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.

0, 0, 1, 4, 23, 53, 87, 593, 5807, 415267, 8758459, 274431867, 12856077691, 905435186299, 481691519113703, 77763074616922439, 3824113551749834107, 1437016892446437662971, 165559472503434318118655, 146602912901791088694069887, 200050146291129782743679367167

Offset: 1

Author

Amiram Eldar, Dec 04 2020

Keywords

• nonn
• frac

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).

Examples

			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).
		

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.

Crossrefs

Cf. A000045, A001791, A084623, A234951, A243398, A339393 (denominators).

Programs

Mathematica

f = Table[k/(Fibonacci[k + 2] - 1), {k, 2, 20}]; Numerator[1 - FoldList[Times, 1, f]]

Formula

a(n) = numerator(1 - Product_{k=2..n} k/(Fibonacci(k+2)-1)).

Lim_{n->oo} a(n)/A339393(n) = 1.