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A129268 Slowest increasing sequence: the sum of three consecutive terms shares no digit with any of the summands. 1
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 (list; graph; refs; listen; history; text; internal format)
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 A330220 A039169
KEYWORD
base,easy,fini,nonn
AUTHOR
Eric Angelini, May 25 2007
STATUS
approved

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Last modified April 24 10:11 EDT 2024. Contains 371935 sequences. (Running on oeis4.)