## A Function for Folding Odd-Numbered Grids?

General discussion about Origami, Papers, Diagramming, ...

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

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.
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Baltorigamist
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### Re: A Function for Folding Odd-Numbered Grids?

I really enjoy number theory problems like this. I'll preface this post by saying that I may be overcomplicating things here.

It sounds like your problem involves partitions of an integer. These course notes explain partitions better than I can: https://math.berkeley.edu/~mhaiman/math172-spring10/partitions.pdf

When you write out the sequences:

5: {1, 4}, {2, 3}
The folding sequence: {1, 4}, {2, 3}, {1, 4}, {2, 3}

11: {1, 10}, {2, 9}, {3, 8}, {4, 7}, {5, 6}
The folding sequence: {3, 8}, {4, 7}, {2, 9}, {1, 10}, {5, 6}, {3, 8}, {4, 7}, {2, 9}, {1, 10}, {5, 6}

13: {1, 12}, {2, 11}, {3, 10}, {4, 9}, {5, 8}, {6, 7}
The folding sequence: {1, 12}, {6, 7}, {3, 10}, {5, 8}, {4, 9}, {2, 11}, {1, 12}, {6, 7}, {3, 10}, {5, 8}, {4, 9}, {2, 11}

For the numbers that work, the pattern repeats itself halfway through the sequence, and you need N - 1 elements (folds in this case). All the 2 part partitions are also used. 7 and 9 don't work because no matter where you start you can't find a sequence that uses all the 2 part partitions.

I can't immediately see a pattern off the top of my head, but I suspect it will be something that doesn't have a closed form solution. I might try to code this up in python and see if I can see a pattern when I do it for more numbers.
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unknownfolder
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### Re: A Function for Folding Odd-Numbered Grids?

I don't really understand the partition part, but otherwise your explanation makes a lot of sense. If it helps, I'm fairly sure that any number that's the product of two primes (excluding 2) won't work, as well as (most) numbers of the form (2^n)+/-1. The latter function doesn't work for any lower values, but I'm not sure if it generalizes to values above n=3.
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Baltorigamist
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### Re: A Function for Folding Odd-Numbered Grids?

I got curious with this since I fold odd grids sometimes and also noticed this peculiarity. I tried up to 31 manually on Oripa and here are some results:

Work: 2, 3, 5, 11, 13, 19, 29
Doesn't work: 7, 9, 15, 17, 21, 23, 25, 27, 31

The numbers in bold are not prime numbers. Those will never work as they have an independant grid inside them. For instance, when folding 9ths, you will always be left with the 3rds to fold separately. With 25ths, same thing but with 5ths.

The problem then seems to be relevant to prime numbers only (but the numbers in bold above might still prove useful to find the solution). I'm not sure how to proceed from there other than trying some simple functions and try to figure out some pattern. My intuition tells me it has to do with the pair (n+-1)/2.

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

I decided to give this a go in python. The stuff I mentioned about partitions was me over-complicating things.

I wrote a program that automatically checks if there is a folding sequence for whatever N you specify. These were my results for N = 199: 3, 5, 11, 13, 19, 29, 37, 53, 59, 61, 67, 83, 101, 107, 131, 139, 149, 163, 173, 179, 181, 197

I didn't see a pattern, but what Sunburst said about it being relevant to prime numbers only was correct! I decided to see what google would tell me when I searched for a few numbers of the sequence, and I found this wikipedia page as the top result: https://en.wikipedia.org/wiki/Full_reptend_prime If you scroll down to the section for "Binary full reptend primes", the sequence is there.

I ran my program for N=300, and the sequence I get seems to line up!
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unknownfolder
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### Re: A Function for Folding Odd-Numbered Grids?

«Binary full reptend primes»... interesting. That wiki definition is confusing as heck though.

Sunburst
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### Re: A Function for Folding Odd-Numbered Grids?

Very interesting. So, if I'm interpreting this correctly, these "binary full reptend primes" are the only numbers that work with the method I described in the OP. I'd had a hunch it had to do with prime numbers, but I never would have found my way to that Wiki page. Thanks!

The "obligatory" next question is what fraction each sequence needs to start at--unless it has to do with the primes' periodicity somehow.
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### Re: A Function for Folding Odd-Numbered Grids?

Baltorigamist wrote:The "obligatory" next question is what fraction each sequence needs to start at--unless it has to do with the primes' periodicity somehow.

Maybe I am mistaken, but here is how I understand it: after the last needed fold is done, you'll get back to the first one if you'd keep the same folding logic[*].

Baltorigamist wrote:For fifths:
1/5 - 1:4
3/5 - 3:2
4/5 - 4:1
2/5 - 2:3

another possibility is:
3/5 - 3:2
4/5 - 4:1
2/5 - 2:3
1/5 - 1:4
(starting with your 2d line, then 3rd one, 4th one, and back to the 1st one).

Any other rotation would also work (i.e., "4/5, 2/5, 1/5, 3/5" and "2/5, 1/5, 3/5, 4/5").

So, if you know where is one of the needed reference on your paper (the 1/5 from your example; the 3/5 from my first example; ...), you can fold the other references using related rotation. No needed start, each and every reference is equivalent.

[*]: Which is why you can fold these proportions using non-exact sequences such as dichotomy.
For example, to fold the fifths:
- Guess-fold the 2/5 proportion.
- Fold precisely the next proportions (based on the guessed 2/5, but precisely paper-edge-to-crease) in the sequence order (1/5, 3/5, 4/5).
- Now, fold precisely 2/5 "again". If your first guess was right, the new 2/5 will be the same as guess-folded 2/5. Otherwise, keep folding the whole sequence once again.
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### Re: A Function for Folding Odd-Numbered Grids?

You’re right about that, actually. So it really doesn’t matter where you start folding an odd grid, then. If it’s a full reptend prime, it will work, but if it’s not, it won’t.
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Baltorigamist
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### Re: A Function for Folding Odd-Numbered Grids?

Yep. My program confirms that as well.

Thank you for the fun problem! It kind of reminded me of the programming problems on Project Euler. Now I'm wondering if any number theorists have stumbled upon this relationship as well. I'm almost tempted to try to find one online and email him/her a link to this forum post.
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