{"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":"<p>La formule de MCC est \\( L_{r,n} = r \\cdot 4^n + \\frac{4^n &#8211; 1}{3} \\)<\/p>\n\n\n[mathjax]\n<p>Voici une formule inline : \\\\( E = mc^2 \\\\)<\/p>\n<p>Et une formule en bloc : \\\n\n\\[ \\int_0^1 x^2 \\, dx \\\\]\n\n<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">\n  MCC \u2014 recentrage sur le 1er 0&nbsp;mod&nbsp;3 (MCC2), racines en 1re colonne, <u>tri par MCC2 croissant<\/u>\n<\/h3>\n\n<div class=\"wp-block-group has-pale-cyan-blue-background-color has-background\" style=\"border-left-color:#2563eb;border-left-width:4px;border-radius:6px;padding:.4rem .9rem\">\n  <p><strong>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 &#8211; 1}{3} \\\\). On place \\\\( L_{r,n_\\star} \\\\) en <strong>MCC2<\/strong>, et les colonnes adjacentes donnent \\\\( L_{r,n_\\star &#8211; 1} \\\\) (MCC1) et \\\\( L_{r,n_\\star &#8211; 2} \\\\) (MCC0) quand elles existent. <u>On n\u2019affiche que des lignes de racines<\/u> (1,3,7 mod 8) en 1re colonne. Les lignes sont <u>tri\u00e9es par la valeur de MCC2 croissante<\/u>.\n  <\/p>\n<\/div>\n\n\n<!-- ===================== CLASSIQUE (IMPAIRS) ===================== -->\n\n<h4 class=\"wp-block-heading\">MCC \u2014 Classique (impairs) \u2014 tri par MCC2<\/h4>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table>\n  <thead>\n    <tr><th>fratrie $latex r$<\/th><th>MCC0<\/th><th>MCC1<\/th><th style=\"background:#fef3c7\">MCC2 (1er 0 mod 3)<\/th><th>MCC3<\/th><th>MCC4<\/th><th>MCC5<\/th><\/tr>\n  <\/thead>\n  <tbody>\n    <!-- n* = 0 : MCC2 = r ; MCC0\/MCC1 vides -->\n    <tr><td>$latex 3$<\/td><td><\/td><td><\/td><td style=\"background:#fef3c7\"><strong>3<\/strong><\/td><td>13<\/td><td>53<\/td><td>213<\/td><\/tr>\n    <tr><td>$latex 9$<\/td><td><\/td><td><\/td><td style=\"background:#fef3c7\"><strong>9<\/strong><\/td><td>37<\/td><td>149<\/td><td>597<\/td><\/tr>\n    <tr><td>$latex 15$<\/td><td><\/td><td><\/td><td style=\"background:#fef3c7\"><strong>15<\/strong><\/td><td>61<\/td><td>245<\/td><td>981<\/td><\/tr>\n\n    <!-- n* = 2 : MCC0 = r ; MCC1 = L_{r,1} ; MCC2 = L_{r,2} -->\n    <tr><td>$latex 1$<\/td><td><strong>1<\/strong><\/td><td>5<\/td><td style=\"background:#fef3c7\">21<\/td><td>85<\/td><td>341<\/td><td>1365<\/td><\/tr>\n\n    <!-- n* = 0 -->\n    <tr><td>$latex 27$<\/td><td><\/td><td><\/td><td style=\"background:#fef3c7\"><strong>27<\/strong><\/td><td>109<\/td><td>437<\/td><td>1749<\/td><\/tr>\n    <!-- *** AJOUTS *** -->\n    <tr><td>$latex 33$<\/td><td><\/td><td><\/td><td style=\"background:#fef3c7\"><strong>33<\/strong><\/td><td>133<\/td><td>533<\/td><td>2133<\/td><\/tr>\n    <tr><td>$latex 39$<\/td><td><\/td><td><\/td><td style=\"background:#fef3c7\"><strong>39<\/strong><\/td><td>157<\/td><td>629<\/td><td>2517<\/td><\/tr>\n    <!-- *** FIN AJOUTS *** -->\n\n    <!