Numerical Triangle Question

Question

If a numerical triangle is constructed such that each number is the sum of three numbers in the previous row: the number above it, the number to the left of the one above it, the number to the right of the one above it, and each number that isn't filled is treated as 0. We get this $$\begin{matrix} &&&&&1\\ &&&&1&1&1\\ &&&1&2&3&2&1\\ &&1&3&6&7&6&3&1\\ &...&...&...&...&...&...&...&...&...\\ \end{matrix}$$

I'm trying to prove that each row after the second row will always contain an even number. I'm not really sure how to go about it.

Answer 1

First, you have to notice the following easily observable properties of your "pyramid of numbers":

• it is symmetric with respect to the vertical axis
• each row has an odd number of elements (1, 3, 5, 7...)

Now suppose that you have a row in your pyramid containing odd numbers only. What would be the composition of the row directly above it? Let us simply denote the odd number with O and even number with E. A simple ananlysis will show that the row with all odd numbers and the row above it must look like this:

$$\begin{matrix} & O & E & E & O & E & E & O & E & E & O & E & E & ...\\ O & O & O & O & O & O & O & O & O & O & O & O & O & ... \end{matrix}\tag{1}$$

Notice the pattern that repeats in the row above. It's basically a group $(O, E, E)$ repeated multiple times. One such group will contain the middle element of the row. That middle element cannot be $E$ because numbers to the left and right of it are not of the same parity and the row has to be symmetric. So the middle element has to be $O$.

You can still construct a row that is symmetric, has odd number of elements and is made of $(O, E, E)$ sequences only (the last sequence does not have to be complete, that was the error in my previous version of the proof). The only possible way to do it is to make the row with an even number of $(O,E,E)$ groups and add one $O$ element to the end of it.

For example: $\color{blue}O,E,E,\color{blue}O,E,E,\color{blue}O,E,E,\color{red}O,E,E,\color{blue}O,E,E,\color{blue}O,E,E,O$

But now repeat the same exercise as we did in (1) and check the composition of the row above it.

You can easily show that the row above must have the following pattern:

$$\begin{matrix} & O & O & E & E & E & E & O & O & E & E & E & E & O & O & E & E & E & E & ...\\ O & E & E & O & E & E & O & E & E & O & E & E & O & E & E & O & E & E & O & ... \end{matrix}\tag{1}$$

This time the pattern $(O,O,E,E,E,E)$ repeats multiple times. And now you are out of luck because you cannot pick any element from it as the central element of the row to make the row symmetric. Each odd element is between odd and even number so it cannot be in the middle of the row. If you pick any even number as the middle element just check a pair of numbers on the left and the right side. These are never symmetric.

So you cannot have a row with all odd numbers.

Answer 2

See A027907 at the Online Encyclopedia of Integer Sequences.

Here's what the rows look like (mod 2).

In Mathematica, the $k$ row is Mod[CoefficientList[(1 + x + x^2)^k, x], 2] .

The number of odd values is maximized when $k$ is a Mersenne numbers $2^n -1$.
That number of odd values is the Jacobsthal sequence A001045.