Take any number x (x>1), with p sets of consecutive integers with k elements adding up to x.
Note: for every k there is only one p

If you multiply any even x by 2 (or 2^n), p raises with the amount of uneven divisors of x.

This is because, as partly shown in the solution for the number 2016, in all situations:
1) consider k uneven, the set has an average value of x/k, being an integer. Multiplying a number by 2^n doesn't give it more (or less) possibilities of multiplying prime factors and getting an uneven number.
2) consider k even. x/k has to be a/2, with a an integer. Thus, to have an even amount of numbers in any p x has to be uneven. So, if x is already even and you multiply it by 2, you don't increase nor decrease the amount of values of k with k being even.

Conclusion: if the prime factorisation of x is 2^m*... with m>1, you can divide it by 2 without changing p.

Since 294912 = 2^15 * 3^2, 294912 has the same value for p as 2 * 3^2 = 18.
If we divide 18 by all integers lower than sqrt(18) (1, 2, 3 and 4), we'll see that k can only be 1 or 3. (fyi: {294912} and {98303,98304,98305}