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 $(n-1)!$.



Combinatorial interpretation of ${i\choose n}$, where $i^2=-1$
**Asymptotics:** Lets look at the quantity 


$$S(n)=(-1)^{n}(n+1)\binom{i}{n+1}=i\prod\_{k=1}^{n+1}\left(1-\frac{i}{k}\right).$$ It's just your binomial coefficient above with the $(-1)^{n+1}$ factored in, and an extra $n+1$ so it factors nicely as a product.



> 
> **Claim:** We have that 
> 
> 
> $$S(n)=\sqrt{\frac{\sinh{\pi}}{\pi}}e^{iC\_{0}}e^{-i\log n}\left(1+O\left(\frac{1}{n}\right)\right),$$ 
>  where
> 
> 
> $$C\_{0}=\frac{-\pi}{2}-1+\int\_0^\infty \frac{\{x\}}{1+x^2}dx =\arg(\Gamma(i))\approx 
> -1.872.$$ 
> 
> 
> 


In particular, the angle moves around the circle like $\log n$. 


**Application to your question:** The above claim shows that 
$$f\_{n+1} = (-1)^{n+1} n! \sqrt{\frac{\sinh{\pi}}{\pi}}\cos(-\log n+C\_0)\left(1+O\left(\frac{1}{n}\right)\right)$$ and $$g\_{n+1} \sim (-1)^{n+1} n! \sqrt{\frac{\sinh{\pi}}{\pi}} \sin(-\log n+C\_0)\left(1+O\left(\frac{1}{n}\right)\right).$$ 


In particular, the ratio $g\_n/f\_n$ can be made arbitrarily large or small.


*Proof of the claim:* We first note that the size is


$$\sqrt{\prod\_{k=1}^{n}\left(1+\frac{1}{k^{2}}\right)}=\sqrt{\prod\_{k=1}^{\infty}\left(1+\frac{1}{k^{2}}\right)}+O\left(\frac{1}{n}\right).$$ To evaluate this product, recall the Weierstrass product for the Gamma function $$\left(\Gamma(z)\right)^{-1}=ze^{\gamma z}\prod\_{k=1}^{\infty}\left(1+\frac{z}{k^{2}}\right)e^{-\frac{z}{r}}.$$ From this it follows that $$\frac{1}{|\Gamma(i)|^{2}}=\frac{1}{\Gamma(i)\Gamma(-i)}=\prod\_{k=1}^{\infty}\left(1+\frac{1}{k^{2}}\right).$$ Using the identity $$\Gamma(x)\Gamma(-x)=-\frac{\pi}{x\sin\left(\pi x\right)},$$ we now have that $$\frac{1}{\Gamma(i)\Gamma(-i)}=\frac{-i\sin(i\pi)}{\pi}=\frac{\sinh(\pi)}{\pi},$$ which gives rise to the $\sqrt{\frac{\sinh(\pi)}{\pi}}$ term. Moving on to the evaluation of the angle, by looking at each triangle, and noting that the argument is additive when multiplied, we get that the argument equals


$$-\sum\_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right).$$ 


The negative sign arises since we are working in the fourth quadrant. By looking at the Taylor series for $\tan^{-1}$ we see that the above is $\log n+O(1)$, however, I would like to compute this argument more precisely, and obtain the constant. Lets compare our $\tan^{-1}$ series to the harmonic series. Rewriting things in terms of a Riemann Stieltjes integral, and using summation by parts, we have that 


$$\sum\_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right)=\int\_{0}^{n}\tan^{-1}\left(\frac{1}{x}\right)d\left[x\right]=[n]\tan^{-1}(1/n)\int\_{0}^{n}\frac{\left[x\right]}{1+x^{2}}dx. $$


Pulling out the main term with the identity $[x]=x-\{x\}$, the above equals 


$$\int\_{0}^{n}\frac{x}{1+x^{2}}dx-\int\_{0}^{n}\frac{\{x\}}{1+x^{2}}dx.$$


Since the first integral evaluates to $\frac{1}{2}\log(1+x^2)$, we have that $$\sum\_{k=1}^{n}\tan^{-1}\left(\frac{1}{k}\right)=\log n +1-\int\_0^\infty \frac{\{x\}}{1+x^2}dx +O\left(\frac{1}{n}\right).$$


**Acknowledgements:** I would like to thank Noam Elkies for pointing out that $$\prod\_{k=1}^\infty \sqrt{1+\frac{1}{k^2}}=\frac{1}{|\Gamma(i)|}=\sqrt{\frac{\sinh(\pi)}{\pi}}$$ in the comments.


**Edit:** Fixed the constants appearing. Interestingly $$\Gamma(i)=\sqrt{\frac{\pi}{\sinh{\pi}}}\exp\left(i\left(\frac{-\pi}{2}-1+\int\_0^\infty \frac{\{x\}}{1+x^2}dx \right)\right).$$


