Folding Exercise

In this exercise, we were given a piece of paper and had to fold it so that the left end was on top of the right end. This was repeated as many times as the paper would permit, always left end on top of right end. After each fold, we had to unravel the paper and record the order of the creases that were vertex up and vertex down.

These were the results we got for up to five folds:

u = vertex up
d = vertex down

1st Fold: d ---> 1 crease
2nd Fold: u, d, d ---> 3 creases
3rd Fold: u, u, d, d, u, d, d ---> 7 creases
4th Fold: u, u, d, u, u, d, d, d, u, u, d, d, u, d, d ---> 15 creases
5th Fold: u, u, d, u, u, d, d, u, u, u, d, d, u, d, d, d, u, u, d, u, u, d, d, d, u, u, d, d, u, d, d ---> 31 creases

By this time we noticed a pattern of a "middle d", as shown above in green.
We observed that everything to the right of this "middle d" was identical to the entire previous fold. Everything to the left of the "middle d" was the reverse of the right side, mirrored by the "middle d".

Further observation showed that from the first crease up to and including the "middle d", there was a pattern:

1st Fold: 1 crease ---> 2^(1-1) = 2^0 = 1 crease + prev creases
2nd Fold: 2 creases ---> 2^(2-1) = 2^1 = 2 creases + prev creases
3rd Fold: 4 creases ---> 2^(3-1) = 2^2 = 4 creases + prev creases
4th Fold: 8 creases ---> 2^(4-1) = 2^3 = 8 creases + prev creases
5th Fold: 16 creases ---> 2^(5-1) = 2^4 = 16 creases + prev creases

Therefore can conclude that on any given fold, the formula for finding the number of creases produced is 2^(fold # - 1) + the number of creases from the previous fold.

However, the number of creases from the previous fold is always 1 fold less than the number of creases up to and including the "middle d". Therefore a more exact formula would be:

2[2^(fold # -1)] - 1

This formula is guaranteed to give you the number of creases in the paper if given the fold number you are on.

Ex: Fold # 5 ----> 2[2^(5-1)] -1
= 2[2^4] -1
= 2[16] - 1
= 32 -1
= 31

There are 31 creases on the 5th fold, as confirmed in the observations above.

Note: I'm aware this formula is a side step away from the actual question of "can you predict the sequence of ups and downs on a specific fold?" To be honest, I'm not sure how to predict the ups and downs other than what was observed above. An idea of a recursion method comes to mind, but I'm not sure how to implement it if that's correct. So I'm satisfied with finding the number of creases for any given fold instead.