Good (asymptotic) upper bound for recurrence over divisors I am looking for a good (asymptotic) upper bound on the following recurrence relation ($T(0) = 1)$:
$$
T(n) = \left[ \sum_{1\leq d<n, d|n} T(d) \right] + 1
$$
Note that the recursion is only for divisors $d < n$ to ensure termination*. 
What I have so far:
An upper bound can be found by that fact that $\sigma_0(n) \leq \sqrt{n}$ and $d \leq n/2$:
$$
T(n) \leq \sqrt{n} T(n/2) + 1 \\
$$
It's easy to see that $T(n) = O(n^{\log_2(n)/2})$ (and this can be further improved by the fact that $\sigma_0(n) \leq n^{1.5379 \log2/\log\log n }$).
When flat out computing $T(n)$ for all $n$ up to $10^6$, I noticed that $T(n) \leq n^2 + 1$. 
However, I fail to see how this bound can be derived (or how one can prove that $T(n) = \omega(n^c)$ for some constant $c$). Obviously, one has to do better than estimating $T(d) \leq  T(n/2)$ for all divisors of $n$, since this is where the $\log n$ in the exponent comes from. 
*Bonus question: I am using Computer Science terminology here. What is correct (mathematical) way? Is the recursion above a finite one and would a recursion with $T(n)$ on the right hand side be an infinite one? 
 A: Here are some observations that might provide a basis for conjectures and further investigation by an expert, somewhat more than what would fit into a comment.
Suppose we have $T(1) = 1$ and
$$T(n) = 1 + \sum_{d|n,\; d<n} T(d).$$
Then we can make a statement about the average order of $T(n)$ using the Wiener-Ikehara Theorem.
To do this we need the closed form of
$$L(s) = \sum_{n\ge 1} \frac{T(n)}{n^s}.$$
The recurrence yields
$$2 T(n) = 1 + \sum_{d|n} T(d)$$
which implies
$$2 L(s) = \zeta(s) + L(s)\zeta(s)$$
or $$ L(s) = \frac{\zeta(s)}{2-\zeta(s)}.$$
Numeric experiments (which require confirmation / refutation) would seem to indicate that the dominant pole of $L(s)$ is at 
$$\rho = 1.72864723899818361813510301029769,$$
that it is simple and that the residue has the value
$$c = \mathrm{Res}\left(L(s); s=\rho\right) = 2\times 0.5500100054.$$
The Wiener-Ikehara theorem then says that
$$\sum_{n\le x} T(n) \sim c \frac{x^\rho}{\rho}
\quad\text{i.e. that the average order is}\quad
\frac{1}{x} \sum_{n\le x} T(n) \sim c \frac{x^{\rho-1}}{\rho}.$$
The precision from this formula is exceptional. For example, when $x=1000$ the exact value is $97.227$ while the formula gives $97.641,$ similarly for $x=6000$ the exact value is $359.9451$ while the formula gives $360.2746.$
BTW this sequence is OEIS A067824.
