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A118247
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a(0)=0. Concatenate onto the end of the sequence (from left to right) the integer m_n converted into binary (with the most significant digit on the left), where m_n is the smallest integer > A118248(n-1) and whose binary representation does not occur anywhere earlier in the sequence (when the concatenated sequence is read from left to right). A118248(n) then equals m_n when written in decimal.
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9
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0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0
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OFFSET
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0,1
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COMMENTS
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Sequence can be regarded as an irregular number triangle containing the binary digits of A118248(n) in row n. - Michael De Vlieger, Aug 19 2017
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LINKS
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EXAMPLE
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The sequence begins 0,1,1,0,1,0,0,1,1,1,1,0,0,0. Now A118248(5) = 8, which is represented by the 1,0,0,0 at the end of the sequence. The binary representation of 9 (1001 in binary) and (decimal) 10 (1010 in binary) both occur earlier in the sequence. But the binary representation of (decimal) 11 (1011 in binary) does not occur earlier in the sequence, so (1,0,1,1) is added to the end of the sequence. And A118248(6) becomes 11.
n A118248(n) Binary digits appended to a(n)
0 0 0
1 1 1
2 2 10
3 4 100
4 7 111
5 8 1000
6 11 1011
7 16 10000
8 18 10010
9 21 10101
10 22 10110
11 25 11001
12 29 11101
(End)
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MATHEMATICA
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Block[{a = {{0}}, b = {0}, k, d}, Do[k = FromDigits[#, 2] &@ Last@ a + 1; While[SequenceCount[Flatten@ a, Set[d, IntegerDigits[k, 2]]] > 0, k++]; AppendTo[a, d]; AppendTo[b, k], {i, 22}]; Flatten@ a] (* Michael De Vlieger, Aug 19 2017 *)
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CROSSREFS
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KEYWORD
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easy,nonn,base,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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