2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ – José Carlos Santos Dec 6 '17 at 15:57
3
$\begingroup$

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 $$\frac{\exp(\frac {n^2}2)}{n^{\frac n2}}=\exp \left( \frac{n^2}2-\frac n2\log n \right)$$

$\endgroup$
  • $\begingroup$ Perfect! Thank you so much. $\endgroup$ – MaxWell Dec 6 '17 at 16:27
  • $\begingroup$ If the growth is dominated by $\exp(\frac{n^2}{2}-\frac{n}{2}\log{}n)$, is it equivalent to say that $a(n)\in\mathcal{O}(\exp(\frac{n^2}{2}-\frac{n}{2}\log{}n))$? $\endgroup$ – MaxWell Dec 7 '17 at 8:33
  • $\begingroup$ @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. $\endgroup$ – Ross Millikan Dec 7 '17 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.