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Search: id:A078188
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| A078188 |
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a(1) = 1; for n > 1 take the digits of a(n-1) one after the other and remove one occurrence of this digit from the digits of m = k*n (k > 0) if possible; a(n) is the smallest multiple m of n which has the least number of remaining digits (counted with multiplicity). |
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+0
1
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| 1, 2, 3, 4, 5, 6, 7, 8, 9, 90, 99, 96, 26, 28, 285, 528, 85, 18, 19, 20, 21, 22, 23, 24, 25, 52, 27, 28, 29, 90, 93, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71
(list; graph; listen)
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OFFSET
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1,2
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EXAMPLE
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For n = 10 we have a(n-1) = a(9) = 9; removing 9 from 9*n = 9*10 = 90 results in one remaining digit (i.e. 0) and for every smaller multiple of 10 (i.e. 10, 20, ..., 80) there are two remaining digits, so a(10) = 90. For n = 20 we have a(n-1) = a(19) = 19; for 1*n = 1*20 = 20 no digit is removed and two digits remain and there is no multiple of 20 which has less than two remaining digits (for 100 twice the digit 0), so a(20) = 20.
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CROSSREFS
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Sequence in context: A032799 A160343 A024664 this_sequence A061805 A061219 A071271
Adjacent sequences: A078185 A078186 A078187 this_sequence A078189 A078190 A078191
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KEYWORD
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base,nonn
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AUTHOR
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Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Nov 21 2002
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EXTENSIONS
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Edited, corrected and extended by Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Dec 03 2002
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