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... -Riley (o0lit3) P.S. - Note the following example as well (to avoid confusion with negative numbers) 27 is divisible by 3, 9 times, so we can write: 27 = 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 and rewrite this to: 27 = (3-4) + (3-3) + (3-2) + (3-1) + 3 + (3+1) + (3+2) + (3+3) + (3+4), which is the same as: 27 = -1 + 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 Removing all non-positive integers is easy because 0 is trivial and numbers bellow 0 will cancel with their corresponding positive conterparts... 27 = 2 + 3 + 4 + 5 + 6 + 7
