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A006066
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Kobon triangles: maximal number of nonoverlapping triangles that can be formed from n lines drawn in the plane.
(Formerly M1334)
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2
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OFFSET
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1,4
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COMMENTS
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The known values a = a(n) and upper bounds U (usually A032765(n)) with name of discoverer of the arrangement when known are as follows:
n a U [Found by]
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1 0 0
2 0 0
3 1 1
4 2 2
5 5 5
6 7 7
7 11 11
8 15 16
9 21 21
10 25? 26 [Grünbaum]
11 32? 33 [See link below]
12 ? 40
13 47 47 [Kabanovitch]
14 >= 53 56 [Bader]
15 65 65 [Suzuki]
16 >=72 74 [Bader]
17 85 85 [Bader]
18 >= 93 96 [Bader]
19 >= 104 107 [Bader]
20 >= 115 120 [Bader]
21 >= 130 133 [Bader]
22 ? 146
23 ? 161
24 ? 176
25 ? 191
26 ? 208
27 ? 225
28 ? 242
29 ? 261
30 ? 280
31 ? 299
32 ? 320
Ed Pegg's web page gives the upper bound for a(6) as 8. But by considering all possible arrangements of 6 lines - the sixth term of A048872 - one can see that 8 is impossible. - N. J. A. Sloane, Nov 11 2007
Although they are somewhat similar, this sequence is strictly different from A084935, since A084935(12) = 48 exceeds the upper bound on a(12) from A032765. - Floor van Lamoen, Nov 16 2005
The name is sometimes incorrectly entered as "Kodon" triangles.
Named after the Japanese puzzle expert and mathematics teacher Kobon Fujimura (1903-1983). - Amiram Eldar, Jun 19 2021
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REFERENCES
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Martin Gardner, Wheels, Life and Other Mathematical Amusements, Freeman, NY, 1983, pp. 170, 171, 178. Mentions that the problem was invented by Kobon Fujimura.
Branko Grünbaum, Convex Polytopes, Wiley, NY, 1967; p. 400 shows that a(10) >= 25.
Viatcheslav Kabanovitch, Kobon Triangle Solutions, Sharada (Charade, by the Russian puzzle club Diogen), pp. 1-2, June 1999.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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Johannes Bader, Kobon Triangles. [Cached copy, with permission, pdf format]
Ed Pegg, Jr., Kobon Triangles, 2006. [Cached copy, with permission, pdf format]
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FORMULA
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An upper bound on this sequence is given by A032765.
For any odd n > 1, if n == 1 (mod 6), a(n) <= (n^2 - (2n + 2))/3; in other odd cases, a(n) <= (n^2 - 2n)/3. For any even n > 0, if n == 4 (mod 6), a(n) <= (n^2 - (2n + 2))/3, otherwise a(n) <= (n^2 - 2n)/3. - Sergey Pavlov, Feb 11 2017
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EXAMPLE
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a(17) = 85 because the a configuration with 85 exists meeting the upper bound.
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CROSSREFS
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KEYWORD
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nonn,hard,more
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AUTHOR
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EXTENSIONS
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A perfect solution for 13 lines was found in 1999 by Kabanovitch. - Ed Pegg Jr, Feb 08 2006
Updated with results from Johannes Bader (johannes.bader(AT)tik.ee.ethz.ch), Dec 06 2007, who says "Acknowledgments and dedication to Corinne Thomet".
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STATUS
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approved
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