Greatest Odd Divisors

Let $d(n)$ be the largest odd divisor of $n$ .

Consider the integers $n+1, n+2, ..., 2n$ . Clearly, none of the integers in the list divide any of the others, as the ratio of any two numbers in the list is less than 2. This means that they must all have different largest odd divisors. Hence, these largest odd divisors are $n$ distinct odd integers in the range $[1,2n-1]$ . However, there are only $n$ positive odd positive integers less than 2n, so our odd parts are all the odd positive integers from $1$ to $2n-1$ , and their sum is $n^2$ .

So we have $d(n+1)+d(n+2)+...+d(2n)=n^2$ .

Note that $d(1)=1$ . Using the above fact, we find:

$d(2)=1$ ,

$d(3)+d(4)=2^2$ ,

$d(5)+d(6)+d(7)+d(8)=2^4$ ,

and so on, until:

$d(2^{2018}+1)+...+d(2^{2019})=2^{4036}$ .

Hence, $d(1)+d(2)+...+d(2^{2019})=1+(1+2^2+2^4+2^6+...+2^{4036})=1+\dfrac{1((2^2)^{2019}-1)}{3} = \dfrac{2^{4038}+2}{3}$ .

Finally, $2+4038+2+3=\boxed{4045}$ .