Partitions into Consecutive Integers

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Partitions into Consecutive Integers

For any positive integer N, let f(N) denote the number of ways in which n can be expressed as the sum of consecutive positive integers.

For example, since 9 can be expressed as 2+3+4 and as 4+5 and as 9, we have f(9)=3.

The sum of the integers from 1 to n is n(n+1)/2, so we're looking for the number of ways in which a given integer N can be expressed in the form

N = n(n+1)/2 - m(m+1)/2 (1) for a positive integer n and non-negative integer m. Solving this for m gives

____________________

-1 + / 1 - 4[2N - n(n+1)]

m = ---------------------------

2 which implies that the quantity inside the square root must be a square integer. Thus there is an integer u such that

4n^2 + 4n + 1 - 8N - u^2 = 0

Clearly u must be odd. Solving this for n gives

_________

-1 + / 8N + u^2

n = ----------------

2

Again, this requires that the quantity inside the square root is a square integer, so we have an integer v such that

8N = v^2 - u^2 = (v-u)(v+u) and since u is odd, v must also be odd. Thus we can write this in the form

2N = [(v-u)/2] [(v+u)/2]

So we are looking for ways of factoring 2N into two factors, A=(v-u)/2 and B=(v+u)/2 with odd integers u,v. We have v=A+B and u=B-A, so A and B must have opposite parity, i.e., one is odd and one is even.

Thus, the integers A,B give a solution if and only if 2N = AB and one of A,B is odd and the other is even. Clearly this occurs for any odd divisor d of N, which let's us take A=d and B=2N/d (or vice versa, which ever give B > A). Consequently, the number of ways of expressing

N as a sum of consecutive positive integers is equal to the number of odd divisors of N.

For example, there are three odd divisors of N=9, namely 1, 3, and 9, so we have f(9) = 3, corresponding to the three solutions

A B v u n m expression

--- --- --- --- --- --- -------------

1 18 19 17 9 8 9

3 6 9 3 4 1 2+3+4

2 9 11 7 5 3 4+5

Incidentally, if we let F(N) denote the number of ways of expressing

N as the sum of consecutive integers, without requiring that they be positive integers, then clearly F(N) = 2f(N), because for any sum of consecutive integers extending from -m to n the total is again given by equation (1), and for each solution that ranges from m+1 to n there corresponds an equivalent solution ranging from -m to n.

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