 2^{x+1} + 1 \equiv 1 \pmod {2^{x+1}}$, and so $p^3 \equiv p$ and inductively we have $p^{2i} + p^{2i+1} \equiv 1 + p \equiv 2^x \pmod{2^{x+1}}$. Thus each individual term of this "reduced" sum has $\beta = x$, and so naively we might assume that 2 such terms divide $2^{x+1}$, three terms divide only $2^x$, four terms divide $2^{x+2}$ and so on, such that with $l=k+1$ terms in the sum, we would have as many extra powers of two as divide $l$, i.e. that $\tau(p^a) = \beta(\sum_{i=0}^{k}({p^{2i}+p^{2i+1})}) = \tau(p) + \beta(l)$. Of course this isn't a stringent argument, for in the two term case, we don't know that in fact the sum does not divide $2^{x+2}$, so that this argument may only conclude that $\tau(p^a) \geq \tau(p) + \beta(l)$. But in fact, in my numerical tests, I have been unable to find a counter example where $\tau(p^a) \gt \tau(p) + \beta(l)$, where the search extended to $p \lt 10^9$, and each prime was tested up to $a = 99$. I believe this should certainly cover any pathological case where the odd parts of the $p^{2i} + p^{2i+1}$ terms add up to more than the specified power of two (which is certainly easy to do with two arbitrary odd numbers); alas, I have been unable to prove the proposition. Defining $w_i$ by the equation $p^{2i} + p^{2i+1} = 2^x \cdot w_i$, it would be sufficient to show that among the $l$ different $w_i$, an even number of them are 1 mod 4 and thus with the other amount of them, also even in quantity, 3 mod 4 (if $l$ is odd then $\sum{w_i}$ is odd anyways) whence we would conclude that $\sum{w_i} \equiv 2 \pmod{4} \implies \beta(\sum{w_i})) = 1 \lt 2$, and thus no extra powers of two to make $tau(p^a) \gt x + \beta(l)$.

Like I said though, despite the numerical evidence to the contrary, I as yet have no such proof. Any thoughts on the matter would be appreciated.