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.
Bonus Question:
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.
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Posted by tomarken
on 2014-02-28 18:49:43 |
Some thoughts on the bonus
