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A277268
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If we call "T" this sequence and consider the k-digit term a(n) of T with digits abcd...k, then a(n+1) = [a(n) + the a-th digit of T + the b-th digit of T + the c-th digit of T + ... + the k-th digit of T]. This is the lexicographically first such infinite sequence containing no duplicate term.
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2
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11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 45, 49, 53, 55, 57, 59, 61, 67, 73, 75, 77, 79, 81, 89, 97, 99, 101, 103, 105, 107, 109, 111, 114, 119, 122, 125, 128, 137, 140, 144, 151, 154, 159, 162, 169, 176, 183, 192, 195, 198, 207, 209, 211, 214, 219, 222, 225, 228, 237, 240, 244, 251, 254, 259, 262, 269, 276, 283, 292, 295, 298, 307
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OFFSET
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1,1
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COMMENTS
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There are 3 lexicographically earlier sequences with this property, but they all stop at some point. The first one is A276514, starting with 5,10,15,25,31,... and halting with a(49) = 300. The second one starts with 9,10,19,29,31,... and stops with a(246) = 3003. The third one starts with 10,11,13,15,17,... and ends with a(8) = 22, as shown here: 10, 11, 13, 15, 17, 19, 21, 22. [To compute a hypothetical a(9), one has to add to 22 the second digit of the sequence (which is zero) and (again) the second digit of the sequence (again zero): 22+0+0 = 22.]
As this sequence, starting with 11, 13, 15, 17, 19, ..., shows no zero digit among its first 10 digits, it will never stop.
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LINKS
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EXAMPLE
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To compute a(2), add to a(1) the 1st digit of T and the 1st digit of T, so a(2) = 11+1+1 = 13;
to compute a(3), add to a(2) the 1st digit of T and the 3rd digit of T, so a(3) = 13+1+1 = 15;
to compute a(4), add to a(3) the 1st digit of T and the 5th digit of T, so a(4) = 15+1+1 = 17;
to compute a(5), add to a(4) the 1st digit of T and the 7th digit of T, so a(5) = 17+1+1 = 19;
to compute a(6), add to a(5) the 1st digit of T and the 9th digit of T, so a(6) = 19+1+1 = 21;
to compute a(7), add to a(6) the 2nd digit of T and the 1st digit of T, so a(7) = 19+1+1 = 23;
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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