Base Polynomials for Schultz Invariants of [620034]
Base Polynomials for Schultz Invariants of
n-Bilinear Straight Pentachain
Abdul Rauf Nizami, Khurram Shabbir and Muhammad Qasim
Abstract. The Schultz and modied Schultz polynomials were intro-
duced by Gutman in 2005 for a connected graph GasH1(G;x) =
1
2P
fv;ug2V(u+v)xd(u;v)andH2(G;x) =1
2P
fv;ug2V(uv)xd(u;v),
whereduis the degree of the vertex uandd(u;v) is the distance between
uandv. In this paper we give lengths of paths in terms of polynomials,
which ultimately serve as bases for Schultz polynomials and indices of
then-bilinear straight pentachain.
Subject Classication (2010) . 05C12; 05C07; 05C31
Keywords . Bilinear straight pentachain; Schultz polynomials; Schultz
indices
1. Introduction
AgraphGis a pair (V;E), whereVis the set of vertices and Ethe set of
edges. The edge ebetween two vertices uandvis denoted by ( u;v). The
degree of a vertex u, denoted by duis the number of edges incident to it. A
path from a vertex vto a vertex wis a sequence of vertices and edges that
starts from vand stops at w. The number of edges in a path is the length of
that path. A graph is said to be connected if there is a path between any two
of its vertices, as you can see in Figure 1.
/0/0 /1/1/0/0 /1/1
/0/0 /1/1/0/0 /1/1 /0/0 /1/1
/0/0 /1/1/0/0 /1/1
/0/0 /1/1 /0/0 /1/1
1 2 345 6 7 8
vv v v v
v v v v9
Figure 1. A connected graph with a highlighted shortest
path fromv1tov5, and withd(v1) = 4 andd(v5) = 3
Amolecular graph is a representation of a chemical compound in terms of
graph theory. Specically, molecular graph is a graph whose vertices corre-
spond to (carbon) atoms of the compound and whose edges correspond to
chemical bonds. For instance, the molecular graph of 1-pentene C5H10is
given in Figure 2:
2 Nizami, Khurram, and Qasim
ab
c
d e
Figure 2. 1-pentene
Denition 1.1. A function Iwhich assigns to every connected graph Ga
unique number I(G) is called a graph invariant . Instead of the function Iit
is custom to say the number I(G) as the invariant. An invariant of a molecular
graph which can be used to determine structure-property or structure-activity
correlation is called the topological index . A topological index is said to be
degree (distance) based if it depends on degrees (distance) of the vertices of
the graph.
In 1989 Harry Schultz introduced the Schultz index in [9] and was further
studied in [10, 5, 3].
Denition 1.2. [9] The Schultz index of Gis dened as
S(G) =1
2X
fv;ug2V(u+v)d(u;v)
Hereduis the degree of the vertex uandu6=v.
In 1997 Klavzar and Gutman introduced the modied Schultz index.
Denition 1.3. [7] The modied Schultz index of Gis dened as
MS(G) =1
2X
fv;ug2V(uv)d(u;v)
Hereduis the degree of the vertex uandu6=v.
The Schultz and modied Schultz polynomials were introduced by Gutman in
2005 and found some relations of these polynomials with Wiener polynomial
of trees [6].
Denition 1.4. [6] The Schultz and modied Schultz polynomials of Gare
dened respectively as:
H1(G;x) =1
2X
fv;ug2V(u+v)xd(u;v)
H2(G;x) =1
2X
fv;ug2V(uv)xd(u;v)
Hereduis the degree of the vertex uandu6=v.
Base Polynomials for Schultz Invariants of n-Bilinear Straight Pentachain 3
In 2006 Sen-Peng et. al. did a similar work for hexagonal chains [8]. In 2008
Eliasi and Taeri gave Schultz polynomials of some composite graphs [1].
Denition 1.5. By ann-linear pentachain we shall mean a concatenation of
npentene molecular graphs along a line. An n-bilinear straight pentachain ,
denoted by G(n;S 1), is dened in Figure 3.
Figure 3. n-bilinear straight pentachain
This article is devoted to study the Schultz invariants of the bilinear
straight pentachain.
2. The Results
ByDp;qwe shall mean the matrix that represents the distances, and by Hp;q
b
we shall mean the base polynomial that counts the paths of dierent lengths
among vertices of degrees pandqof the straight bilinear pentachain G(n;S 1).
