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What's the asymptotic growth of:

$$ a(n)=\prod_{k\in\{1,\dots,n\}}{2^{k+1} \choose 2^k}\quad?$$

So far I was able to find only some lower bounds. For example, $a(n)>\prod_{k\in \{1\dots n\}} 2^{k+1}=\Theta(2^n)$

Also, I was thinking that it would be easy to find it in OEIS (and there to find some data about it, e.g. its asymptotic growth), but unfortuantely I could find it there.

StubbornAtom
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Dudi Frid
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  • Notice that $a_k := \binom{2^{k + 1}}{2^k} = \frac{2^{2^{k + 1}} \Gamma\left(\frac{1}{2} + 2^k\right) }{\sqrt{\pi} \Gamma(1 + 2^k)}$. Therefore, $$ 2^{n 2^{n + 1}} \ge \prod_{k = 1}^{n} a_k \ge 2^{4n}.$$ – ViktorStein Aug 30 '19 at 14:17

1 Answers1

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We have $\binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}}$, hence by considering $n=2^k$ and multiplying both sides for $k\in\{1,2,\ldots,N\}$ we have that

$$\prod_{k=1}^{N}\binom{2^{k+1}}{2^k}\sim \prod_{k=1}^{N}\frac{4^{2^k}}{\sqrt{\pi}\,2^{k/2}}=\frac{4^{2^{N+1}-2}}{\pi^{N/2}2^{N(N+1)/4}} $$ which essentially is $2^{2^{N+2}}$.

Jack D'Aurizio
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