* is close to 1 (why?), gives the following formula.
     */
    double fractionOfBitsSet = (double) bitCount / bitSize;
    return DoubleMath.roundToLong(
        -Math.log1p(-fractionOfBitsSet) * bitSize / numHashFunctions, RoundingMode.HALF_UP);
  }

  /** Returns the number of bits in the underlying bit array. */
  @VisibleForTesting
  long bitSize() {
    return bits.bitSize();

  @GwtIncompatible // TODO
  public void testLog10HalfEven() {
    for (BigInteger x : POSITIVE_BIGINTEGER_CANDIDATES) {
      int halfEven = BigIntegerMath.log10(x, HALF_EVEN);
      // Now figure out what rounding mode we should behave like (it depends if FLOOR was
      // odd/even).
      boolean floorWasEven = (BigIntegerMath.log10(x, FLOOR) & 1) == 0;

     * can narrow the possible floor(log10(x)) values to two. For example, if floor(log2(x)) is 6,
     * then 64 <= x < 128, so floor(log10(x)) is either 1 or 2.
     */
    int y = maxLog10ForLeadingZeros[Long.numberOfLeadingZeros(x)];
    /*
     * y is the higher of the two possible values of floor(log10(x)). If x < 10^y, then we want the

        `tf.math.greater`, `tf.math.greater_equal`, `tf.math.igamma`,
        `tf.math.igammac`, `tf.math.invert_permutation`, `tf.math.less`,
        `tf.math.less_equal`, `tf.math.lgamma`, `tf.math.log`, `tf.math.log1p`,
        `tf.math.logical_and`, `tf.math.logical_not`, `tf.math.logical_or`,
        `tf.math.maximum`, `tf.math.minimum`, `tf.math.not_equal`,
        `tf.math.polygamma`, `tf.math.reciprocal`, `tf.math.rint`,