Richard Williamson proved that all odd numbers are polite: This is because any odd number can be expressed as an even number + 1 and hence is the sum of half that even number, and the next integer. For instance, 13 = 6 + 7, 45 = 22 + 23, etc, where 6 and 22 are half of 12 and 44 respectively. As an odd number can always be expressed as the sum of two consecutive integers, and as this satisfies the definition of politeness, all odd numbers are polite.

Richard also found the consecutive sums for 544 and 434 and the way of characterizing impolite numbers, namely that they are all powers of 2. All other integers are polite. Shu Cao of the Oxford High School for Girls also proved this result.

James Newton of Hills Road Sixth Form College, Cambridge proved this result and also gave a nice generalization defining a 'degree of politeness' of numbers.

The desired sequence of integers has sum $0.5 n (2 a + n - 1)$ where $n$ is the number of terms, $a$ is the initial value. If we are looking for a sequence of integers that sum to 544, $n (2 a + n - 1) = 544 \times 2 = 1088 = 2^{6} \times 17$ . A sensible value for n would therefore be 17. This means that $a = 24$ , so $544 = 24 + 25 + 26 + . . . + 39 + 40$ . Using the same procedure on 424 yields $424 = 19 + 20 + 21 + . . . + 34$ . Hypothesis: all numbers of the form $2^{m}$ (where $m$ is a positive integer) are impolite. Proof: The desired sequence of integers which add up to give a polite number has the sum $0.5 n (2 a + n - 1)$ where $n ≧ 2$ and one of the factors $n$ or $(2 a + n - 1)$ must be odd and the other even. Thus every polite number has an odd factor greater than or equal to 3. If the number is a power of 2 then it does not have an odd factor so it is impolite. Hypothesis: all numbers not of the form $2^{m}$ are polite If $k$ is a polite number then $2 k = n^{2} + (2 a - 1) n$ is equal to $2^{j} \times r$ where $r$ is odd (as above). Set whichever is the smaller of $2^{j}$ and $r$ equal to $n$ . Then either (1) $2^{j} + 2 a - 1 = r$ if $2^{j} < r$ or (2) $r + 2 a - 1 = 2^{j}$ if $2^{j} > r$ . In case (1) $2 a = r + 1 - 2^{j}$ . This is greater than 0, as $2^{j} < r$ , and is also even, as $2^{j} - 1$ is an odd number and $r$ is an odd number. (The difference of two odd numbers is even). There therefore exists an integer value of $a$ . Similarly in case (2) $2 a = 2^{j} + 1 - r$ which is a positive even number. This proves that all numbers not of the form $2^{m}$ are polite, and all those of the form $2^{m}$ are impolite. It is interesting to note that numbers in which $r$ is not prime may have two or more different sequences of integers which sum to them. For example

$\begin{matrix} 140 & = 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 \\ = 26 + 27 + 28 + 29 + 30 \\ = 17 + 18 + 19 + 20 + 21 + 22 + 23 . \end{matrix}$

An extension of the above problem is to find a general formula for the 'degree of politeness' of a number. 140 has politeness 3, as it can be broken down into 3 different sums of consecutive integers. In general the politeness of a number is $2^{k} - 1$ where $k$ is the number of prime factors of $r$ if $r$ is less than $2^{j}$ in the context used above. Repeated prime factors are counted repeatedly, for example $3^{5}$ contributes 5 to the number of prime factors. The number of different values for n, with 1 excluded, is given by the number of possible divisors of $r$ . The situation is more complicated if some of the factors of $r$ are greater than $2^{j}$ and some less, but the politeness will still be equal to the number of different divisors of $r$ which are less than $2^{j}$ . If $2^{j}$ is less than all the factors of $r$ then the politeness of the number will be equal to 1.