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 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 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) + ... + x + ... + (x + c'2) + (x + c'1)
>
> An example demonstrates this more clearly:
> 35 is divisible by 7, 5 times, so we can write:
>
> 35 = 7 + 7 + 7 + 7 + 7
>
> and rewrite this to:
>
> 35 = (7-2) + (7-1) + 7 + (7+1) + (7+2),
>
> which is the same as:
>
> 35 = 5 + 6 + 7 + 8 + 9
>
> For every odd divisor, there will be a corresponding summation.
> The reason even divisors don't allow this should be obvious: we
> need an odd number of components in order to be able to add
> the same amounts on the right that we subtracted from the left...
>
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
=> (7-0.5)+(7+0.5) => not integer, oops
The other split:
14 = 2+2...2 (7 times) => (2-3)+(2-2)+(2-1)+2+(2+1)+(2+2)+(2+3) =>
2-3 is smaller than one, even smaller than zero, oops
