Recursive formula for the number of connected labelled graphs with n vertices and k edges

Let $n,k$ be integers such that $n - 1 \le k \le {n \choose 2}$. Let $T(n, k)$ be the number of connected, simple graphs without self-loops on $n$ labelled vertices having exactly $k$ edges. Can we give an expression for $T(n,k)$ in terms of $T(m,h)$ for $m < n$ or $h < k$ (that is, a recursive formula)?

$T(n, k)$ is sequence A123527 on the OEIS: http://oeis.org/A123527.

Variations on this question have been asked before on this site (for instance: here), but I wasn't able to piece together a recursive formula from their answers. My motivation is to write a program that computes it for small $n$, for which a recursive formula can be used.

So far I've noticed that a few base cases are easy to compute. In fact, if $k = n - 1$ then $T(n, k)$ is counting the number of trees on $n$ vertices, which, by Cayley's formula, is $$T(n, n - 1) = n ^ {n - 2}\text{,}$$ while if $k \ge {n - 1 \choose 2} + 1$ then every graph is surely connected, therefore $$T(n, k) = {{n \choose 2} \choose k}\text{.}$$

I know I am late but I had the same question and came up with following recurrence. I hope this helps somebody who comes across this.

def numberOfedges(n,k):
    t = combination(n,2)
    ret = combination(t,k)
    if k < combination(n - 1,2) + 1:
        for i in range(1,n):
            ni = combination(n - 1, i - 1) 
            x = combination(i,2)
            for j in range(i - 1, min(x,k) + 1): 
                ret -= ni * combination(combination(n - i,2),k - j) * numberOfedges(i, j) 
    return ret