Posterior coin count after exactly one head ($P(N=n\mid H=1)=n(1-r{)}^2{r}^{n-1}$)

Let $N$ have the positive-integer geometric distribution with success parameter $p\in (0,1)$, and conditionally toss $N$ independent coins with head probability $\theta \in (0,1)$. Exactly one head has conditional probability $n\theta (1-\theta {)}^{n-1}$. Combining this with the prior by Bayes' theorem gives the displayed posterior with $r=(1-p)(1-\theta )$; its normalization uses ${\sum }_{n\ge 1}n{r}^{n-1}=(1-r{)}^{-2}$. For $p=\theta =1/2$, the posterior is $9n/4^{n+1}$. It is also a shifted negative binomial distribution, with $N-1$ equal in law to the number of failures before two successes of probability $1-r$.