Proof of a conjecture about class III ME

Abstract -
It appears that the number 12 is exceptional also in respect to the notion of class III Maximally Even sets introduced by Ian Quinn in his dissertation: nobody could find an instance of a c (not prime) pc-universe greater than 12 without instances of class III ME. We prove this is indeed impossible. The modulo 12 is yet again very special in this respect.

Lemma 1
For all composite c > 12 there exists a divisor k | c and a prime number p,  p < k-1  and p is not a factor of k.

Note this is false for c = 12 : even for k = 6 all numbers < 5 have a common factor with k.

Proof : it is known since Bertrand and Tchebitchiev (who proved this using 17 lemmas) that there is always a prime number between n and 2  n for n ≥ 2.
Let k be the greatest divisor of c (k ≠ c): we have either
• k = 2  n + 1 or 
• k = 2  n + 2. 
For c > 24 and c composite we can claim k ≥ 5 and hence n ≥ 2 (see by hand for c = 14,  15,  16…24). We take our prime p∈] n, 
      2  n[ or more precisely p ∈ [n + 1, 2  n - 1]: then k/2 < p < k - 1.

Lemma 2 : there is a ME of cardinality p in (Subscript[ℤ, k]).

Proof: this is well known, a way to produce it is to take {0, p, 2  p, … (p - 1) p} modulo k. It is generated (G) and its cardinality p being coprime with the modulo k it is not a LTM (Limited Transposition Mode): hence it is class I.

Theorem :
For c > 12, there exists a ME of class III in (Subscript[ℤ, c]), with cardinality d =  p (c/k)with notations as above. It is the set {⌊((k n)/p)⌋  |  k = 
          0 … d - 1}.

Proof: We produce this ME by double generation, concatenating (c/k) copies of the ME provided by the last Lemma. Another way to put it is to exhibit the generating function

\(TraditionalForm\`J : 
          n -> ⌊\(n\ c\)\/\(\(c\ p\)\/k\)⌋ = ⌊\(k\ n\)\/p⌋\)

Clearly, the first p images are spread between 0 and k - 1; equally clearly, the set is k periodic as J(n + p)  =  J(n) + k.
These generating functions provide ME (Clough & , thm 1.5).
It only remains to verify this is a class III set.
\[MathematicaIcon] it has transpositional symmetry (k-periodic).
\[MathematicaIcon] we must check that Gcd(c, d) ≠ c - d. Notice d= (c/k) p. There are two cases to consider, as (c/k) obviously divides c:
	✶ the gcd is (c/k). But 
	(c/k) = c-(c/k) p means p = k - 1, which is forbidden by construction.
	✶ the gcd is (c/k) p. But 
	(c/k) p = c-(c/k) p means p = k/2, which is forbidden by construction too.
This ends the proof.

Implementations

Class 3

findPK[
        c_] := 
Module[{k, p, 
          i = 1}, 
k  =  
          Max[Most[ Divisors[c]]]; 
While[
          Mod[k,  p = Prime[i]] == 
            0,  i++]; 
{k, 
          p}]

buildClassIIIME[
        c_] := 
Module[{k, p,  
          d}, 
{k, p} = 
          findPK[c]; 
d  =  (c/k) 
            p; 
Table[Floor[n k/p],  {n, 0,  d - 1}] // 
          Sort]

horloge[set_,  c_] := 
      Show[Graphics[{RGBColor[0.789502,  1,  
              0.413352], 
Disk[{0, 0},  
              1], 
PointSize[
              0.02], 
RGBColor[0.208896,  0.149676,  
              1], 
Map[
              Point[{Sin[2 (π/c) #],  Cos[2 (π/c) #]}] &,  
              Range[c]], 
RGBColor[0.841627,  0.0914931,  
              0.0617227], 
            PointSize[0.035], 
Circle[{0, 0}, 
              1], 
Map[
              Point[{Sin[2 (π/c) #],  Cos[2 (π/c) #]}] &,  
              set]} ],  
Background ->  
          RGBColor[0.204913,  0.0961624,  
            0.280537], 
AspectRatio -> 
          Automatic, 
PlotLabel -> 
          StyleForm["modulo " <> 
              ToString[c], 
FontFamily -> "Times", 
            FontSize -> 14, 
            FontWeight -> "Bold"]
]

With[{r = 1.5}, 
      Line[{{r, r}, {r, -r}, {-r, -r}, {-r, r}, {r, 
            r}}]], 
PlotRange -> {{-1.3,  
          1.3},  {-1.3,  1.3}},

horloge[buildClassIIIME[#],  #] &  @@  
        Select[Range[14,  60], Not[ PrimeQ[#]] &];

Derivation of all ME

Union[
      With[{c = 12}, 
Table[
            Floor[n c/d],  {d, c - 1},  {n, 0,  d - 1}] // 
          Sort]]

horloge[#,  12] &  /@  Rest@%;

Show[GraphicsArray[Partition[%,  5]]];

Notice for instance that collection (0 1 6 7) verifies
		(0 6)⊂(0 1 6 7) ⊂ (0 1 3 4 6 7 9 A) 
which pretty well describes its position among the ME species.

Derivation of all prototypes

prototype(c_, k_, d_):=Sort[Table[(((c n)/k) + ⌊(n/k)⌋) mod c,{n, 0, d - 1}]]

horloge[prototype[12, 4, 6],  12];

SetOptions[Graphics,  
        DisplayFunction -> Identity]; 
\(\(Table[
        horloge[prototype[12, a, \ b], \ 12], \ {b, 2, 
          12}, 
	{a, 6}];\)\)

Show[GraphicsArray[%],  
        DisplayFunction -> $DisplayFunction];
SetOptions[Graphics,  
        DisplayFunction -> $DisplayFunction];

Little movie

\(\(\(Do[
      If\ [Not[PrimeQ[n]], 
	
        horloge[buildClassIIIME[n], \ n]], \ {n, 6, 45}]\)\(\ \)\)\)