I noticed something curious when I was working on my summer project last year. Basically, you consider the set of binary strings (strings of length where each symbol is either a or a and ‘s are non-adjacent). So we construct a table like

Length |
1 |
2 |
3 |
4 |
5 |
6 |

Strings |
0;1 | 00;01;10 | 000;001;010;100;101 | 0000;0001;0010;0100;0101;1000;1001;1010 | … | … |

And then count the number of elements of length with ones.

Number of ones [m] (below) |
Number of strings (below) |
|||||

Length of string [n] (right) |
1 |
2 |
3 |
4 |
5 |
6 |

0 |
1 | 1 | 1 | 1 | 1 | 1 |

1 |
1 | 2 | 3 | 4 | 5 | 6 |

2 |
0 | 0 | 1 | 3 | 6 | 10 |

3 |
0 | 0 | 1 | 4 | ||

4 |
0 | 0 |

Do you see Pascal’s Triangle hiding inside? Starting from the top left and diagonally down to the left. First one is 1,2,1, then is 1,3,3,1 etcThe explanation is simple enough. We are using the fact that the subset of composed of strings with ones (lets call this ) can be made by taking an element in and appending a zero to it or an element in and appending a zero-one to it. We end up with the recurrence . If we consider as the element in the diagonal, then we realize that follows the same recurrence. This reccurence is well documented in the literature for calculating , here we are just using this idea for subsets of with exactly ones to connect it with Pascal’s Triangle.

The adventure continues in Part 2!