-- n* = 1 : MCC1 = r ; MCC2 = L_{r,1} -->\n    <tr><td>$latex 11$<\/td><td><\/td><td><strong>11<\/strong><\/td><td style=\"background:#fef3c7\">45<\/td><td>181<\/td><td>725<\/td><td>2901<\/td><\/tr>\n    <tr><td>$latex 17$<\/td><td><\/td><td><strong>17<\/strong><\/td><td style=\"background:#fef3c7\">69<\/td><td>277<\/td><td>1109<\/td><td>4437<\/td><\/tr>\n\n    <!-- n* = 2 -->\n    <tr><td>$latex 7$<\/td><td><strong>7<\/strong><\/td><td>29<\/td><td style=\"background:#fef3c7\">117<\/td><td>469<\/td><td>1877<\/td><td>7509<\/td><\/tr>\n  <\/tbody>\n<\/table><\/figure>\n\n\n<!-- ===================== VARIANTE W (PAIRS) ===================== -->\n\n<h4 class=\"wp-block-heading\">MCC \u2014 Variante W (pairs) \u2014 tri par MCC2 (valeurs \u00d72)<\/h4>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table>\n  <thead>\n    <tr><th>fratrie $latex 2r$<\/th><th>MCC0<\/th><th>MCC1<\/th><th style=\"background:#ecfeff\">MCC2<\/th><th>MCC3<\/th><th>MCC4<\/th><th>MCC5<\/th><\/tr>\n  <\/thead>\n  <tbody>\n    <tr><td>$latex 6$<\/td><td><\/td><td><\/td><td style=\"background:#ecfeff\"><strong>6<\/strong><\/td><td>26<\/td><td>106<\/td><td>426<\/td><\/tr>\n    <tr><td>$latex 18$<\/td><td><\/td><td><\/td><td style=\"background:#ecfeff\"><strong>18<\/strong><\/td><td>74<\/td><td>298<\/td><td>1194<\/td><\/tr>\n    <tr><td>$latex 30$<\/td><td><\/td><td><\/td><td style=\"background:#ecfeff\"><strong>30<\/strong><\/td><td>122<\/td><td>490<\/td><td>1962<\/td><\/tr>\n\n    <tr><td>$latex 2$<\/td><td><strong>2<\/strong><\/td><td>10<\/td><td style=\"background:#ecfeff\">42<\/td><td>170<\/td><td>682<\/td><td>2730<\/td><\/tr>\n\n    <tr><td>$latex 54$<\/td><td><\/td><td><\/td><td style=\"background:#ecfeff\"><strong>54<\/strong><\/td><td>218<\/td><td>874<\/td><td>3498<\/td><\/tr>\n    <!-- *** AJOUTS (doublage de 33 et 39) *** -->\n    <tr><td>$latex 66$<\/td><td><\/td><td><\/td><td style=\"background:#ecfeff\"><strong>66<\/strong><\/td><td>266<\/td><td>1066<\/td><td>4266<\/td><\/tr>\n    <tr><td>$latex 78$<\/td><td><\/td><td><\/td><td style=\"background:#ecfeff\"><strong>78<\/strong><\/td><td>314<\/td><td>1258<\/td><td>5034<\/td><\/tr>\n    <!-- *** FIN AJOUTS *** -->\n\n    <tr><td>$latex 22$<\/td><td><\/td><td><strong>22<\/strong><\/td><td style=\"background:#ecfeff\">90<\/td><td>362<\/td><td>1450<\/td><td>5802<\/td><\/tr>\n    <tr><td>$latex 34$<\/td><td><\/td><td><strong>34<\/strong><\/td><td style=\"background:#ecfeff\">138<\/td><td>554<\/td><td>2218<\/td><td>8874<\/td><\/tr>\n    <tr><td>$latex 14$<\/td><td><strong>14<\/strong><\/td><td>58<\/td><td style=\"background:#ecfeff\">234<\/td><td>938<\/td><td>3754<\/td><td>15018<\/td><\/tr>\n  <\/tbody>\n<\/table><\/figure>\n\n\n\n<!-- ===================== DIAGONALES MCC \u2014 FORMULES R\u00c9VIS\u00c9ES ===================== -->\n\n<h4 class=\"wp-block-heading\">Diagonales MCC \u2014 formules r\u00e9vis\u00e9es (avec tri par MCC2)<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n  <li><strong>Indexation par la colonne MCC.