TITLE: Combinatorial interpretation of ${i\choose n}$, where $i^2=-1$
QUESTION [45 upvotes]: At MIT all departments have numbers, and math is 18. Last year MIT
math majors produced a tee shirt that said ${i\choose 18}$ ("I choose
18") on the front, and on the back
   $$ \frac{34376687+1499084559i}{14485008384}. $$
With the more natural denominator $18!$ this is
   $$ \frac{15194495654000+662595375078000i}{18!}. $$
This suggests the question: for any $n\geq 1$ find a "nice"
combinatorial interpretation of the real and imaginary parts of
$i(i-1)(i-2)\cdots (i-n+1)=f_n+ig_n$. It is easy to express $f_n$ and
$g_n$ as certain alternating sums of Stirling numbers of the first
kind, but I don't consider this "nice." The $g_n$'s seem to alternate
in sign beginning with $n=5$. The $f_n$'s alternate in sign up to
$n=17$ and then seem to alternate in sign beginning with $n=18$. It is
curious that $i(i-1)(i-2)(i-3)=-10$, a real number. One could ask the
same question with $i$ replaced by any Gaussian integer $a+bi$. One
can also ask about the asymptotic rate of growth of $f_n$ and
$g_n$. Clearly $f_n^2+g_n^2\sim C\cdot (n-1)!^2$, so one would expect
$f_n$ and $g_n$ to be roughly of the size of $(n-1)!$.

REPLY [24 votes]: There is no need to reinvent the wheel by estimating
$\prod_{k<n}(1+\frac1{k^2})$.
The asymptotic formula for $f_n + i g_n$  follows readily from
Stirling's approximation (as I already noted in my comment to
the original question), and indeed the same is true for
the asymptotics as $n \rightarrow \infty$ of $w \choose n$
for any $w \in {\bf C}$; the answer is simply
$$
\phantom{*0000000000000000000}
{w \choose n} =
 \Bigl(1+O\bigl(\frac1n\bigr)\Bigr) \frac{(-1)^n}{\Gamma(-w)} n^{-w-1}
\phantom{0000000000000000000}(*)
$$
(and the $O(1/n)$ can be refined to an asymptotic series in powers of $1/n$).
Note that this gives zero precisely for the values $w=0,1,2,3,\ldots$
for which $-w$ is a pole of $\Gamma$, which are also the $w$ for which
${w \choose n} = 0$ for sufficiently large $n$.
For $w=i$, we recover the observed behavior:
$\Gamma(i)$ is a complex number of absolute value
$(\pi / \sinh \pi)^{1/2}$ [in general
$$|\Gamma(it)| = (\Gamma(it)\Gamma(-it))^{-1/2} = 
\left(\frac \pi {t \phantom. \sinh \pi t} \right)^{1/2}
$$
for real $t \neq 0$],
and $n^{-w-1}$ is a complex number of absolute value $1/n$ that goes
once around the origin when $n$ increases by a factor $e^{2\pi}$.
Thus each of $\lbrace f_n \rbrace$ and $\lbrace g_n \rbrace$
alternates in sign outside an infinite sequence of exceptions
that's asymptotically a geometric sequence with common ratio $e^\pi$.
To prove $(*)$, write
$$
{w \choose n} = \frac{(-1)^n}{n!} \prod_{k=0}^{n-1} (k-w)
= \frac{(-1)^n}{n!} \frac{\Gamma(n-w)}{\Gamma(-w)}
= \frac{(-1)^n}{n\Gamma(-w)} \frac{\Gamma(n-w)}{\Gamma(n)}.
$$
Now we understand $(-1)^n/n$,
and the factor $1 / \Gamma(-w)$ is constant,
so we're left with $\Gamma(n-w) / \Gamma(n)$.
We apply the following form of Stirling's formula:
there exists a constant $\varpi>0$ (known to equal $2\pi$,
but we shall not need this) such that
$$
\Gamma(z) = \bigl(1 + O(|z|^{-1}\bigr) z^z e^{-z} \sqrt{\varpi/z}
$$
holds as $|z| \rightarrow \infty$ in the right half-plane,
where $z^z = \exp (z \log z)$ and $\sqrt{\varpi/z}$
are defined using the principal branches of $\log z$ and $\sqrt z$.
This gives
$$
\frac{\Gamma(n-w)}{\Gamma(n)} = \Bigl(1+O\bigl(\frac1n\bigr)\Bigr)
  \frac{(n-w)^{n-w} e^{-(n-w)} (\varpi/(n-w))^{1/2}}
    {n^n e^{-n} (\varpi/n)^{1/2}}
$$ $$
 = \Bigl(1+O\bigl(\frac1n\bigr)\Bigr)
\frac{(n-w)^{n-w}}{n^n} e^w \left(1-\frac{w}{n}\right)^{-1/2}.
$$
Now the last factor is $1 + O(1/n)$; the factor $e^w$ is constant;
and
$$
(n-w)^{n-w} = (n-w)^{-w} (n-w)^n
= \Bigl(1+O\bigl(\frac1n\bigr)\Bigr) n^{-w} (n-w)^n.
$$
So we're left with
$$
\frac{\Gamma(n-w)}{\Gamma(n)} =
\Bigl(1+O\bigl(\frac1n\bigr)\Bigr) n^{-w} e^{-w} \left(1 - \frac{w}{n}\right)^n
= \Bigl(1+O\bigl(\frac1n\bigr)\Bigr) n^{-w}.
$$
This completes the proof of $(*)$ (and the cancellation in the last step
leads me to suspect that even this use of Stirling is more complicated
than necessary).