Consecutive Sums (2)

Note: all the considerations below are referred to the specific case (integers from 1 to 1.000.000).

When you sum two consecutive numbers you obtain only odd numbers, greater or equal to 3:

$n + (n+1) = 2n + 1 = [3, 5, 7, 9, 11, ....]$

when you sum three consecutive numbers you obtain multiples of 3, greater or equal to 6:

$n + (n + 1) + (n + 2) = 3n + 3 = 3(n +1) = [6, 9, 12, 15, .....]$

When you sum four consecutive numbers you obtain even numbers at a distance of four each other, greater or equal to 10:

$n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 = 2(2n + 3) = [10, 14, 18, 22, ....]$

When you sum five consecutive numbers you obtain only multiples of 5 greater or equal to 15:

$n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 5n + 10 = 5(n + 2) = [15, 20, 25, 30, 35, ....]$

In general, when you sum x consecutive numbers you obtain:

$n + (n + 1) + (n + 2) + (n + 3) + ... + (n + x - 1) = xn + \frac {(x - 1) \times (x)} {2} = xn + \frac {(x^2 - x)} {2}$

In words it's "x" times the first number of the sequence plus the sum of the first (x - 1) integers

It's easy to see that none of the possible sequences can contain any power of 2.

So those are the only numbers you cannot obtain as a sum of two or more consecutive positive integers. In this case they are the ones from $2^0 [= 1]$ to $2^{19} [= 524288]$ for a total amount of 20 numbers