<\/strong> Pour une ligne $latex r$ donn\u00e9e, posons $latex n_\\star(r)$ son 1er z\u00e9ro mod&nbsp;3. En notant $latex c\\in\\{0,1,2,\\dots\\}$ l\u2019indice de colonne MCC (MCC0\/MCC1\/MCC2\/MCC3\u2026 = 0\/1\/2\/3\u2026), on a la formule <em>locale<\/em> :\n    <div style=\"text-align:center;margin:.4rem 0\">$latex \\displaystyle \\mathrm{MCC}[r,c]\\;=\\;L_{r,\\;n_\\star(r)+c-2}\\,.$<\/div>\n    (La cellule existe ssi $latex n_\\star(r)+c-2\\ge0$.)<\/li>\n\n  <li><strong>Diagonale NE <em>sur une ligne<\/em>.<\/strong> Passer de $latex c$ \u00e0 $latex c{+}1$ :\n    <div style=\"text-align:center;margin:.4rem 0\">$latex \\mathrm{MCC}[r,c{+}1]=4\\,\\mathrm{MCC}[r,c]+1$.<\/div>\n    <em>Valuation :<\/em> $latex \\nu_2\\!\\bigl(3\\,\\mathrm{MCC}[r,c{+}1]+1\\bigr)=\\nu_2\\!\\bigl(3\\,\\mathrm{MCC}[r,c]+1\\bigr)+2$ (et c\u00f4t\u00e9 W : $latex \\nu_2(3x{+}2)-1$ subit le m\u00eame $latex +2$).<\/li>\n\n  <li><strong>P\u00e9riodicit\u00e9 $latex \\bmod 3$ (ancr\u00e9e en MCC2).<\/strong> Comme $latex \\mathrm{MCC}[r,2]\\equiv0\\ (\\bmod 3)$ et $latex \\mathrm{MCC}[r,c{+}1]\\equiv\\mathrm{MCC}[r,c]+1\\ (\\bmod 3)$, on a :\n    <div style=\"text-align:center;margin:.4rem 0\"><u>$latex \\mathrm{MCC}[r,c]\\equiv0\\ (\\bmod 3)$ ssi $latex c\\equiv2\\ (\\bmod 3)$<\/u> (z\u00e9ros tous les 3 pas).<\/div><\/li>\n\n  <li><strong>Barri\u00e8re $latex \\bmod 64$ (classique).<\/strong> Pour $latex c\\ge5$ (i.e. $latex c{-}2\\ge3$), $latex 4^{\\,c-2}\\equiv0\\ (\\bmod 64)$ et\n    <div style=\"text-align:center;margin:.4rem 0\">$latex \\displaystyle \\mathrm{MCC}[r,c]\\equiv\\frac{4^{\\,c-2}-1}{3}\\equiv\\frac{-1}{3}\\equiv21\\pmod{64}$,<\/div>\n    car $latex 3^{-1}\\equiv43\\ (\\bmod 64)$ et $latex -43\\equiv21$.<\/li>\n\n  <li><strong>Barri\u00e8re W $latex \\bmod 128$.<\/strong> $latex \\mathrm{MCC}_W[r,c]=2\\,\\mathrm{MCC}[r,c]$, donc pour $latex c\\ge5$ :\n    <div style=\"text-align:center;margin:.4rem 0\">$latex \\mathrm{MCC}_W[r,c]\\equiv 42\\pmod{128}$.<\/div><\/li>\n\n  <li><strong>Effet du tri par MCC2 (propri\u00e9t\u00e9 de monotonie verticale).<\/strong> Comme $latex \\mathrm{MCC}[r,c]=4^{\\,c-2}\\,\\mathrm{MCC}[r,2]+\\tfrac{4^{\\,c-2}-1}{3}$ est croissante en $latex \\mathrm{MCC}[r,2]$, trier les lignes par MCC2 rend chaque <em>colonne<\/em> (c fix\u00e9e) croissante de haut en bas. Les formules NE\/SE ci-dessus restent <em>identiques<\/em> (elles sont locales \u00e0 la ligne), mais la lecture verticale est maintenant ordonn\u00e9e.<\/li>\n\n  <li><strong>SE (inverse local).<\/strong> Quand d\u00e9fini : $latex \\mathrm{MCC}[r,c{-}1]=\\dfrac{\\mathrm{MCC}[r,c]-1}{4}$ (et c\u00f4t\u00e9 W : idem avec les valeurs doubl\u00e9es).