I've found myself puzzled by the following simple discrete distribution: open-ended Bernoulli distribution, which I will now define.
The distribution has 2 parameters: $p$, the success probability, and $q$, the repeat probability. I will define it recursively as follows: \begin{equation} X = \begin{cases} 0 & (1-p)(1-q)\\ 1 & p(1-q)\\ X' & (1-p)q\\ 1+X' & pq \end{cases} \end{equation} where the right column of the "cases" environment represents probabilities (it sums to 1) and the left column represents the "result" of $X$, where $X'$ is an independent random variable sharing the same distribution as $X$ (with the same parameters).
Of course, $X$ can be infinite, but this happens with probability 0 (at least when $q<1$); also, it seems that the mean will always be finite (again, when $q<1$). However, I am not only interested in the mean, but rather in the full pmf; I am interested in its shape, in the moments, etc.
Some effort
Here's some effort I've made: I tried to understand some edge particular cases. Here it is:
The case $p=0$
This is easy, as in this case $X\equiv 0$.
The case $p=1$
In this case this is a geometric distribution (the version whose support does not include 0).
The case $q=0$
In this case $X$ is a Bernoulli trial.
The case $q=1$
In this case, unless $p=0$, $X$ is always $\infty$.
More generality
In fact, in a more general and more interesting case, the parameters $a,b,a^+,b^+$ replace $p,q$ as follows: \begin{equation} X = \begin{cases} 0 & a\\ 1 & b\\ X' & a^+\\ 1+X' & b^+ \end{cases} \end{equation}
Can you find the pmf?
Instead of $a^+$ and $b^+$ I use $c$ and $d$ respectively. Note that $a+b+c+d=1$.
$P\left\{ X=0\right\} =a+cP\left\{ X=0\right\} $ so: $$P\left\{ X=0\right\} =\frac{a}{1-c}$$
$P\left\{ X=1\right\} =b+cP\left\{ X=1\right\} +dP\left\{ X=0\right\} $ so: $$P\left\{ X=1\right\} =\frac{b}{1-c}+\frac{ad}{\left(1-c\right)^{2}}$$
If $n>0$ then $P\left\{ X=n+1\right\} =cP\left\{ X=n+1\right\} +dP\left\{ X=n\right\} $ so: $$P\left\{ X=n+1\right\} =\frac{d}{1-c}P\left\{ X=n\right\} $$
This makes it possible to prove by induction that:
$$P\left\{ X=n\right\} =\frac{bd^{n-1}}{\left(1-c\right)^{n}}+\frac{ad^{n}}{\left(1-c\right)^{n+1}}$$ for $n\geq1$ and under condition $a+b>0$.