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A129268
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Slowest increasing sequence: the sum of three consecutive terms shares no digit with any of the summands.
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1
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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; internal format)
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OFFSET
| 0,3
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COMMENTS
| The sequence is finite and has 112 terms. Max Alekseyev (who computed the sequence together with Peter Pein) proved that 175414854, 415748410 and1631058958 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)
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LINKS
| Eric Angelini (eric.angelini(AT)kntv.be), May 25 2007, Table of n, a(n) for n = 0..111
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CROSSREFS
| Sequence in context: A175020 A050728 A180590 * A039169 A156068 A039263
Adjacent sequences: A129265 A129266 A129267 * A129269 A129270 A129271
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KEYWORD
| base,easy,fini,nonn
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AUTHOR
| Eric Angelini (eric.angelini(AT)kntv.be), May 25 2007
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