A360190 Starting from 1, successively take the smallest "Choix de Bruxelles" with factor 13 which is not already in the sequence.
1, 13, 133, 1333, 13333, 133333, 1333333, 125641, 1256413, 12564133, 1197241, 117481, 9037, 90391, 9031, 90313, 903133, 90241, 902413, 9024133, 90241333, 6941641, 693241, 6932413, 69324133, 6717241, 671557, 65557, 5557, 55591, 5431, 54313, 543133, 54241
Offset: 0
Examples
Below, square brackets [] represent multiplication by 13 (e.g., [4] = 52); curly brackets {} represent division by 13 (e.g., {26} = 2); digits outside the brackets are not affected by the multiplication or division (e.g., 1[3] = 139 and 1{169} = 113). We begin with 1 and, at each step, we go to the smallest number possible that hasn't yet appeared in the sequence: 1 --> [1] = 13 13 --> [1]3 = 133 133 --> [1]33 = 1333 1333 --> [1]333 = 13333 13333 --> [1]3333 = 133333 133333 --> [1]33333 = 1333333 1333333 --> 1{333333} = 125641 ... and so on.
Links
- Alon Vinkler, Table of n, a(n) for n = 0..6851
- Eric Angelini, Lars Blomberg, Charlie Neder, Remy Sigrist, and N. J. A. Sloane, "Choix de Bruxelles": A New Operation on Positive Integers, arXiv:1902.01444 [math.NT], Feb 2019; Fib. Quart. 57:3 (2019), 195-200.
- Eric Angelini, Lars Blomberg, Charlie Neder, Remy Sigrist, and N. J. A. Sloane,, "Choix de Bruxelles": A New Operation on Positive Integers, Local copy.
- Alon Vinkler, C# Program
Comments