cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A232890 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.

Original entry on oeis.org

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
Offset: 1

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Author

Clark Kimberling, Dec 02 2013

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    Off[Power::infy]; x = {0}; Do[x = DeleteDuplicates[Flatten[Transpose[{x, x + 1, -1/(x + 1)} /. ComplexInfinity -> 0]]], {8}]; x
    On[Power::infy]; Denominator[x]  (* Peter J. C. Moses, Nov 29 2013 *)