1. An origami-inspired adventure in number theory…and on limits
Jeanine Meyer
Purchase College/SUNY

2. The adventure begins
… Mark Kennedy led people coming together to wait for free tickets to Shakespeare in the Park in NYC and fold. He taught the dollar bill rosette.

3. Overview [Warm-up] Making estimates Dollar bill rosette
Estimating thirds
Dollar bill rosette
Procedure for making off sections. What does procedure do?
Iterates to improve original estimate: E  0
Marks off ALL of intermediate sections
For what numbers besides 11 does the procedure work?
Python Program generating numbers (under 1000) that work
Used this set to identify class of primes
Proof

4. Kissy fish http://faculty. purchase. edu/jeanine

5. Method to divide into thirds
"S" method (actually a Z). Form a S or Z shape looking at it sideways and keep adjusting parts until the same size.
Variant: Make an estimate and fold to the estimate (dividing the 2/3 side into 2 part). Suppose the initial fold is off by error E: Say it should be A, but instead is A+E. (Can assume it is on one side). This divides the bill into two parts measuring A+E and Length – (A+E)
Assuming that folding a section in half is pretty accurate, the next fold produces a mark at ( Length-(A+E) )/2
So this new mark is off by E/2. The error has been halved!
Keep going

6. Dollar bill rosette
Directions (crude drawings) at

7. Dollar bill rosette first step
Fan fold of 22 parts starts with need to divide bill into 11ths.
Start with guess for 1/11. Dollar bill is divided into 1 part and 10 parts, even part and odd part. Divide the even part in half to get 6 parts and 5 parts. Continue.
1 and 10
6 and 5
3 and 8
7 and 4
9 and 2
10 and 1
Keeping going. Get back to 1 and 10.
So now have mark at each place 1/11, 2/11, etc.

8. Repeating argument on error
Suppose first estimate is off by error E: Say it should be A, but instead is A+E. (Can assume it is on one side). This divides the bill into two parts measuring A+E and Length – (A+E)
Assuming that folding a section in half is pretty accurate, the next fold produces a mark at ( Length-(A+E) )/2
So this new mark is off by E/2. The error has been halved!
Continuing through the 10 steps, when back to (around) the first mark, the new mark is off by E/210

9. Continuing folding
With improved estimate, goes through one more time making complete folds, all valleys
Divide each of the 11 sections by 2, making mountain fold, to get 22 sections in a fan fold
Divide into two
Fold out 3 panels on each side
Fold at the midpoint
Make tabs to fold in to attach the two parts.
Pop open to be a rosette

10. Rosette procedure
Besides improving the estimate, the procedure gets back to the start (1, N-1) hitting all the divisions in-between.
For what numbers does the procedure work?
Must be odd number
Must be prime
Proof?
What kind of prime

11. Suggestion that it must be prime
Suppose P is not prime, and equal to F * G, F and G each >1.
Note: neither F or G can be even.
In the rosette procedure, at some point, the pair of numbers is G and (F-1)*G. What is the next step? G+((F-1)/2)*G and ((F-1)/2)*G
Claim: never get to 1 and (F*G) -1.

12. All primes? Does procedure hit all numbers for 17 NO 1 and 16 9 and 8

13. What primes? Adventure continues…
After learning fold, a long time ago, a mathematician colleague told me that it probably was certain category of numbers, but I forgot…
So I wrote Python program to generate numbers that work

14. Experiment Wrote program to invoke the testing program for all numbers
May have checked for primes first
… from 3 to 1000

15. def tryProcedure(N): count = 1 # start with 1 and N-1 currentpos = 1 remainder = N-currentpos while True: if (isEven(currentpos)): # decide which side is even currentpos = currentpos//2 # integer division remainder = N - currentpos else: currentpos = currentpos + remainder//2 # integer division count = count + 1 if (currentpos==1): break if (count==N): print(" ",N,end="") #good return

16. Numbers from 3 to 1000

17. Experimental computational mathematics
Generated list of the numbers that work up to 1000.
Took those numbers and put into Google….
Got to

18. What is reptend prime base 2?
Numbers P for which 2 is a primitive root modulo P
This is a procedure, also.
…. Number P for which 2 raised to the power N, N going from 0 to P-2, produces the numbers 1 to P-1, modulo P

19. What is mod
Two numbers are equal mod X if the difference is a multiple of X.
M = N mod P if and only if (this is the definition) M-N is a multiple of P. Putting it another way: M = N + a*P where a is an integer
This sometimes is called clock arithmetic. For clock arithmetic, the P is 12.

