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
Amiram Eldar, Table of n, a(n) for n = 1..100
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
KEYWORD
nonn,frac
AUTHOR
Amiram Eldar, Dec 04 2020
STATUS
approved