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() {

        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 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

        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

  int mod(int reps) {
    int tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += LongMath.mod(longs[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 += LongMath.mod(nonnegative[j], positive[j]);
    }
    return tmp;
  }

  @Benchmark

// all the ellipses sizes.
func getDivisibleSize(totalSizes []uint64) (result uint64) {
	gcd := func(x, y uint64) uint64 {
		for y != 0 {
			x, y = y, x%y
		}
		return x
	}
	result = totalSizes[0]
	for i := 1; i < len(totalSizes); i++ {
		result = gcd(result, totalSizes[i])
	}
	return result
}

// isValidSetSize - checks whether given count is a valid set size for erasure coding.

    //     moved at that point. Otherwise, we can rotate the cycle a[1], a[1 + d], a[1 + 2d], etc,
    //     then a[2] etc, and so on until we have rotated all elements. There are gcd(d, n) cycles
    //     in all.
    // (3) "Successive". We can consider that we are exchanging a block of size d (a[0..d-1]) with a
    //     block of size n-d (a[d..n-1]), where in general these blocks have different sizes. If we

57785ca45b8873032f17 GCD = 42 A = 0 B = 42 LCM = 0 GCD = 42 A = 42 B = 0 LCM = 0 GCD = 42 A = 42 B = 42 LCM = 42 GCD = f60d A = ef7886c3391407529d5c B = d1d3ec32fa3103911830 LCM = cc376ed2dc362c38a45a GCD = 9370 A = 1ee02fb1c02100d1937f B = 67432fd1482d19c4a1c2 LCM = 159ff177bdb0ffbd09e2 GCD = c5f A = 5a3a2088b5c759420ed0 B = 1b1eb33b006a98178bb3 LCM = c5cbbbe9532d30d2a7dd GCD = e052 A = 67429f79b2ec3847cfc7 B = 39faa7cbdeb78f9028c1 LCM = 1ab071fb733ef142e94d GCD = 3523 A = 0 B = 3523 LCM = 0 GCD = 3523 A = 3523...