Trying to solve Project Euler problem 119:

The number 512 is interesting because it is equal to the sum of its digits raised to some power: 5 + 1 + 2 = 8, and 8^3 = 512. Another example of a number with this property is 614656 = 28^4.

We shall define an to be the nth term of this sequence and insist that a number must contain at least two digits to have a sum.

You are given that a2 = 512 and a10 = 614656.

Find a30.

Question: Is there a more efficient way to find the answer than just checking every number until a30 is found?

My Code

    int currentNum = 0;
    long value = 0;
    for (long a = 11; currentNum != 30; a++){ //maybe a++ is inefficient
        int test = Util.sumDigits(a);
        if (isPower(a, test)) {
            currentNum++;
            value = a;
            System.out.println(value + ":" + currentNum);
        }
    }
    System.out.println(value);

isPower checks if a is a power of test. Util.sumDigits:

    public static int sumDigits(long n){
        int sum = 0;
        String s = "" + n;
        while (!s.equals("")){
            sum += Integer.parseInt("" + s.charAt(0));
            s = s.substring(1);
        }
        return sum;
    }

program has been running for about 30 minutes (might be overflow on the long). Output (so far):

81:1

512:2

2401:3

4913:4

5832:5

17576:6

19683:7

234256:8

390625:9

614656:10

1679616:11

17210368:12

34012224:13

52521875:14

60466176:15

205962976:16

612220032:17