

A339392


Numerators of the probability that when a stick is broken up at n1 points independently and uniformly chosen at random along its length there exist 3 of the n pieces that can form a triangle.


3



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

Martin Gardner, Probability and Ambiguity, The Colossal Book of Mathematics, W. W. Norton, New York, 2001, chapter 21, pp. 273285.


FORMULA

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



KEYWORD

nonn,frac


AUTHOR



STATUS

approved



