{"id":56428,"date":"2025-10-02T14:30:18","date_gmt":"2025-10-02T13:30:18","guid":{"rendered":"https:\/\/4.exstyle.fr\/le-blog-photo\/?p=56428"},"modified":"2025-10-04T13:26:58","modified_gmt":"2025-10-04T12:26:58","slug":"fr-mcc","status":"publish","type":"post","link":"https:\/\/4.exstyle.fr\/le-blog-photo\/fr-mcc\/","title":{"rendered":"FR > decouverte variante (cas simple)"},"content":{"rendered":"

La formule de MCC est \\( L_{r,n} = r \\cdot 4^n + \\frac{4^n – 1}{3} \\)<\/p>\n\n\n\n

Voici une formule inline : \\\\( E = mc^2 \\\\)<\/p>\n

Et une formule en bloc : \\\n\n\\[ \\int_0^1 x^2 \\, dx \\\\]\n\n<\/p>\n\n\n\n

\n MCC \u2014 recentrage sur le 1er 0 mod 3 (MCC2), racines en 1re colonne, tri par MCC2 croissant<\/u>\n<\/h3>\n\n
\n

R\u00e8gle MCC.<\/strong> Pour une fratrie \\\\( r \\\\) (racine minimale \\\\( r \\equiv 1,3,7 \\pmod{8} \\\\)), soit \\\\( n_\\star \\\\) le plus petit rang avec \\\\( L_{r,n_\\star} \\equiv 0 \\pmod{3} \\\\), o\u00f9 \\\\( L_{r,n} = r \\cdot 4^n + \\frac{4^n – 1}{3} \\\\). On place \\\\( L_{r,n_\\star} \\\\) en MCC2<\/strong>, et les colonnes adjacentes donnent \\\\( L_{r,n_\\star – 1} \\\\) (MCC1) et \\\\( L_{r,n_\\star – 2} \\\\) (MCC0) quand elles existent. On n\u2019affiche que des lignes de racines<\/u> (1,3,7 mod 8) en 1re colonne. Les lignes sont tri\u00e9es par la valeur de MCC2 croissante<\/u>.\n <\/p>\n<\/div>\n\n\n\n\n

MCC \u2014 Classique (impairs) \u2014 tri par MCC2<\/h4>\n\n\n\n
\n \n \n \n
fratrie \\(r\\)<\/th>MCC0<\/th>MCC1<\/th>MCC2 (1er 0 mod 3)<\/th>MCC3<\/th>MCC4<\/th>MCC5<\/th><\/tr>\n <\/thead>\n
\\(3\\)<\/td><\/td><\/td>3<\/strong><\/td>13<\/td>53<\/td>213<\/td><\/tr>\n
\\(9\\)<\/td><\/td><\/td>9<\/strong><\/td>37<\/td>149<\/td>597<\/td><\/tr>\n
\\(15\\)<\/td><\/td><\/td>15<\/strong><\/td>61<\/td>245<\/td>981<\/td><\/tr>\n\n \n
\\(1\\)<\/td>1<\/strong><\/td>5<\/td>21<\/td>85<\/td>341<\/td>1365<\/td><\/tr>\n\n \n
\\(27\\)<\/td><\/td><\/td>27<\/strong><\/td>109<\/td>437<\/td>1749<\/td><\/tr>\n \n
\\(33\\)<\/td><\/td><\/td>33<\/strong><\/td>133<\/td>533<\/td>2133<\/td><\/tr>\n
\\(39\\)<\/td><\/td><\/td>39<\/strong><\/td>157<\/td>629<\/td>2517<\/td><\/tr>\n \n\n \n
\\(11\\)<\/td><\/td>11<\/strong><\/td>45<\/td>181<\/td>725<\/td>2901<\/td><\/tr>\n
\\(17\\)<\/td><\/td>17<\/strong><\/td>69<\/td>277<\/td>1109<\/td>4437<\/td><\/tr>\n\n \n
\\(7\\)<\/td>7<\/strong><\/td>29<\/td>117<\/td>469<\/td>1877<\/td>7509<\/td><\/tr>\n <\/tbody>\n<\/table><\/figure>\n\n\n\n\n

