yichao chen, jonathan l. gross and toufik mansour- genus distributions of circular ladders: a...

8/3/2019 Yichao Chen, Jonathan L. Gross and Toufik Mansour- Genus Distributions of Circular Ladders: A Website Supplement

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GENUS DISTRIBUTIONS OF CIRCULAR LADDERS:

A WEBSITE SUPPLEMENT

YICHAO CHEN, JONATHAN L. GROSS, AND TOUFIK MANSOUR

Abstract. The genus distribution of the circular ladder CLn is re-derived here, using overlap

matrices and Chebyshev polynomials.

1. The genus distributions of circular ladders

We define two sets of matrices and two polynomials:

(1) An+2 as the set of all matrices over Z2 of the form Mc,x,y,X,Y,Z

n+2 ;(2) An+2(z) =

n+2j=0 An+2(j)z

j as the rank-distribution polynomial of the set An+2, thatis, of the overlap matrices for general rotation systems for the circular ladder CLn;

(3) Bn+2 as the set of all matrices of the form Mc,x,y,0,Y,Zn+2 ; and

(4) Bn+2(z) =n+2

j=0 Bn+2(j)zj as the rank-distribution polynomial of the set Bn+2, that

is, of the overlap matrices for pure rotation systems for the circular ladder CLn.

In a matrix of the form Mc,x,y,X,Y,Zn+2 , suppose that we first add the second row to the first rowand next add the second column to the first column. Without changing the rank of the matrix,this produces a matrix of the following form:

xe c + xf x 0 0 0 yc + xf xf z1 z2

zn2 zn1 zn

x z1 x1 y1

0 z2 y1 x2. . . 0

......

. . .. . .

0 zn2 0 yn20 zn1 0 yn2 xn1 yn1y zn yn1 xn

For xy {00, 01, 10, 11}, we define two more sets and two more polynomials:(1) A

xyn+2 as the set of all matrices over Z2 of the form M

c,x,y,X,Y,Zn+2 ;

(2) Axyn+2(z) =

n+2j=0 A

xyn+2(j)z

j as the rank-distribution polynomial of the set Axyn+2;

(3) Bxyn+2 as the set of all matrices of the form Mc,x,y,0,Y,Zn+2 ; and

(4) Bxy

n+2(z) =n+2

j=0 Bxy

n+2(j)zj

as the rank-distribution polynomial of the set Bxy

n+2.

2000 Mathematics Subject Classification. Primary: 05C10; Secondary: 30B70, 42C05.Key words and phrases. graph embedding; total embedding distribution; circular ladders; overlap matrix;

Chebyshev polynomials.

1


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2 YICHAO CHEN, JONATHAN L. GROSS, AND TOUFIK MANSOUR

Clearly, we have the following property.

Property 1.1. For all n 1,An+2(z) =

xy=00,01,10,11

Axyn+2(z) andBn+2(z) =

xy=00,01,10,11

Bxyn+2(z).

For the vectors X = (x0, x1, . . . , xn), Y = (y1, y2, . . . , yn1), and Z = (z1, z2, . . . , zn), withxi, yj , zk Z2, we define the matrices

(1) MX,Yn =

x1 y1y1 x2 y2 0

y2 x3 y3. . .

. . .. . .

0 yn2 xn1 yn1yn1 xn

and

(2) MX,Y,Zn+1 =

x0 z1 z2 z3 . . . zn1 znz1 x1 y1z2 y1 x2 y2 0

z3 y2 x3. . .

.... . .

. . . yn2zn1 0 yn2 xn1 yn1

zn yn1 xn

.

As described by [1] and [3], every overlap matrix of the closed-end ladder Ln1 has the form

MX,Yn+1 , and every overlap matrix of the Ringel ladder Rn1 has the form MX,Y,Zn+1 . (Note that

the subscripts ofRn1 and MX,Y,Zn+1 differ by two.) Now we further define

(1) Pn as the set of all matrices over Z2 of the form M0,Y,Zn ;

(2) Cn(j) as the number of overlap matrices for the Ringel ladder Rn that are of rank j;

(3) Pn(z) =n+1

j=0 Cn(j)zj as the rank-distribution polynomialof the set Pn; and

(4) On(z) as the rank-distribution polynomial of the overlap matrices of the closed-endladder Ln1 over the set of matrices of the form M

0,Yn .

