Challenge #12 solution, part III

And now for the solution to problem #3 from Challenge #12, which asked: how many ways are there to write a positive integer n as a sum of powers of two, with no restrictions on how many powers of two may be used?

In the comments to Challenge #12, Jonathan posted some good thoughts, and Anand posted a thorough analysis, so if you want more details you can go read those.

The main idea is to break our analysis down into two cases based on whether n is even or odd, like we did in our analysis of the solution to problem #1. If n = 2k + 1 is odd, then any way of writing n must have at least one copy of 2^0 (since otherwise the sum would be even). If we just remove one copy of 2^0, we get a way to write n - 1 = 2k; conversely, any way to write 2k can be made into a way to write 2k + 1 by adding 2^0. Therefore, letting p(n) denote the number of ways to write n as a sum of powers of two, we have p(2k + 1) = p(2k).

On the other hand, suppose n = 2k is even. For any particular way to write n, we consider two cases: either it includes at least one 2^0, in which case we can remove it to give a way to write n - 1; or it doesn’t contain 2^0, in which case we can divide everything else by 2 to yield a way to write n/2 = k. So, we have p(2k) = p(2k - 1) + p(k).

Oh, and of course, for a base case, p(1) = 1. Let’s calculate a few values of p(2k) (omitting odd values since they are simply equal to the previous value (boring!)):

\{p(2k)\}_{k=1,2,\dots} = \{ 2, 4, 6, 10, 14, 20, 26, 36, 46, \dots \}

Notice how the first two differences (4 - 2 and 6 - 4) are equal to 2; the next two differences are 4; then 6, then 10… and yes, then 14, and so on. Interesting! This self-similar structure is a direct result of the recurrence p(2k) = p(2k-1) + p(k). You can see more of the sequence on the Online Encyclopedia of Integer Sequences (one of my favorite web sites!).

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