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Exemples guid\u00e9s \u2014 calcul des diagonales en MCC<\/h3>\n\n\n\n<p><strong>R\u00e8gle locale (rappel).<\/strong> Pour une fratrie racine $latex r\\equiv1,3,7\\ (\\mathrm{mod}\\ 8)$, note $latex n_\\star(r)$ le plus petit rang tel que $latex L_{r,n_\\star}\\equiv0\\ (\\mathrm{mod}\\ 3)$ avec $latex L_{r,n}=r4^n+\\frac{4^n-1}{3}$. En MCC : \n<span style=\"white-space:nowrap\">$latex \\mathrm{MCC}[r,2]=L_{r,n_\\star}$<\/span>, et pour toute colonne $latex c\\ge0$ :\n<\/p>\n\n\n\n<p class=\"has-text-align-center\">$latex \\displaystyle \\mathrm{MCC}[r,c]=L_{r,\\;n_\\star(r)+c-2}$ &nbsp;&nbsp; (la cellule existe ssi $latex n_\\star+c-2\\ge0$).<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n  <li><strong>Pas NE<\/strong> (une colonne vers la droite) : $latex \\mathrm{MCC}[r,c{+}1]=4\\,\\mathrm{MCC}[r,c]+1$.<\/li>\n  <li><strong>Pas SE<\/strong> (inverse, une colonne vers la gauche quand d\u00e9fini) : $latex \\mathrm{MCC}[r,c{-}1]=\\dfrac{\\mathrm{MCC}[r,c]-1}{4}$.<\/li>\n  <li><strong>Valuation<\/strong> : $latex k_c=\\nu_2\\!\\bigl(3\\,\\mathrm{MCC}[r,c]+1\\bigr)$ (et c\u00f4t\u00e9 variante W : $latex k&rsquo;_c=\\nu_2(3\\,\\mathrm{MCC}_W[r,c]+2)-1$). \u00c0 chaque pas NE, $latex k$ gagne syst\u00e9matiquement <strong>+2<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">Exemple 1 \u2014 racine $latex r=27$ (cas $latex r\\equiv0\\ (\\mathrm{mod}\\ 3)$, donc $latex n_\\star=0$)<\/h4>\n\n\n\n<p><strong>Point d\u2019ancrage.<\/strong> $latex \\mathrm{MCC}[27,2]=L_{27,0}=27$ (MCC0\/MCC1 vides).<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n  <li><em>NE vers MCC3<\/em> : $latex 4\\cdot 27+1=109$.<\/li>\n  <li><em>NE vers MCC4<\/em> : $latex 4\\cdot 109+1=437$.<\/li>\n  <li><em>Valuations<\/em> : $latex \\nu_2(3\\cdot27+1)=\\nu_2(82)=1$, puis $latex \\nu_2(3\\cdot109+1)=\\nu_2(328)=3$, puis $latex \\nu_2(3\\cdot437+1)=\\nu_2(1312)=5$ : on gagne bien +2 \u00e0 chaque pas.<\/li>\n<\/ul>\n\n\n\n<p><strong>Formule ferm\u00e9e (depuis MCC2).<\/strong> Pour $latex c\\ge2$, $latex \\mathrm{MCC}[27,c]=4^{c-2}\\cdot 27+\\dfrac{4^{c-2}-1}{3}$. Par ex. pour $latex c=5$ : $latex 4^3\\cdot27+\\frac{64-1}{3}=64\\cdot27+21=1749$.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">Exemple 2 \u2014 racine $latex r=11$ (cas $latex r\\equiv2\\ (\\mathrm{mod}\\ 3)$, donc $latex n_\\star=1$)<\/h4>\n\n\n\n<p><strong>Points d\u2019ancrage.<\/strong> $latex \\mathrm{MCC}[11,1]=L_{11,0}=11$, $latex \\mathrm{MCC}[11,2]=L_{11,1}=4\\cdot11+1=45$.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n  <li><em>NE (MCC2 \u2192 MCC3)<\/em> : $latex 4\\cdot45+1=181$.<\/li>\n  <li><em>NE (MCC3 \u2192 MCC4)<\/em> : $latex 4\\cdot181+1=725$.<\/li>\n  <li><em>Valuations<\/em> : $latex \\nu_2(3\\cdot11+1)=\\nu_2(34)=1$, puis $latex \\nu_2(3\\cdot45+1)=\\nu_2(136)=3$, puis $latex \\nu_2(3\\cdot181+1)=\\nu_2(544)=5$.