# What is the growth rate of the products of binomial coefficients?

Claim: Experimental data seems to suggest that

$${n \choose 1^a b}{n \choose 2^a b}{n \choose 3^a b}\cdots {n \choose m^a b} \sim \exp\bigg(\frac{2n^{1 + \frac{1}{a}}}{ab+3b}\bigg)$$ where $$a$$ and $$b$$ are a fixed positive integers and $$m$$ is the largest positive integer such that $$m^a b \le n$$.

Note that the above asymptotic are supported by the data even if we relax the condition that $$a,b$$ are integers and allow them to be reals $$a > 0, b > 0$$ and replace $$k^a b$$ with $$\lfloor k^a b\rfloor$$.

As an illustration, for $$a = 3, b = 1$$, the $$\%$$ error between the asymptotic and the actual product is shown below.

Note: Posted in MO since in it unanswered in MSE

Update 19-Dec-19: Combined the two individual claims into a single claim based on experimental data.

• i would play with Stirling's approximation, may make things easier – gt6989b Jul 24 '19 at 8:11
• I would also expect the denominator to be something like $a+e$ instead of $a+3$... – gt6989b Jul 24 '19 at 8:13
• @gt6989b I have a proof using Stirling's approximation for the simpler case $a = 1$ and the RHS is $e^{n^2}/2$ so the only constant that can fit in the generalization is 3 – Nilotpal Sinha Jul 24 '19 at 8:16
• Great, happy to be wrong :-) – gt6989b Jul 24 '19 at 11:06
• Presumably when one takes the logarithm of the left-hand sides and applies Stirling's approximation, the resulting expression is a Riemann sum for some integral.... – Greg Martin Jul 25 '19 at 4:47