MCC \u2014 Variante W (pairs) \u2014 tri par MCC2 (valeurs \u00d72)<\/h4>\n\n\n\n
\n \n \n
fratrie \\(2r\\)<\/th>MCC0<\/th>MCC1<\/th>MCC2<\/th>MCC3<\/th>MCC4<\/th>MCC5<\/th><\/tr>\n <\/thead>\n
\\(6\\)<\/td><\/td><\/td>6<\/strong><\/td>26<\/td>106<\/td>426<\/td><\/tr>\n
\\(18\\)<\/td><\/td><\/td>18<\/strong><\/td>74<\/td>298<\/td>1194<\/td><\/tr>\n
\\(30\\)<\/td><\/td><\/td>30<\/strong><\/td>122<\/td>490<\/td>1962<\/td><\/tr>\n\n
\\(2\\)<\/td>2<\/strong><\/td>10<\/td>42<\/td>170<\/td>682<\/td>2730<\/td><\/tr>\n\n
\\(54\\)<\/td><\/td><\/td>54<\/strong><\/td>218<\/td>874<\/td>3498<\/td><\/tr>\n \n
\\(66\\)<\/td><\/td><\/td>66<\/strong><\/td>266<\/td>1066<\/td>4266<\/td><\/tr>\n
\\(78\\)<\/td><\/td><\/td>78<\/strong><\/td>314<\/td>1258<\/td>5034<\/td><\/tr>\n \n\n
\\(22\\)<\/td><\/td>22<\/strong><\/td>90<\/td>362<\/td>1450<\/td>5802<\/td><\/tr>\n
\\(34\\)<\/td><\/td>34<\/strong><\/td>138<\/td>554<\/td>2218<\/td>8874<\/td><\/tr>\n
\\(14\\)<\/td>14<\/strong><\/td>58<\/td>234<\/td>938<\/td>3754<\/td>15018<\/td><\/tr>\n <\/tbody>\n<\/table><\/figure>\n\n\n\n\n\n