Theorem 1.2. (see [2]) The rank-distribution polynomialOn(z) of the overlap matrix for theclosed-end ladder Ln1 satisfies the recurrence relation

On(z) = On1(z) + 2z2On2(z)

with the initial conditions O1(z) = 1 andO2(z) = z2 + 1. Moreover, the generating function

O(t; z) =

n1 On(z)tn is given by

O(t; z) =t + z2t2

1 t 2z2t2 .


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GENUS DISTRIBUTIONS OF CIRCULAR LADDERS: A WEBSITE SUPPLEMENT 3

Theorem 1.3. [3] LetOn1(z) be the rank-distribution polynomial of the closed-end ladderLn2. Then the polynomialPn(z) (n 3) satisfies the recurrence relation

Pn+1(z) = Pn(z) + 8z2Pn1(z) + 2

n1z2On1(z).(3)

with the initial conditionP2(z) = z2 +1, P3(z) = 7z

2 +1 andP4(z) = 12z4 +19z2 +1. Moreover,

the generating functionP(t; z) =

n2 Pn(z)tn is given by

P(t; z) =t2(1 + z2 2(1 2z2)t 4z2(4 + z2)t2 32z4t3)

(1 2t 8z2t2)(1 t 8z2t2) .

1.1. Rank-distribution polynomial for circular ladders. Let MO,Y,Zevenn+1 be the number

of matrices such that the number of elements of Z equals to 1 is even and MO,Y,Zoddn+1 be the

number of matrices such that the number of elements of Z equals to 1 is odd. Let POn+1(z) be

the rank distribution polynomial over the set MO,Y,Zoddn+1 and PEn+1(z) be the rank distribution

polynomial over the set MO,Y,Zeven

n+1 . By the definition, we have Pn(z) = PO

n (z) + PE

n (z).

Lemma 1.4. The polynomialPEn (z) (n 4) satisfies the recurrence relation

PEn+1(z) = P

En (z) + 4z

2Pn1(z) + 2

n2z2On1(z).

with the initial conditionPE3 (z) = 3z2 + 1 andPE4 (z) = 6z

4 + 9z2 + 1, where On1(z) andPn(z)are the rank-distribution polynomials of the closed-end ladder Ln2 and the Ringel ladder Rn2,respectively.

Proof. The proof has two cases.

Case 1: yn1 = 0. If zn = 0, the matrix contributes a term PEn (z); otherwise, if zn = 1,it contributes a term 2n2z2On1(z).

Case 2: yn1 = 1. Ifzn = 0, we add the n +1-st column and row to the first and n1-stcolumn and row, which transforms the above matrix (2) to the following form:

0 z1 z2 zn2 0 0z1 0 y1

z2 y1 0. . . 0

.... . .

. . .

zn20 0 0 1

0 1 0

,

Depending whether zn1 = 0 or 1, this matrix contributes either 2z2P

En1(z) or 2z

2POn1(z)

to PEn+1(z). Otherwise zn = 1, and we add the last column and row to the first column


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and row, and thereby obtain the following matrix.

0 z1 z2 zn2 0 1z1 0 y1

z2 y1 0

. .. 0...

. . .. . .

zn2 0 yn20 0 yn2 0 11 1 0

,

(1) yn2 = 0. In this subcase, according to the values of zn1, it contributes to PEn+1(z)

by a term z2POn1(z) or z2PEn1(z)

(2) yn2 = 1. We add the last row to the n 1-st row in this subcase. Depending onthe value of zn1, it contributes either z

2POn1(z) or z

2PEn1(z):

0 z1 z2 zn2 + 1 0 1z1 0 y1

z2 y1 0 . . . 0...

. . .. . .

zn2 + 10 0 0 11 1 0

,

Proposition 1.5. For all n 3,

PEn (z) =

1

2(Pn(z) + 1 z2).

Moreover, the generating functionPE(t; z) = n3

PE

n

(z)tn is given by

t3(1 + 3z2 3(1 z2)(1 + 2z2)t + 2(1 10z2 13z4)t2 + 8z2(3 2z2 4z4)t3 + 64z4t4)(1 2t 8z2t2)(1 t 8z2t2)(1 t) .

