### Sums of Consecutive Counting Numbers

There is a cool little proof by Wai Yan Pong titled Sums of Consecutive IntegersHere is a slightly longer way of saying the same thing (this is basically the thought process I went through to understand the Pong proof).

## Terminology

We will call a consecutive series of counting numbers that sums to n a "decomposition" of n. If the decomposition has m members, we will say that the decomposition is of length m.

E.g. the series 2, 3, 4 is a decomposition of 9, and it has a length of 3.

We will call the number n itself the "trivial decomposition" of n. That is, n = n, and n is a series of counting numbers of length 1.

## Question

What counting numbers have no decomposition other than the trivial decomposition?

## Approach

We will proceed by showing that (1) a decomposition of n can be constructed using a procedure that starts with an odd factor of n. Next, we will prove that (2) every decomposition of n can be constructed this way.  Finally, we will observe that (3) for odd factors greater than 1, the constructed decomposition will have a length greater than 1. Therefore, (4) the counting numbers that have no decomposition other than the trivial decomposition are the counting numbers that have no odd factors other than 1, a.k.a. the powers of 2.

## By design, the sum of this series is mk = n. This means that if m-(k-1)/2 is positive, the series is a consecutive series of counting numbers that sums to n, which would make the series a decomposition of n. For example, if we are looking for a decomposition of 75, we might consider the odd factor 5, and then construct the series of 5 consecutive integers centered on 15, which is 13, 14, 15, 16, 17. This series sums to 75, and all of the numbers in the series are counting numbers, hence the series is a decomposition of 75.

So this suggests that odd factors are an interesting place to start when looking for decompositions. But sometimes m-(k-1)/2 will not be positive. For example, let us again assume we are looking for a decomposition of 75. Instead of considering the odd factor 5, we might consider the odd factor 15. We then would construct a series of 15 consecutive integers centered on 5, which is -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. This series sums to 75, but it has a lot of numbers that aren't counting numbers.

It turns out that there is a way to fix this problem: We can eliminate the first several members of the series that together sum to 0, which is to say, eliminate the negative numbers, the 0, and the additive inverses of the eliminated negative numbers, leaving 3, 4, 5, 6, 7, 8, 9, 10, 11, 12:

-2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Because the eliminated numbers sum to 0, the remaining series still sums to 75, and all of the numbers in this series are counting numbers, so we still have a decomposition of 75. Generalizing, if m-(k-1)/2 is not positive, we remove this sub-series:

## which leaves us with:

(k-1)/2-m+1, (k-1)/2-m+2, . . . m+(k-1)/2

which is a decomposition of n. The removed sub-series has -2 (m-(k-1)/2) + 1 terms, which simplifies to -2m+k, and thus the decomposition has a length of 2m, which is also 2n/k

Either way, starting with an odd factor, we have constructed a decomposition of n, which we will call the decomposition of associated with k.

## (2) There is no decomposition of n that is not associated with an odd factor k.

It is interesting (and hopefully now fairly intuitive) that we can construct a decomposition of n using the odd factors of n. But might there are some decompositions of n that cannot be created this way? As it turns out the answer is "no." We will show this to be the case by starting with an arbitrary decomposition of n and then will show that there is an odd factor of k with which this decomposition is associated.

We will call our arbitrary decomposition D. This decomposition starts with some counting number a and is of length j:

a, a+1, . . . a+j-1

We will show that this is precisely the decomposition of n associated with some odd factor k.

Because D is a decomposition of n, this means:

n = a + a+1 + . . . + a+j-1

The right side of that equation has terms, and the average of the terms is (a a+j-1)/2. Thus:

n = j(a + (a+j-1))/2
= j(2a+j-1)/2

If j is odd, this means j is an odd factor of nIf j is even, then j/2 is a whole number, which means (2a+j-1) is a factor of n. And because is even, it follows that (2a+j-1) is odd

Let us consider both possibilities, and verify that in either case, the decomposition of n associated with whichever of those is an odd factor of n is in fact the decomposition D.

(2a) If is odd, then the decomposition of n associated with is D.

Consider the case where is odd. Let us construct the decomposition associated with j. Thus, = j, and m = (2a+j-1)/2. We then construct this series:

## m-(k-1)/2, m-(k-1)/2+1, . . . m-(k-1)/2

Is the first term of that series positive?

## m-(k-1)/2 = (2a+j-1)/2 - (j-1)/2    = a+j/2-1/2 - j/2+1/2    = a

Not only is the first term positive, it is a. Thus, if is odd, the decomposition of associated with starts with a and has length j, which is D.

(2b) If j is not odd, then (2a+j-1) is odd, and then the decomposition of associated with (2a+j-1) is D.

Now consider the other case, where j is even. If so, then (2a+j-1) is odd. Let us construct the decomposition associated with (2a+j-1)Thus, k = (2a+j-1)and m= j/2We then construct this series:

## m-(k-1)/2, m-(k-1)/2+1, . . . m-(k-1)/2

Is the first term of that series positive?

m-(k-1)/2 = j/2-((2a+j-1)-1)/2
= j/2-(2a+j-2)/2
= j/2-(a+j/2-1)
= j/2-a-j/2+1
= -a+1

Because a is a counting number, this means -a+1 is not positive. We therefore know that the decomposition of associated with (2a+j-1) begins with the term (k-1)/2-m+1:

(k-1)/2-m+1 =  ( (2a+j-1) -1)/2 - j/2 + 1
=  (2a+j-1)/2 -1/2 - j/2 + 1
= a+j/2-1/2 -1/2 - j/2 + 1
= a

We also know that the decomposition of associated with (2a+j-1) has a length of 2n/(2a+j-1). Because = j(2a+j-1)/2, it follows that 2n/j = 2a+j-1, and thus that 2n/j-j+1 = 2a. Substituting for 2a yields:

2n/(2a+j-1) = 2n/(2n/j-j+1+j-1)
= 2n/(2n/j)
= j

Therefore, if j is even, the decomposition of associated with (2a-j-1) starts with and has length jwhich is D.

Thus, not only is it true that every odd factor of n can be used to construct a decomposition of n, it also is true that every decomposition of n is associated with an odd factor of n.

## (3) A counting number n that has at least one odd factor other than 1 has a non-trivial decomposition.

Recall that the decomposition of associated with k is either of length k or 2m. This means that the decomposition of always has a length > 1 if k > 1. That is, any counting number n with an odd factor other than 1 has a non-trivial decomposition!

And the counting numbers with no odd factors other than 1 are the powers of 2 (because all prime numbers other than 2 are odd, and thus all counting numbers other than powers of 2 have odd factors).

Moreover, for n = a power of 2, n has only one odd factor, 1, and that odd factor is associated with the trivial decomposition. Thus, powers of 2 never have a non-trivial decomposition.

(4) Therefore:

The counting numbers with no decomposition other than the trivial decomposition are the powers of 2.