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A232890
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Denominator of n-th term of sequence (or tree) S of all rational numbers generated by these rules: 0 is in S; if x is in S then x + 1 is in S, and if x + 1 is nonzero, then -1/(x + 1) is in S; duplicates are deleted as they occur.
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2
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1, 1, 1, 1, 2, 1, 3, 2, 1, 1, 4, 3, 2, 2, 3, 1, 5, 4, 3, 3, 5, 2, 5, 3, 1, 1, 6, 5, 4, 4, 7, 3, 8, 5, 2, 2, 7, 5, 3, 3, 4, 1, 7, 6, 5, 5, 9, 4, 11, 7, 3, 3, 11, 8, 5, 5, 7, 2, 9, 7, 5, 5, 8, 3, 7, 4, 1, 1, 8, 7, 6, 6, 11, 5, 14, 9, 4, 4, 15, 11, 7, 7, 10, 3
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OFFSET
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1,5
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COMMENTS
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Let S be the sequence (or tree) of numbers generated by these rules: 0 is in S; if x is in S then x + 1 is in S, and if x + 1 is nonzero, then -1/(x + 1) is in S. Deleting duplicates as they occur, the generations of S are given by g(1) = (0), g(2) = (1,-1), g(3) = (2,-1/2), g(4) = (3, -1/3, 1/2, -2), ... Concatenating gives 0, 1, -1, 2, -1/2, 3, -1/3, 1/2, -2, 4, -1/4, ...
Conjectures: If b/c is a positive rational number, the position of n + b/c for n >= 0 forms a linear recurrence sequence with signature (1,1), and the position of -n - b/c forms a linear recurrence sequence with signature (4, -4, 1). For n>=1, the numbers -(1 + 1/n) are terminal nodes in the tree, and their positions are linearly recurrent with signature (2,0,-1). For n >=3, the n-th generation g(n) consists of F(n-1) positive numbers and F(n-1) negative numbers, where F = A000045, the Fibonacci numbers.
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LINKS
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EXAMPLE
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To generate S, the number 0 begets (1,-1), whence 1 begets 2 and -1/2, whereas -1 begets 0 and -1/2, both of which are (deleted )duplicates, so that g(3) = (2, -1/2). The resulting concatenation of all the generations g(n) begins with 0, 1, -1, 2, -1/2, 3, -1/3, 1/2, -2, 4, -1/4, so that A232890 begins with 1,1,1,1,2,1,3,2,1,1,4.
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MATHEMATICA
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Off[Power::infy]; x = {0}; Do[x = DeleteDuplicates[Flatten[Transpose[{x, x + 1, -1/(x + 1)} /. ComplexInfinity -> 0]]], {8}]; x
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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