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Le Tripalin se présente ici . Voici les solutions des palindromes du septième et dernier fichier ( Boutique fermée ).  #119 CASTOR + TROT + SAC #120 REVUE + JEU + VER #121 RUE + OSSÈTES + S Œ UR #122 LESSIVE + LEVI'S + SEL #123 RU + ELFE + FLEUR #124 ÉCU + PELLE + PUCE #125 ZÉNANA + NANA + NEZ #126 AIL + ERRO + BORRELIA #127 AILE + BOL + LOBELIA #128 AS + LABO + BALSA #129 DÉ + ÎLE + LIED #130 BOLÉRO + TORE + LOB #131 CIPPE + CEP + PIC #132 LAC + ARA + CARACAL #133 CÂBLE + GEL + BAC #134 SAPEUR + GRUE + PAS #135 TUF + FANON + AFFÛT #136 AILE + MAC + CAMÉLIA #137 COR + CÈNE + CROC #138 FIN + ACE + CANIF #139 CABRI + TIR + BAC #140 LAC + OBÈSE + BOCAL #141 ABSIDE + LED + ISBA #142 OUD + SAS + DUO #143 LAVE + SORT + ROSEVAL #144 ÂGE + MOËLLE + OMEGA #145 CABRIOLE + LOIR + BAC #146 CIRCÉE + FÉE + CRIC #147 COIN + AIDE + MANIOC #148 CRAPS + ANANAS + PARC #149 SERF + FOC + COFFRES #150 RIS + IOLE + LOISIR #151 RÔT + AIDE + MÉDIATOR #152 SERRE + TRAPÈZE + PARTERRES #153 SEL + CRÉ...

We start S with a(1) = 1 and a(2) = 2 . S is always extended with the integer k by obeying these two simple rules: 1/ if the last two terms of S have the same length, they are concatenated to form k ; 2/ else the longest term is divided by the shortest one and k is the integer part. Example As the last two terms (1 and 2) of S have the same length, they are concatenated to form a(3): S = 1, 2, 12 , … As the last two terms of S have not the same length, we divide the longest one by the shortest one (this is 12/2 = 6) and extend S with 6: S = 1, 2, 12, 6 , … As the last two terms of S have not the same length, we make 12/6 = 2 and extend S accordingly: S = 1, 2, 12, 6, 2 , … As the last two terms (6 and 2) of S have the same length, we extend S with the concatenation: S = 1, 2, 12, 6, 2, 62 , … As the last two terms of S have not the same length, we make 62/2 = 31 and extend S: S = 1, 2, 12, 6, 2, 62, 31 , … As the last two terms (62 and 31) of S have the same length, we extend S ...

  https://www.centrepompidou.fr/fr/programme/agenda/evenement/9htHbj4 Gotlib , Rubrique-à-brac tome 5 , 1974  Art Spiegelman , Comics as a Medium for Self-Expression , 1981 Gir/ Moebius , Starwatcher , 1986 Marc-Antoine Mathieu ,  Julius Corentin Acquefacques, tome 4 : L’Épaisseur du miroir, 1995 Yokoyama Yuichi , Travaux publics (Doboku), vers 2005 Winshluss , Pinocchio , 2008 Eric Lambé , La Saison des vendanges , 2016 Magritte , Souvenir de voyage , 1926 Gabriella Giandelli , Prenazi , 2013 Christian Schad , Graf St. Genois d ’ Anneaucourt (Comte St-Genois d ’ Anneaucourt), 1927 Joann Sfar , Pacsin , 2005 Jules Pacsin , La belle Anglaise , 1916 Anna Sommer , Barbara , 2020 Francis Picabia , Dresseur d ’a nimaux , 1923 Philippe Dupuy , Mon papa dessine des femmes nues , 2020 Robert Combas , Egyptian Reggae , 1979 Chris Ware , Jimmy Corrigan , 1988 Theo Van Doesburg , Composition X , 1918 Roy Lichtenstein , Knock, Knock Poster , 1975 Winsor McCay , Little Nemo in Slumb...

L ’ A venturier , Roland Topor The palindromes 66 and 101, for instance, are balanced numbers : if you put them on a graduated ruler which is placed on a fulcrum, the ruler will remain horizontal. But there are balanced numbers which are not palindromes – like 102 or 2024: Why are 102 and 2024 balanced numbers? B ecause the distance which separates a digit from the  fulcrum  comes here into play: the more this distance increases, the heavier the digit in question is. We see above that the digit 1 of  1 02 weighs 2 in reality (weight*distance =  1 *2 = 2 ). This quantity balances the influence of the 2 of 10 2  (for which weight*distance =  2 *1 = 2 too). We have the same kind of balance with the number 2024 if we put the fulcrum exactly under the rightmost 2. I am 99% sure that those balanced numbers are in the OEIS — but I could not find them: where are they? And, a last little puzzle : can you find the first four consecutive balanced numbers >...