我觉得你的结果没什么错误.其结果可以进一步简化为:

(4^n-1)/3

结果是一个关于n的函数。

Let’s use Q(1,2,3,..,2^n) to represent 各个正整数1, 2, 3, ..., 2^n 的最大的奇因数的和.

So

Q(2^0) = 1

Q(1,2^1) = 2

Q(1,2,3, 2^2) = 6

and

Q(1,2,3,..,2^n) = Q(1,,3,5,..,(2^n-1)) + Q(2,4,..,2^n)

Q(1,3,5,..,(2^n-1)) = 1+3+5+…+2^n-1 = 2^(2(n-1))

since 2,4,6,…2^n = 2(1,2,3,…, 2^(n-1)),

Q(2,4,..,2^n) = Q(1,2,3,..,2^(n-1))

So, Q(1,2,3,..,2^n) = 2^(2(n-1))+ Q(1,2,3,..,2^(n-1)) = 2^(2(n-1))+ 2^(2(n-2)) +….+1