### A Function for Folding Odd-Numbered Grids?

Posted:

**May 22nd, 2020, 5:35 pm**This question has been on my mind for a long time now, but I don't remember posting it here before. It's kind of tricky to explain, so bear with me.

For a few odd-numbered grids, you can make all the creases simply by folding one of the edges to the last crease you made. In other words, each successive crease divides the paper into (2^n) divisions and an odd number of divisions such that each (2^n)th crease has yet to be made. The examples below state the position of the crease from left to right and then the ratio of the number of divisions on either side of the crease.

For fifths:

1/5 - 1:4

3/5 - 3:2

4/5 - 4:1

2/5 - 2:3

And for elevenths:

3/11 - 3:8

7/11 - 7:4

9/11 - 9:2

10/11 - 10:1

5/11 - 5:6

8/11 - 8:3

...and then you can rotate the paper and repeat for the remaining creases.

However, this process doesn't work for most other odd-numbered divisions, such as sevenths or ninths; you almost always have to align two existing creases in order to get all the divisions.

My question is whether there's a single function that will tell whether a given odd number is possible without aligning two creases.

I initially considered that (2(2^n))+/-1 might give numbers that cannot be constructed by the method above (e.g., 7 and 9), but 21, 23, and 25 aren't outputs of that function. It's not (n(2^n))+/-1 either, since 21 is still not an output.

I'd appreciate any help.

For a few odd-numbered grids, you can make all the creases simply by folding one of the edges to the last crease you made. In other words, each successive crease divides the paper into (2^n) divisions and an odd number of divisions such that each (2^n)th crease has yet to be made. The examples below state the position of the crease from left to right and then the ratio of the number of divisions on either side of the crease.

For fifths:

1/5 - 1:4

3/5 - 3:2

4/5 - 4:1

2/5 - 2:3

And for elevenths:

3/11 - 3:8

7/11 - 7:4

9/11 - 9:2

10/11 - 10:1

5/11 - 5:6

8/11 - 8:3

...and then you can rotate the paper and repeat for the remaining creases.

However, this process doesn't work for most other odd-numbered divisions, such as sevenths or ninths; you almost always have to align two existing creases in order to get all the divisions.

My question is whether there's a single function that will tell whether a given odd number is possible without aligning two creases.

I initially considered that (2(2^n))+/-1 might give numbers that cannot be constructed by the method above (e.g., 7 and 9), but 21, 23, and 25 aren't outputs of that function. It's not (n(2^n))+/-1 either, since 21 is still not an output.

I'd appreciate any help.