20. Let's try 11
20 is 1 = 1 mod is 2 = 2 mod is 4 = 4 mod is 8 = 8 mod is 16 = 5 mod = 5 + 1*11 25 is 32 = 10 mod = *11 26 is 64 = 9 mod 11 ? 27 is 128 = 7 mod is 256 = 3 mod is 512 = 6 mod 11

21. Interesting….
Not just that 11 works: reptend process hits all the intermediate numbers AND
the numbers appear to be the numbers in the dollar bill rosette procedure going backwards…..

22. Plan
To prove that the numbers that can work using the folding procedures are the reptend primes base 2, one needs to prove that the numbers for which the folding procedure hits all the intermediate numbers are the same as the numbers for which the reptend process, raising 2 to powers from 0 to number -1 and then taking mod, hits all the intermediate numbers. Instead, I decided to try for a stronger result: the two procedures are the same procedure, though in opposite order.

23. Reverse folding process
Need to determine: if F and P-F go to G and P-G, starting at G, how to get back to F
Consider cases of if F is (was) odd or even
The pair G and P-G came from F and P-F either by halving F or halving P-F, so Consider 2*G. If 2*G > P, then need to consider 2*(P-G)
If 2*G > P, then 2*(P-G) <P. Why?
Both ways "work". I use the second way.
Two numbers adding up to P. They are different (note P is odd so they can't be equal) so one must be less than ½ P and the other must be more.

24. Stronger result
the two procedures are the same procedure, though need to take folding in reverse order.
Need to prove that for k, the first number of the M, P-M pair by (reverse) folding is 2k mod P Note: both processes can continue past P-2.
Proof by induction
Initial case
Induction case
Assume true for k, prove for k+1

25. Initial case
Reptend and reverse folding start out with 1 Reptend: 20 =1, so 20 =1 mod P Reverse folding starts with 1.

26. Induction step
Can assume G = 2k mod P which means G = 2k + a*P where a is an integer
What is the next case for reverse folding?
Two cases: 2*G< P and 2*G > P. Case 1: 2*G< P. So F = 2*G. Expand F F=2*(2k + a*P) = 2k+1 + 2*a*P so F = 2k+1 mod P

27. Continuing with 2nd case
Case 2: 2*G>P. So F and P-F produced next step by halving the P-F term. So the P-F term is equal to 2*(P-G). F = P-2*(P-G) F = P – 2*P + 2*G rearranging terms F = 2*G – P now substitute G F = 2*(2k + a*P) – P F = 2k+1 + 2*a*P – P F = 2k+1 + (2*a-1)*P F = 2k+1 mod P

28. Conclusion of proof
The reptend process of raising 2 to a power and then calculating mod P produces the same numbers as the reverse folding process.
Because both processes yield the same results at each step, both either satisfy criteria of hitting all the intermediate points between 1 and P-1 or neither does.
Note: the fact that folding resembled reptend for 11 and that the numbers under 1000 matches are not proofs.
What I meant by a stronger result

29. Recap Origami inspired adventure folding improving approximations
programming
online investigation
proof

30. References
My books frequently contain origami-related material
Programming 101 (Processing), Apress, June, 2018
Chapter 6: Origami flower; Chapter 8 (extra) Directions.
HTML5 and JavaScript Projects, Apress, 2nd Edition scheduled for publication very soon
Chapter 7: Using HTML & JavaScript for Origami Directions