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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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David Petry

Posts: 1,097
Registered: 12/8/04
Re: Reciprocals of integers summing to 1
Posted: Nov 22, 2012 9:08 PM
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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
>
>
>
> (the gaps are where I didn't need them for the above work).
>
>
>
> This table at least is easy to do by computer.
>
> It seems that #(p) = 2 for prime p, and in general
>
> 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.





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