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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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Posts: 821
Registered: 9/1/10
Re: Reciprocals of integers summing to 1
Posted: Nov 23, 2012 6:06 PM
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On Nov 23, 4:16 am, Bill Taylor <wfc.tay...@gmail.com> wrote:
> 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" !


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