Distribution of population size $n$ given binomial sampled quantity $k$ and selection probability $\pi$ Given a drawn (without replacement) sample size $k$ from a binomial distribution with known probability parameter $\pi$, is there a function which gives distribution of likely population size $n$ from which these $k$ were sampled?  For instance, let's say we have $k=315$ items randomly selected with known probability $\pi=0.34$ from a population of $n$ items.  Here most likely value is $\hat{n}=926$ but what is probability distribution for $n$. Is there a distribution which gives $p(n)$?
I know that $p(\pi | k,n)$ is given by the beta distribution and that $p(k |\pi, n)$ is the binomial distribution.  I'm looking for that third creature, $p(n |\pi, k)$, properly normalized of course such that $\sum_{n=k}^{\infty} p(n)=1$
first "attempt" at this, given the normal approximation to binomial distribution is $p(k|\pi, n)=\mathcal{N}(k/\pi,k\pi(1-\pi))$, is that $p(n|\pi,k)\approx\mathcal{N}(k/\pi,k\pi(1-\pi))$?
 A: Let's start with this equation.
$\sum_{n=k}^{\infty}{{n \choose k}\pi^k(1-\pi)^{n-k}} $
Hopefully this is self-explanatory, but as just an intuition you can see this as a brute-force way of calculating your distribution by finding the probability that n pulls came from each possible binomial function. You would then need to divide by some constant $C$ such that: $\sum_{n=k}^{\infty}{\frac{{n \choose k}\pi^k(1-\pi)^{n-k}}{C}} = 1 $ to get a PMF. So if we can figure out what C is, we've got our distribution (even if we don't know the (EV or Varience).
The internet (wolframalpha) seems to be suggesting to me that C is equal to $1/\pi$. Feel free to confirm, but I'm sure that's correct. With that constant, the model simplifies to:
${n \choose k}\pi^{k+1}(1-\pi)^{n-k} $
I ran a simulation with this equation where we have observed n=1 with a probability of .5.
> pmf<-function(n,k,pi){
+   choose(n,k)*pi^(k+1)*(1-pi)^(n-k)
+ }
> 
> 
> graph<-1:100
> for(i in 1:100){
+ graph[i]<-pmf(i,1,.5)
+ }  
> 
> plot(graph)
> sum(graph)
[1] 1


Hopefully this distribution "rings true." 25% of the time we observe 1 success with 50% chance of success, it was after n=1, 25% it was n=2, and then it trails of exponentially from then. The expected value of our distribution is given by:
$\sum_{n=k}^{\infty}{n{n \choose k}\pi^k(1-\pi)^{n-k}} $
and variance:
$ E((X-E(n)^{2}) $
Unfortunately I don't currently have the time to solve those, but I challenge someone here to do so.
Edit: others have suggested a Bayesian solution for this problem. My bone to pick with those is that they assume that n HAS a distribution. Your question seems to assume that N is only distributed in as far as it is dependent on the binomial distribution(s) of $\pi$.
A: Bayes' theorem tells us:
$$
p(n \mid k, \pi) = \frac{p(k \mid n, \pi) p(n \mid \pi)}{\sum_{m=k}^\infty p(k \mid m, \pi) p(m \mid \pi)}
$$
We know $p(k \mid n, \pi)$. That's the Binomial distribution. However, we don't know the form of the prior, $p(n \mid \pi).$ But any possible prior for $n$ can give you an answer to the question. Therefore this question is underspecified. You would get a different answer if $n$ was Poisson distributed (with different parameters), or Negative Binomial distributed, or even distributed from a different Binomial distribution. But regardless, if you knew the distribution you would simply calculate the above expression.
As to your comment that we have $p(\pi \mid k,n)$ in the form of the Beta distribution, that has a similar issue to $n$. That is, we require a prior over $\pi$ to determine a posterior distribution for $\pi.$ For example, a uniform distribution prior gives a posterior of $\text{Beta}(k+1, n-k+1)$ with an expected value $(k+1)/(n+2)$. But a frequentist would estimate $\pi$ as having an EV of $k/n.$ Neither of them are "wrong", it just depends on your assumptions.
$k$ is different from $n$ or $\pi$ in that knowing the other two quantities gives you the information you need to know the distribution over $k$ exactly. The same is not true for either $n$ or $\pi.$

If you want to try an example for an assumed prior of $n$, let's suppose $n$ is Poisson distributed with mean $\lambda.$ Just keep in mind that this is an example, but the above answer still holds; that there's no definite answer to your question, it depends on the prior distribution of $n$.
If $n$ is distributed as $\text{Poisson}(\lambda)$ then the evidence function (the denominator) is,
$$
\begin{split}
p(k) &= \sum_{m=k}^\infty {m \choose k} \pi^k (1-\pi)^{m-k} \frac{\lambda^m e^{-\lambda}}{m!} & \\
&= \frac{ \left( \pi \lambda \right)^k } {k!} e^{- \lambda} \sum_{m=k}^\infty \frac{ \left[ \lambda (1-\pi) \right]^{m-k} } {(m-k)!} \\
&= \frac{ \left( \pi \lambda \right)^k } {k!} e^{-\lambda} e^{\lambda (1-\pi)}\\
&= \frac{ \left( \pi \lambda \right)^k } {k!} e^{-\lambda \pi}.
\end{split}
$$
It's interesting to note that the evidence function, i.e. $p(k)$ is the Poisson distribution with mean $\lambda \pi.$ This seems intuitively obvious when you think about it.
The numerator would simply be the summand of the first line of the above expression, with $n$ replacing $m$. Taking that and dividing the evidence, we get,
$$
p(n \mid k, \pi) = \frac{ \left[ \lambda (1-\pi) \right]^{n-k} } {(n-k)!} e^{-\lambda (1-\pi)}.
$$
Thus $n$ would be distributed such that $n-k$ is Poisson distributed with mean $\lambda (1 - \pi).$
