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    <p>Based on the requirements, you can do it in different ways. If I was doing it, I would calculate the prime numbers using 'Sieve of Eratosthenes' in the required range (A to (9*2)*B.length), cache them (again, depending on your setup, you can use in-memory or disk cache) and use it for the next run.</p> <p>I just coded a fast solution (Java), as below (<strong>NOTE</strong>: Integer overflow is not checked. Just a fast example. Also, my code is not optimized.):</p> <pre><code>import java.util.ArrayList; import java.util.Arrays; public class LuckyNumbers { public static void main(String[] args) { int a = 0, b = 1000; LuckyNumbers luckyNums = new LuckyNumbers(); ArrayList&lt;Integer&gt; luckyList = luckyNums.findLuckyNums(a, b); System.out.println(luckyList); } private ArrayList&lt;Integer&gt; findLuckyNums(int a, int b) { ArrayList&lt;Integer&gt; luckyList = new ArrayList&lt;Integer&gt;(); int size = ("" + b).length(); int maxNum = 81 * 4; //9*2*b.length() - 9 is used, coz it's the max digit System.out.println("Size : " + size + " MaxNum : " + maxNum); boolean[] primeArray = sieve(maxNum); for(int i=a;i&lt;=b;i++) { String num = "" + i; int sumDigits = 0; int sumSquareDigits = 0; for(int j=0;j&lt;num.length();j++) { int digit = Integer.valueOf("" + num.charAt(j)); sumDigits += digit; sumSquareDigits += Math.pow(digit, 2); } if(primeArray[sumDigits] &amp;&amp; primeArray[sumSquareDigits]) { luckyList.add(i); } } return luckyList; } private boolean[] sieve(int n) { boolean[] prime = new boolean[n + 1]; Arrays.fill(prime, true); prime[0] = false; prime[1] = false; int m = (int) Math.sqrt(n); for (int i = 2; i &lt;= m; i++) { if (prime[i]) { for (int k = i * i; k &lt;= n; k += i) { prime[k] = false; } } } return prime; } } </code></pre> <p>And the output was:</p> <p>[11, 12, 14, 16, 21, 23, 25, 32, 38, 41, 49, 52, 56, 58, 61, 65, 83, 85, 94, 101, 102, 104, 106, 110, 111, 113, 119, 120, 131, 133, 137, 140, 146, 160, 164, 166, 173, 179, 191, 197, 199, 201, 203, 205, 210, 223, 229, 230, 232, 250, 289, 292, 298, 302, 308, 311, 313, 317, 320, 322, 331, 335, 337, 344, 346, 353, 355, 364, 368, 371, 373, 377, 379, 380, 386, 388, 397, 401, 409, 410, 416, 434, 436, 443, 449, 461, 463, 467, 476, 490, 494, 502, 506, 508, 520, 533, 535, 553, 559, 560, 566, 580, 595, 601, 605, 610, 614, 616, 634, 638, 641, 643, 647, 650, 656, 661, 665, 674, 683, 689, 698, 713, 719, 731, 733, 737, 739, 746, 764, 773, 779, 791, 793, 797, 803, 805, 829, 830, 836, 838, 850, 863, 869, 883, 892, 896, 904, 911, 917, 919, 922, 928, 937, 940, 944, 955, 968, 971, 973, 977, 982, 986, 991]</p>
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    1. POAlgorithm to find Lucky Numbers
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    1. COUnfortunately, your algorithm will be still slow where (B - A) -> 10^18, cause it is still linear
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    2. COIn my implementation I have used Eratosthenes method. That's trivial. The main thing to be optimized is traversing all numbers and calculating sum of their digits because the numbers range from 1 to 10^18.
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    3. COthe range of primes is actually much smaller
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