New upper bounds for the forgotten index among bicyclic graphs
A. Jahanbani, L. Shahbazi, S.M. Sheikholeslami, R. Rasi, J. Rodriguez
aa r X i v : . [ m a t h . C O ] F e b New upper bounds for the forgotten indexamong bicyclic graphs
A. Jahanbani a ∗ , L. Shahbazi a , S.M. Sheikholeslami a ,R. Rasi a , and J. Rodr´ıguez b a Department of Mathematics, Azarbaijan Shahid Madani UniversityTabriz, Iran b Departamento de Matem´aticas, Facultad de Ciencias B´asicas, Universidad deAntofagasta, Av Angamos 601, Antofagasta, Chile.
Abstract
The forgotten topological index of a graph G , denoted by F ( G ), is defined as the sumof weights d ( u ) + d ( v ) over all edges uv of G , where d ( u ) denotes the degree of a vertex u . In this paper, we give sharp upper bounds of the F-index (forgotten topological index)over bicyclic graphs, in terms of the order and maximum degree. Keywords:
Forgotten index, Bicyclic graph, Molecular graph, Maximum degree
AMS subject classification: 05C50, 92E10
A topological index is a numeric quantity associated with a molecular graph that remainsinvariant under graph isomorphism and encodes at least one physical or chemical prop-erty of the underlying organic molecule. Topological indices play an important role inpredicting the physical as well as the chemical properties (boiling point, volatility, sta-bility, solubility, connectivity, chirality and melting point) of chemical compounds. Formore information we refer to [12, 13] and the references cited therein.Let G be a simple graph with vertex set V = V ( G ) and edge set E = E ( G ). Theintegers n = n ( G ) = | V ( G ) | and m = m ( G ) = | E ( G ) | are the order and the size of thegraph G , respectively. If m = n + 1 then we say that G is a bicyclic graph. The openneighborhood of vertex v is defined as N ( v ) = N G ( v ) = { u ∈ V ( G ) : uv ∈ E ( G ) } andthe degree of v is d G ( v ) = d v = | N ( v ) | . The maximum degree of a graph G is denoted by∆ = ∆( G ). ∗ Corresponding author.
E-mail addresses:
[email protected] (A. Jahanbani), [email protected] (S.M. Sheikholeslami), [email protected] (L.Shahbazi), [email protected] (R. Rasi), [email protected] (J. Rodr´ıguez) F ( G ) for a graph G and defined as F = F ( G ) = X uv ∈ E [ d ( u ) + d ( v ) ] = X v ∈ V d ( v ) . This topological index was named and first studied by Furtula and Gutman [6] in 2015,but it first appeared in 1972 [8] within the study of structure-dependency of the total π -electron energy. For recent results in the F-index of graphs we refer to [1, 4, 5, 7, 9–11].In this paper, we established sharp upper bounds for the F-index among bicyclic graphs,in terms of the order and the maximum degree. In this section, we establish new upper bounds for he forgotten topological index of ofbicyclic graphs. Now we present some known results that will be needed in this section.If n is a positive integer, then an integer partition of n is a non-increasing sequenceof positive integers y = ( x , x , . . . , x t ), such that n = P ti =1 x i . If ∆ ≥ x ≥ x ≥ . . . ≥ x t ≥
1, then ( x , x , . . . , x t ) is called a ∆-partition or an integer partition of n on N ∆ = { , , . . . , ∆ } .A ∆-partition y = ( y , y , . . . , y t ) of n is called an integer ∆-dominant sequence if thenumber ∆ in this partition is as large as possible. In other words, if n = t ∆, then y = (∆ , . . . , ∆) is the integer ∆-dominant sequence and if n = t ∆ + b where 0 < b < ∆then y = (∆ , . . . , ∆ , b ) is the integer ∆-partition.Let B be a bicyclic graph of order n and maximum degree ∆. For each i ∈ { , , . . . , ∆ } ,let n i denote the number of vertices of degree i . Then n + n + · · · + n ∆ = n (1)and n + 2 n + . . . + ∆ n ∆ = 2 m = 2 n. (2)Subtracting (1) from (2), yields n + 2 n + . . . + (∆ − n ∆ = n. (3)2y (3), we obtain the (∆ − n as follows:( ∆ − , . . . , ∆ − | {z } n ∆ , . . . , , . . . , | {z } n , , . . . , | {z } n ) . (4)Next result is an immediate consequence of the above discussion. Corollary 1.
