Mathematics desk
< January 30	<< Dec \| January \| Feb >>	February 1 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

January 31

Harmonic mean problem

The sequence 20, 12, 6, 5, 2, 1 arose in a newspaper puzzle. It has the property that the cumulative HM from L to R is integral, with the specific values 20, 15, 10, 8, 5, 3. Is it possible to construct an increasing sequence without end with the same property?31.54.246.96 (talk) 23:30, 31 January 2015 (UTC)[reply]

According to my calculations, 1⋅2, 2⋅3, 3⋅4, 4⋅5, ... has cumulative harmonic means 2, 3, 4, 5 ... . --RDBury (talk) 03:44, 1 February 2015 (UTC)[reply]

True, analogous to 1, 3, 5, 7, ... for AM and 1, 4, 9, 16, ... for GM. I wonder if there is an HM sequence for all starting integers apart from 1.→31.54.246.96 (talk) 09:47, 1 February 2015 (UTC)[reply]

That GM sequence doesn't work, try 1, 4, 16, 64,... . -- Meni Rosenfeld (talk) 13:09, 1 February 2015 (UTC)[reply]

Actually there is one starting with 1: The sequence 1, 3, 5, ... are the cumulative HM's, of 1, -1⋅3, -3⋅5, -5⋅7, -7⋅9. If you're assuming positive integers in the starting sequence then 1 cannot be the first HM, since if the first harmonic mean is 1 then the second is at most 2, preventing it from being an integer. In fact, if the starting sequence is positive and the cumulative harmonic means are a₁, a₂, a₃, ... then a₂ < 2a₁, 2a₃ < 3a₂, 3a₄ < 4a₃, ... . The sequence a, a(2a-1), (2a-1)(3a-2), (3a-2)(4a-3) has cumulative HM's a, 2a-1, 3a-2, ... , so any starting point a > 1 is possible.

A more interesting question (meaning I don't know the answer) is what starting pairs a, b are possible. --RDBury (talk) 14:08, 1 February 2015 (UTC)[reply]