A129268 Slowest increasing sequence: the sum of three consecutive terms shares no digit with any of the summands.

0, 1, 2, 3, 4, 5, 7, 8, 9, 10, 14, 15, 31, 32, 33, 34, 35, 37, 38, 39, 40, 43, 44, 63, 64, 65, 68, 69, 73, 76, 79, 80, 83, 86, 88, 96, 116, 118, 119, 120, 124, 125, 128, 140, 267, 426, 440, 445, 446, 447, 460, 474, 604, 733, 774, 775, 777, 778, 779, 785, 797, 818, 819

Offset

0,3

Comments

The sequence is finite and has 112 terms. Max Alekseyev (who computed the sequence together with Peter Pein) proved that 175414854, 415748410, and 1631058958 are the last three terms: (Quoting Max Alekseyev) Suppose that the next term is x, then the sum s = 415748410 + 1631058958 + x = 2046807368 + x may contain only decimal digit 2. Therefore all solutions are given by the formula x(k) = 2*(10^k-1)/9 - 2046807368 where k>=10. It is easy to see that while x(10)=175414854 is smaller than 1631058958 (hence it cannot be an element of our sequence), all other x(k) contain a decimal digit 2 which is not allowed: x(11) = 20175414854, x(12) = 220175414854, x(13) = 2220175414854, ... Therefore there is no next term in this sequence. QED. (End of quote)

Links

Eric Angelini, May 25 2007, Table of n, a(n) for n = 0..111

Crossrefs

Sequence in context: A289342 A286302 A299767 * A271317 A374759 A330220

Adjacent sequences: A129265 A129266 A129267 * A129269 A129270 A129271

Keyword

base,easy,fini,full,nonn

Author

Eric Angelini, May 25 2007

Status

approved