Cinquante signes

  To define the surface S of a number N we form two substrings of digits b and h with a single cut through N; we then compute the product b * h (base*height = S).   The surface of 8 is zero (as b or h are missing). The surface of 10 is also zero (as b = 1 and h = 0, for instance). The surface of 11 is 1 (as 1*1 = 1). The surface of 39 is 27 (as 3*9 = 27). 2021 has two distinct surfaces, depending on where you cut:   the cut 202 / 1 will produce S = 202*1 = 202 and the cut 20 / 21 will produce S = 20*21 = 420 (no substring can start with a leading zero, except the substring 0 itself). 1234567890 has 9 distinct surfaces (9 possible cuts), etc.   The interesting part is that every number is the surface of at least two other numbers; 17 (a prime number) is one of the surfaces of 117 (1*17) and one of the surfaces of 171 (17*1); 2021 is one of the surfaces of 4347 (as 43*47 = 2021) but also  one of the surfaces   of 12021 and 20211 (as 2021 = 1*2021 = 20...

  This is the cover of Weird Science #22, pencils and ink by Wally Wood . But, wait, this #22 reminds us something! (HD image visible here – on the Wayback Machine ) (the full story of those 22 panels is  there ) (YouTube screen capture – the full 6'42'' are visible here ) (Many thanks to the wonderful Arnold Zwicky's blog )

  Squares for Scott Hi Scott, Our latest idea, inspired by your spirals (except we don’t spiral at all here !-)   We want to lexico-fill a quarter of an infinite board with distinct integers such that the [a(n)]^2 terms belonging to any square with a(n) in the top/left corner always sum up to a prime. We fill a(n)’s square line by line, top to bottom and left to right.   The first square is thus:   2  3   4  8 with prime sum 17 The next square has 3 in its upper/left corner:  2 . 3  5  6  4 . 8  7  9     10 11 12 with prime sum 71 The next square has 4 in its upper/left corner:  2  3  5  6 . 4  8  7  9 13 10 11 12 14 15 16 17 18 19 20 30 with prime sum 223 The next square has 5 in its upper/left corner:  2  3 . 5  6 21 22 23  4  8 . 7  9 24 25 26 13 10 11 12 27 28 29 14 15 16 17 31 32 33 18 19 20 30 34 35 40 with prime sum 563 (if I’m...

Suite des K- phénix On commence par un entier K > 9 (10 par exemple), on multiplie les deux chiffres les plus à gauche, on colle le résultat tout à droite derrière K, on efface le chiffre le plus à gauche, on itère (on trouvera dans l ’ OEIS le même protocole concernant l ’ addition — alors qu ’ il s ’ agit ici de multiplier à gauche). Nous nous intéressons aux K- phéni x qui réapparaissent dans la concaténation ainsi générée. La suite des K- phénix semble commencer par : K = 12, 21, 22, 23, 24, 26, 28, … (mais nous ne sommes sûrs de rien) Aucun K- phénix ne se termine en 0 ou 5 (mais 53 est dans la suite K car 53 ->5315->31 53 ). 10 100 1000 10000... K = 10 finit en point fixe sur 0 11 111 1111 11111... K = 11 finit en point fixe sur 1 12 1 2 2 12 2 4 122 4 8 1224 8 3 2 12248 3 2 24 122483 2 2 4 6 1224832 2 4 6 4 12248322 4 6 4 8 122483224 6 4 8 24 1224832246 4 8 24 24 12248322464 8 2 424 32 122483224648 2 4 243...