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How can we approximate the median of the numbers in the first $n$ rows of Pascal's triangle? (The top row is the $0$th row.)

Using Excel, I made a graph of the natural log of the median against $n$, for $n=6$ to $n=140$. (For $n=0$ to $n=5$, the median is $1$.)

enter image description here

I used the points from $n=101$ to $n=140$ to calculate a trendline, and I got $\text{median}\approx 0.126\times1.380^n$ Do the constants have asymptotic closed forms? That $1.380$ is very close to $\ln4\approx1.386$.

Context: I asked my high school students to find the mean of the numbers in the first $n$ rows of Pascal's triangle (the answer is $\frac{2(2^n-1)}{n(n+1)}$). Then I wondered, what about the median? (The mode of the entire triangle, if we ignore the $1$'s, is conjectured to be $3003$.)

EDIT:

If the median of the first $n$ rows is indeed approximately $c\times (\ln4)^n$ for some constant $c$, then we would have the following result:

The median of the first $n$ rows is approximately equal to the central binomial coefficient of row $2k$, where $2k\approx \left(1+\frac{\ln \ln 2}{\ln 2}\right)n\approx 0.47123n$. (So the median is approximately in the middle of the triangle.)

Explanation: The central binomial coefficient of row $2k$ is approximately $\frac{4^k}{\sqrt{\pi k}}$. Setting this equal to $c\times(\ln4)^n$, we get $2k=\frac{ \ln c+\frac12 \ln \pi+\frac12 \ln k +n \ln \ln 4 }{\ln 2}$. So for large $n$, we have $2k\approx \left(1+\frac{\ln \ln 2}{\ln 2}\right)n$.

But I don't know how to show that the median is approximately $c\times (\ln4)^n$ for some constant $c$.

Dan
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  • Could $1.38$ be $2\log 2$? – Gary Nov 20 '23 at 06:43
  • @Gary I wondered that also. To get a better approximation, I would have to use many more rows of Pascal's triangle, but Excel is quite limited. – Dan Nov 20 '23 at 06:47
  • What do you get if you compute the median and divide it with $(2\log 2)^n$? – Gary Nov 20 '23 at 06:50
  • @Gary The graph of $\frac{\text{median}}{(2\log 2)^2}$ against $n$ appears to converge to about $0.09$, but the curve is a bit "bumpy", I guess due to the discrete nature of the binomial coefficients. – Dan Nov 20 '23 at 06:58

1 Answers1

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Suppose we have Pascal's triangle with $n$ rows, where $n$ is large. From this answer to another question, we have that the proportion of the numbers in the triangle that are less than the central binomial coefficient of row $2k$, is

$$P=2 \left[\mu_0 + \left(\frac{2k}{n}\ln 2\right)^2 \int_{\mu_0}^{1/2} \frac{1}{g(\mu)^2}\mathrm d\mu \right]$$

where $g(\mu)=-\mu\ln \mu-(1-\mu)\ln (1-\mu)$ and $\mu_0=g^{-1}\left(\frac{2k}{n}\ln 2\right)$.

Setting $P=0.5$ and working numerically, we get $k\approx 0.238752608n$ (and $\mu_0\approx 0.10270281182$).

So the median is approximately equal to the central binomial coefficient of row $2k$, where $k\approx0.238752608n$.

The central binomial coefficient of row $2k$ is approximately $\frac{4^k}{\sqrt{\pi k}}$.

So the median is approximately $\dfrac{4^{an}}{\sqrt{\pi an}}$ where $a=0.238752608$.

Dan
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