Diagonales MCC \u2014 formules r\u00e9vis\u00e9es (avec tri par MCC2)<\/h4>\n\n\n\n
    \n
  • Indexation par la colonne MCC.<\/strong> Pour une ligne \\(r\\) donn\u00e9e, posons \\(n_\\star(r)\\) son 1er z\u00e9ro mod 3. En notant \\(c\\in\\{0,1,2,\\dots\\}\\) l\u2019indice de colonne MCC (MCC0\/MCC1\/MCC2\/MCC3\u2026 = 0\/1\/2\/3\u2026), on a la formule locale<\/em> :\n
    \\(\\displaystyle \\mathrm{MCC}[r,c]\\;=\\;L_{r,\\;n_\\star(r)+c-2}\\,.\\)<\/div>\n (La cellule existe ssi \\(n_\\star(r)+c-2\\ge0\\).)<\/li>\n\n
  • Diagonale NE sur une ligne<\/em>.<\/strong> Passer de \\(c\\) \u00e0 \\(c{+}1\\) :\n
    \\(\\mathrm{MCC}[r,c{+}1]=4\\,\\mathrm{MCC}[r,c]+1\\).<\/div>\n Valuation :<\/em> \\(\\nu_2\\!\\bigl(3\\,\\mathrm{MCC}[r,c{+}1]+1\\bigr)=\\nu_2\\!\\bigl(3\\,\\mathrm{MCC}[r,c]+1\\bigr)+2\\) (et c\u00f4t\u00e9 W : \\(\\nu_2(3x{+}2)-1\\) subit le m\u00eame \\(+2\\)).<\/li>\n\n
  • P\u00e9riodicit\u00e9 \\(\\bmod 3\\) (ancr\u00e9e en MCC2).<\/strong> Comme \\(\\mathrm{MCC}[r,2]\\equiv0\\ (\\bmod 3)\\) et \\(\\mathrm{MCC}[r,c{+}1]\\equiv\\mathrm{MCC}[r,c]+1\\ (\\bmod 3)\\), on a :\n
    \\(\\mathrm{MCC}[r,c]\\equiv0\\ (\\bmod 3)\\) ssi \\(c\\equiv2\\ (\\bmod 3)\\)<\/u> (z\u00e9ros tous les 3 pas).<\/div><\/li>\n\n
  • Barri\u00e8re \\(\\bmod 64\\) (classique).<\/strong> Pour \\(c\\ge5\\) (i.e. \\(c{-}2\\ge3\\)), \\(4^{\\,c-2}\\equiv0\\ (\\bmod 64)\\) et\n
    \\(\\displaystyle \\mathrm{MCC}[r,c]\\equiv\\frac{4^{\\,c-2}-1}{3}\\equiv\\frac{-1}{3}\\equiv21\\pmod{64}\\),<\/div>\n car \\(3^{-1}\\equiv43\\ (\\bmod 64)\\) et \\(-43\\equiv21\\).<\/li>\n\n
  • Barri\u00e8re W \\(\\bmod 128\\).<\/strong> \\(\\mathrm{MCC}_W[r,c]=2\\,\\mathrm{MCC}[r,c]\\), donc pour \\(c\\ge5\\) :\n
    \\(\\mathrm{MCC}_W[r,c]\\equiv 42\\pmod{128}\\).<\/div><\/li>\n\n
  • Effet du tri par MCC2 (propri\u00e9t\u00e9 de monotonie verticale).<\/strong> Comme \\(\\mathrm{MCC}[r,c]=4^{\\,c-2}\\,\\mathrm{MCC}[r,2]+\\tfrac{4^{\\,c-2}-1}{3}\\) est croissante en \\(\\mathrm{MCC}[r,2]\\), trier les lignes par MCC2 rend chaque colonne<\/em> (c fix\u00e9e) croissante de haut en bas. Les formules NE\/SE ci-dessus restent identiques<\/em> (elles sont locales \u00e0 la ligne), mais la lecture verticale est maintenant ordonn\u00e9e.<\/li>\n\n
  • SE (inverse local).<\/strong> Quand d\u00e9fini : \\(\\mathrm{MCC}[r,c{-}1]=\\dfrac{\\mathrm{MCC}[r,c]-1}{4}\\) (et c\u00f4t\u00e9 W : idem avec les valeurs doubl\u00e9es).<\/li>\n<\/ul>\n\n\n\n

    Exemples guid\u00e9s \u2014 calcul des diagonales en MCC<\/h3>\n\n\n\n

    R\u00e8gle locale (rappel).<\/strong> Pour une fratrie racine \\(r\\equiv1,3,7\\ (\\mathrm{mod}\\ 8)\\), note \\(n_\\star(r)\\) le plus petit rang tel que \\(L_{r,n_\\star}\\equiv0\\ (\\mathrm{mod}\\ 3)\\) avec \\(L_{r,n}=r4^n+\\frac{4^n-1}{3}\\). En MCC : \n\\(\\mathrm{MCC}[r,2]=L_{r,n_\\star}\\)<\/span>, et pour toute colonne \\(c\\ge0\\) :\n<\/p>\n\n\n\n

    \\(\\displaystyle \\mathrm{MCC}[r,c]=L_{r,\\;n_\\star(r)+c-2}\\)    (la cellule existe ssi \\(n_\\star+c-2\\ge0\\)).<\/p>\n\n\n\n

