Consider the preliminary question of getting a sequence of $$N$$ heads out of $$k$$ throws, with probability $$1-p(N,k)$$. This is given by the recurrence formula $$p(N,k) = \begin{cases} 1 &\text{if } k Indeed, my reasoning is that no consecutive $$N$$ heads out of $$k$$ draws can be decomposed according to the first occurrence of a tail out of the first $$N$$ throws. Conditioning on whether this first tail occurs at the first, second, ..., $$N$$th draw leads to this recurrence relation.
Next, the probability of getting the first consecutive N heads in $$m\ge N$$ throws is $$q(N,m) =\begin{cases} \dfrac{1}{2^N} &\text{if }m=N\  p(N,m-N-1) \dfrac{1}{2^{N+1}} &\text{if } N<m<2N+1 \end{cases} $$ The first case is self-explanatory. the second case corresponds to a tail occuring at the $m-N-1$th draw, followed by $N$ heads, and the last case prohibits $N$ consecutive heads prior to the $m-N-1$th draw. (The two last cases could be condensed into one, granted!)
Now, the probability to get $$M$$ heads first and the first consecutive $$N$$ heads in exactly $$m\ge N$$ throws (and no less) is $$r(M,N,m) = \begin{cases} 1/2^N &\text{if }m=N\  0 &\text{if } N<m\le N+M\\ \dfrac{1}{2^{M}}\sum_{r=M+1}^{N}\dfrac{1}{2^{r-M}}q(N,m-r)&\text{if } N+M<m  \end{cases} $$Hence the conditional probability of waiting m steps to get N consecutive heads given the first M consecutive heads is$$ s(M,N,m) = \begin{cases} 1/{2^{N-M}} &\text{if }m=N\ 0 &\text{if } N \sum_{r=M+1}^{N}\dfrac{q(N,m-r)}{2^{r-M}}&\text{if } N+M \end{cases} $$The expected number of draws can then be derived by$$ \mathfrak{E}(M,N)= \sum_{m=N}^\infty m\, s(M,N,m)$$ or $\mathfrak{E}(M,N)-M$ for the number of additional steps...