Cinquante signes

I've read a Quanta paper last week about tiling patterns and thought: "How could I tile a sequence?" The first idea was to use distinct integers whose digits would always sum up to 10 – the "tiles".  The hereunder sequence  A  is already in the OEIS , of course – and so are  A011557 (1), A052216 (2), A052217 (3), A052218 (4), A052219 (5), A052220 (6), A052221 (7), A052222 (8), A052223 (9), A166311 (11), A235151 (12), A143164 (13), A235225 (14), A235226 (15), A235227 (16), A166370 (17), A235228 (18), A166459 (19), A235229 (20) : A = 19, 28, 37, 46, 55, 64, 73, 82, 91, 109, 118, 127, 136, 145, 154, 163, 172, 181, 190, 208, 217, 226, 235, 244, 253, 262, 271, 280, 307, 316, 325, 334, 343, 352, 361, 370, 406, 415, 424, 433, 442, 451, 460, 505, 514, 523, 532, 541, 550, 604, 613, 622, 631, 640,... Then I thought, ok, what about distinct pairs of integers whose sum would always be 10?  B  is not in the OEIS: B = 0, 19 , 1, 9 , 2, 8 , 3, 7 , 4, 6...

(Anni Albers) Lexicographically earliest sequence S of distinct nonnegative integers such that 9 out of the last 10 digits of S always sum up to a square. S = 1000000000, 0, 1, 2, 3, 4, 6, 5, 7, 8, 9, 10, 11, 13, 12, 15, 14, 17, 16, 21, 18, 19, 20, 27, 22, 25, 23, 30, 24, 26, 31, 28, 29, 32, 34, 33, 36, 38, 35, 37, 40, 39, 41, 42, 43, 44, 47, 45, 50, 46, 51, 48, 52, 54, 55, 49, 57, 53, 56, 58, 59, 60, 64, 61, 62, 63 (terms computed by Michael S. Branicky , see the draft/seq here ) The smallest available 10-digit integer to start S with is a(1) = 1000000000. The sequence is infinite as there are infinitely many integers whose 9 of the last 10 digits sum up to a square. We start with a(1) = 1000000000: S = 1000000000, The last 10 digits of S are [ 100000000 0 ]; we discard a 0 and the sum of the 9 remaining digits = 1 (a square). We try to extend S with 0: S = 1000000000,0, The last 10 digits of S are [ 000000000 0 ]; we discard a 0 and the sum of the 9 remaining digits = 0 (a square). W...

Polygones binaires (sur des idées de Gilles Esposito-Farèse et Nicolas Graner ) Dans la série « Comment transformer un entier N en polygone » faisons appel aujourd’hui au système binaire. Premier cas Les entiers N dont la représentation binaire B ne comprend qu’un seul chiffre « 1 » sont représentés par un rectangle horizontal R de côté-unité 1. La longueur de ce rectangle (en côtés-unité) est donnée par le nombre de chiffres de  B . Ci-dessous, les nombres binaires   1,   10,   100,   1000   et   10000   avec les rectangles associés de longueur 1, 2, 3, 4 et 5 (les périmètres et surfaces de tous ces rectangles sont des nombres entiers) :     Deuxième cas Les entiers N dont la représentation binaire  B  comprend exactement deux chiffres « 1 » (suite A018900 de l’OEIS) sont représentés par deux rectangles  R  formant un  L majuscule. La case de coin du  L , commune aux deux rectangles, n ’ ap...

  The above submissions can be seen here and there . The hereunder three seqs ( not submitted yet to the OEIS because they are too difficult to calculate by hand ) deal with patterns of primes and nonprimes (terms & digits). Jean-Marc Falcoz was quick to help, check, correct , extend  A ,  B  and  C  with their respective nice graphs: A ] Terms as well as digits fit the nonprime/prime pattern; this is the lexicographically earliest injective sequence with this property [this is now A375326 ]: A ]  =  0,2,1,3,4,5,6,7,8,29,20,31,21,59,24,71,26,79,28,263,9,283,12,13,15,17,42,43,45,47,62,67,63,83,65,97,82,131,30,293,85,139,34,307,87,151,36,313,92,179,38,317,93,421,39,347,95,431,50,367,120,383,121,397,124,503,126,547,128,563,129,587,130,593,134,743,136,787,138,797,150,2029,51,2039,54,2129,56,2131,58,2179,70,2459,74,2621,76,2659,78,2671,202,439,203,479,205,631,207,659,212,821,213,829,215,839,217,859,242,929,243,971,245,1213...  B ] Terms as we...

  We will use the hereunder Digit wheel to illustrate our iteration idea of the day Say we start our iteration with « 12 ». To go (on the wheel) from 1 to 2 we have to travel a single arc which is produced by an angle of 36 degrees. We have thus 12 —> 36 (which is the result of 1 arc x 36 degrees) To go (on the wheel) from 3 to 6 we have to travel 3 arcs which are produced by an angle of 108 degrees. We have thus 12 —> 36 —> 108 (the result of 3 x 36 degrees) To go (on the wheel) from 1 to 0 we have to travel 9 arcs (we always travel clockwise) and to go from 0 to 8 we have to travel 8 arcs, for a total of 9 + 8 = 17 arcs or 612 degrees. We have thus 12 —> 36 —> 108 —> 612 (the result of 17 x 36 degrees) To go from 6 to 1 we have to travel 5 arcs and to go from 1 to 2 we have to travel 1 arc, for a total of 5 + 1 = 6 arcs or 216 degrees. We have thus 12 —> 36 —> 108 —> 612 —> 216 (the result of 6 x 36) To go from 2 to 1 (clockwise) we have to trav...