Theorem 2.1. Letn8. Then
H2;2
b(G(n;S 1)) = (n+ 5)x2+ 4×3+ (n+ 2)x4+ 4×5+n 9X
k=0(n k+ 1)x6+k
+9xn 2+ 8xn 1+ 9xn+ 6xn+1+ 3xn+2:
Proof. In this case we have D2;2
S1=0
@A0A1A2
AT
1A3A4
AT
2AT
4A01
A;whereA0;A1;A2;A3,
andA4are submatrices of orders 3 3, 3n 2, 33,n 2n 2, and
n 23, respectively, and are: A0=0
@0 2 2
2 0 2
2 2 01
A;
A1=0
@3 4 5 6n 1n
3 4 5 6n 1n
2 4 6 7n n + 11
A;
A2=0
@n+ 1n n + 2
n+ 1n+ 2n
n+ 2n+ 1n+ 11
A;
4 Nizami, Khurram, and Qasim
A3=0
BBBBBBBBBBB@0 2 4 6 7 8 n 2n 1n
0 2 4 6 7 n 3n 2n 1
0 2 4 6n 4n 3n 2
………
0 2 4 6
0 2 4
0 2
01
CCCCCCCCCCCA;
AT
4=0
@n+ 1n n 1 7 6 4 2
n n 1n 2 6 5 4 3
n n 1n 2 6 5 4 31
A
Now we give the counts of distinct paths among vertices of degree 2 in
G(n;S 1):
c2= (no. of 2 in A0)(no. ofA0)+(no. of 2 in A1)(no. ofA1)+(no. of 2 in A3)
(no. ofA3) + (no. of 2 in A4)(no. ofA4) = (3)(2) + (1)(1) + ( n 3)(1) +
(1)(1) =n+ 5
c3= (no. of 3 in A1)(no. ofA1)+(no. of 3 in A4)(no. ofA4) = (2)(1)+
(2)(1) = 4
c4= (no. of 4 in A1)(no. ofA1)+(no. of 4 in A3)(no. ofA3)+(no. of 4 in A4)
(no. ofA4) = (3)(1) + ( n 4)(1) + (3)(1) = n+ 2
c5= (no. of 5 in A1)(no. ofA1) + no. of 5 in A4)(no. ofA4) = (2)(1) +
(2)(1) = 4
c6+k= (no. of 6 + kinA1)(no. ofA1)+(no. of 6 + kinA4)(no. ofA4)+
(no. of 6 + kinA3)(no. ofA3) = (3)(1)+(3)(1)+( n k+5)(1) =n k+1
cn 2= (no. ofn 2 inA1)(no. ofA1s)+(no. ofn 2 inA4)(no. ofA4s)+
(no. ofn 2 inA3)(no. ofA3s) = (3)(1) + (3)(1) + (3)(1) = 9
cn 1= (no. ofn 1 inA1)(no. ofA1s)+(no. ofm 1 inA4)(no. ofA4s)+
(no. ofn 1 inA3)(no. ofA3s) = (3)(1) + (3)(1) + (2)(1) = 8
cn= (no. ofninA1)(no. ofA1s)+(no. ofninA2)(no. ofA2s)+(no. ofninA3)
(no. ofA3s) + (no. of ninA4)(no. ofA4s) = (3)(1) + (2)(1) + (1)(1) +
(3)(1) = 9
cn+1= (no. ofn+ 1 inA1)(no. ofA1s)+(no. ofn+ 1 inA2)(no. ofA2s)+
(no. ofn+ 1 inA4)(no. ofA4s) = (1)(1) + (4)(1) + (1)(1) = 6
cn+2= (no. ofn+ 2 inA2)(no. ofA2s) = (3)(1) = 3
Base Polynomials for Schultz Invariants of n-Bilinear Straight Pentachain 5
Theorem 2.2. Forn5, we have
H2;3
b(G(n;S 1)) = 10×1+ (4n)x2+ (4n+ 2)x3+ (4n 4)x4
+n+1X
k=5(4n 4k+ 14)xk+ 2xn+2:
