* 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

     * 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();

pkg math, func Jn(int, float64) float64
pkg math, func Ldexp(float64, int) float64
pkg math, func Lgamma(float64) (float64, int)
pkg math, func Log(float64) float64
pkg math, func Log10(float64) float64
pkg math, func Log1p(float64) float64
pkg math, func Log2(float64) float64
pkg math, func Logb(float64) float64
pkg math, func Max(float64, float64) float64
pkg math, func Min(float64, float64) float64

        `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`,

pkg math, const Ln2 = 0.693147  // 693147180559945309417232121458176568075500134360255254120680009/1000000000000000000000000000000000000000000000000000000000000000
pkg math, const Log10E = 0.434294  // 10000000000000000000000000000000000000000000000000000000000000/23025850929940456840179914546843642076011014886287729760333279