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

1,4

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

Table of n, a(n) for n=1..21.

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

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 * A030653 A085505 A065972

Adjacent sequences: A339389 A339390 A339391 * A339393 A339394 A339395

Keyword

nonn,frac

Author

Amiram Eldar, Dec 04 2020

Status

approved