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URL: https://oeis.org/A339392

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A339392
Numerators of the probability that when a stick is broken up at n-1 points independently and uniformly chosen at random along its length there exist 3 of the n pieces that can form a triangle.
4
0, 0, 1, 4, 23, 53, 87, 593, 5807, 415267, 8758459, 274431867, 12856077691, 905435186299, 481691519113703, 77763074616922439, 3824113551749834107, 1437016892446437662971, 165559472503434318118655, 146602912901791088694069887, 200050146291129782743679367167
OFFSET
1,4
COMMENTS
For the corresponding probability that any triple of pieces can form a triangle, see A001791. The probabilities for these two cases were found by Kong et al. (2013).
LINKS
Alexander Bogomolny, Stick Broken Into Three Pieces (Trilinear Coordinates), Interactive Mathematics Miscellany and Puzzles, Cut the Knot website.
P. A. Crowdmath, The Broken Stick Project, arXiv:1805.06512 [math.HO], 2018.
Martin Gardner, Probability and Ambiguity, The Colossal Book of Mathematics, W. W. Norton, New York, 2001, chapter 21, pp. 273-285.
Lingyi Kong, Luvsandondov Lkhamsuren, Abigail Turner, Aananya Uppal and A. J. Hildebrand, Random Points, Broken Sticks, and Triangles, Project Report, Illinois Geometry Lab, 2013.
FORMULA
a(n) = numerator(1 - Product_{k=2..n} k/(Fibonacci(k+2)-1)).
Lim_{n->oo} a(n)/A339393(n) = 1.
EXAMPLE
Fractions begin with 0, 0, 1/4, 4/7, 23/28, 53/56, 87/88, 593/594, 5807/5808, 415267/415272, 8758459/8758464, 274431867/274431872, ...
For n = 1 or 2 the number of pieces is less than 3, so the probability is 0.
For n = 3, the stick is being broken into 3 pieces and the probability that they can form a triangle is 1/4, the solution to the classical broken stick problem (see, e.g., Gardner, 2001).
MATHEMATICA
f = Table[k/(Fibonacci[k + 2] - 1), {k, 2, 20}]; Numerator[1 - FoldList[Times, 1, f]]
CROSSREFS
Cf. A000045, A001791, A084623, A234951, A243398, A339393 (denominators).
Sequence in context: A106684 A239624 A179628 * A383464 A030653 A085505
KEYWORD
nonn,frac
AUTHOR
Amiram Eldar, Dec 04 2020
STATUS
approved