%I #14 Oct 16 2023 01:48:07
%S 0,0,1,4,23,53,87,593,5807,415267,8758459,274431867,12856077691,
%T 905435186299,481691519113703,77763074616922439,3824113551749834107,
%U 1437016892446437662971,165559472503434318118655,146602912901791088694069887,200050146291129782743679367167
%N 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.
%C 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).
%H Amiram Eldar, <a href="/A339392/b339392.txt">Table of n, a(n) for n = 1..100</a>
%H Alexander Bogomolny, <a href="https://www.cut-the-knot.org/Curriculum/Probability/TriProbability.shtml">Stick Broken Into Three Pieces (Trilinear Coordinates)</a>, Interactive Mathematics Miscellany and Puzzles, Cut the Knot website.
%H P. A. Crowdmath, <a href="https://arxiv.org/abs/1805.06512">The Broken Stick Project</a>, arXiv:1805.06512 [math.HO], 2018.
%H Martin Gardner, <a href="https://archive.org/details/martingardnerthecolossalbookofmathematics/page/n289/mode/2up">Probability and Ambiguity</a>, The Colossal Book of Mathematics, W. W. Norton, New York, 2001, chapter 21, pp. 273-285.
%H Lingyi Kong, Luvsandondov Lkhamsuren, Abigail Turner, Aananya Uppal and A. J. Hildebrand, <a href="https://faculty.math.illinois.edu/~hildebr/ugresearch/integrals-spring2013report-brokenstick.pdf">Random Points, Broken Sticks, and Triangles, Project Report</a>, Illinois Geometry Lab, 2013.
%F a(n) = numerator(1 - Product_{k=2..n} k/(Fibonacci(k+2)-1)).
%F Lim_{n->oo} a(n)/A339393(n) = 1.
%e 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, ...
%e For n = 1 or 2 the number of pieces is less than 3, so the probability is 0.
%e 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).
%t f = Table[k/(Fibonacci[k + 2] - 1), {k, 2, 20}]; Numerator[1 - FoldList[Times, 1, f]]
%Y Cf. A000045, A001791, A084623, A234951, A243398, A339393 (denominators).
%K nonn,frac
%O 1,4
%A _Amiram Eldar_, Dec 04 2020