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A302656
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Replacing each term of this sequence S with its digitsum produces a new sequence S' such that S' and S share the same succession of digits.
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4
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1, 2, 3, 4, 5, 6, 7, 8, 9, 109, 18, 10, 17, 19, 89, 100, 27, 26, 36, 199999999999, 11, 16, 20, 15, 12, 24, 199, 45, 54, 63, 72, 81, 90, 108, 117, 126, 135, 29, 79, 299, 69, 39, 101, 13, 289, 144, 22, 14, 23, 31, 33, 21, 25, 110, 35, 1000, 9999999999, 28, 44, 38, 34, 48, 42, 49, 32, 200, 153, 43
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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 present that doesn't lead to a contradiction.
Huge jumps appear in this sequence: a(96) = 41, a(97) = 2*10^111-1, and a(98) = 234.
Records after a(97) are:
a(176) = 2*10^1111-1
a(396) = 2*10^11111-1
a(463) = 2*10^111111-1
a(1918) = 2*10^1111111-1
a(1984) = 2*10^11111111-1
a(2279) = 2*10^111111111-1
...
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LINKS
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EXAMPLE
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The first nine terms are replaced by themselves;
109 = a(10) is replaced by the digitsum 1 + 0 + 9 = 10;
18 = a(11) is replaced by the digitsum 1 + 8 = 9;
10 = a(12) is replaced by the digitsum 1 + 0 = 1;
17 = a(13) is replaced by the digitsum 1 + 7 = 8;
19 = a(14) is replaced by the digitsum 1 + 9 = 10;
89 = a(15) is replaced by the digitsum 8 + 9 = 17;
100 = a(16) is replaced by the digitsum 1 + 0 + 0 = 1;
27 = a(17) is replaced by the digitsum 2 + 7 = 9;
26 = a(18) is replaced by the digitsum 2 + 6 = 8;
36 = a(19) is replaced by the digitsum 3 + 6 = 9;
199999999999 = a(20) is replaced by the digitsum 1 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 = 100; etc.
We see that the first and the last column here (the terms of S and S') share the same succession of digits:
1, 0, 9, 1, 8, 1, 0, 1, 7, 1, 9, 8, 9, 1, 0, 0, 2, 7, 2, 6, 3, 6, 1, 9, 9, 9, 9, ...
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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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