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A274606
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Pick any two successive integers in the sequence; there is a larger one (L) and a smaller one (S); the last digit of L is the remainder of L/S.
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1
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1, 10, 2, 11, 5, 12, 60, 3, 30, 6, 31, 15, 32, 160, 4, 20, 21, 210, 7, 70, 14, 71, 35, 72, 360, 8, 40, 41, 410, 82, 16, 80, 81, 810, 9, 90, 18, 91, 45, 92, 460, 23, 230, 46, 231, 115, 22, 110, 55, 25, 50, 51, 510, 17, 170, 34, 171, 85, 172, 860, 43, 430, 86, 431, 215, 42, 211, 105, 100, 101, 1010, 202, 200, 201, 2010, 67
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OFFSET
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1,2
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COMMENTS
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The sequence starts with a(1)=1 and is always extended with the smallest integer not yet in the sequence and not leading to a contradiction.
This sequence is probably a permutation of the positive integers.
After 10000 terms, the smallest integer not yet present in the sequence is 913 and the largest one present is 1144810.
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LINKS
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EXAMPLE
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The sequence starts with 1,10,2,11,5,12,60,3,30,6,31,15,32,160,... and indeed, 10/1 has 0 as remainder [the last digit of "10", which is the biggest term of the pair (1,10)]; 10/2 has 0 as remainder (again, the last digit of "10"); 11/2 has 1 as remainder; 11/5 has 1 as remainder; 12/5 has 2 as remainder; 60/12 has 0 as remainder; etc.
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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