How can I solve the following:
$$ T(n) = \sum_{i = 1}^{d(n) - 2}T(v_i) + \sum_{i = d(n) - 1}^{n - 1}c + c' $$
Where $d(n)$ is the Tau function, and v is the set of values dividing n.
e.g. $d(18) = 6$, and $v_0 = 1, v_1 = 2, v_2 = 3, v_3 = 6, v_4 = 9, v_5 = 18$.
Edit: From the discussion in the comments, it looks like the expression sought is: $$ T(n) = \sum_{\substack{d \mid n \\ d < n}} T(d) + c (n - \tau(n)) + c' $$