Robert Eisele
Engineer, Systems Architect and DBA

Project Euler 27: Quadratic primes

Problem 27

Euler discovered the remarkable quadratic formula:

\[n^2 + n + 41\]

It turns out that the formula will produce 40 primes for the consecutive integer values \(0 \le n \le 39\). However, when \(n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41\) is divisible by 41, and certainly when \(n = 41, 41^2 + 41 + 41\) is clearly divisible by 41.

The incredible formula \(n^2 - 79n + 1601\) was discovered, which produces 80 primes for the consecutive values \(0 \le n \le 79\). The product of the coefficients, -79 and 1601, is -126479.

Considering quadratics of the form:

\[n^2 + an + b,\] where \(|a| < 1000\) and \(|b| \le 1000\) and where \(|n|\) is the modulus/absolute value of \(n\), e.g. \(|11| = 11\) and \(|-4| = 4\)

Find the product of the coefficients, \(a\) and \(b\), for the quadratic expression that produces the maximum number of primes for consecutive values of \(n\), starting with \(n = 0\).

Solution

Based on the problem statement, a very simple brute force algorithm can be stated, using the isPrime function from Problem 7:

function solution() {

  var maxC = 0;
  var maxAB = 0;
  for (var a = -999; a <= 999; a++) {
    for (var b = -1000; b <= 1000; b++) {

      var c = countConsecutivePrimes(a, b);
      if (c > maxC) {
        maxC = c;
        maxAB = a * b;
      }
    }
  }
  return maxAB;
}

function countConsecutivePrimes(a, b) {

  for (var n = 0; ; n++) {
    var t = n * n + a * n + b;
    if (!isPrime(t)) {
      return n;
    }
  }
}

This sledgehammer method is quite fast already, but far from satisfying. Since \(n^2 + an + b\) must generate a series of primes, we can easily find optimizations to algorithm above. Starting with \(n_0:= 0\), the first element becomes

\[n_0^2+an_0+b = b\]

Since the quadratic equation must be prime for every element, it follows that \(b\) must be prime as well. For the next element, a similar step can be performed with \(n_1=1\):

\[n_1^2+an_1+b = 1 + a + b\]

Since \(b\) is prime, and all primes but 2 are odd, it follows that \(a\) must be odd for all \(b\neq 2\) and even for \(b= 2\). This lets us improve the algorithm as follows:

var primes = [
   2,   3,   5,   7,  11,  13,  17,  19,  23,  29,  31,  37,  41,
  43,  47,  53,  59,  61,  67,  71,  73,  79,  83,  89,  97, 101,
 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167,
 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239,
 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313,
 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397,
 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467,
 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569,
 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643,
 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733,
 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823,
 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911,
 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997];    

function solution() {

  var maxC = 0;
  var maxAB = 0;
  for (var a = -999; a <= 1001; a+= 2) {
    for (var i = 0; i < primes.length; i++) {

      var b = primes[i];
      var c = countConsecutivePrimes(a - (i == 0 ? 1 : 0), b);
      if (c > maxC) {
        maxC = c;
        maxAB = a * b;
      }
    }
  }
  return maxAB;
}

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