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On Pascal's triangle with any number of rows, draw the three medians.

enter image description here

For each median, calculate the product of the numbers that zig-zag along that median.

The three products are always equal! (proof below)

Note that, on the vertical median, every other number is either just to the left, or just to the right of the median; it makes no difference. On the oblique medians, every other number is just below the median (this rule applies when the number of rows is even or odd).

Here is an example with a triangle with $10$ rows (the top $1$ is row $0$):

enter image description here

$1\cdot 1\cdot 2\cdot 3\cdot 6\cdot 10\cdot 20\cdot 35\cdot 70\cdot 126\cdot 252=560105280000$

enter image description here

$1\cdot 6\cdot 15\cdot 35\cdot 35\cdot 56\cdot 28\cdot 36\cdot 9\cdot 10\cdot 1=560105280000$

This is such a simple result, so I wonder if there is an intuitive explanation. My question is:

Is there an intuitive explanation for this result?

To give an idea of what I mean by "intuitive explanation", here are some examples:

  • The sum of the numbers in the $n$th row is $2^n$, because $\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{n}=(1+1)^n=2^n$.
  • $\sum_{k=0}^n \binom n k ^2 = \binom {2n} n$ has an intuitive combinatorial proof.
  • In the left-justified Pascal's triangle, any square matrix that shares a border with the left edge of the triangle has determinant $1$; this has an intuitive proof.

My non-intuitive proof

I prove the case when the number of rows is even, $2n$. The proof with an odd number of rows is very similar.

$P_v=$ product along vertical median $=\prod\limits_{k=1}^n\binom{2k-1}{k-1}\binom{2k}{k}$

$P_o=$ product along oblique median $=\prod\limits_{k=1}^n\binom{n+k}{2k-1}\binom{n+k}{2k}$

$\begin{align} \dfrac{P_v}{P_o}&=\prod\limits_{k=1}^n\dfrac{(2k-1)!^2 (2k)!^2 (n+1-k)!(n-k)!}{(k-1)!k!^3 (n+k)!^2}\\ &=\prod\limits_{k=1}^n \dfrac{(2k-1)!^2(2k)!^2}{k!^2 (n+k)!^2}\\ &=\prod\limits_{k=1}^n \dfrac{(2k)!^4}{k!^2 (n+k)!^2 (2k)^2}\\ &=\dfrac{(2!4!6!\dots (2n)!)^4}{(1!2!3!\dots n!)^2((n+1)!(n+2)!(n+3)!\dots (2n)!)^2(2\cdot 4\cdot 6\dots (2n))^2}\\ &=\dfrac{(2!4!6!\dots (2n)!)^2}{(1!3!5!\dots (2n-1)!)^2 (2\cdot 4\cdot 6\dots(2n))^2}\\ &=1 \end{align}$

Context

I have been playing with Pascal's triangle, investigating some of its mysterious, geometrical and humourous properties.