For any bicyclic graph B of order n with maximum degree ∆ , the F-index F ( B ) = P v ∈ V d v is maximum if and only if the (∆ − -partition (4) is a (∆ − -dominantsequence of n . Remark 2.
In other words, with regard to the (∆ − -dominant sequence of n , n ∆ (numberof vertices with degree ∆ − ) must be maximum. In this case, sequence ( n , n , . . . , n ∆ ) is called a major sequence for B . Theorem 3.
Let B be a bicyclic graph of order n and maximum degree △ with n ≡ mod ( △ − . Then F ( B ) ≤ (∆ + ∆ + 2) n + 26 Proof.
Without loss of generality, assume that n = ( △ − k .By equality in (4), we have n ∆ = n + 2 − ( n + 2 n + . . . + (∆ − n ∆ − )∆ − k − r where r = n + 2 n + · · · + ( △ − n △− − △ − . then − ≤ r ≤ k − ≤ n △ ≤ k. Thus, consider the following cases.
Case 0. r = − n △ = k + 1. It follows that n + 2 n + . . . + (∆ − n ∆ − + (∆ − k + 1) = (∆ − k + 2and so n + 2 n + . . . + (∆ − n ∆ − = − (∆ −
1) + 2that it is not possible. so r = 0, △ > . ase 1. r = 0 . Thus, n △ = k , n = 1, n = . . . = n △− = 0. Since n + n + . . . + n △ = n , we concludethat n = ( △ − k −
1. By Corollary 1, we obtain( n , n , n , . . . , n ∆ − , n ∆ ) = ((∆ − k − , , , , . . . , , k )which is the optimal solution and so F ( B ) is maximum. Therefore, F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + . . . + (∆ − n ∆ − + ∆ n ∆ = (∆ − k − + ∆ ( k )= (∆ + ∆ − k + 26= (∆ + ∆ + 2)(∆ − k + 26= (∆ + ∆ + 2) n + 26 . Case 2. r = 1 . Since n △ = k −
1, it follows from (4) that n + 2 n + · · · + ( △ − n △− = ( △ −
1) + 2 = ( △ −
2) + 3 . First let △ >
4, so, n = 1, n △− = 1, n = . . . = n △− = 0. Since n + n + . . . + n △ = n ,we conclude that n = ( △ − k −
1. By Corollary 1,( n , n , n , n . . . , n ∆ − , n ∆ ) = ((∆ − k − , , , , , . . . , , k − F ( B ) is maximum. Therefore, F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + . . . + (∆ − n ∆ − + ∆ n ∆ = (∆ − k − + ( △ − + ∆ ( k − + ∆ − k + 63 − △ +3 △ −
1= (∆ + ∆ − k − △ +3 △ +62= (∆ + ∆ + 2)( △ − k − △ ( △ −
1) + 62 ≤ (∆ + ∆ + 2) n − △ −
1) + 62= (∆ + ∆ + 2) n − △ +77 ≤ (∆ + ∆ + 2) n + 2 < (∆ + ∆ + 2) n + 26 . Now, if let △ = 4, by Corollary 1 4 n , n , n , n ) = (2 k − , , , k − F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + 3 n + 4 n = (2 k −
2) + 8 + 3 (2) + 4 ( k − + 4 − k −
4= ( △ + △ − k −
4= ( △ + △ + 2)( △ − k −
4= ( △ + △ + 2) n − < ( △ + △ + 2) n + 26 . Also, if △ = 3, by Corollary 1( n , n , n ) = ( k − , , k − F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + 3 n = ( k −