<\/li>\n  <li><em>SE (inverse)<\/em> : on revient de MCC2 \u00e0 MCC1 par $latex (45-1)\/4=11$.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">Exemple 3 \u2014 racine $latex r=7$ (cas $latex r\\equiv1\\ (\\mathrm{mod}\\ 3)$, donc $latex n_\\star=2$)<\/h4>\n\n\n\n<p><strong>Points d\u2019ancrage.<\/strong> $latex \\mathrm{MCC}[7,2]=L_{7,2}=117$. Comme $latex n_\\star=2$, les colonnes de gauche existent : $latex \\mathrm{MCC}[7,1]=L_{7,1}=29$, $latex \\mathrm{MCC}[7,0]=L_{7,0}=7$.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n  <li><em>NE (MCC2 \u2192 MCC3)<\/em> : $latex 4\\cdot117+1=469$.<\/li>\n  <li><em>NE (MCC3 \u2192 MCC4)<\/em> : $latex 4\\cdot469+1=1877$.<\/li>\n  <li><em>Valuations<\/em> : $latex \\nu_2(3\\cdot117+1)=\\nu_2(352)=5$, puis $latex \\nu_2(3\\cdot469+1)=\\nu_2(1408)=7$.<\/li>\n  <li><em>SE (inverse)<\/em> : $latex \\mathrm{MCC}[7,1]=(117-1)\/4=29$, puis $latex \\mathrm{MCC}[7,0]=(29-1)\/4=7$.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">Exemple 4 \u2014 calcul direct par la formule ferm\u00e9e (depuis MCC2)<\/h4>\n\n\n\n<p>Pour toute ligne, si l\u2019on conna\u00eet $latex \\mathrm{MCC}[r,2]=M_2(r)$, alors pour $latex c\\ge2$ :<\/p>\n\n\n\n<p class=\"has-text-align-center\">$latex \\displaystyle \\mathrm{MCC}[r,c]=4^{\\,c-2}\\,M_2(r)\\;+\\;\\frac{4^{\\,c-2}-1}{3}\\,.$<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n  <li><em>Ex.<\/em> $latex r=1$ : $latex \\mathrm{MCC}[1,2]=21$. Pour $latex c=5$ : $latex 4^3\\cdot21+\\frac{64-1}{3}=64\\cdot21+21=1344+21=1365$ (c\u2019est bien MCC5 de la ligne $latex r=1$).<\/li>\n  <li><em>Ex.<\/em> $latex r=7$ : $latex \\mathrm{MCC}[7,2]=117$. Pour $latex c=5$ : $latex 4^3\\cdot117+\\frac{64-1}{3}=64\\cdot117+21=7488+21=7509$.<\/li>\n<\/ul>\n\n\n\n<div class=\"wp-block-group has-luminous-vivid-amber-background-color has-background\" style=\"border-left-color:#16a34a;border-left-width:4px;border-radius:4px;padding:.4rem .9rem;margin:.6rem 0 1rem\"><div class=\"wp-block-group__inner-container is-layout-flow wp-block-group-is-layout-flow\">\n  <p><strong>Cons\u00e9quences rapides.<\/strong> (i) <em>P\u00e9riodicit\u00e9 mod 3<\/em> : $latex \\mathrm{MCC}[r,c]\\equiv0$ ssi $latex c\\equiv2\\ (\\mathrm{mod}\\ 3)$ ; (ii) <em>barri\u00e8re mod 64<\/em> : pour $latex c\\ge5$, toutes les lignes tombent en $latex 21\\ (\\mathrm{mod}\\ 64)$ ; (iii) <em>valutions<\/em> : \u00e0 chaque pas NE, $latex k$ augmente de $latex +2$.<\/p>\n<\/div><\/div>\n\n\n\n<h4 class=\"wp-block-heading\">Variante W (pairs) \u2014 m\u00eame diagonale, valeurs \u00d72<\/h4>\n\n\n\n<p>On double toutes les valeurs : $latex \\mathrm{MCC}_W[r,c]=2\\,\\mathrm{MCC}[r,c]$ et $latex k&rsquo;_c=\\nu_2(3\\,\\mathrm{MCC}_W[r,c]+2)-1=\\nu_2(3\\,\\mathrm{MCC}[r,c]+1)=k_c$.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n  <li><em>Ex.