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
ee that the number of ways
> to write a number as the sum of sequences of consecutive positive
> integers the same as the number of odd divisors of a number.
>
> If a number, y, is divisible by x, z times, then y can be expressed
> as x + x + ... + x, z times.
>
> We can
ailto:[EMAIL PROTECTED]
Sent: Wednesday, February 11, 2004 8:16 PM
To: [EMAIL PROTECTED]
Subject: Number of odd divisors
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
number of odd divisors of a number.
If a number, y, is divisible by x, z times, then y can be expressed
as x + x + ... + x, z times.
We can subtract the appropriate amount from the left values, and
increment the respective right values accordingly to rewrite this as
(x - c'1) + (x - c'2)