3) + 8(4) + 3 ( k − + 3 − k + 2= ( △ + △ − k + 2= ( △ + △ + 2)( △ − k + 2= ( △ + △ + 2) n + 2 < ( △ + △ + 2) n + 26 . Case 3. ≤ r < △ − . As above, n + 2 n + ... + ( △ − n △− = ( △ − r + 2 = ( △ − r + r + 2 . since r + 2 < △ −
1, it follows from Corollary 1 that( n , n , . . . , n r +3 , . . . , n ∆ − , n ∆ − , n ∆ ) = ((∆ − k − , , . . . , , . . . , , r, k − r )which is the optimal solution. Thus F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + . . . + (∆ − n ∆ − + ∆ n ∆ − k − r + 3) + (∆ − r + ∆ ( k − r )= (∆ + ∆ − k − r + 3) − r + 3∆ r − r = (∆ + ∆ + 2) n + ( r + 3) − r + 3∆ r − r −
1= (∆ + ∆ + 2) n + ( r + 3) + r ( − + 3∆ − − < (∆ + ∆ + 2) n + △ + ( △ − − + 3∆ − −
1= (∆ + ∆ + 2) n − △ +12 △ − △ +2= (∆ + ∆ + 2) n − △ (2 △ − − △ +2 < (∆ + ∆ + 2) n − △ − −
50 + 2= (∆ + ∆ + 2) n − △ +252 < (∆ + ∆ + 2) n + 2 < (∆ + ∆ + 2) n + 26 . Because 5 < △ . Case 4. △ − ≤ r ≤ k − . Then n + 2 n + . . . + (∆ − n ∆ − = (∆ − r + r + 2 . Thereby, there are non-negative integers t, s such that r + 2 = t ( △ −
2) + s with0 ≤ s < △ −
2. Hence n + 2 n + . . . + ( △ − n △− = ( △ − r + t ) + s. If 0 < s < △ −
2, then( n , n , . . . , n s , n s +1 , n s +2 , . . . , n ∆ − , n ∆ − , n ∆ ) = ((∆ − k − ( t +1) , , . . . , , , , . . . , , , t + r, k − r )which is optimal solution and since s < △ − − r ≤ − △ and △ >
5, we obtain F ( B ) ≤ n + 2 n + . . . + (∆ − n ∆ − + ∆ n ∆ = (∆ − k − ( t + 1) + ( s + 1) + (∆ − ( t + r ) + ∆ ( k − r )= (∆ + ∆ − k − t − s + 1) + ∆ t − t + 3∆ t − t − r + 3∆ r − r = (∆ + ∆ − k + ( s + 1) − − t ( − ∆ + 3∆ −
3∆ + 2) − r (3∆ + 3∆ + 1) < (∆ + ∆ − k + (∆ − − − − ∆ + 3∆ −
3∆ + 2) + (3 − ∆)(3∆ + 3∆ + 1)= (∆ + ∆ + 2) n − ∆ + 6∆ − −
16 (∆ + ∆ + 2) n − ∆ (∆ − − − < (∆ + ∆ + 2) n − − −
21= (∆ + ∆ + 2) n −
25∆ + 129 < (∆ + ∆ + 2) n + 4 < (∆ + ∆ + 2) n + 26 . If s = 0, then the optimal solution is( n , n , . . . , n ∆ − , n ∆ − , n ∆ ) = ((∆ − k − t, , . . . , , r + t, k − r ) . Since − r < (4 − △ ), △ >
3, we conclude that F ( B ) ≤ n + 2 n + . . . + (∆ − n ∆ − + ∆ n ∆ = (∆ − k − t + ∆ t − t + 3∆ t − t − r + 3∆ r − r = (∆ + ∆ − k − t ( − ∆ + 3∆ −
3∆ + 2) − r (3∆ −
3∆ + 1) < (∆ + ∆ − k − − ∆ + 3∆ −
3∆ + 2) + (4 − ∆)(3∆ −
3∆ + 1)= (∆ + ∆ + 2) n − + 12∆ −
10∆ + 2= (∆ + ∆ + 2) n − (∆ − −
10∆ + 2 < (∆ + ∆ + 2) n − − −
48= (∆ + ∆ + 2) n −
50∆ + 252 < (∆ + ∆ + 2) n − < (∆ + ∆ + 2) n + 26 . Theorem 4.
