Cinquante signes

  ( Because it's there ) ( El Cap par Honnold ici )

We would like to extend the recent idea expressed below (and  here ): NAME Lexicographically earliest sequence of distinct terms > 0 such that the concatenation of the rightmost digit of a(n) and the leftmost digit of a(n+1) forms a prime number. The rightmost digit of a(n) cannot be 0. DATA 1, 3, 7, 9, 71, 11, 12, 31, 13, 14, 15, 32, 33, 16, 17, 18, 34, 19, 72, 35, 36, 73, 74, 37, 38, 39, 75, 91, 76, 77, 92, 93, 78, 94, 79, 701, 95, 96, 101, 97, 98, 99, 702, 301, 102, 302, 303, 103, 104, 105, 304, 106, 107, 108, ... EXAMPLE a(1) = 1 and a(2) = 3 form 13, a prime number; a(2) = 3 and a(3) = 7 form 37, a prime number; a(3) = 7 and a(4) = 9 form 79, a prime number; a(4) = 9 and the leftmost digit of a(5) = 71 form 97, a prime number; a(5) = 71 and its rightmost digit, concatenated to the leftmost digit of a(6) = 11, form 11, a prime number; etc. _____________ What we will call now a "prime weld" is a link K between two integers X and Y  (say X =  2023  and Y = 1951)...

Hello Nicolas , Si tu es sur une plage, ne mets ni ton grain de sel ici, ni tes grains de sable ! Cool, vacances ! Sinon, regarde cette idée : La suite S est composée de termes N qui sont la concaténation de trois entiers : A, B et C. A est la position de N dans S ; B est la quantité de chiffres pairs présents dans N ; C est la quantité de chiffres impairs dans N. Ainsi N = 10323 affirme-t-il trois choses : – que N est le 103 e terme de la suite S, – que N comporte 2 chiffres pairs (oui, le 0 et le 2), – que N comporte 3 chiffres impairs (oui, le 1, le 3 central et le 3 final). Ma question  : Ai-je raison d’affirmer que la lecture de N, quelle que soit la quantité de chiffres qu’il contienne, n’est jamais ambiguë ? Qu’il y a donc toujours moyen de reconstituer A, B et C, pour tout N ? Voici S (modulo erreurs dues à mon travail à la main) : « Lexico-première suite de termes N distincts formés par la concaténation...

 ... en mai 2023

NAME Lexicographically earliest sequence of distinct terms > 0 such that the number of odd digits present in each term is described by another term, or itself if possible (see how in the Comments section). DATA 20, 40, 10, 60, 31, 80, 52, 200, 71, 92, 101, 112, 122, 131, 143, 152, 162, 171, 183, 220, 192, 240, 212, 260, 231, 280, 252, 400, 271, 292, 301, 312, 322, 331, 343, 352, 362, 371, 383, 420, 392, 440, 412, 460, 431, 480, 452, 600, 471, 492, 501, 512, 522, 531, 543, 552, … COMMENTS All integers N of the sequence are the concatenation of two parts, X and Y.  X is the position of N in the sequence; Y is the number of odd digits present in N.  The term a(5) = 31 means that the 3rd term of the sequence contains 1 odd digit: this is true as a(3) = 10.  The first term a(1) = 20 means that the 2nd term contains 0 odd digit: this is true as a(2) = 40. Etc.  ____________________ Possible variants of this sequence might be: NAME =  Lexicographically earlies...

Take the integer K = 418918636212793150 for instance. This is an unrestricted pandigital number. Is K the smallest of its kind?  The said kind being that if you replace each digit of K by its triple, K reemerges in the result R: If we forget about the pandigital restriction, we can list all such "triple-my-successive-digits" reemerging integers. I guess the list begins with: K =  0, 5, 15, 50, 150, 315, 500, ... (not in the OEIS). ____________________ Sunday, August 13th  update Gilles  E.-F. has extended the above sequence K (and computed much more nice things): K =  0, 5, 15, 50, 150, 315, 500, 1500, 3150, 5000, 9315, 15000, 31500, 50000, 79315, 93150, 150000, 279315, 315000, 500000, 793150, 931500, ... The substitution hereunder of each digit d by its sevenfold would produce the sequence L: L = 0, 5, 35, 50, 135, 350, 500, 1350, 2135, 3500, 5000, 13500, 21350, 35000, 50000, 72135, 135000, 213500, 350000, 472135, 500000, 721350, ... And if d --> ...

Hi Jon (...) your last remark about the Collatz name made me think: «Perhaps Jon and Alois are right, the Collatz name shouldn’t be present in my  submission – as the Redux rules and the original Collatz problem are indeed a bit different»…  I then said to myself «Well, it’s too late now to change what you’ve written in your blog, Eric, or to cancel your submission to the OEIS – but remember the lesson for the next time!» I then thought – «What could be a  rule   closer to the original Collatz problem – slighly different but interesting and new?» The idea of the Ztalloc rule popped up quickly. Orbit of 53 under the Ztalloc rule, which asks to divide N by 2 if N is even and to compute (N + 1)*3 if N is odd. Z = 53-162-81-246-123-372-186-93-282-141-426-213-642-321-966-483-1452-726-363-1092-546-273-822-411-1236-618-309-930-465-1398-699-2100-1050-525-1578-789-2370-1185-3558-1779-5340-2670-1335-4008-2004-1002-501-1506-753-2262-1131-3396-1698-849-2550-1275-3828-1914-9...