Proof. By Lemma 1.4 and together with induction on n, we have

PEn (z) = P

E4 (z) +

n2j=3

(4z2Pj(z) + 2j1z2Oj(z)).

Thus, by (3) we obtain

PEn (z) = P

E4 (z) +

1

2

n2

j=3

(Pj+2(z) Pj+1(z)) = PE4 (z) +1

2(Pn(z) P4(z))

=1

2(Pn(z) + 1 z2),

as claimed. The rest follows immediately from Theorem 1.3.


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Lemma 1.6. The polynomialB00n (z) (n 4) equals

B00n+2(z) = P

En+1(z) + 2

n1z2On(z) =1

2(Pn+1(z) + 1 z2) + 2n1z2On(z).(4)

where On1(z) is the rank-distribution polynomial of closed-end ladders Ln2 andPEn (z) is therank-distribution polynomial over the set MO,Y,Zevenn .

Proof. If c = 0, in this case, it contributes the term PEn+1(z) to B00n+2(z). Otherwise c = 1,

and since there are 2n1 possible choices of z1, z2, , zn such that the number of variablesin {z1, z2, , zn} equal to 1 is odd, it contributes 2n1z2On(z) to B00n+2(z). The rest followsimmediately by Proposition 1.5.

Lemma 1.7. The polynomialB10n (z) (n 4) is given by the formula

B10n+2(z) = 4z

2Pn(z),

where Pn(z) is the rank-distribution polynomial of the Ringel ladder Rn2.

Proof. Multiply row 3 by the constant c and add to the second row, Similarly, Multiply column3 by the constant c and add the result to the second row. If z1 = 1, we add the first row tothe third row and then add the first column to the third column. There is a similar situationfor y1. We interchange row 2 with row 3 and column 2 with column 3, and thereby transformMc,1,0,0,Y,Zn+2 to the following form:

0 11 0 0

0 z2 z3 zn1 znz2 0 y2

0 z3 y2 0 . . . 0...

. . .. . .

zn1 0 0 yn1zn yn1 0

.

Note that the lower-right n n submatrix has the form M0,Y,Zn . There are 22 different as-signments of the variables y1 and z1 in this case, so it contributes 4z

2Pn(z) to the polynomialB10n+2(z).

By symmetry, we have the following lemma.

Lemma 1.8. The polynomialB01n

(z) (n

4) satisfies this formula:

B01n+2(z) = 4z

2Pn(z),

where Pn(z) is the rank-distribution polynomial of the Ringel ladder Rn2.


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Let

MX,Y,Z,1n+1 =

x0 z1 z2 z3 . . . zn1 znz1 x1 y1 0 . . . 0 1z2 y1 x2 y2

z3 0 y2 x3 . . . 0...

.... . .

. . . yn2zn1 0 yn2 xn1 yn1

zn 1 yn1 xn

, where

X = (x0, x1, . . . , xn) Zn+12 , Y = (y1, y2, . . . , yn1) Zn12 , and Z = (z1, z2, . . . , zn) Zn2 .

We let Qn+1 be the set of all matrices over Z2 of the form MO,Y,Z,1n+1 , and we let

Qn+1(z) =n+1j=0

Qn+1(j)zj

be the rank-distribution polynomialof the set Qn+1.

We now introduce some further notations:

MO,Y,Zeven,1n+1 is the number of matrices with evenly many elements of Z equal to 1; MO,Y,Zodd,1n+1 is the number of matrices with oddly many elements of Z equal to 1; QOn+1(z) is the rank-distribution polynomial over the set MO,Y,Zoddn+1 ; QEn+1(z) is the rank-distribution polynomial over the set MO,Y,Zeven,1n+1 .

Its clear that Qn+1(z) = QEn+1(z) + Q

On+1(z).

