Let $\Gamma$ be the Gamma function and abbreviate $x!:=\Gamma(x+1)$, $x>-1$. For $\alpha>0$ let us generalize the binomial coefficients in the following way: $$\binom{n+m}{n}_\alpha:=\frac{(\alpha n+\alpha m)!}{(\alpha n)!(\alpha m)!}$$

Of course for $\alpha=1$ this reduces to the ordinary binomial coefficients. My question is:

How can one show that this is really the only case where they coincide?

That is, $\displaystyle\binom{n+m}{n}=\binom{n+m}{n}_\alpha$ for all $n,m\in\mathbb{N}$ implies $\alpha=1$.

Or even more general (but currently not of interest to me):

$\displaystyle\binom{n+m}{n}_\alpha=\binom{n+m}{n}_\beta$ implies $\alpha=\beta$.

I tried to use Stirling's formula: $\Gamma(z+1)\sim\sqrt{2\pi z}\bigl(\frac{z}{e}\bigr)^z$ as $z\to\infty$, but didn't get very far.

Thanks in advance for any help.


1 Answer 1


Set, for example, $m=1$ and consider the limit $n\rightarrow\infty$. Then $$ {\alpha n+\alpha\choose \alpha n}\sim \frac{(\alpha n)^{\alpha}}{\alpha!},\qquad {n+1 \choose n}\sim n.$$ It is clear that the only possibility for both asymptotics to agree is $\alpha=1$. In the more general situation, the same argument shows that $\alpha=\beta$.


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