(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.1' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 59554, 1234]*) (*NotebookOutlinePosition[ 60243, 1258]*) (* CellTagsIndexPosition[ 60199, 1254]*) (*WindowFrame->Normal*) Notebook[{ Cell["Proof of a conjecture about class III ME", "Title"], Cell[TextData[{ StyleBox["Abstract", FontVariations->{"Underline"->True}], " -\nIt 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 ", Cell[BoxData[ \(TraditionalForm\`c\)]], " (not prime) pc-universe greater than 12 without instances of class III \ ME. We prove this is indeed impossible. 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number ", Cell[BoxData[ \(TraditionalForm\`p, \ p < \(\(k\)\(-\)\(1\)\(\ \)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`p\)]], " is not a factor of ", Cell[BoxData[ \(TraditionalForm\`k\)]], "." }], "Section"], Cell[TextData[{ "Note this is ", StyleBox["false", FontWeight->"Bold"], " for ", Cell[BoxData[ \(TraditionalForm\`c = 12\)]], " : even for ", Cell[BoxData[ \(TraditionalForm\`k = 6\)]], " all numbers < 5 have a common factor with ", Cell[BoxData[ \(TraditionalForm\`k\)]], "." }], "Text"], Cell[TextData[{ "Proof : it is known since Bertrand and Tchebitchiev (who proved this using \ 17 lemmas) that there is always a prime number between ", Cell[BoxData[ \(TraditionalForm\`n\)]], " and ", Cell[BoxData[ \(TraditionalForm\`2 n\)]], " for ", Cell[BoxData[ \(TraditionalForm\`n \[GreaterEqual] 2\)]], ".\nLet ", Cell[BoxData[ \(TraditionalForm\`k\)]], " be the greatest divisor of ", Cell[BoxData[ \(TraditionalForm\`c\ \((k \[NotEqual] c)\)\)]], ": we have either\n\[Bullet] ", Cell[BoxData[ \(TraditionalForm\`k = 2 n + 1\)]], " or \n\[Bullet] ", Cell[BoxData[ \(TraditionalForm\`k = 2 n + 2\)]], ". \nFor ", Cell[BoxData[ \(TraditionalForm\`c > 24\)]], " and ", Cell[BoxData[ \(TraditionalForm\`c\)]], " composite we can claim ", Cell[BoxData[ \(TraditionalForm\`k \[GreaterEqual] 5\)]], " and hence ", Cell[BoxData[ \(TraditionalForm\`n \[GreaterEqual] 2\)]], " (see by hand for ", Cell[BoxData[ \(TraditionalForm\`c = 14, \ 15, \ 16\)]], "\[Ellipsis]24). We take our prime ", Cell[BoxData[ \(TraditionalForm\`\(\(\(p\)\(\[Element]\)\)\(]\)\) n, 2 \( \(n\)\([\)\)\)]], " or more precisely ", Cell[BoxData[ \(TraditionalForm\`p \[Element] \([n + 1, 2 n - 1]\)\)]], ": then ", Cell[BoxData[ \(TraditionalForm\`k/2 < p < k - 1\)]], "." }], "Exercise"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Lemma 2 : there is a ME of cardinality ", Cell[BoxData[ \(TraditionalForm\`p\)]], " in ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\_k\)]], "." }], "Section"], Cell[TextData[{ "Proof: this is well known, a way to produce it is to take ", Cell[BoxData[ \(TraditionalForm\`{0, p, 2 p, \[Ellipsis]\ \((p - 1)\) p}\)]], " modulo ", Cell[BoxData[ \(TraditionalForm\`k\)]], ". It is generated (G) and its cardinality ", Cell[BoxData[ \(TraditionalForm\`p\)]], " being coprime with the modulo ", Cell[BoxData[ \(TraditionalForm\`k\)]], " it is not a LTM (Limited Transposition Mode): hence it is class I." }], "Exercise"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Theorem :\nFor ", Cell[BoxData[ \(TraditionalForm\`c > 12\)]], ", there exists a ME of class III in ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalZ]\_c\)]], ", with cardinality ", Cell[BoxData[ \(TraditionalForm\`d = \ p\ c\/k\)]], "with notations as above. It is the set ", Cell[BoxData[ \(TraditionalForm\`{\[LeftFloor]\(k\ n\)\/p\[RightFloor]\ | \ k = 0\ \[Ellipsis]\ d - 1}\)]], "." }], "Section"], Cell[CellGroupData[{ Cell[TextData[{ "Proof: We produce this ME by double generation, concatenating ", Cell[BoxData[ \(TraditionalForm\`c\/k\)]], " copies of the ME provided by the last Lemma. Another way to put it is to \ exhibit the generating function" }], "Exercise"], Cell[BoxData[ \(TraditionalForm\`J : n \[Rule] \[LeftFloor]\(n\ c\)\/\(\(c\ p\)\/k\)\[RightFloor] = \ \[LeftFloor]\(k\ n\)\/p\[RightFloor]\)], "DisplayFormula"] }, Open ]], Cell[TextData[{ "Clearly, the first ", Cell[BoxData[ \(TraditionalForm\`p\)]], " images are spread between 0 and ", Cell[BoxData[ \(TraditionalForm\`k - 1\)]], "; equally clearly, the set is ", Cell[BoxData[ \(TraditionalForm\`k\)]], " periodic as ", Cell[BoxData[ \(TraditionalForm\`J(n + p)\ = \ J(n) + k\)]], ".\nThese generating functions provide ME (Clough & , thm 1.5).\nIt only \ remains to verify this is a class III set.\n\[MathematicaIcon] it has \ transpositional symmetry (", Cell[BoxData[ \(TraditionalForm\`k\)]], "-periodic).\n\[MathematicaIcon] we must check that ", Cell[BoxData[ \(TraditionalForm\`Gcd(c, d) \[NotEqual] c - d\)]], ". Notice ", Cell[BoxData[ \(TraditionalForm\`\(\(d\)\(=\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`\(c\/k\) \(\(p\)\(.\)\)\)]], " There are two cases to consider, as ", Cell[BoxData[ \(TraditionalForm\`c\/k\)]], " obviously divides ", Cell[BoxData[ \(TraditionalForm\`c\)]], ":\n\t\[SixPointedStar] the gcd is ", Cell[BoxData[ \(TraditionalForm\`c\/k\)]], ". But \n\t", Cell[BoxData[ \(TraditionalForm\`c\/k = \(\(c\)\(-\)\)\)]], Cell[BoxData[ \(TraditionalForm\`\(c\/k\) p\)]], " means ", Cell[BoxData[ \(TraditionalForm\`p = k - 1\)]], ", which is forbidden by construction.\n\t\[SixPointedStar] the gcd is ", Cell[BoxData[ \(TraditionalForm\`\(c\/k\) p\)]], ". 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