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January 31[edit]

Harmonic mean problem[edit]

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]