Xander Henderson
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Dan
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    Don't know if it might help, but it seems that the equality can be divided in two: $\prod\limits_{k=1}^n\binom{2k}{k}=\prod\limits_{k=1}^n\binom{n+k}{2k-1}$ and $\prod\limits_{k=1}^n\binom{2k-1}{k-1}=\prod\limits_{k=1}^n\binom{n+k}{2k}$ – Fabius Wiesner Dec 05 '23 at 17:37
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    Related conjecture: Let "strict-median" numbers be along lines that don't wiggle. Then, for a $2n$ triangle, $$\frac{\text{vertical strict-median product}}{\text{oblique strict-median product}}=2^{n}$$ For instance, with your $n=5$ example, $$\frac{1\cdot2\cdot6\cdot20\cdot70\cdot252}{1\cdot15\cdot35\cdot28\cdot9\cdot1}=2^5$$ And, note: the oblique off-median numbers (eg, your blue $6,35,56,35,10$), along with a "$1$", are the strict-median numbers for a $2n+1$ triangle. (Verticals are slightly weirder to describe.) – Blue Dec 05 '23 at 17:40
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    The algebraic form of @Blue conjecture is $\frac{\prod\limits_{k=1}^n\binom{2k}{k}}{\prod\limits_{k=1}^n\binom{n+k}{2k}}=\frac{\prod\limits_{k=1}^n\binom{n+k}{2k-1}}{\prod\limits_{k=1}^n\binom{2k-1}{k-1}}=2^n$. – Fabius Wiesner Dec 05 '23 at 17:55
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    Proof of Blue's conjecture: $\dfrac{\prod\limits_{k=0}^n\binom{2k}{k}}{\prod\limits_{k=0}^n\binom{n+k}{2k}}=\prod\limits_{k=0}^n \dfrac{(2k)!^2(n-k)!}{k!^2(n+k)!}=\prod\limits_{k=0}^n \dfrac{(2k)!^2}{k!(n+k)!}=\dfrac{(2!4!6!\dots (2n)!)^2}{(1!2!3!\dots n!)(n!(n+1)!(n+2)!\dots (2n)!)}=\dfrac{2!4!6!\dots (2n)!}{1!3!5!\dots (2n-1)!n!}=\dfrac{2\cdot 4\cdot 6\cdot \dots (2n)}{n!}=2^n$ – Dan Dec 05 '23 at 21:29

1 Answers1

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Self-answering. I found what I consider to be an intuitive proof.

It looks long, but after you understand what it's saying, it's fairly intuitive. In fact, I thought of this proof while riding my bicycle, without writing anything.

Define:
$P_o(a)=$ product along oblique median in a triangle with $a$ rows.
$P_v(a)=$ product along vertical median in a triangle with $a$ rows.

The result in the OP is equivalent to $\dfrac{P_o(a+1)}{P_o(a)}=\dfrac{P_v(a+1)}{P_v(a)}$ for all $a\in\mathbb{N}$.

In the diagram,
$P_o(10)$ is the product of the numbers in blue and green.
$P_o(11)$ is the product of the numbers in yellow and green.

enter image description here

$\begin{align} \dfrac{P_o(11)}{P_o(10)}&=\dfrac{\binom{6}{0}}{\binom{5}{0}}\cdot\dfrac{\binom{7}{2}}{\binom{6}{2}}\cdot\dfrac{\binom{8}{4}}{\binom{7}{4}}\cdot\dfrac{\binom{9}{6}}{\binom{8}{6}}\cdot\dfrac{\binom{10}{8}}{\binom{9}{8}}\cdot\dfrac{\binom{11}{10}}{\binom{10}{10}}\\ &=\dfrac{6}{6}\cdot\dfrac{7}{5}\cdot\dfrac{8}{4}\cdot\dfrac{9}{3}\cdot\dfrac{10}{2}\cdot\dfrac{11}{1}\text{ because }\dfrac{\binom{p}{r}}{\binom{p-1}{r}}=\dfrac{p}{p-r}\\ &=\binom{11}{5}\\ &=\dfrac{P_v(11)}{P_v(10)}\\ \end{align}$

Or in general,
$\begin{align} \dfrac{P_o(2n+1)}{P_o(2n)}&=\prod\limits_{k=0}^n \dfrac{\binom{n+1+k}{2k}}{\binom{n+k}{2k}}\\ &=\prod\limits_{k=0}^n \dfrac{n+1+k}{n+1-k}\\ &=\binom{2n+1}{n}\\ &=\dfrac{P_v(2n+1)}{P_v(2n)}\\ \end{align}$

A similar argument shows that $\dfrac{P_o(2n)}{P_o(2n-1)}=\dfrac{P_v(2n)}{P_v(2n-1)}$.

$\therefore\dfrac{P_o(a+1)}{P_o(a)}=\dfrac{P_v(a+1)}{P_v(a)}$ for all $a\in\mathbb{N}$.

Dan
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