greatest divisors!

Let $m \epsilon A$ are three cases:

1. $m=2^n \Rightarrow$ the greatest odd divisor is $1$

2. $m=2k \Rightarrow$ the greatest odd divisor is $k$

3. $m=2k-1 \Rightarrow$ the greatest odd divisor is $2k-1$

Then $S(1,2,3...2^{2015}$ sum of the greatest odd divisor each elemte of $A$

Have that

$S_{2105}=S(1,2,3...2^{2015})=S(1,3,5...2^{2015}-1)+S(2,4,6...2^{2015})$

$S(1,2,3...2^{2015})=(2^{2014})^2+S(1,2,3...2^{2014})$

$S(1,2,3...2^{2015})=(2^{2014})^2+S_{2014}$

Then $S_{2014}=S(1,3,5,..2^{2014}-1)+S(2,4,6...2^{2014})$

$S_{2014}=(2^{2013})^2+S(2,4,6...2^{2014})$

$S_{2014}=(2^{2013})^2+S_{2013}$

and so successively.

Later

$S_{2015}-S_{2014}=(2^{2014})^{2}=4^{2014}$

$S_{2014}-S_{2013}=(2^{2013})^{2}=4^{2013}$

$\vdots$

$S_{3}-S_{2}=(2^{2})^{2}=4^{2}$

$S_{2}-S_{1}=(2^{1})^{2}=4^{1}$

$\Rightarrow (S_{2015}-S_{2014})+(S_{2014}-S_{2013})+...+(S_{3}-S_{2})+(S_{2}-S_{1})=S_{2015}-S_{1}$

$S_{2015}-S_{1}=4^{2014}+4^{2013}+4^{2012}...4^{3}+4^{2}+4=\frac{4^{2015}-4}{4-1}$

$S_{2015}-2=\frac{4^{2015}-4}{3}$

$S_{2015}=\frac{4^{2015}-4}{3}+2$

$S_{2015}=\frac{4^{2015}+2}{3}$

So

$a=2015$

$b=3$

$\boxed{a+b=4+2015+3=2022}$

We have to "extract" the highest power of $2$ from each element of $A$ to get the highest odd divisor of each of the elements.

For the odd elements the highest power of $2$ is of course $0$ , so we add all the odd numbers, $2^{2014}$ of them in total, to get

$1 + 3 + 5 + ..... + (2^{2015} - 1) = (2^{2014})^{2} = 4^{2014}$ ,

since the sum of the first $n$ odd natural numbers is $n^{2}$ .

Next, divide by $2$ those elements of $A$ for which the highest power of $2$ in their prime factorization is $1$ . What remains is

$1 + 3 + 5 + ..... + (2^{2014} - 1) = (2^{2013})^{2} = 4^{2013}$ .

We then deal with the elements of $A$ for which the highest power of $2$ in their prime factorizations is $2$ . What remains is

$1 + 3 + 5 + ..... + (2^{2013} - 1) = (2^{2012})^{2} = 4^{2012}$ .

We continue this process for successively higher powers $k$ of $2$ , in each case the sum of highest odd divisors being $4^{2014 - k}$ for $0 \le k \le 2014$ . We thus end up with the geometric series

$4^{2014} + 4^{2013} + 4^{2012} + ..... + 4^{2} + 4^{1} + 4^{0} = \dfrac{4^{2015} - 1}{3}.$

Finally we need to consider $2^{2015}$ itself, the highest odd divisor being of course $1$ . So the desired sum is

$\dfrac{4^{2015} - 1}{3} + 1 = \dfrac{4^{2015} + 2}{3}$ ,

and so $a + b + c = 4 + 2015 + 3 = \boxed{2022}$ .