Finding the total number of set-bits from 1 to n

Question

Write an algorithm to find F(n) the number of bits set to 1, in all numbers from 1 to n for any given value of n.

Complexity should be O(log n)

Answer 1

this is coded in java...
logic: say number is 34, binary equal-ant is 10010, which can be written as 10000 + 10. 10000 has 4 zeros, so count of all 1's before this number is 2^4(reason below). so count is 2^4 + 2^1 + 1(number it self). so answer is 35.
*for binary number 10000. total combinations of filling 4 places is 2*2*2*2x2(one or zero). So total combinations of ones is 2*2*2*2.

public static int getOnesCount(int number) {
    String binary = Integer.toBinaryString(number);
    return getOnesCount(binary);
}

private static int getOnesCount(String binary) {
    int i = binary.length();

    if (i>0 && binary.charAt(0) == '1') {
        return gePowerof2(i) + getOnesCount(binary.substring(1));
    }else if(i==0)
        return 1;
    else
        return getOnesCount(binary.substring(1));

}
//can be replaced with any default power function
private static int gePowerof2(int i){
    int count = 1;
    while(i>1){
        count*=2;
        i--;
    }
    return count;
}