Is there a way to analytically compute the recurrence time of a finite Markov process? Let $X_t$ be an ergodic (time-homogeneous) Markov process (in discrete or continuous time) on a finite state space $\{1,\dots,n\}$. Let $T(X_0)$ be the stopping time given by the infimum of times such that $X$ has covered the space (i.e., for all $j$ with $1 \le j \le n$ there exists some $t_j \le T(X_0)$ s.t. $X_{t_j} = j$) and $X_{T(X_0)} = X_0$. Clearly $T(X_0)$ dominates the cover time of $X$. I would expect it to be dominated in turn by the sum of the cover time and the expected hitting time of $X_0$ starting from a state chosen w/r/t the invariant distribution $p$.
Define the recurrence time as $\sum_{X_0} p(X_0) \cdot \mathbb{E}_{X_0} T(X_0)$, where again the first term is the invariant distribution of $X$. 
Now it has been quite a while (early 2000s) since I looked at cover and hitting times, but I recall that while the fundamental matrix (in discrete time) or the "deviation matrix" (in continuous time) give lots of information about hitting times, computing the cover time is hard. I am aware of the Matthews bound, but I do not know of a simpler way to compute the cover time than by simulating the chain. In particular, I don't know of an analytical approach to this quantity. 
I am in the same situation w/r/t the recurrence time, and it is this quantity that interests me much more than the cover time per se. But both are of some interest/utility to me.
So my questions are: 

  
*
  
*Has the recurrence time (or a similar quantity besides the cover time or first return time) been treated anywhere?
  
*Are there known analytical results on computing or at least (besides the Matthews
  bound) bounding cover times or
  recurrence times?
  

 A: This is a response to a comment.
The coupon collector's problem is elementary. I don't have a particular scholarly reference in mind, but rather the technique of the proofs. There are a few proofs of the $n H_n$ expected time to collect all coupons. One possibility is that you can compute the expected time to collect the $k$th new coupon, 
$n/(n-k+1)$. That uses a lot of symmetry you don't have for a general Markov process. Here, you have transition probabilities and times on (current location, subset visited so far). 
Analogous to what I did here, you can use inclusion-exclusion. The expected time to cover everything (with discrete time) is the sum of the probability that you haven't covered everything by time $t-1$, which you can express as 
$$\sum_t \sum_{S\subset V} -1^{|S|+1}Prob(\{X_i\}_{i\lt t}\cap S = \emptyset) $$
where $V$ = $\{1,...,n\}$. You can switch the order of summation to get about $2^n$ analytically solvable problems about avoiding particular subsets.
$$\sum_{S\subset V} -1^{|S|+1} A(S)$$
where $A(S)$ = expected time before you first enter $S$.
The same holds for continuous time. 
A: http://arxiv.org/abs/1004.4371 might be useful.
A note regarding your remark:
"Clearly T(X0) dominates the cover time of X. I would expect it to be dominated in turn by the sum of the cover time and the expected hitting time of X0 starting from a state chosen w/r/t the invariant distribution p."
This is false (unless I'm misreading it). If you start from 0 on the discrete interval {0,1,..,n}, then after covering you have to return to 0 from n, which takes longer than returning to 0 from a uniformly random point.
