for (int b : POSITIVE_INTEGER_CANDIDATES) {
        assertEquals(valueOf(a).gcd(valueOf(b)), valueOf(IntMath.gcd(a, b)));
      }
    }
  }

  public void testGCDZero() {
    for (int a : POSITIVE_INTEGER_CANDIDATES) {
      assertEquals(a, IntMath.gcd(a, 0));
      assertEquals(a, IntMath.gcd(0, a));
    }
    assertEquals(0, IntMath.gcd(0, 0));
  }

  public void testGCDNegativePositiveThrows() {

    int bTwos = Integer.numberOfTrailingZeros(b);
    b >>= bTwos; // divide out all 2s
    while (a != b) { // both a, b are odd
      // The key to the binary GCD algorithm is as follows:
      // Both a and b are odd. Assume a > b; then gcd(a - b, b) = gcd(a, b).
      // But in gcd(a - b, b), a - b is even and b is odd, so we can divide out powers of two.

      // We bend over backwards to avoid branching, adapting a technique from

        assertEquals(valueOf(a).gcd(valueOf(b)), valueOf(LongMath.gcd(a, b)));
      }
    }
  }

  @GwtIncompatible // TODO
  public void testGCDZero() {
    for (long a : POSITIVE_LONG_CANDIDATES) {
      assertEquals(a, LongMath.gcd(a, 0));
      assertEquals(a, LongMath.gcd(0, a));
    }
    assertEquals(0, LongMath.gcd(0, 0));
  }

  @GwtIncompatible // TODO

      int j = i & ARRAY_MASK;
      tmp += IntMath.mod(ints[j], positive[j]);
    }
    return tmp;
  }

  @Benchmark
  int gCD(int reps) {
    int tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += IntMath.gcd(nonnegative[j], positive[j]);
    }
    return tmp;
  }

  @Benchmark
  int factorial(int reps) {
    int tmp = 0;

      int j = i & ARRAY_MASK;
      tmp += IntMath.mod(ints[j], positive[j]);
    }
    return tmp;
  }

  @Benchmark
  int gCD(int reps) {
    int tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += IntMath.gcd(nonnegative[j], positive[j]);
    }
    return tmp;
  }

  @Benchmark
  int factorial(int reps) {
    int tmp = 0;

1024 drives. In this scenario 16 becomes the erasure set size. This is decided based on the greatest common divisor (GCD) of acceptable erasure set sizes ranging from *4 to 16*.

- *If total drives has many common divisors the algorithm chooses the minimum amounts of erasure sets possible for a erasure set size of any N*.  In the example with 1024 drives - 4, 8, 16 are GCD factors. With 16 drives we get a total of 64 possible sets, with 8 drives we get a total of 128 possible sets, with 4...

      for (int b : POSITIVE_INTEGER_CANDIDATES) {
        assertEquals(valueOf(a).gcd(valueOf(b)), valueOf(IntMath.gcd(a, b)));
      }
    }
  }

  public void testGCDZero() {
    for (int a : POSITIVE_INTEGER_CANDIDATES) {
      assertEquals(a, IntMath.gcd(a, 0));
      assertEquals(a, IntMath.gcd(0, a));
    }
    assertEquals(0, IntMath.gcd(0, 0));
  }

  public void testGCDNegativePositiveThrows() {

    int bTwos = Long.numberOfTrailingZeros(b);
    b >>= bTwos; // divide out all 2s
    while (a != b) { // both a, b are odd
      // The key to the binary GCD algorithm is as follows:
      // Both a and b are odd. Assume a > b; then gcd(a - b, b) = gcd(a, b).
      // But in gcd(a - b, b), a - b is even and b is odd, so we can divide out powers of two.

      // We bend over backwards to avoid branching, adapting a technique from

        assertEquals(valueOf(a).gcd(valueOf(b)), valueOf(LongMath.gcd(a, b)));
      }
    }
  }

  @GwtIncompatible // TODO
  public void testGCDZero() {
    for (long a : POSITIVE_LONG_CANDIDATES) {
      assertEquals(a, LongMath.gcd(a, 0));
      assertEquals(a, LongMath.gcd(0, a));
    }
    assertEquals(0, LongMath.gcd(0, 0));
  }

  @GwtIncompatible // TODO

        return DoubleMath.factorial(n);
      }

      @Override
      public int gcdInt(int a, int b) {
        return IntMath.gcd(a, b);
      }

      @Override
      public long gcdLong(long a, long b) {
        return LongMath.gcd(a, b);
      }

      @Override
      public long binomialCoefficient(int n, int k) {
        return LongMath.binomial(n, k);
      }

      @Override