- N is an integer
- The value of N > 2 and is always even
- The maximum unique twist value is at max(T) = floor((N+2)/4)
- Since negative twists have the same surface counts as positive twists we only need evaluate abs(T)
- Let us use the rule: if 2N mod abs(2T)-1 = 0 then CS = abs(T) - 1
- We need to show that CS > 0 for all cases when N/2 is odd and N > 2
- For N/2 is odd and N > 2, max(T) will be the sequence (2, 3, 4, 5, 6, ...)
- If T is set to the value max(T), using the rule, CS will be the sequence (1, 2 ,3, 4, 5, ...)
- Therefore there is one value of T for each N/2 is odd and N > 2 where CS > 0
- Let us use the rule: if 2N mod abs(2T)-1 = 0 then CS = abs(T) - 1
- We need to show that CS > 0 for all cases when N/2 is even and N > 2, except for the powers of 2
- Let L = max(T)
- Let C = ceil(L/2)
- if abs(T) > C then T = L - abs(T) + 1 (Surface Count Reflection Property)
- Because of this property only values of T up to ceil(max(T)/2) need be considered
- For N/2 is even and N > 2, max(T) will be the sequence (1, 2, 3, 4, 5, 6, ...)
- For N/2 is even and N > 2, ceil(max(T)/2) will be the sequence (1, 1, 2, 2, 3, 3, ...)
- Divide the problem into two cases, when N/4 is odd, and when N/4 is even
- When N/4 is odd, (4, 12, 20, 28, 36, ...), ceil(max(T)/2) will be the sequence (1, 2, 3, 4, 5, ...)
- If T is set to the value ceil(max(T)/2), using the rule, CS will be the sequence (0, 1 ,2, 3, 4, ...)
**CS = 0 when N = 4, a power of 2**- When N/4 is even, (8, 16, 24, 32, 40, ...), ceil(max(T)/2) will be the sequence (1, 2, 3, 4, 5, ...)
- Note in the rule for the sequence of T (1, 2, 3, 4, 5, ...), the value of abs(2T)-1 is the sequence (1, 3, 5 ,7, 9, ...), the odd integers.
- For the powers of 2, 2N will have only have 2 as its only factor
**Therefore 2N mod abs(2T)-1 != 0, for any T when N is a power of 2**- N is a positive integer
- N/4 is even, and N is not a power of 2, (24, 40, 48, 56, 72, 80, ...)
- The maximum of T at max(T) = floor((N+2)/4), (6, 10, 12, 14, 18, 20, ...)
- If T is is a value from 1 to ceil(max(T)/2), (3, 5, 6, 7, 9, 10, ...) then:
- Premise: There exists at least one case where 2N mod abs(2T)-1 = 0
- For N, 2N is the sequence (48, 80, 96, 112, 144, 160, ...)
- The term abs(2max(T))-1 is the sequence(5, 9, 11, 13, 17, 19, ...)
- Since 2N is not prime there will be at least one factor less than sqrt(2N)
- If sqrt(2N) <= abs(2max(T))-1 then all candidate factors are checked and at least one will be a factor
- Restated: If N <= square(abs(2max(T))-1)/2 then it is proven
- Replacing max(T), abs no longer needed since N is positive: If N <= square(2*floor((N+2)/4)-1)/2
- floor no longer needed after numerator multiplied by 2: If N <= square(((2N+4)/4)-1)/2
- Simplify: If N <= square(((N/2)+1)-1)/2
- Simplify: If N <= square(N/2)/2
- Simplify: If N <= (square(N)/4)/2
- Simplify: If N <= square(N)/8
**N <= square(N)/8 for any N >= 8**

Computational analysis indicates that Hybrid N-icons, for each value of N there is at least one twist resulting in more than one surface with the exception of when N is a power of 2. In the case of Hybrid N-icons when N is a power of 2, there is no twist position where there is more than the one implicit discontinuous surface. Can these statements be proven?

Given, for Hybrid N-icons:

The case when N/2 is odd, (6, 10, 14, 18, 22, ...):

The case when N/2 is even, (4, 8, 12, 16, 20, ...):

Still needed is to prove is that all other cases when N/4 is even have at lease one twist such that CS > 0

Question or comments about the web page should be directed to **polyhedra@bigfoot.com**.

The generation of OFF, VRML, and Live3D files was done with Antiprism. The Hedron application by Jim McNeill was used to generate VRML Switch files.

History:

2007-09-06 Initial Release

2007-10-19 Note for rule 2 deleted since it no longer applies

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