Let B be a bicyclic graph of order n and maximum degree △ with n ≡ mod ( △ − . Then F ( B ) ≤ ( △ + △ + 2) n − ( △ + △ − . Proof.
Let n = ( △ − k + 1. Set r = n + 2 n + ... + ( △ − n △− − △ − . By equality in ( ), we have n ∆ = k − ( n + 2 n + . . . + (∆ − n ∆ − − − k − r. − ≤ r ≤ k − ≤ n △ ≤ k. We consider the following cases:
Case 0. r = − n △ = k + 1. It follow that n + 2 n + . . . + (∆ − n ∆ − + (∆ − k + 1) = (∆ − k + 3and so n + 2 n + . . . + (∆ − n ∆ − = − (∆ −
1) + 3that it is not possible. so r = 0, △ > . Case 1. r = 0.Since r = n + 2 n + ... + ( △ − n △− − △ − , it follows that n = 1, n = ... = n △− = 0 and n △ = k . From (2), we have n = ( △ − k .Now, by Corollary 1, we have( n , n , n , n . . . , n ∆ − , n ∆ − , n ∆ ) = ((∆ − k, , , , . . . , , k )which is the optimal solution. Thus F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + . . . + (∆ − n ∆ − + ∆ n ∆ = (∆ − k + 8 + ∆ ( k )= (∆ + ∆ − k + 8= (∆ + ∆ + 2)(∆ − k + 8= (∆ + ∆ + 2)( n −
1) + 8= (∆ + ∆ + 2) n − (∆ + ∆ − . Case 2. r = 1Since n △ = k −
1, it follows from (4) that n + 2 n + . . . + (∆ − n ∆ − = (∆ −
1) = (∆ −
2) + 4 . First let △ >
5. By Corollary 1,( n , n , n , n , . . . , n ∆ − , n ∆ − , n ∆ ) = ((∆ − k − , , , , . . . , , , k − F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + . . . + (∆ − n ∆ − + ∆ n ∆ = (∆ − k + 53 + (∆ + 1) + ∆ ( k − + ∆ − k + 52 − + 3∆= (∆ + ∆ + 2)( n −
1) + 52 − + 3∆= (∆ + ∆ + 2) n − + 2∆ + 50= (∆ + ∆ + 2) n − + ∆ −
6) + 6∆ + 26 < (∆ + ∆ + 2) n − (∆ + ∆ − . Now, if let △ = 5, by Corollary 1,( n , n , n , n , n ) = (3 k − , , , , k − F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + 3 n + 4 n + 5 n = 3 k − ( k − + 5 − k −
8= (∆ + ∆ − k −
8= (∆ + ∆ + 2)(∆ − k −
8= (∆ + ∆ + 2)( n − −
8= (∆ + ∆ + 2) n − (∆ + ∆ + 10)= (∆ + ∆ + 2) n − (∆ + ∆ − − < (∆ + ∆ + 2) n − (∆ + ∆ − . Also, if let △ = 4. By Corollary 1,( n , n , n , n ) = (2 k − , , , k − F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + 3 n + 4 n = 2 k − ( k − + 4 − k + 49 (∆ + ∆ − n −
1) + 4= (∆ + ∆ + 2) n − (∆ + ∆ − + ∆ + 2) n − (∆ + ∆ − − < (∆ + ∆ + 2) n − (∆ + ∆ − . Case 3. ≤ r < △ − ≤ r < △ −
4, thus △ >
6. In this case:
