References on binomial coefficients of the form $\binom{-1/m}{k} $

Solution 1:

Consider the formal power series $y(x)$ satisfying $y^m = \frac{1}{1 - mx}$, which gives

$$y(x) = (1 - mx)^{- \frac 1 m} = \sum_{n \ge 0} (-1)^n m^n {-\frac 1 m \choose n} x^n.$$

Taking logs and exponentiating again gives

$$y(x) = \exp \left( \frac{1}{m} \log \frac{1}{1 - mx} \right) = \exp \left( \sum_{k \ge 1} \frac{m^{n-1} x^n}{n} \right).$$

This gives us that $n! [x^n] y(x)$ counts permutations but where a permutation with cycles of length $\ell_1, \dots \ell_n$ is weighted by $m^{\sum (\ell_i - 1)}$. The quantity $\sum (\ell_i - 1)$ is sometimes called the length of a permutation; it gives the minimal $\ell$ such that a permutation can be expressed as a product of $\ell$ transpositions. It's equal to $n$ minus the number of cycles, which means we get the Stirling numbers of the first kind but in reverse order:

$$\boxed{ (-1)^n m^n {-\frac 1 m \choose n} = \frac{1}{n!} \sum_{i=0}^n \left[ {n \atop i} \right] m^{n-i} }.$$

Compare to the generating function for the Stirling numbers in the usual order, which is

$$(-1)^n {-m \choose n} = {n+m-1 \choose n} = \frac{1}{n!} \sum_{i=0}^n \left[ {n \atop i} \right] m^i$$

and which corresponds to looking at $(1 - x)^{-m}$.

What you might've been hoping for was an expression in terms of ${mn \choose n}$ generalizing the case $m = 2$, and it turns out that some nice things can be said about this sequence. Using Lagrange inversion or the more explicit combinatorial argument in this blog post we can show that the formal power series $z(x)$ satisfying $z = 1 + xz^m$ has coefficients

$$z(x) = \sum_{n \ge 0} \frac{1}{(m-1)n + 1} {mn \choose n} x^n$$

which generalizes the well-known generating function for the Catalan numbers, to which it reduces when $m = 2$ (and in fact these coefficients count $m$-ary trees in a way that generalizes the way the Catalan numbers count binary trees). This is Example 6.2.6 in Stanley's Enumerative Combinatorics, Vol. II. Example 6.2.7 uses this to show that the formal power series $w(x) = \sum_{n \ge 0} {mn \choose n} x^n$ satisfies

$$\frac{w - 1}{1 + (m-1)w} = x \left( \frac{mw}{1 + (m-1)w} \right)^m$$

so like $y$ and $z$ it is also algebraic, but its minimal polynomial appears to be more complicated. For example, setting $m = 3$ and clearing denominators gives that $w = \sum_{n \ge 0} {3n \choose n} x^n$ satisfies

$$(w - 1)(1 + 2w)^2 = 27x w^3$$

and expanding and rearranging gives

$$(27x - 4) w^3 + 3w + 1 = 0.$$

So more complicated than $(1 - 3x) y^3 = 1$, unfortunately. Stanley is probably your best bet for learning more about this sort of thing; it's extremely comprehensive, especially the exercises. The series $w$ appears again in Exercise 6.13.

Solution 2:

A simple identity:

$${(1+x)^{1/m}}=\sum_{k\geq 0}{\binom{1/m}{k}}x^k.$$

converting it to the binomial form $$\binom{km}{k}$$ is far harder though

For m=3, it arises when finding the roots of a cubic equation [1] and certain enumerative combinatorics problems in which the characteristic equation is a cubic , such as random walks involving two steps forward and one step back. The 1 step forward, one step back process involves the usual m=2, but m=3 case is far harder and nice analytic expressions as far as I know do not exist ,except for simple cases in which there is a single absorbing barrier.

see:

[1] Find the second real root for cubic $x^3+1-x/b=0$