Similarly, we have further notations:

Hn+1 is the set of all matrices over Z2 tof the form M

X,Y,Z,1n+1 ;

Hn+1(z) = n+1j=0 Hn+1(j)zj is the rank-distribution polynomialof the set Qn+1; MX,Y,Zeven,1n+1 is the number of matrices with evenly many elements of Z equal to 1; QEn+1(z) is the rank-distribution polynomial of the set MX,Y,Zeven,1n+1 ; MX,Y,Zodd,1n+1 is the number of matrices such that oddly many elements of Z equal 1; and HOn+1(z) is the rank-distribution polynomial of the set MX,Y,Zoddn+1 .

Lemma 1.9. The polynomialQEn (z) (n 4) satisfies the recurrence relation

QEn+1(z) = 6z

2Pn1(z) + 4z

2QEn1(z)

with the initial conditions QE2 (z) = 1 andQE3 (z) = 1+3z

2, where Pn1(z) is the rank-distributionpolynomial of the Ringel ladder Rn3.

Proof. There are four cases.

(1) Case 1: zn = 0, yn1 = 0. Ifz1 = 1, we first add the last row to the first row, and thenadd the last column to the first column. By a similar discussion for y1, we can transform


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MO,Y,Z,1n+1 into the following forms:

0 z1 z2 z3 . . . zn1 0z1 0 y1 0 . . . 0 1z2 y1 0 y2

z3 y2 0 . . . 0... 0

. . .. . . yn2

zn1 yn2 0 00 1 0 . . . 0 0 0

and

0 0 z2 z3 . . . zn1 00 0 0 0 . . . 0 1

z2 0 0 y2

z3 y2 0. . . 0

... 0. . .

. . . yn2

zn1 yn2 0 00 1 0 . . . 0 0 0

.

Note that there are four different ways to assign the variables y1 and z1 in the matrixMO,Y,Z,1n+1 . If z1 = 0, the matrix contributes 2z

2PEn1(z) to the polynomial QEn+1(z).

Otherwise z1 = 1, and it contributes a term 2z2POn1(z). Since Pn1(z) = P

En1(z) +

POn1(z), it contributes 2z2Pn1(z) to the polynomial Q

En+1(z).

(2) Case 2: zn = 1, yn1 = 0. We first add the second row to the first row, and then add the

second column to the first column. This transforms MO,Y,Z,1n+1 into the following form:

0 z1 z2 + y1 z3 . . . zn1 0z1 0 y1 0 . . . 0 1

z2 + y1 y1 0 y2

z3 y2 0. . . 0

... 0. . .

. . . yn2zn1 yn2 0 0

0 1 0 . . . 0 0 0

.

If z1 = 1, we first add the last row to the first row, and then add the last column tothe first column. By similar discussion regarding y1, we can transform the above matrixinto the following form:

0 0 z2 + y1 z3 . . . zn1 00 0 0 0 . . . 0 1

z2 + y1 0 0 y2

z3 y2 0. . . 0

..

. 0

. ..

. .. yn2

zn1 yn2 0 00 1 0 . . . 0 0 0

.

Accordingly, this case contributes 2z2Pn1(z) to the polynomial QEn+1(z).


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(3) Case 3: zn = 0, yn1 = 1.

0 z1 z2 z3 . . . zn2 zn1 0z1 0 y1 0 . . . 0 0 1z2 y1 0 y2 0

z3 0 y2 0. . . 0 ...

......

. . .. . .

zn2 0 0 yn2 0zn1 0 0 yn2 0 1

0 1 0 . . . 0 1 0

,

We add the n-th row to the second row, then add the n-th column to the second column,the resulted matrix has the following form:

0 z1 + zn1 z2 z3 . . . zn2 zn1 0z1 + zn1 0 y1 0 . . . yn2 0 0

z2 y1 0 y2 0

z3 0 y2 0.. . 0

.

.....

.... . .

. . .

zn2 yn2 0 yn2 0zn1 0 0 yn2 0 1

0 0 0 . . . 0 1 0

,

No matter what the assignments of the values of zn1 and yn2, we can transfer theabove matrix to the following form:

0 z1 + zn1 z2 z3 . . . zn2 0 0z1 + zn1 0 y1 0 . . . yn2 0 0

z2 y1 0 y2 0

z3 0 y2 0. . . 0

...

......