Hypothesis : n = (∆ − k + 12 ≤ r = n +2 n + ... +(∆ − n ∆ − − − < ∆ − n + 2 n + . . . + (∆ − n ∆ − = (∆ − r + ( r + 3) r + 1 ≤ ∆ − n ∆ = k − rn ∆ − = rn r +4 = 1 n = (∆ − k and from Corollary 1 we obtain( n , n , . . . , n r +3 , n r +4 , n r +5 , . . . , n ∆ − , n ∆ − , n ∆ ) = ((∆ − k, , . . . , , , , . . . , , r, k − r )which which is the optimal optimization. Then F ( U ) ≤ F max ( U n, ∆ ) = n + 2 n + . . . + (∆ − n ∆ − + ∆ n ∆ = (∆ − k + ( r + 4) + (∆ − r + ∆ ( k − r ) < (∆ + ∆ − k + △ + r ( − + 3∆ − + ∆ + 2)( n −
1) + ( r + 2) + r ( − + 3∆ − < (∆ + ∆ + 2) n − ∆ − ∆ − △ − − + 3∆ − + ∆ + 2) n − + 14∆ − △ +4= (∆ + ∆ + 2) n − (2 △ − △ + △ − − △ +100 < (∆ + ∆ + 2) n − ( △ + △ − . Since ∆ > Case 4. △ − ≤ rk − . Then, there are non-negative integers t, s such that r + 3 = t ( △ −
2) + s , t ≥ ≤ s < △ −
2. By substituting in (4), we have n + 2 n + . . . + (∆ − n ∆ − = (∆ − r + t ) + s. ypothesis : n = (∆ − k − ( t + 1)∆ − ≤ r = n +2 n + ... +(∆ − n ∆ − − − ≤ k − n + 2 n + . . . + (∆ − n ∆ − = (∆ − r + t ) + ss + 1 ≤ ∆ − n ∆ = k − rn ∆ − = r + tn s +1 = 1 n = (∆ − k − ( t + 1)First let 0 < s . From Corollary 1, we have( n , n , . . . , n s , n s +1 , n s +2 , . . . , n ∆ − , n ∆ − , n ∆ ) = ((∆ − k − ( t +1) , , . . . , , , , . . . , , , r + t, k − r )which is the optimal solution. Thus F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + · · · + (∆ − n ∆ − + ∆ n ∆ = (∆ − k − ( t + 1) + ( s + 1) + (∆ − ( t + r ) + ∆ ( k − r )= (∆ + ∆ − k − s + 1) − t (∆ + 3∆ −
3∆ + 2) − r (3∆ −
3∆ + 1) < (∆ + ∆ − k − △ − − + 3∆ −
3∆ + 2) + (4 − △ )(3∆ −
3∆ + 1)= (∆ + ∆ + 2) n − △ + 8 △ − △ −
2= (∆ + ∆ + 2) n + ( △ − − △ − △ +6) − △ +52 < (∆ + ∆ + 2) n + ( △ + △ − . Now let s = 0, then the optimal solution is( n , n , . . . , n ∆ − , n ∆ − , n ∆ ) = ((∆ − k − t + 1 , , . . . , , r + t, k − r ) . where we have that Hypothesis : n = (∆ − k + 1∆ − ≤ r = n +2 n + ... +(∆ − n ∆ − − − < k − n + 2 n + . . . + (∆ − n ∆ − = (∆ − r + t ) n ∆ = k − rn ∆ − = r + tn = n = . . . = n ∆ − = 0 n = (∆ − k − t + 1 . Therefore F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + . . . + (∆ − n ∆ − + ∆ n ∆ = (∆ − k − t + 1 + (∆ − ( t + r ) + ∆ ( k − r )= (∆ + ∆ − k − t (∆ + 3∆ −
3∆ + 2) + 1 − r (3∆ −
3∆ + 1)11 (∆ + ∆ − k − + 3∆ −
3∆ + 2) + 1 + (4 − △ )(3∆ −
3∆ + 1)= (∆ + ∆ + 2) n − △ +11 △ − △ +1= (∆ + ∆ + 2) n + (2 △ − − △ − △ +6) + 36 △ +79 < (∆ + ∆ + 2) n − ( △ + △ − . Theorem 5.
