Ways of making polite numbers? Given that $n$ is a polite number, meaning that it can be expressed as the sum of two or more consecutive positive integers, how many different ways are there to express $n$ as the sum of at least two consecutive positive integers? Can you prove that the method you describe works?
 A: The page you linked to answers your question:

The politeness of a positive number is defined as the number of ways it can be expressed as the sum of consecutive integers. For every x, the politeness of x equals the number of odd divisors of x that are greater than one.

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An easy way of calculating the politeness of a positive number is that of decomposing the number into its prime factors, taking the powers of all prime factors greater than 2, adding 1 to all of them, multiplying the numbers thus obtained with each other and subtracting 1.

I guess I'll say a word or two about why looking at odd divisors works.  If $x$ is a sum of an odd number $k$ of consecutive positive integers with average term $d$, then $x=kd$.  Conversely, if $k>1$ is an odd divisor of $x$, and $x=kd$ with $d-\frac{k-1}{2} > 0$, then $x$ is the sum of $k$ consecutive integers with average term $d$.  For example, if $x=6$, then this process relates the odd divisor $k=3$ of $x$ with the three-term sum centered at $d=\frac{6}{3}=2$, namely $6=1+2+3$.
It seems like we need a different bijection for sums of an even number of consecutive positive integers.  But actually, we can force the same bijection to work anyway, just by canceling all the negative terms.  For example, if $x=27$ and $k=9$, then we naively get a nine-term sum centered at $d=\frac{27}{9}=3$, namely $27 = (-1)+0+1+2+3+4+5+6+7$.  But throwing away the first three terms gives us $27=2+3+4+5+6+7$.  So this even-length sum corresponds to the odd divisor $9$ of $27$.
A: Check out the section Politeness here Politeness