Proof. In this case we have D2;3
S1=0
@A11A12
A21A22
A12A111
A;whereA11;A12,A21and
A22are submatrices of order 3 n, 3n,n 2nandn 2nrespectively,
and are:
A11=0
@1 2 3 4n 2n 1n+ 1
2 3 4 5n 1n n + 2
3 4 5 6n n + 1n+ 11
A
A12=0
@n+ 1n n 1 4 3 1
n n 1n 2 3 2 1
n 1n 2n 3 2 1 11
A
A21=0
BBBBBBBBB@2 2 3 4 5 n 2n 1n+ 1
3 2 2 3 4 n 3n 2n
4 3 2 2 3 n 4n 3n 1
………
n 3n 4n 5n 6n 7 3 4 6
n 2n 3n 4n 5n 6 2 3 5
n 1n 2n 3n 4n 5 2 2 31
CCCCCCCCCA
A22=0
BBBBBBBBB@n 1n 2n 3n 4n 5 2 2 3
n 2n 3n 4n 5n 6 2 3 5
n 3n 4n 5n 6n 7 3 4 6
………
4 3 2 2 3 n 4n 3n 1
3 2 2 3 4 n 3n 2n
2 2 3 4 5 n 2n 1n+ 11
CCCCCCCCCA
Now we go for paths:
c1= (no. of 1 in A11)(no. ofA11s) + (no. of 1 in A12)(no. ofA12s) =
(1)(2) + (4)(2) = 10
c2= (no. of 2 in A11)(no. ofA11s) + (no. of 2 in A12)(no. ofA12s) +
(no. of 2 in A21)(no. ofA21s)+(no. of 2 in A22)(no. ofA22s) = (2)(2)+
(2)(2) + (2n 4)(1) + (2n 4)(1) = 4n
c3= (no. of 3 in A11)(no. ofA11s) + (no. of 3 in A12)(no. ofA12s) +
(no. of 3 in A21)(no. ofA21s)+(no. of 3 in A22)(no. ofA22s) = (3)(2)+
6 Nizami, Khurram, and Qasim
(3)(2) + (2n 5)(1) + (2n 5)(1) = 4n+ 2
c4= (no. of 4 in A11)(no. ofA11s) + (no. of 4 in A12)(no. ofA12s) +
(no. of 4 in A21)(no. ofA21s)+(no. of 4 in A22)(no. ofA22s) = (3)(2)+
(3)(2) + (2n 8)(1) + (2n 8)(1) = 4n 4
ck= (no. ofkinA11)(no. ofA11s) + (no. of kinA12)(no. ofA12s) +
(no. ofkinA21)(no. ofA21s)+(no. ofkinA22)(no. ofA22s) = (3)(2)+
(3)(2) + (2n 2k+ 1)(1) + (2 n 2k+ 1)(1) = 4 n 4k+ 14
cn+2= (no. ofn+ 2 inA11)(no. ofA11s) = (1)(2) = 2
Theorem 2.3. Forn5we have
H2;4
b(G(n;S 1)) = (2n 2)x1+ 4×2+ (2n)x3+ 4×4
+nX
k=5(2n 2k+ 8)xk+ 2xn+1:
Proof. Here the distance matrix is D2;4
S1=0
@A1
A2
A31
A;whereA1;A2, andA3
are submatrices of orders 3 n 1,n 2n 1, and 3n 1, respectively,
and are:
A1=0
@2 3 4 5n 2n 1n
2 3 4 5n 2n 1n
1 3 5 6n 1n n + 11
A
A2=0
BBBBBBBBB@1 1 3 5 6 n 2n 1n
3 1 1 3 5 n 3n 2n 1
5 3 1 1 3 n 4n 3n 2
…………
n 2n 3n 4 3 1 1 3 5
n 1n 2n 3 5 3 1 1 3
n n 1n 2 6 5 3 1 11
CCCCCCCCCA;
A3=0
@n+ 1n n 1 6 5 3 1
n n 1n 2 5 4 3 2
n n 1n 2 5 4 3 21
A;
The following are counts of distinct paths in G(n;S 1).
