Genera and crossing numbers of $2$-bridge knots

In this paper, we determine the average genus of all the $2$-bridge knots with a given crossing number. As a consequence, we obtain the oblique asymptote of this value as the crossing number grows.


Introduction
Dunfield provided experimental data in [6] which suggests that the genus of any knot grows linearly with respect to its crossing number.It is known that the family of 2-bridge knots is a fundamental class of knots.Many knot properties are first studied for 2-bridge knots before for the general case.
Let g c be the average genus of all the 2-bridge knots with c crossings.Motivated by Dunfield's result, Cohen in [2] [3] gave lower and upper bounds on g c in terms of c.In this paper, we will determine the explicit value of g c .As a consequence, we can obtain the oblique asymptote of g c .That is to say, the following is the main theorem of this paper.
In particular, g c ∼ This result supports Dunfield's observation affirmatively.Cohen and Lowrance [4] obtained the same asymptote independently.

Preliminary
In this section, we review some known facts about 2-bridge knots from [8], [5], and show some fundamental formulas about binomial coefficients.
It is known that a 2-bridge knot corresponds to a rational number.We denote by K([a 1 , a 2 , . . ., a n ]) the 2-bridge knot determined by a continued fraction [a 1 , a 2 , . . ., a n ] = 1 a 1 + The genus of the 2-bridge knot corresponding to a rational number is a half of the length of the even continued fraction.Namely, if we take the even continued fraction [2a 1 , 2a 2 , . . ., 2a 2m ] of a rational number where a i = 0, then the genus of K = K([2a 1 , 2a 2 , . . ., 2a 2m ]) is m.Remark that the lengths of the even continued fractions for 2-bridge knots are always even.Moreover, the crossing number c of K is given by where ℓ is the number of sign changes in the sequence (2a 1 , 2a 2 , . . ., 2a 2m ) (See [9], [11], and [10]).Since ),ℓ be the set of all the 2-bridge knots K([2a 1 , 2a 2 , . . ., 2a 2m ]) such that a i = ±b i and that the number of sign changes in the sequence (a 1 , a 2 , . . ., a 2m ) is exactly ℓ.For example, we have ).These two knots are distinct if and only if (2a 1 , 2a 2 , . . ., 2a 2m ) is not symmetric.
In [7], Ernst and Sumners determined the number of 2-bridge knots with respect to crossing number.Theorem 2.1 (Ernst-Sumners [7]).For c ≥ 3, the number of all the 2-bridge knots with c crossings is given by .
In this paper, we make use of several identities on binomial coefficients in order to prove Theorem 1.In addition to well-known formulas: we prepare the following.
Lemma 2.3.We have . Then their derivatives with respect to x are a a−b , the derivative of α n with respect to x is given by , we obtain the first desired equality.Since β n = α n+1 − xα n , we have By n q=0 q 2 q 2n−q q = 2β ′ n | x=2 , we also obtain the second desired equality.

Total genus
In this section, we compute the total genus of 2-bridge knots with a given crossing number, which is denoted by T G(c) for a given crossing number c.By definition, we have where c(K) and g(K) are the crossing number and the genus of a knot K respectively.Proposition 3.1.For c ≥ 3, we have .
Combining Theorem 2.1 and Proposition 3.1, we obtain Theorem 1.We will divide the proof of Proposition 3.1 into the cases of even crossing number and odd crossing number.).The number of ways of choosing a i = ±b i such that the number of sign changes among (a 1 , a 2 , . . ., a 2m ) is exactly ℓ = 2l is equal to twice the number of solutions (d 1 , d 2 , . . ., d ℓ+1 ) ∈ (Z ≥1 ) ℓ+1 of the equation and so it is equal to ) is symmetric (this only occurs when k + l is even and the number of symmetric cases is equal to (k+l)/2−1 m−1 ) also belong to S b,ℓ .We need to consider the two cases where (2a and the total genus, namely, the total number of the genera of these 2-bridge knots is If k + l is even, then there are both cases where (b 1 , b 2 , . . ., b 2m ) is symmetric and not symmetric.Therefore the number of 2-bridge knots with crossing number c = 2k and number of sign changes ℓ = 2l is equal to and the total genus is As a consequence, A l and B l are the same regardless of the parity of k + l.
In order to simplify B l , we compute 2B l − (2l + 1)A l and A l as follows: When we take the sum of B l with respect to l, B l depends on the three cases, 0 By Lemma 2.2, Lemma 2.3 and c = 2k, we conclude Therefore Proposition 3.1 for the case of even crossing number is proved.).Note that (−2a 2m , −2a 2m−1 , . . ., −2a 1 ) = (2a 1 , 2a 2 , . . ., 2a 2m ) if and only if (2a 1 , 2a 2 , . . ., ) is not symmetric.Therefore the number of 2-bridge knots with crossing number c = 2k + 1 and number of sign changes ℓ = 2l + 1 is equal to and the total genus is If k + l is odd, then there are both cases where (b 1 , b 2 , . . ., b 2m ) is symmetric and not symmetric.Therefore the number of 2-bridge knots with crossing number c = 2k + 1 and the number of sign changes ℓ = 2l + 1 is equal to and the total genus is Hence we obtain the total genus Moreover, we simplify this expression by using binomial coefficient formulas.When we take the sum of B l with respect to l of the first term, B l depends on the three cases, 0 Similarly, if l + k + 1 ≡ 0 (mod 2), then Then we obtain Furthermore, the first term of the above expression can be simplified by using the following: The third term depends on the parity of k.In both cases, Lemma 2.2 and Lemma 2. It is easy to see that g c = g * c for odd crossing numbers, since T K * (2k+1) = 1 2 T K(2k+1) and T G * (2k + 1) = 1  2 T G(2k + 1).As a final example, we show Table 1 on the values which are provided in this paper.

Table 1 .
g c and g *