      \n
    • Pas NE<\/strong> (une colonne vers la droite) : \\(\\mathrm{MCC}[r,c{+}1]=4\\,\\mathrm{MCC}[r,c]+1\\).<\/li>\n
    • Pas SE<\/strong> (inverse, une colonne vers la gauche quand d\u00e9fini) : \\(\\mathrm{MCC}[r,c{-}1]=\\dfrac{\\mathrm{MCC}[r,c]-1}{4}\\).<\/li>\n
    • Valuation<\/strong> : \\(k_c=\\nu_2\\!\\bigl(3\\,\\mathrm{MCC}[r,c]+1\\bigr)\\) (et c\u00f4t\u00e9 variante W : \\(k’_c=\\nu_2(3\\,\\mathrm{MCC}_W[r,c]+2)-1\\)). \u00c0 chaque pas NE, \\(k\\) gagne syst\u00e9matiquement +2<\/strong>.<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Exemple 1 \u2014 racine \\(r=27\\) (cas \\(r\\equiv0\\ (\\mathrm{mod}\\ 3)\\), donc \\(n_\\star=0\\))<\/h4>\n\n\n\n

      Point d\u2019ancrage.<\/strong> \\(\\mathrm{MCC}[27,2]=L_{27,0}=27\\) (MCC0\/MCC1 vides).<\/p>\n\n\n\n

        \n
      • NE vers MCC3<\/em> : \\(4\\cdot 27+1=109\\).<\/li>\n
      • NE vers MCC4<\/em> : \\(4\\cdot 109+1=437\\).<\/li>\n
      • Valuations<\/em> : \\(\\nu_2(3\\cdot27+1)=\\nu_2(82)=1\\), puis \\(\\nu_2(3\\cdot109+1)=\\nu_2(328)=3\\), puis \\(\\nu_2(3\\cdot437+1)=\\nu_2(1312)=5\\) : on gagne bien +2 \u00e0 chaque pas.<\/li>\n<\/ul>\n\n\n\n

        Formule ferm\u00e9e (depuis MCC2).<\/strong> Pour \\(c\\ge2\\), \\(\\mathrm{MCC}[27,c]=4^{c-2}\\cdot 27+\\dfrac{4^{c-2}-1}{3}\\). Par ex. pour \\(c=5\\) : \\(4^3\\cdot27+\\frac{64-1}{3}=64\\cdot27+21=1749\\).<\/p>\n\n\n\n

        \n\n\n\n

        Exemple 2 \u2014 racine \\(r=11\\) (cas \\(r\\equiv2\\ (\\mathrm{mod}\\ 3)\\), donc \\(n_\\star=1\\))<\/h4>\n\n\n\n

        Points d\u2019ancrage.<\/strong> \\(\\mathrm{MCC}[11,1]=L_{11,0}=11\\), \\(\\mathrm{MCC}[11,2]=L_{11,1}=4\\cdot11+1=45\\).<\/p>\n\n\n\n

          \n
        • NE (MCC2 \u2192 MCC3)<\/em> : \\(4\\cdot45+1=181\\).<\/li>\n
        • NE (MCC3 \u2192 MCC4)<\/em> : \\(4\\cdot181+1=725\\).<\/li>\n
        • Valuations<\/em> : \\(\\nu_2(3\\cdot11+1)=\\nu_2(34)=1\\), puis \\(\\nu_2(3\\cdot45+1)=\\nu_2(136)=3\\), puis \\(\\nu_2(3\\cdot181+1)=\\nu_2(544)=5\\).<\/li>\n
        • SE (inverse)<\/em> : on revient de MCC2 \u00e0 MCC1 par \\((45-1)\/4=11\\).<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          Exemple 3 \u2014 racine \\(r=7\\) (cas \\(r\\equiv1\\ (\\mathrm{mod}\\ 3)\\), donc \\(n_\\star=2\\))<\/h4>\n\n\n\n

          Points d\u2019ancrage.<\/strong> \\(\\mathrm{MCC}[7,2]=L_{7,2}=117\\). Comme \\(n_\\star=2\\), les colonnes de gauche existent : \\(\\mathrm{MCC}[7,1]=L_{7,1}=29\\), \\(\\mathrm{MCC}[7,0]=L_{7,0}=7\\).<\/p>\n\n\n\n

