

A057114


Permutation of N induced by the orderpreserving permutation of the rational numbers (x > x+1); positions in SternBrocot 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
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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) 343366.


LINKS

Table of n, a(n) for n=1..87.
A. Bogomolny, About the SternBrocot tree
P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
Index entries for sequences related to Stern's sequences
Index entries for sequences that are permutations of the natural numbers


FORMULA

a(n) = frac2position_in_whole_SB_tree (sbtree_perm_1_1_right (SternBrocotTreeNum(n) / SternBrocotTreeDen(n))).


EXAMPLE

Consider the following "extended" SternBrocot 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 breadthfirst (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" SternBrocot 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" SternBrocot 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/(1x)), (`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



