Digit sum of powers in bases

CJam, 13 bytes

q~,f#\fb::+$p

I think this can be improved, but lets get this started.

The input is in the following order : b a n

Example input:

3 2 4

Output:

[1 2 2 4]

Code expansion:

q~             "Read the input and parse it. Now we have on stack, b a and n"
  ,            "Get an array from 0 to n-1";
   f#          "Get all the powers of a present in the array, i.e. 0 to n-1";
     \         "Swap the array of a^i and bring the base on top";
      fb       "For each number in the array, convert it into its base representation array";
        ::+    "Calculate the sum of each of the sub arrays in the bigger array";
           $p  "Sort and print the sum of digits of base representation of n powers of a";

Try it online here

Pyth, 11

Smsj^vzdQvw

Note that this requires the inputs separated by newlines in the order a,b,n.

Try it online here

Pyth has built-ins that do each step for us. We map over each value from 0 to n-1, raising a to that power. Then we use j to convert each of these numbers to base b. We sum the resulting lists, and the final list of numbers is Sorted.

The input is read in the order z, Q then w. z and w have to be evaled in order to be used as numbers.

JavaScript (ES6) 170 175 196 98 100

Edit New version, handles bigger numbers using digits array. Score almost doubled :-(

Multiplying digit by digit is more difficult, summing the digits is simpler.

F=(a,b,n)=>
  (t=>{
    for(r=t;--n;r[n]=s)
      for(k=1,y=t,z=a;z;z=z/b|0)
        t=[...t.map(v=>(v=c+(k?w*~~y[i++]+v:w*v),s+=d=v%b,c=v/b|0,d),
           w=z%b,s=c=0,i=--k),c]       
  })([1])||r.sort((a,b)=>a-b)

Less golfed

F=(a, b, n) =>
{
  var r=[1]; // power 0 in position 0

  for(w=[]; w.push(a%b),a=a/b|0; ); // a in base b

  for(t = [1]; --n; )
  {
    // calc next power, meanwhile compute digits sum
    y=[...t]
    k=1
    w.map(w=>
      (t=[...t.map(v=>(
        v = c + (k?w*~~y[i++]+v:w*v), 
        d = v % b, // current digit
        s += d, // digits sum
        c = (v-d)/b, // carry
        d)
       ,s=c=0,i=--k),c]
      )
    ); 
    r[n] = s // store sum in result array (positions n-1..1)
  }
  return r.sort((a,b)=>a-b) // sort result in numeric order
}

My first attempt using doubles, fails for big numbers. JS doubles have 52 mantissa bits when for instance 15^14 needs 55 bits.

F=(a,b,n,r=[1])=>(x=>{for(;--n;r[n]=s)for(t=x*=a,s=0;t;t=(t-d)/b)s+=d=t%b})(1)
||r.sort((a,b)=>a-b)

Less golfed

F=(a, b, n) =>
{
  var r=[1]; // power 0 in position 0
  for(x = 1; --n; )
  {
    x *= a; // calc next power starting at 1
    for(t = x, s = 0; t; t = (t-d) / b) // loop to find digits
    {
      d = t % b;
      s += d;
    }
    r[n] = s // store sum in result array (positions n-1..1)
  }
  return r.sort((a,b)=>a-b) // sort result in numeric order
}

Test In Firefox/FireBug console

;[[2,5,10],[3,2,20],[5,6,1],[6,6,11],[6,8,20],[8,2,10],[15,17,18]]
.forEach(t=>console.log(...t,F(...t)))

2 5 10 [1, 2, 4, 4, 4, 4, 4, 4, 8, 8]
3 2 20 [1, 2, 2, 3, 4, 5, 6, 6, 6, 8, 9, 10, 11, 11, 13, 14, 14, 14, 15, 17]
5 6 1 [1]
6 6 11 [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
6 8 20 [1, 6, 6, 8, 8, 8, 13, 13, 15, 20, 22, 22, 22, 27, 29, 34, 34, 34, 36, 41]
8 2 10 [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
15 17 18 [1, 15, 17, 31, 31, 33, 49, 49, 63, 65, 81, 95, 95, 95, 95, 113, 127, 129]