Which numbers were used in this sum ?

converting 1627622 to binary 110001101010111100110 , now expanding this binary form in powers of 2 , we get it in same form as given question where
a1 =20
a2 =19
a3 =15
a4 =14
a5 =12
a6 = 10
a7 = 8
a8 = 7
a9 = 6
a10 = 5
a11 = 2
a12 = 1
hence ΣAn + n = 119 + 12 = 131

We observe that $( 2^1 + 2^2 + 2^3 ... 2^{n-1} ) + 2 = 2^{n}$ .
Thus $2^{n}$ is more than sum of 2's of all powers less than $n$ .
Here $2^{a_x}$ where $a_x$ is maximum of all $a$ 's will be more than $\frac {1627622}{2}$ .

Now $2^{21} = 2097152$ and $2^{20} = 1048576$ . $2^{21}$ is more than the sum i.e. $1627622$ but $2^{20}$ is the maximum power of 2 less than the sum .

So there exists some $a_x = 20$ for x in 1 to n .

Now we know $2^{20}$ exists so we will minus that.
$1627622 - 1048576 = 579046$

Similarly now we will check the max power of 2 that can be accommodated in $579046$ . It is $2^{19} = 524288$ .
Minus that.
Now we have $54758$ . The max power of 2 that can be accommodated is $2^{15} = 32768$ .
.....
The same process can continue and all powers of 2 that were used can be extracted.
In the sum 1627622 , the powers of 2 ( $a_x$ ) that were used are -- 20 19 15 14 12 10 8 7 6 5 2 1 .

The $n$ is 12 and the sum of $a_x$ where $x=1..12$ is 119 .
So answer = 131

Here is the code in AutoHotkey which I wrote to solve this problem.

msgbox % getParams(l := 1627622, 0)
msgbox % z := getParams(l, 1)
f := 0
loop, parse, z, %A_Space%
    f+=A_LoopField+1
msgbox % f
return

getParams(sum, r=0){
    static a := 1
    while sum>0
        loop
        {
            a*=2
            if (a>sum)
            {
                a/=2,p.= (r ? Round(log(a)/log(2)) : Round(a)) " ",sum-=a,a:=1
                break
            }
        }
    return Substr(p,1,-1)
}

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import java.util.ArrayList;
import java.util.List;
import java.util.Collections;
public class x
{
    public static void main(String[] args)
    {
        ArrayList<Integer> x = new ArrayList<Integer>();
        int i = 1627622;
        int k = 50;
        while(i>0)
        {
            if(Math.pow(2,k) <= i)
            {
                i-=Math.pow(2,k);
                x.add(k);
            }
            k--;
        }
        System.out.println(x);
    }
}

Adding, we get $20+19+15+14+12+10+8+7+6+5+2+1+12=\boxed{131}$

Here is my elegant C code to solve this problem :

int main(){    
    int x=1627622,i=0,s=0;
    int i=0;
    while (x!=0){
        i++;
        if (x%2==1)
            s+=i;
        x/=2;
    }
    printf("%d",s);
}

Explanation : We need to convert 1627622 to binary , then we have to count the sum of ones in its binary representation added to the sum of their positions in the representation , in other words : their powers . Since each time we find a one we add to our sum its position plus one for its presence : s+=i+1 , or we can , as I wrote in my code , initiate the first value of i to 0 and increment it as we enter the while loop , so the counting starts with 1.

We are being asked to express this number as a sum of base two logarithms. It's easy to find that ${ 2 }^{ 22 }$ exceeds the number. Therefore, what we need to do is to start with ${ 2 }^{ 22 }$ and remove successively smaller powers of two, if present in the number, and noting their presence. Here is one possible way of doing that.

>>> indices = [ ]

>>> r=1627622

>>> for k in range ( 22, -1, -1 ) :

... term = 2 ** k

... if term < r :

... r -= term

... indices. append(k)

... elif term == r :

... indices.append(k)

... break

... else :

... pass

...

>>> indices

[20, 19, 15, 14, 12, 10, 8, 7, 6, 5, 2, 1]

>>> sum+len(indices)

131