A different kind of divisor sums Define, for a number $n$ the function $\mathcal{T}(n)$ which equals the number of different sums $d_1 + d_2,$ where $d_{1, 2}$ are divisors of $n.$ Now, one assumes that $\mathcal{T}(n)$ is approximately $\tau^2(n)/2$ for most $n$ but is there any analytic machinery to analyse the average/maximal behavior of something like this?
 A: The sequence $1, 3, 3, 6, 3, 9, 3, 10, 6, 10, 3, 16, 3, 10, \ldots$ does not seem to be in the OEIS yet.  That may be about to change...
Plotting $T(n)$ against $\tau(n)^2/2$ for $n \le 20000$, I get

which doesn't appear to support your conjecture.  
EDIT:
Some more data points: the medians of $T(n)/\tau(n)^2$ for 
$n$ in the intervals $[2^d, 2^{d+1})$ for $d = 1, \ldots, 19$ are
$$ \frac{3}{4},\; \frac{17}{24},\; \frac{5}{8},\; \frac{5}{8},\; \frac{5}{8},\; \frac{5}{8},\; \frac{7}{12},\; \frac{7}{12},\; \frac{7}{12},\; \frac{9}{16},\; \frac{9}{16},\; \frac{9}{16},\; \frac{9}{16},\; \frac{9}{16},\; \frac{9}{16},\; \frac{9}{16},\; \frac{9}{16},\; \frac{9}{16},\; \frac{9}{16}
$$
Well, maybe the limit is in fact $1/2$, it's hard to say.  But it looks like my graph above may be misleading, overemphasizing those $n$ that have unusually large numbers of divisors, and these may have unusually small $T(n)/\tau(n)^2$.
EDIT: The sequence is now in the OEIS as A260152.
A: One thing you can try to use to get estimates for $\mathcal T$ is bounds for the number of divisors $2^k\leq d\leq 2^{k+1}$. Since pairs of divisors from different dyadic ranges have distinct sums, you can get a rough idea for a lower bound of $\mathcal T$. 
The maximum $$\Delta(n)=\max_{k}|\{d|n:2^k\leq d\leq 2^{k+1}\}|$$ has been estimated by Hooley [On a new technique and its application to the theory of numbers, 1977] - something like $$\sum_{n\leq x}\Delta(n)\ll x(\log x)^{4/\pi-1}.$$ Thus, the typical number $n$ has $\Delta(n)=O((\log x)^{4/\pi-1})$ and so has about $(\log n)^{2/3}$ divisors with each from distinct dyadic ranges. Any pair of such divisors has a distinct sum so you get $\mathcal T(n)\gg (\log n)^{4/3}$. This is some way away from your conjecture however...
