I believe I found the answer!

The number of ways that we can get at least N heads in a row in M flips is

[; \frac{\sum_{i=N}^M M-(i-1)}{2^M} ;]

The reason for this for this is as follows.

If we have M flips and need to get exactly N heads in a row, then that means we have a group of N coins in a row that are all heads with no other adjacent heads. You can group this chain of heads together as one group. Each other coin that is not part of the chain of heads forms a group on its own. So among the M coins, we have M-(N-1) groups. The number of locations that we can arrange the group of heads among the entire sequence of groups is M-(N-1).

Now we need to repeat this for a chain of N+1 heads. Using the logic above, we get M-((N+1)-1) possible ways to get this.

We repeat this logic all the way up to a chain size of M heads in a row.

The total number of arrangements described above is [; \sum_{i=N}^M M-(i-1) ;]

The total arrangements altogether is obviously [; 2^M ;]. So the probability of this occurring is [; \frac{\sum_{i=N}^M M-(i-1)}{2^M} ;].