c1= (no. of 1 in A1)(no. ofA1s)+(no. of 1 in A2)(no. ofA2s)+(no. of 1 in A3)
(no. ofA3s) = (1)(1) + (2 n 4)(1) + (1)(1) = 2 n 2
Base Polynomials for Schultz Invariants of n-Bilinear Straight Pentachain 7
c2= (no. of 2 in A1)(no. ofA1s)+(no. of 2 in A3)(no. ofA3s) = (2)(1)+
(2)(1) = 4
c3= (no. of 3 in A1)(no. ofA1s)+(no. of 3 in A2)(no. ofA2s)+(no. of 3 in A3)
(no. ofA3s) = (3)(1) + (2 n 6)(1) + (3)(1) = 2 n
c4= (no. of 4 in A1)(no. ofA1s)+(no. of 4 in A3)(no. ofA3s) = (2)(1)+
(2)(1) = 4
ck= (no. ofkinA1)(no. ofA1s)+(no. ofkinA2)(no. ofA2s)+(no. ofkinA3)
(no. ofA3s) = (3)(1) + (2 n 2k+ 2)(1) + (3)(1) = 2 n 2k+ 8
cn+1= (no. ofn+ 1 inA1)(no. ofA1s)+(no. ofn+ 1 inA3)(no. ofA3s) =
(1)(1) + (1)(1) = 2
Theorem 2.4. Forn4we have
H3;3
b(G(n;S 1)) = (2n 4)x1+ (3n 3)x2+n 1X
k=3(4n 4k+ 4)xk+ 6xn+xn+2:
Proof. Here the distance matrix is D3;3
S1=A1A2
AT
2A1
;whereA1andA2
are submatrices of order nn, and are:
A1=0
BBBBBBBBBBB@0 1 2 3 4 n 2n
0 2 4 5n 3n 1
0 2 4n 4n 2
…………
0 1 2 4
0 1 3
0 2
01
CCCCCCCCCCCA
A2=0
BBBBBBBBBBB@n n 1n 2 4 3 2 2
n 1n 2n 3 3 2 3 3
n 2n 3n 4 2 3 4 4
n 3n 4n 5 3 4 5 5
…………
3 2 3 n 3n 2n 1n 1
2 3 4 n 2n 1n n
2 3 4 n 2n 1n n + 21
CCCCCCCCCCCA;
The following are counts of distinct paths in G(n;S 1).
c1= (no. of 1 in A1)(no. ofA1s= (n 2)(2) = 2n 4
8 Nizami, Khurram, and Qasim
c2= (no. of 2 in A1)(no. ofA1s) + (no. of 2 in A2)(no. ofA2s) = (n
2)(2) + (n+ 1)(1) = 3 n 3
ck= (no. ofkinA1)(no. ofA1s) + (no. ofkinA2)(no. ofA2s) = (n
k)(2) + (2n 2k+ 4)(1) = 4 n 4k+ 4
cn 1= (no. ofn 1 inA1)(no. ofA1s)+(no. ofn 1 inA2)(no. ofA2s) =
(1)(2) + (6)(1) = 8
cn= (no. ofninA1)(no. ofA1s)+(no. ofninA2)(no. ofA2s) = (1)(2)+
(4)(1) = 6
cn+2= (no. ofninA2)(no. ofA2s) = (1)(1) = 1
Theorem 2.5. Forn6we have
H3;4
b(G(n;S 1)) = (2n 2)x1+ (4n 6)x2+ (3n 8)x3+ (4n 14)x4
+nX
k=5(4n 4k+ 3)xk+ 2xn+1:
Proof. Here the distance matrix is D3;4
S1=A1
A2
;whereA1andA2are
submatrices of order nn, and are:
A1=0
BBBBBBBBBBB@1 2 3 4 n 2n 1
2 1 2 3 n 3n 2
3 2 1 2 n 4n 3
4 3 2 1 n 5n 4
……
n 1n 2 4 2 1 2
n n 1 5 4 2 1
n+ 1n 6 5 4 21
CCCCCCCCCCCA
A2=0
BBBBBBBBB@n 1n 2n 3 4 3 2 1
n 2n 3n 4 3 2 1 2
n 3n 4n 5 2 1 2 3
…………
2 1 2 n 3n 2n 1n 1
1 2 3 n 2n 1n n
2 4 5 n 2n 1n n + 11
CCCCCCCCCA
The following are counts of distinct paths in G(n;S 1).
c1= (no. of 1 in A1)(no. ofA1s) + (no. of 1 in A2)(no. ofA2s) = (n
1)(1) + (n 1)(1) = 2n 2
Base Polynomials for Schultz Invariants of n-Bilinear Straight Pentachain 9
c2= (no. of 2 in A1)(no. ofA1s) + (no. of 2 in A2)(no. ofA2s) = (2n
3)(1) + (2n 3)(1) = 4n 6
c3= (no. of 3 in A1)(no. ofA1s) + (no. of 3 in A2)(no. ofA2s) = (n
2)(1) + (2n 6)(1) = 3n 8
ck= (no. ofkinA1)(no. ofA1s)+(no. ofkinA2)(no. ofA2s) = (2n
2k+ 2)(1) + (2 n 2k+ 1)(1) = 4 n 4k+ 3
cn+1= (no. ofn+ 1 inA1)(no. ofA1s)+(no. ofn+ 1 inA2)(no. ofA2s) =
(1)(1) + (1)(1) = 2
Theorem 2.6. Forn6we have
H4;4
b(G(n;S 1)) = (n 2)x2+ (n 3)x4+n+1X
k=6(n k+ 2)xk:
Proof. Here the distance matrix is
D4;4
S1=0
BBBBBBBBBBB@0 2 4 6 7 8 n n + 1
0 2 4 6 7 n 1n
0 2 4 6n 2n 1
…………
0 2 4 6
0 2 4
0 2
01
CCCCCCCCCCCA
The counts for this matrix are obvious.
