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.

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  • $\begingroup$ i would play with Stirling's approximation, may make things easier $\endgroup$ – gt6989b Jul 24 '19 at 8:11
  • $\begingroup$ I would also expect the denominator to be something like $a+e$ instead of $a+3$... $\endgroup$ – gt6989b Jul 24 '19 at 8:13
  • $\begingroup$ @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 $\endgroup$ – Nilotpal Sinha Jul 24 '19 at 8:16
  • $\begingroup$ Great, happy to be wrong :-) $\endgroup$ – gt6989b Jul 24 '19 at 11:06
  • $\begingroup$ 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.... $\endgroup$ – Greg Martin Jul 25 '19 at 4:47

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