2010
DOI: 10.1016/j.ejc.2007.10.005
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Locating terms in the Stern–Brocot tree

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Cited by 18 publications
(22 citation statements)
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“…Example 43. For m = 3, the sequence counting admissible subwords associated with the Tribonacci numeration system starts with 1,2,3,3,4,5,5,5,7,8,6,7,7,6,9,11,9,11,12,10,9,11,11,9,7,11,14,12,15,17,15,14,18,19,15,14,14,11,15,17,15,15,17,15,8,13,17,15,19,22, . .…”
Section: Discussionmentioning
confidence: 99%
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“…Example 43. For m = 3, the sequence counting admissible subwords associated with the Tribonacci numeration system starts with 1,2,3,3,4,5,5,5,7,8,6,7,7,6,9,11,9,11,12,10,9,11,11,9,7,11,14,12,15,17,15,14,18,19,15,14,14,11,15,17,15,15,17,15,8,13,17,15,19,22, . .…”
Section: Discussionmentioning
confidence: 99%
“…Plugging in the first few terms of (S(n)) n≥0 in Sloane's On-Line Encyclopedia of Integer Sequences [33], it seems to be a shifted version of the sequence A007306 of the denominators occurring in the Farey tree (left subtree of the full Stern-Brocot tree), which contains every (reduced) positive rational less than 1 exactly once. Many papers deal with this tree; for instance, see [4,18]. The Farey tree is an infinite binary tree made up of mediants.…”
Section: The Sequence A007306 Farey Tree and Stern-brocot Sequencementioning
confidence: 99%
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“…Some of the recent developments include the following. In [1], B. Bates et al gave a simple method of identifying both the level and the position within the level of each fraction in the Stern-Brocot tree. In [4], K. Dilcher and K. B. Stolarsky discovered an interesting polynomial analogue to the Stern sequence.…”
Section: Introductionmentioning
confidence: 99%
“…Bates et al [5] have shown that when expressed in the shortest form of their simple continued fractions: i) Left mediants possess continued fractions of the form [a 0 , a 1 , . .…”
Section: Definition 11 (Left and Right Mediants)mentioning
confidence: 99%