Posts:
821
Registered:
9/1/10
|
|
Re: Reciprocals of integers summing to 1
Posted:
Nov 23, 2012 6:06 PM
|
|
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" !
0
|
|