<\/em> Ligne $latex r=1$ : $latex \\mathrm{MCC}[1,2]=21 \\mapsto \\mathrm{MCC}_W[2,2]=42$. Valuation : $latex \\nu_2(3\\cdot42+2)-1=\\nu_2(128)-1=7-1=6$, identique \u00e0 $latex \\nu_2(3\\cdot21+1)=6$.<\/li>\n  <li><em>Ex.<\/em> Puis NE : $latex 42\\mapsto 4\\cdot42+2=170$ ; $latex \\nu_2(3\\cdot170+2)-1=\\nu_2(512)-1=9-1=8$ (on a bien gagn\u00e9 +2, comme c\u00f4t\u00e9 impair).<\/li>\n<\/ul>\n\n\n\n <h3 class=\"wp-block-heading\">Fratries &nbsp;L<sub>r,n<\/sub> (n = 0\u20266)<\/h3>   <table class=\"wp-block-table is-style-regular\"> <thead> <tr> <th>r \\ n<\/th> <th>0<\/th><th>1<\/th><th>2<\/th><th>3<\/th><th>4<\/th><th>5<\/th><th>6<\/th> <\/tr> <\/thead> <tbody> <tr> <td><strong>r = 1<\/strong><\/td> <td>1&nbsp;<em>(C)<\/em><\/td> <td>5&nbsp;<strong>(P)<\/strong><\/td> <td>21&nbsp;<em>(C)<\/em><\/td> <td>85&nbsp;<em>(C)<\/em><\/td> <td>341&nbsp;<em>(C)<\/em><\/td> <td>1365&nbsp;<em>(C)<\/em><\/td> <td>5461&nbsp;<em>(C)<\/em><\/td> <\/tr> <tr> <td><strong>r = 3<\/strong><\/td> <td>3&nbsp;<strong>(P)<\/strong><\/td> <td>13&nbsp;<strong>(P)<\/strong><\/td> <td>53&nbsp;<strong>(P)<\/strong><\/td> <td>213&nbsp;<em>(C)<\/em><\/td> <td>853&nbsp;<strong>(P)<\/strong><\/td> <td>3413&nbsp;<strong>(P)<\/strong><\/td> <td>13653&nbsp;<em>(C)<\/em><\/td> <\/tr> <tr> <td><strong>r = 7<\/strong><\/td> <td>7&nbsp;<strong>(P)<\/strong><\/td> <td>29&nbsp;<strong>(P)<\/strong><\/td> <td>117&nbsp;<em>(C)<\/em><\/td> <td>469&nbsp;<em>(C)<\/em><\/td> <td>1877&nbsp;<strong>(P)<\/strong><\/td> <td>7509&nbsp;<em>(C)<\/em><\/td> <td>30037&nbsp;<em>(C)<\/em><\/td> <\/tr> <\/tbody> <\/table>   <h3 class=\"wp-block-heading\">Image en 2 pas : G<sub>2<\/sub>(L<sub>r,n<\/sub>)<\/h3>   <p>Pour lecture rapide : P &rarr; \\((3y+1)\/2\\), C &rarr; \\(9y+2\\). Les valeurs ci-dessous sont des <em>entiers<\/em> (pairs ou impairs) atteints en <strong>deux pas<\/strong> depuis chaque cellule du tableau ci-dessus.<\/p>   <table class=\"wp-block-table is-style-regular\"> <thead> <tr> <th>r \\ n<\/th> <th>0<\/th><th>1<\/th><th>2<\/th><th>3<\/th><th>4<\/th><th>5<\/th><th>6<\/th> <\/tr> <\/thead> <tbody> <tr> <td><strong>r = 1<\/strong><\/td> <td>11<\/td><td>8<\/td><td>191<\/td><td>767<\/td><td>3071<\/td><td>12287<\/td><td>49151<\/td> <\/tr> <tr> <td><strong>r = 3<\/strong><\/td> <td>5<\/td><td>20<\/td><td>80<\/td><td>1919<\/td><td>1280<\/td><td>5120<\/td><td>122879<\/td> <\/tr> <tr> <td><strong>r = 7<\/strong><\/td> <td>11<\/td><td>44<\/td><td>1055<\/td><td>4223<\/td><td>2816<\/td><td>67583<\/td><td>270335<\/td> <\/tr> <\/tbody> <\/table> \n","protected":false},"excerpt":{"rendered":"<p>La formule de MCC est \\( L_{r,n} = r \\cdot 4^n + \\frac{4^n &#8211; 1}{3} \\) [mathjax] 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&nbsp;mod&nbsp;3 (MCC2), racines en 1re colonne, tri par MCC2 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"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}]}}