Listen to the bass line. It's a timeline of 2-3-4 (even though the downbeat is on 3, so it would be 3-4-2, but for our purposes 2-3-4 is fine).
Suspicion: It's divisible by 3 because it's 3 consecutive integers. How about 3, 4, 5: adds up to 12. 4, 5, 6: adds up to 15.
However, it only seems to work for an odd count of consecutive integers, e.g. 2 + 3 + 4 + 5 = 14, not divisible by 4.
If we regard the median as the symmetry axis:
2 + 3 + 4
n-1 + n + n+1; -1 +1 cancel each other out, we're left with n + n + n = 3n.
3 + 4 + 5 + 6 + 7
n-2 + n-1 + n + n+1 + n+2; -2 +2, -1 +1, cancel each other out, we're left with n + n + n + n + n = 5n.
Conjecture: The sum of k consecutive integers with median n is kn if k is odd.
Proof below.
4 5 6 => k = 3, n = 5, kn = 15.
| k | 3 | 3 | 3 | 3 | 3 | ||||||||||
| n | 5 | 5 | 5 | ||||||||||||
| kn | 4 | 5 | 6 |
| k | 3 | 3 | 3 | 3 | 3 | 3 | 3 | ||||||||||||||
| n | 7 | 7 | 7 | ||||||||||||||||||
| kn | 6 | 7 | 8 |
2 3 4 5 6 => k = 5, n = 4, kn = 20.
| k | 5 | 5 | 5 | 5 | ||||||||||||||||
| n | 4 | 4 | 4 | 4 | 4 | |||||||||||||||
| kn | 2 | 3 | 4 | 5 | 6 |
The sum of the first n integers is m(m+1)/2.
So the sum of k consecutive integers is
m(m+1)/2 - (m-k)(m-k+1)/2, simplified:
((2m+1)k - k2)/2. According to the conjecture this is supposed to equal kn if k is odd.
((2m+1)k - k2)/2 = kn.
(2m+1 - k)/2 = n.
n must be integer, so (2m+1 - k)/2 must be integer, too, hence (2m+1 - k) must be even.
2m+1 is guaranteed to be odd, hence k must be odd in order for the result to be even.
q.e.d.
© Bernhard Wagner, May 28th, 2012.