            \n
          • NE (MCC2 \u2192 MCC3)<\/em> : \\(4\\cdot117+1=469\\).<\/li>\n
          • NE (MCC3 \u2192 MCC4)<\/em> : \\(4\\cdot469+1=1877\\).<\/li>\n
          • Valuations<\/em> : \\(\\nu_2(3\\cdot117+1)=\\nu_2(352)=5\\), puis \\(\\nu_2(3\\cdot469+1)=\\nu_2(1408)=7\\).<\/li>\n
          • SE (inverse)<\/em> : \\(\\mathrm{MCC}[7,1]=(117-1)\/4=29\\), puis \\(\\mathrm{MCC}[7,0]=(29-1)\/4=7\\).<\/li>\n<\/ul>\n\n\n\n
            \n\n\n\n

            Exemple 4 \u2014 calcul direct par la formule ferm\u00e9e (depuis MCC2)<\/h4>\n\n\n\n

            Pour toute ligne, si l\u2019on conna\u00eet \\(\\mathrm{MCC}[r,2]=M_2(r)\\), alors pour \\(c\\ge2\\) :<\/p>\n\n\n\n

            \\(\\displaystyle \\mathrm{MCC}[r,c]=4^{\\,c-2}\\,M_2(r)\\;+\\;\\frac{4^{\\,c-2}-1}{3}\\,.\\)<\/p>\n\n\n\n

              \n
            • Ex.<\/em> \\(r=1\\) : \\(\\mathrm{MCC}[1,2]=21\\). Pour \\(c=5\\) : \\(4^3\\cdot21+\\frac{64-1}{3}=64\\cdot21+21=1344+21=1365\\) (c\u2019est bien MCC5 de la ligne \\(r=1\\)).<\/li>\n
            • Ex.<\/em> \\(r=7\\) : \\(\\mathrm{MCC}[7,2]=117\\). Pour \\(c=5\\) : \\(4^3\\cdot117+\\frac{64-1}{3}=64\\cdot117+21=7488+21=7509\\).<\/li>\n<\/ul>\n\n\n\n
              \n

              Cons\u00e9quences rapides.<\/strong> (i) P\u00e9riodicit\u00e9 mod 3<\/em> : \\(\\mathrm{MCC}[r,c]\\equiv0\\) ssi \\(c\\equiv2\\ (\\mathrm{mod}\\ 3)\\) ; (ii) barri\u00e8re mod 64<\/em> : pour \\(c\\ge5\\), toutes les lignes tombent en \\(21\\ (\\mathrm{mod}\\ 64)\\) ; (iii) valutions<\/em> : \u00e0 chaque pas NE, \\(k\\) augmente de \\(+2\\).<\/p>\n<\/div><\/div>\n\n\n\n

              Variante W (pairs) \u2014 m\u00eame diagonale, valeurs \u00d72<\/h4>\n\n\n\n

              On double toutes les valeurs : \\(\\mathrm{MCC}_W[r,c]=2\\,\\mathrm{MCC}[r,c]\\) et \\(k’_c=\\nu_2(3\\,\\mathrm{MCC}_W[r,c]+2)-1=\\nu_2(3\\,\\mathrm{MCC}[r,c]+1)=k_c\\).<\/p>\n\n\n\n

                \n
              • Ex.<\/em> Ligne \\(r=1\\) : \\(\\mathrm{MCC}[1,2]=21 \\mapsto \\mathrm{MCC}_W[2,2]=42\\). Valuation : \\(\\nu_2(3\\cdot42+2)-1=\\nu_2(128)-1=7-1=6\\), identique \u00e0 \\(\\nu_2(3\\cdot21+1)=6\\).<\/li>\n
              • Ex.<\/em> Puis NE : \\(42\\mapsto 4\\cdot42+2=170\\) ; \\(\\nu_2(3\\cdot170+2)-1=\\nu_2(512)-1=9-1=8\\) (on a bien gagn\u00e9 +2, comme c\u00f4t\u00e9 impair).<\/li>\n<\/ul>\n\n\n\n