Example. Forn= 8 the base polynomials are:
H2;2
b(G(8;S1)) = 3×10+ 6×9+ 9×8+ 8×7+ 9×6+ 4×5+ 10×4+ 4×3+ 13×2,
H2;3
b(G(8;S1)) = 2×10+ 10×9+ 14×8+ 18×7+ 22×6+ 26×5+ 28×4+ 34×3+
32×2+ 10x
H2;4
b(G(8;S1)) = 2×9+ 8×8+ 10×7+ 12×6+ 14×5+ 4×4+ 16×3+ 4×2+ 14x,
H3;3
b(G(8;S1)) =x10+ 6×8+ 8×7+ 12×6+ 16×5+ 20×4+ 24×3+ 21×2+ 12x,
H3;4
b(G(8;S1)) = 2×9+ 3×8+ 7×7+ 11×6+ 15×5+ 18×4+ 16×3+ 26×2+ 14x
H4;4
b(G(8;S1)) =x9+ 2×8+ 3×7+ 4×6+ 5×4+ 6×2. It now follows that the
Schultz polynomial, modied Schultz polynomial, Schultz index, and modied
Schultz index are:
1.H1(G(8;S1)) = 4H2;2
b+ 5H2;3
b+ 6H2;4
b+ 6H3;3
b+ 7H3;4
b+ 8H4;4
b=
4
3×10+ 6×9+ 9×8+ 8×7+ 9×6+ 4×5+ 10×4+ 4×3+ 13×2
+ 5
2×10+
10×9+ 14×8+ 18×7+ 22×6+ 26×5+ 28×4+ 34×3+ 32×2+ 10x
+ 6
2×9+
8×8+10×7+12×6+14×5+4×4+16×3+4×2+14x
+6
x10+6×8+8×7+
12×6+16×5+20×4+24×3+21×2+12x
+7
2×9+3×8+7×7+11×6+15×5+
18×4+16×3+26×2+14x
+8
x9+2×8+3×7+4×6+5×4+6×2
= 28×10+
10 Nizami, Khurram, and Qasim
108×9+ 227×8+ 303×7+ 399×6+ 431×5+ 490×4+ 538×3+ 592×2+ 304x
2.H2(G(8;S1)) = 4H2;2
b+ 6H2;3
b+ 8H2;4
b+ 9H3;3
b+ 12H3;4
b+ 16H4;4
b=
33×10+140×9+306×8+424×7+568×6+608×5+716×4+756×3+873×2+
448x
3.S(G(8;S1)) =d
dxH1(G(8;S1))jx=1= 14800
4.MS(G(8;S1)) =d
dxH2(G(8;S1))jx=1= 20780
3. Conclusions
In this paper we followed the divide and conquer rule to nd Schultz in-
variants of the bilinear straight pentachain. It was observed that obtaining
the general closed forms of the Schultz invariants directly by denition is ex-
tremely dicult. In order to handle the situation we counted the paths among
vertices of degrees 2-2, 2-3, 2-4, 3-3, 3-4, and 4-4 separately and represented
them in terms of polynomials. These polynomials ultimately served as bases
for computing Schultz invariants as one can directly nd Schultz polynomial,
modied Schultz polynomial, Schultz index, and modied Schultz index from
them. Finally, we gave an examples to show how these bases actually work.
References
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Base Polynomials for Schultz Invariants of n-Bilinear Straight Pentachain 11
[9] H. P. Schultz, Topological Organic Chemistry. 1. Graph theory and topological
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Abdul Rauf Nizami
Faculty of Information Technology, University of Central Punjab, Lahore-Pakistan
e-mail: arnizami@ucp.edu.pk
Khurram Shabbir
Department of Mathematics, GC University, Lahore-Pakistan
e-mail: dr.khurramshabbir@gcu.edu.pk
Muhammad Qasim
Department of Mathematics, GC University, Lahore-Pakistan
e-mail: mqasimyahya@gmail.com
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