. . . . . .

zn2 yn2 0 0 00 0 0 0 0 10 0 0 . . . 0 1 0

,

If yn2 = 0, it contributes to the polynomial QEn+1(z) by a term 2z

2PEn1(z). Otherwise

yn2 = 1, this case contributes to the polynomial QEn+1(z) by a term 2z

2QEn1(z).

(4) Case 4: zn = 1, yn1 = 1.

0 z1 z2 z3 . . . zn1 1z1 0 y1 0 . . . 0 1z2 y1 0 y2

z3 y2 0

. .. 0... 0

. . .. . . yn2

zn1 yn2 0 11 1 0 . . . 0 1 0

.


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We first add the nth row to the first and second row, then add the nth column tothe first and second column, the resulting matrix has the following form:

0 z1 + zn1 z2 z3 . . . zn2 + yn2 zn1 0

z1 + zn1 0 y1 0 . . . yn2 0 0z2 y1 0 y2 0

z3 0 y2 0. . . 0

......

.... . .

. . .

zn2 + yn2 yn2 0 yn2 0zn1 0 0 yn2 0 1

0 0 0 . . . 0 1 0

.

If yn2 = 0, it contributes 2z2POn1(z) to the polynomial Q

En+1(z). Otherwise yn2 = 1,

this case contributes 2z2QEn1(z) to the polynomial QEn+1(z).

Proposition 1.10. The generating functionQE

(t; z) =

n2 Q

E

n (z)tn

is given by

t2(1 (2 3z2)t (1 + 19z2 6z4)t2 + 2(1 + 4z2)(1 3z2)t3 + 8z2(3 + 5z2 3z4)t4 + 64z4t5)(1 2t 8z2t2)(1 t 8z2t2)(1 4z2t2) .

Proof. Multiplying the recurrence relation in the statement of Lemma 1.9 by tn+1 and summingover all n 3 with using Theorem 1.3, we attain the result.

Lemma 1.11. The polynomialB11n (z) (n 4) equals

B11n+2(z) = 2z

2Pn(z) + 4z

2QEn (z),

where Pn(z) is the rank-distribution polynomial of the Ringel ladder Rn2, and where QEn (z) is

the rank distribution polynomial over the set MO,Y,Zeven,1n .

Proof. In this case, the matrix has the following form:

0 c 1 0 1c 0 z1 z2 zn2 zn1 zn1 z1 0 y1

z2 y1 0. . . 0

0...

. . .. . .

zn2 0 yn2zn1 0 yn2 0 yn1

1 zn yn1 0

.

Multiply the last row constant c and add to the second row. Similarly, multiply the last columnconstant c and add to the second column. A similar discussion to the row (column) 3. Theabove matrix is transformed into this:


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0 0 0 0 0 0 10 0 z1 + zn z2 zn2 zn1 + cyn1 zn0 z1 + zn 0 y1 0 yn1 0

0 z2 y1 0.. . 0

......

.... . .

. . .

0 zn2 0 0 yn20 zn1 + cyn1 yn1 0 yn2 0 yn11 zn 0 yn1 0

.

(1) Case 1: c = 0. There are four subcases. Subcase 1: zn = 0, yn1 = 0. Note that the down-right submatrix is the matrix

the form MO,Y,Zevenn . In this case, it contributes to the polynomial B11n+2(z) by a

term z2PEn (z). Subcase 2: zn = 1, yn1 = 0. We first add the first row to the second row, then

add the first column to the second column. Note that the down-right submatrixis the matrix the form MO,Y,Zevenn . In this case, it contributes to the polynomialB11n+2(z) by a term z2PEn (z).

Subcase 3: zn = 0, yn1 = 1. We first delete the first row and the last row, thendelete the first column and the last column. At last we obtain a matrix of the formMO,Y,Zeven,1n1 . In this case, it contributes to the polynomial B

11n+2(z) by a term

z2QEn1(z).

0 0 0 0 0 0 10 0 z1 z2 zn2 zn1 00 z1 0 y1 0 1 00 z2 y1 0

. . . 0...

......

. . .. . .

0 zn2 0 0 yn2

0 zn1 1 0 yn2 0 11 0 0 1 0

.

