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Topic: Reciprocals of integers summing to 1
Replies: 21   Last Post: Nov 23, 2012 8:44 PM

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 Bill Taylor Posts: 186 Registered: 11/17/10
Re: Reciprocals of integers summing to 1
Posted: Nov 23, 2012 7:16 AM

On Nov 23, 3:08 pm, david petry <david_lawrence_pe...@yahoo.com>
wrote:
> On Thursday, November 22, 2012 4:07:09 AM UTC-8, Bill Taylor wrote:
>

> > If anyone wants to check my hand work, the number
> > of ways of splitting  1/m  is (I suggest)

>
> > m:   1  2  3  4  5  6  7  8  9 10  12  15  18  20  24  42
> > #(m) 1  2  2  3  2  5  2  4  3  5   4   5   8   8  11  14

>
> > It seems that  #(p) = 2  for prime p, and in general

In fact this is simply routinely proved, unsurprisingly.

> > the # function bears a close resemblance to d, the number
> > of factors of m; though it tends to be a bit bigger,
> > and is not multiplicative like d.

>
> The formula  1/(a*b) = 1/(a*(a+b))  +  1/(b*(a+b))   shows why
> there is a close relation between your function and
> the number of ways the number can be factored into two factors.

Of course! Neat triviality. I wonder if the surplus ways
are also simply characterizable. Must investigate.

-- Bumbling Bill

** There is no moral right to "not be offended".
** There should be no legal right to "not be offended" !

Date Subject Author
11/16/12 Charlie-Boo
11/16/12 William Elliot
11/16/12 Charlie-Boo
11/16/12 William Elliot
11/17/12 Charlie-Boo
11/18/12 Bill Taylor
11/21/12 David Petry
11/22/12 Bill Taylor
11/22/12 Luis A. Rodriguez
11/22/12 David Petry
11/22/12 David Petry
11/23/12 Bill Taylor
11/23/12
11/23/12 Bill Taylor
11/16/12 Don Redmond
11/16/12 gus gassmann
11/16/12 billh04
11/17/12 Luis A. Rodriguez
11/17/12 Charlie-Boo
11/19/12 Luis A. Rodriguez
11/20/12 doumin
11/22/12 Bill Taylor