                Fratries  Lr,n<\/sub> (n = 0\u20266)<\/h3>
                r \\ n<\/th> 0<\/th>1<\/th>2<\/th>3<\/th>4<\/th>5<\/th>6<\/th> <\/tr> <\/thead>
                r = 1<\/strong><\/td> 1 (C)<\/em><\/td> 5 (P)<\/strong><\/td> 21 (C)<\/em><\/td> 85 (C)<\/em><\/td> 341 (C)<\/em><\/td> 1365 (C)<\/em><\/td> 5461 (C)<\/em><\/td> <\/tr>
                r = 3<\/strong><\/td> 3 (P)<\/strong><\/td> 13 (P)<\/strong><\/td> 53 (P)<\/strong><\/td> 213 (C)<\/em><\/td> 853 (P)<\/strong><\/td> 3413 (P)<\/strong><\/td> 13653 (C)<\/em><\/td> <\/tr>
                r = 7<\/strong><\/td> 7 (P)<\/strong><\/td> 29 (P)<\/strong><\/td> 117 (C)<\/em><\/td> 469 (C)<\/em><\/td> 1877 (P)<\/strong><\/td> 7509 (C)<\/em><\/td> 30037 (C)<\/em><\/td> <\/tr> <\/tbody> <\/table>

                Image en 2 pas : G2<\/sub>(Lr,n<\/sub>)<\/h3>

                Pour lecture rapide : P → \\((3y+1)\/2\\), C → \\(9y+2\\). Les valeurs ci-dessous sont des entiers<\/em> (pairs ou impairs) atteints en deux pas<\/strong> depuis chaque cellule du tableau ci-dessus.<\/p>
                r \\ n<\/th> 0<\/th>1<\/th>2<\/th>3<\/th>4<\/th>5<\/th>6<\/th> <\/tr> <\/thead>
                r = 1<\/strong><\/td> 11<\/td>8<\/td>191<\/td>767<\/td>3071<\/td>12287<\/td>49151<\/td> <\/tr>
                r = 3<\/strong><\/td> 5<\/td>20<\/td>80<\/td>1919<\/td>1280<\/td>5120<\/td>122879<\/td> <\/tr>
                r = 7<\/strong><\/td> 11<\/td>44<\/td>1055<\/td>4223<\/td>2816<\/td>67583<\/td>270335<\/td> <\/tr> <\/tbody> <\/table> \n","protected":false},"excerpt":{"rendered":"

                La formule de MCC est \\( L_{r,n} = r \\cdot 4^n + \\frac{4^n – 1}{3} \\) Voici une formule inline : \\\\( E = mc^2 \\\\) Et une formule en bloc : \\ \\[ \\int_0^1 x^2 \\, dx \\\\] MCC \u2014 recentrage sur le 1er 0 mod 3 (MCC2), racines en 1re colonne, tri par MCC2 croissant […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"saved_in_kubio":false,"footnotes":""},"categories":[1],"tags":[436],"class_list":["post-56428","post","type-post","status-publish","format-standard","hentry","category-non-classe","tag-math"],"_links":{"self":[{"href":"https:\/\/4.exstyle.fr\/le-blog-photo\/wp-json\/wp\/v2\/posts\/56428","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/4.exstyle.fr\/le-blog-photo\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/4.exstyle.fr\/le-blog-photo\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/4.exstyle.fr\/le-blog-photo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/4.exstyle.fr\/le-blog-photo\/wp-json\/wp\/v2\/comments?post=56428"}],"version-history":[{"count":8,"href":"https:\/\/4.exstyle.fr\/le-blog-photo\/wp-json\/wp\/v2\/posts\/56428\/revisions"}],"predecessor-version":[{"id":56469,"href":"https:\/\/4.exstyle.fr\/le-blog-photo\/wp-json\/wp\/v2\/posts\/56428\/revisions\/56469"}],"wp:attachment":[{"href":"https:\/\/4.exstyle.fr\/le-blog-photo\/wp-json\/wp\/v2\/media?parent=56428"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/4.exstyle.fr\/le-blog-photo\/wp-json\/wp\/v2\/categories?post=56428"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/4.exstyle.fr\/le-blog-photo\/wp-json\/wp\/v2\/tags?post=56428"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}