Subcase 4: zn = 1, yn1 = 1. We first delete the first row and the last row, thendelete the first column and the last column. At last we obtain a matrix of theform MO,Y,Zeven,1n1 . In this case, it contributes to the polynomial B

11n+2(z) by a term

z2QEn1(z).(2) Case 2: c = 1. In this case also, we have four subcases:

0 0 0 0 0 0 10 0 z1 + zn z2 zn2 zn1 + cyn1 zn0 z1 + zn 0 y1 0 yn1 00 z2 y1 0

. . . 0

... ... ... . . . . . .0 zn2 0 0 yn20 zn1 + cyn1 yn1 0 yn2 0 yn11 zn 0 yn1 0

.


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Subcase 1: zn = 0, yn1 = 0. Note that the down-right submatrix is the matrix theform MO,Y,Zoddn . In this case, it contributes to the polynomial B

11n+2(z) by a term

z2POn (z). Subcase 2: zn = 1, yn1 = 0. We first add the first row to the second row, then add

the first column to the second column. Note that the down-right submatrix is thematrix the form MO,Y,Zoddn . In this case, it contributes to the polynomial B11n+2(z)

by a term z2POn (z). Subcase 3: zn = 0, yn1 = 1. We first delete the first row and the last row, then

delete the first column and the last column. At last we obtain a matrix of the formMO,Y,Zeven,1n1 . In this case, it contributes to the polynomial B

11n+2(z) by a term

z2QEn1(z):

0 0 0 0 0 0 10 0 z1 z2 zn2 zn1 + 1 00 z1 0 y1 0 1 00 z2 y1 0

. . . 0...

.

.....

.. .

.. .

0 zn2 0 0 yn20 zn1 + 1 1 0 yn2 0 11 0 0 1 0

.

Subcase 4: zn = 1, yn1 = 1. We first delete the first row and the last row, thendelete the first column and the last column. At last we obtain a matrix of the formMO,Y,Zeven,1n1 . In this case, it contributes to the polynomial B

11n+2(z) by a term

z2QEn1(z):

0 0 0 0 0 0 10 0 z1 + 1 z2 zn2 zn1 + 1 10 z1 + 1 0 y1

0 1 0

0 z2 y1 0 . . . 0...

......

. . .. . .

0 zn2 0 0 yn20 zn1 + 1 1 0 yn2 0 11 1 0 1 0

.

Theorem 1.12. For all n 5,

Bn(z) =1

2(Pn(z) + 1 z2) + QEn (z).

Moreover, the generating functionB(t; z) =

n5 Bn(z)tn is given by

t5f(t; z)

(1 2t 8z2t2)(1 t 8z2t2)(1 t)(1 4z2t2) ,


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where

f(t; z) = 100z4 + 27z2 + 1 + (z2 1)(160z4 + 76z2 + 3)t + (30z2 640z4 + 2 1312z6)t2 4z2(213z2 7 + 60z4 + 352z6)t3 + 8z2(15z2 + 448z4 + 528z6 1)t4

+ 32z4(96z6 3 + 68z4 65z2)t5 256z6(1 + 19z2 + 12z4)t6.

Proof. By Property 1.1 and Lemmas 1.6, 1.7, 1.8 and 1.11, we obtain Bn(z) = PEn (z) + QEn (z),

and by Proposition 1.5, we obtain Bn(z) =12 (Pn(z) + 1 z2) + QEn (z). Multiplying by tn

and summing over all n 5 with using Theorem 1.3 and Proposition 1.10, we complete theproof.

1.2. The genus polynomial of circular ladders.

Theorem 1.13. We have

n4 gCLn(x)tn = 2

tB(t;

x). Moreover, for all n 4, the number

of distinct cellular imbeddings of CLn in a surface of genus j is

7n+jj

23j3nj1j1

+ n4j+2

j23j3

nj+1j1

+ n

j1 2n+j1

njj2

+2n1n,2j+2 + 2

nn,2j+1 3 2n1n,2j j 2,2n + 8n 2 + 8n,4 j = 1,2 j = 0.

