Posts:
822
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 UTC8, 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

