TITLE: Is every positive integer the permanent of some 0-1 matrix?
QUESTION [24 upvotes]: In the course of discussing another MO question we realized that we did not know the answer to a more basic question, namely:

Is it true that for every positive integer $k$ there exists a balanced bipartite graph with exactly $k$ perfect matchings?

Equivalently, as stated in the title, is every positive integer the permanent of some 0-1 matrix?
The answer is surely yes, but it is not clear to me how to prove it.  Entry A089479 of the OEIS reports the number $T(n,k)$ of times the permanent of a real $n\times n$ zero-one matrix takes the value $k$ but does not address the question of whether, for every $k$, there exists $n$ such that $T(n,k)\ne 0$.  Assuming the answer is yes, the followup question is, what else can we say about the values of $n$ for which $T(n,k)\ne 0$ (e.g., upper and lower bounds)?

REPLY [2 votes]: If an order n (0,1) matrix has r rows with all ones, its permanent is a multiple of r! by an easy argument.  It follows that the largest odd number which is a permanent of an order n matrix is q = der(n) + der(n-1), which is a sum of numbers of derangements, and is a little larger than $n!/e$.  I suspect that the smallest positive number not expressible as a permanent of this size matrix is an odd integer which is not much smaller, perhaps about half, of q.  If so, then we can shave a little bit off of n=log(k) in the construction in Timothy Chow's post.  For some motivation for this problem, look at Will Orrick's talk mentioned (for determinants)  at https://mathoverflow.net/a/271273 .
Gerhard "Also A Potential Polymath Project" Paseman, 2017.08.29.