Three positive integers are chosen at random without replacement from 1,2,....,72. What is the probability that the numbers chosen are in harmonic sequence
Order of choice doesn't matter. For example, 6-3-2 would qualify as numbers in harmonic sequence.
Generalize this result (in terms of n) covering the situation where three positive integers are chosen at random without replacement from 1,2,...,n.
Note from Charlie's simulation the values of n at which the (unreduced) numerator of the probability changes. These are 6, 12, 15, 18, 20...
A little algebra shows that given integers a < b < c, then they are in harmonic sequence if c = ab / (2a - b). The denominator implies that a < b < 2a.
I haven't had any time to take this train of thought further but I obviously wonder if there's a connection, and if it would shed any light on how to compute the probability for a given n.
e.g. we can always calculate the denominator of the probability, it's just n choose 3. And the numerator only increases at the values of n on the integer sequence linked above... but as I said, that's as far as I've been able to go and won't be able to come back to it until later. Perhaps someone else can run with it.
Posted by tomarken
on 2014-02-28 18:49:43