Pascal's Triangle on a Tube

May 6, 2017

The construction rules for Pascal's triangle applied to the surface of a cylinder.

Imagine Pascal's triangle, but on a tube. Each entry is still equal to the sum of the two entries above, but the rows are now rings. Shown above is the physical realization of this idea in the form of a paper tube. Shown below are the entries printed on the tube.

$$\begin{matrix} & 0 & & 1 & & -1 & & 0 & & 0 \\ 0 & & 1 & & 0 & & -1 & & 0 & \\ & 1 & & 1 & & -1 & & -1 & & 0 \\ 1 & & 2 & & 0 & & -2 & & -1 & \\ & 3 & & 2 & & -2 & & -3 & & 0 \\ 3 & & 5 & & 0 & & -5 & & -3 & \\ & 8 & & 5 & & -5 & & -8 & & 0 \\ 8 & & 13 & & 0 & & -13 & & -8 & \\ & 21 & & 13 & & -13 & & -21 & & 0 \\ 21 & & 34 & & 0 & & -34 & & -21 & \\ & 55 & & 34 & & -34 & & -55 & & 0 \\ 55 & & 89 & & 0 & & -89 & & -55 & \\ & 144 & & 89 & & -89 & & -144 & & 0 \\ 144 & & 233 & & 0 & & -233 & & -144 & \\ & 377 & & 233 & & -233 & & -377 & & 0 \\ 377 & & 610 & & 0 & & -610 & & -377 & \\ & 987 & & 610 & & -610 & & -987 & & 0 \\ 987 & & 1597 & & 0 & & -1597 & & -987 & \\ & 2584 & & 1597 & & -1597 & & -2584 & & 0 \\ 2584 & & 4181 & & 0 & & -4181 & & -2584 & \\ & 6765 & & 4181 & & -4181 & & -6765 & & 0 \\ … & & … & & … & & … & & … & \\ & … & & … & & … & & … & & … \\ \end{matrix}$$

This tube has 5 entries per row and generates the fibonacci sequence.

One question you might ask is, given an arbitrary row, can you derive the contents of the row above? The answer is simple enough to find out. Just write out the system of equations that map one row to another in matrix form and see if the coefficient matrix is invertible.

$$\begin{bmatrix} 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_{n1} \\ x_{n2} \\ x_{n3} \\ x_{n4} \\ x_{n5} \\ \end{bmatrix} = \begin{bmatrix} x_{(n+1)1} \\ x_{(n+1)2} \\ x_{(n+1)3} \\ x_{(n+1)4} \\ x_{(n+1)5} \\ \end{bmatrix} $$

Matrix equation mapping one row to the next for a tube having 5 entries per row.

We can evaluate the determinant of the coefficient matrix using the first column. The submatrix formed by removing the first row and first column is the transpose of the submatrix formed by removing the last row and first column. The determinant of each of these submatrices is 1. So the determinant of the entire matrix is \(1 + 1 = 2\) for odd-sized matrices and \(1-1 = 0\) for even-sized matrices. So the coefficient matrix is invertible if and only if it is odd-sized.

So, given an arbitrary row, you can uniquely find the contents of the previous row if and only if the number of entries per row is odd.