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 A057114 Permutation of N induced by the order-preserving permutation of the rational numbers (x -> x+1); positions in Stern-Brocot tree. 19
 3, 1, 7, 2, 6, 14, 15, 4, 5, 12, 13, 28, 29, 30, 31, 8, 9, 10, 11, 24, 25, 26, 27, 56, 57, 58, 59, 60, 61, 62, 63, 16, 17, 18, 19, 20, 21, 22, 23, 48, 49, 50, 51, 52, 53, 54, 55, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 96, 97, 98, 99, 100, 101, 102, 103 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS The "unbalancing operation" used here is what is usually called "rotation of binary trees" (e.g. in Lucas, Ruskey et al. article) REFERENCES Joan Lucas, Dominique Roelants van Baronaigien and Frank Ruskey, On Rotations and the Generation of Binary Trees, Journal of Algorithms, 15 (1993) 343-366. LINKS A. Bogomolny, About the Stern-Brocot tree P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5. FORMULA a(n) = frac2position_in_whole_SB_tree (sbtree_perm_1_1_right (SternBrocotTreeNum(n) / SternBrocotTreeDen(n))). EXAMPLE Consider the following "extended" Stern-Brocot tree (on interval ]-inf,inf[): ....................................0/1 .................-1/1.................................1/1 ......-2/1................-1/2...............1/2...............2/1 .-3/1......-3/2......-2/3......-1/3.....1/3.......2/3.....3/2.......3/1 Enumerate the fractions breadth-first (0/1 = 1, -1/1 = 2, 1/1 = 3, -2/1 = 4, -1/2 = 5, etc.) then use this sequence to pick third, first, 7th, 2nd, etc. fractions. We get a bijection (0/1 -> 1/1, -1/1 -> 0/1, 1/1 -> 2/1, -2/1 -> -1/1, -1/2 -> 1/2, etc.) which is the function x -> x+1. In other words, we cut the edge between 1/1 and 1/2, make 1/1 the new root and create a new edge between 0/1 and 1/2 to get an "unbalanced" Stern-Brocot tree. If we instead make a similar change to subtree 1/1 (cut {2/1,3/2}, create {1/1,3/2} and make 2/1 the new root of the positive side, leaving the negative side as it is), we get the function given in Maple procedure sbtree_perm_1_1_right. Both mappings belong to Cameron's group "A" of permutations of the rational numbers which preserve their linear order and by applying such unbalancing operations successively (possibly infinitely many times) to the "extended" Stern-Brocot tree given above, the whole group "A" can be generated. MAPLE sbtree_perm_1_1_right := x -> (`if`((x <= 0), x, (`if`((x < (1/2)), (x/(1-x)), (`if`((x < 1), (3-(1/x)), (x+1))))))); CROSSREFS SternBrocotNum given in A007305, SternBrocotDen in A047679, frac2position_in_whole_SB_tree in A054424. Inverse permutation: A057115. Cf. also A065249 and A065250. A065259(n) = A059893(A057114(A059893(n))) The first row of A065625, i.e. a(n) = RotateNodeRight(1, n). Sequence in context: A073874 A065287 A065263 * A235800 A065259 A065289 Adjacent sequences:  A057111 A057112 A057113 * A057115 A057116 A057117 KEYWORD nonn AUTHOR Antti Karttunen, Aug 09 2000 STATUS approved

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Last modified October 18 01:04 EDT 2019. Contains 328135 sequences. (Running on oeis4.)