Cinquante signes

Haie des  matheuses œdémateuses et des  matheux œdémateux aima Rébus

CULTURE MUSIQUES L’ eurodance en crise d’identité Hier très europhiles, les DJ italiens se rapprochent des franges nationalistes, à l’instar du fondateur du groupe Corona , Francesco Bontempi. Par Aureliano Tonet Publié aujourd’hui 30 mai 2020 dans le Monde à 00h25, mis à jour à 06h02 « Année bissextile, année funeste », dit un dicton italien. Le 29 février, histoire de conjurer le sort, Dan et Lucie organisent une fête dans leur appartement parisien. Il donne sur la rue des Couronnes, près de Belleville. Or couronne se dit « corona » dans la langue de Dante ; Dan, qui a effectué son Erasmus à Rome, le sait bien. Baptisée « Corona Sound System », la soirée fera sa fête à ce Covid-19 qui tape l’incruste sur le continent. Dernière surprise-partie avant l’apocalypse : « Le coronavirus sera fourni mais si vous apportez le vôtre, ce sera grandement apprécié ! », ironise l’invitation, sur Facebook. La playlist est raccord – le tube du groupe Corona, The Rhythm of the Night ,...

Hello Math-Fun , The idea, here, is to play with the notion of "digit-average". If we take the set {0,1,2,3,4,5,6,7,8,9}, make the sum and divide by the number of elements, we get 45/10 which gives 4.5 as the average value of a digit belonging to the set. What about re-inserting this value back in the set? We would then have the new set  {0,1,2,3,4,4,5,5,6,7,8,9} with sum 54 for 12 elements and "digit-average" 54/12 = 4.5 (the same as before). Re-inserting this value would produce the 3rd set  {0,1,2,3,4,4,4,5,5,5,6,7,8,9} with sum 63 for 14 elements and "digit-average" of 63/14 = 4.5 again. The successive sets show an obvious pattern. Instead of  {0,1,2,3,4,5,6,7,8,9}, let's start with  {1,3}. Sum is 4 with 2 digits; average = 2; new set is  {1,2,3}; sum is 6 with 3 digits; average = 2; new set is  {1,2,2,3}, etc. Similar pattern. But this gets weird sometimes. Start with  {1,10}. Sum is 2 for 3 digits; average = 0.66666666... ( CAVEA...

Hello Math-Fun , The idea here is to always extend the sequence E with the sum of the last two terms – and to apply the "eraser rule" when needed: E = 1, 2, 3, 5, 8, 13, 2 1 , 15, 1 7, ... The eraser rule says that the last term of E and the coming sum cannot share any digit. The duplicate digit (in yellow) are thus erased from the sum. (When needed, the remaining digits are concatenated to form a new integer). Example : We see that after 8 + 13 = 21 we must erase the 1 from 21 and proceed with 2; this 2 will be added to 13 to produce 15; but as 2 + 15 = 17 we will erase 1 again from 17 and extend E with 7; etc. The above start stops quickly: E = 1, 2, 3, 5, 8, 13, 2 1 , 15, 1 7, 22, 2 9, 31, 40, 71, 111 stop. What about leading zeros in the "erased" sum? Let's  see: E = 598, 414, 1 0 1 2,... Well, let's erase any leading zero – we will thus proceed here with: E = 598, 414,  2,  416, 41 8, 424, etc. Questions : What is the lexicograp...

Hello SeqFans, Here is what we call a "poor sandwich": say there is a pair of adjacent integers in S like [1951, 2020]. The sandwich is made of the rightmost digit of a(n), the leftmost digit of a(n+1) and, in between, the absolute difference of those two digits. The pair [1951, 2020] would then produce the sandwich 112 . (Why "poor"? Because a "rich" sandwich would insert the sum of the digits instead of their absolute difference – that is 132 ). Please note that the pair [2020, 1951] would produce the poor sandwich 011 (we keep the leading zero – these are sandwiches after all, not integers!-) Now we want S to be the lexicographically earliest sequence of distinct positive terms such that the successive sandwiches emerging from S rebuild S – digit after digit: S = 2, 1, 110, 10, 1101, 11010, 3, 330, 30, 3303, 33030, 4, 440, 40, 4404, 44040, 5, 550, 50, 5505, 55050, 6, 660, 60, 6606, 66060, 7, 770, 70, 7707, 77070, 8, 880, 80, 8808, 88080, 9, 990, 9...