Proof. Let GCL(t; x) =

n4 gCLn(x)tn. Then by Theorem 1.12 we have

GCL(t; x) =2t4f(t; x)

(1 2t 8xt2)(1 t 8xt2)(1 t)(1 4xt2) ,

where

f(t; z) = 100x2 + 27x + 1 + (x

1)(160x2 + 76x + 3)t + (30x

640x2 + 2

1312x3)t2

4x(213x 7 + 60x2 + 352x3)t3 + 8x(15x + 448x2 + 528x3 1)t4+ 32x2(96x3 3 + 68x2 65x)t5 256x3(1 + 19x + 12x2)t6.

By several algebraic operations, we calculate that the coefficient of xj in GCL(t; x) is given by

(8t315t23t+2)23j3t2j2(1t)j+1 + (13t+2t2)23j2t2j2

(12t)j+1 + (2t2 + 2t + 3/2)22jt2j j 2,

2t4(27+50t28t373t)(12t)(1t)2

j = 1,2t4

1tj = 0.

Hence, the coefficient of xjtn in GCL(t; x) is given by

23j3 8nj1j + 15njj + 3nj+1j

2nj+2j +2n+j1

2nj+2

j

3nj+1j

+njj

+ 2n1n,2j+2 + 2

nn,2j+1 3 2n1n,2j j 2,2n + 8n 2 + 8n,4 j = 1,2 j = 0.


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which is equivalent to

7n+jj

23j3nj1j1

+ n4j+2

j23j3

nj+1j1

+ n

j1 2n+j1

njj2

+2n1n,2j+2 + 2

nn,2j+1 3 2n1n,2j j 2,

2

n

+ 8n 2 + 8n,4 j = 1,2 j = 0.as claimed.

Note that it is not hard to check that our formula in the above theorem is equivalent to theformula in [4, Theorem 3.10].

As a corollary of the above theorem we can obtain the following result.

Corollary 1.14. For all n 4,

gCLn(x) = 1 x +1 3x 2x

4x(2x)n + 1 3x + 2

x

4x(2

x)n

+ 2nx(i2x)n

Un

12i

2x

Un2

1

2i

2x

+ (1 x)(2i

2x)n

Un

1

4i

2x

Un2

1

4i

2x

,

where Us is the s-th Chebyshev polynomial of the second kind and i2 = 1

Proof. By Theorem 1.13 we infer that the generating function f =

n4 gCLn(x)tn can be

written as

f = 2(1 + 19x + 12x2)t3 4(1 + 3x)t2 4t 12

x(1 + 3x 2x2) + 1 3x + 4xt2x(1 4xt2)

+ 2x(1 t)1 2t 8xt2 + 2 2x + xt t1 t 8xt2 + 1 x1 t .

Let n 4, so the coefficient gCLn(x) of tn in the generating function f is given by

gCLn(x) = 1 x +1 3x 2x

4x(2x)n + 1 3x + 2

x

4x(2

x)n

+ 2nx(i

2x)n

Un

1

2i

2x

Un2

1

2i

2x

+ (1 x)(2i

2x)n

Un

1

4i

2x

Un2

1

4i

2x

,

where the first, second and third lines are the coefficients of tn in the generating functions

1 x1 t +

1 3x + 4xt2x(1 4xt2) ,

2x(1 t)1 2t 8xt2 , and

2 2x + xt t1 t 8xt2 ,

respectively.


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References

[1] J. Chen, J. L. Gross, and R. G. Rieper, Overlap matrices and total embeddings, Discrete Math. 128 (1994)

7394.[2] Y. Chen, T. Mansour, and Q. Zou, Embedding distributions and Chebyshev polynomial, Graphs and Com-

bin., DOI: 10.1007/s00373-011-1075-5.[3] Y. Chen, L. Ou, and Q. Zou, Total embedding distributions of Ringel ladders, Discrete Math. 311 (2011)

24632474.[4] L. A. McGeoch, Genus distribution for circular and Mobius ladders, Technical report extracted from PhD

thesis, Carnegie-Mellon University, 1987.

College of Mathematics and Econometrics, Hunan University, 410082 Changsha, China

E-mail address: [email protected]

Department of Computer Science, Columbia University, New York, NY 10027 USA


Department of Mathematics, University of Haifa, 31905 Haifa, Israel


yichao chen, jonathan l. gross and toufik mansour- genus distributions of circular ladders: a...

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