# Product of the entries in a row of Pascal's triangle

The sum of the $n$-th row in Pascal's triangle \begin{equation} \sum_{k=0}^{n}\binom{n}{k} \end{equation} has the well-known value of $2^n$. Now, I'm looking for the value of the product of the $n$-th row in Pascal's triangle, given through \begin{equation} s_n=\prod_{k=0}^{n}\binom{n}{k}. \end{equation} Any ideas how to calculate this value? Is it even possible?

I found some papers (e.g. Finding e in Pascals Triangle) dealing with the growth of this sequence, and it seems to be that the ratio of the ratios $\frac{s_{n+1}/s_n}{s_n/s_{n-1}}$ has the limiting value of \begin{equation} \lim_{n\rightarrow\infty}\frac{s_{n+1}s_{n-1}}{(s_n)^2}=e. \end{equation} Is this helpful for calculating the value of $s_n$? So far, it is not clear to me how the growth rate of a sequence relate to its value.

• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos Dec 6 '17 at 15:57

## 1 Answer

This is OEIS A001142 which begins $$1, 1, 2, 9, 96, 2500, 162000, 26471025, 11014635520, 11759522374656, 32406091200000000, 231627686043080250000, 4311500661703860387840000, 209706417310526095716965894400, 26729809777664965932590782608648192$$ An approximate formula is given $a(n) \approx A^2 * \exp(n^2/2 + n - 1/12) / (n^{(n/2 + 1/3)} * (2*\pi)^{((n+1)/2))}$, where $A = A074962 = 1.2824271291\ldots$ is the Glaisher-Kinkelin constant.

The growth rate is dominated by the term $$\frac{\exp(\frac {n^2}2)}{n^{\frac n2}}=\exp \left( \frac{n^2}2-\frac n2\log n \right)$$

• Perfect! Thank you so much. – MaxWell Dec 6 '17 at 16:27
• If the growth is dominated by $\exp(\frac{n^2}{2}-\frac{n}{2}\log{}n)$, is it equivalent to say that $a(n)\in\mathcal{O}(\exp(\frac{n^2}{2}-\frac{n}{2}\log{}n))$? – MaxWell Dec 7 '17 at 8:33
• @MaxWell: No. Big-O requires you be within a constant factor. The next term here is a factor $(2\pi)^{(n+1)/2}$, which is not constant. Big-O is hard for complicated fast-growing things like this. I found the first couple terms and stopped. – Ross Millikan Dec 7 '17 at 14:52