A more formal writeup of Riley's method:
Consider an odd divisor o of the target number N. Divide the target number
up in that many pieces (n=N/o). Subtract (n-1)/2 from the first one, (n-1)/2-1
from the second etc (one less for each next number). So for example
12 = 3*4 = 4+4+4 = 3+4+5
0r:
14 =
On Thu, 12 Feb 2004, Allen, Greg wrote:
> This proves that the number of sums is at least the same at the number of
> divisors, but it doesn't prove the equivalence. I.e. could there be other
> sums...? I think you need to be more rigorous in demonstrating the
> reversibility of each step.
>
You
In article <[EMAIL PROTECTED]>,
[EMAIL PROTECTED] (Ton Hospel) writes:
> which corresponds to the bottom case in my drawing, so I have that
> case covered. It's not enough however. Consider the top case,
> and try to handle the 7 divisor:
>
> 14 = 7+7 => (7-1)+(7+1) => missing 7, oops
In article <[EMAIL PROTECTED]>,
"Riley" <[EMAIL PROTECTED]> writes:
> Hello All!
>
> After reading the kwiki on this subject provided by ton at
> http://terje2.perlgolf.org/~golf-info/a1227-equivalence.html,
> I think there is a more intuitive way to see that the number of ways
> to write
This proves that the number of sums is at least the same at the number of
divisors, but it doesn't prove the equivalence. I.e. could there be other
sums...? I think you need to be more rigorous in demonstrating the
reversibility of each step.
Greg
-Original Message-
From: Riley [mailto:[E
