Already two computer solutions and neither one even tried anything other than dumb brute force. There is a massive amount of simplification that can be done before getting the number crunching involved.
D(n) can be defined as a two-part recursive relation:
D(2k+1) = 2k+1 and D(2k)=D(k)
Then let S(n) be the sum of D(1) through D(n), then the problem asks for S(2048). S(1)=1 and to get larger values we can do some simple algebra.
Split the sum for S(n) into its odd terms and even terms:
{D(1)+D(3)+...+D(n or n-1 depending on parity)}
+ {D(2)+D(4)+...+D(n or n-1 depending on parity)}
The first half will have ceil[n/2] terms and the second half will have floor[n/2] terms
After applying D(2k+1) = 2k+1, the first half is the sum of the first ceil[n/2] odd numbers, which is simply ceil[n/2]^2.
Then on the second half applying D(2k)=D(k) reduces that down to the summation for S(floor[n/2])
We now have a recursive definition for S(n): S(n) = ceil[n/2]^2 + S(floor[n/2]).
Now to just evaluate at n=2048:
S(2048) = 1024^2 + S(1024)
= 1024^2 + 512^2 + S(512)
= 1024^2 + 512^2 + 256^2 + S(256)
...
= 1024^2 + 512^2 + 256^2 + 128^2 + 64^2 + 32^2 + 16^2 + 8^2 + 4^2 + 2^2 + 1^2 + 1
= 1398102.
Solution