Let B be a bicyclic graph of order n and maximum degree △ with n ≡ pmod ( △ −
1) where 2 ≤ p < △ −
3. Then F ( B ) ≤ ( △ + △ + 2) − p ( △ + △ + 2) + p + 9 p + 28 p + 26 . Proof.
Let n = (∆ − k + p . Suppose that r = n + 2 n + . . . + (∆ − n ∆ − − p − − . By equality in (4), we have n ∆ = k − ( n + 2 n + . . . + (∆ − n ∆ − − p − − k − r. Then clearly − ≤ r ≤ k − ≤ n ∆ ≤ k . We consider the following cases. Case 0. r = − n △ = k + 1. It follow that n + 2 n + . . . + (∆ − n ∆ − + (∆ − k + 1) = (∆ − k + p + 2and so n + 2 n + . . . + (∆ − n ∆ − = − (∆ −
1) + ( p + 2)that it is not possible. So r = 0, △ − ≥ p + 2 implies that △ ≥ p + 3 . We consider2 ≤ p ≤ △ − . Case 1. r = 0.Then n ∆ = k and we by (4) we have n + 2 n + · · · + (∆ − n ∆ − = (∆ − k + p + 2 − ( △ − k = p + 2 . it follows from Corollary (6) that( n , n , . . . , n p +3 , . . . , n ∆ − , n ∆ ) = (( △ − k + p − , , . . . , , , , . . . , , k )12hich is the optimal solution and so F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + . . . + ∆ n ∆ = (∆ − k + p − p + 3) + ∆ ( k )= (∆ + ∆ − k + p + 9 p + 28 p + 26= (∆ + ∆ + 2) n − p (∆ + ∆ + 2) + p + 9 p + 28 p + 26 . Case 2. r = 1.Then n ∆ = k − n + 2 n + . . . + ( △ − n △− = ( △ −
1) + p + 2 = ( △ −
2) + ( p + 3) . Since 5 ≤ p + 3 < △ , we consider three subcases: Subcase 2.1 p + 3 = △ − n + 2 n + . . . + ( △ − n △− = 2( △ −
2) + 1 . Therefore ( n , n , . . . , n △− , n △− , n ∆ ) = ((∆ − k + p − , , , . . . , , , k − p = △ − F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + . . . + ∆ n ∆ = (∆ − k + p − − + ∆ ( k − + ∆ − k + p + ∆ − + 6∆ + 4= (∆ + ∆ + 2)( n − p ) + p (∆ − −
2) + p + ∆ − + 6∆ + 4= (∆ + ∆ + 2)( n − p ) + p (∆ − −
2) + p + (∆ − − − + ∆ + 2)( n − p ) + p (∆ − −
2) + p + ( p + 2)(∆ − − + ∆ + 2)( n − p ) + p (∆ − −
2) + 2∆ − − + ∆ + 2) n − p (5 △ +3) + 2∆ − − < (∆ + ∆ + 2) n − p (∆ + ∆ + 2) + p + 9 p + 28 p + 26 , since − p (5 △ +3) + 2∆ − − < p + 9 p + 28 p + 26 . ubcase 2.2 p + 3 = △ − n + 2 n + . . . + ( △ − n △− = 2( △ − . Therefore( n , n , . . . , n △− , n △− , n ∆ ) = ((∆ − k + p − , , . . . , , , k − p = △ − F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + . . . + ∆ n ∆ = (∆ − k + p − − + ∆ ( k − + ∆ − k + p + △ − △ +6 △ −
3= (∆ + ∆ − k + p + ( △ − − − −
7= (∆ + ∆ − k + p + ( p + 3)(∆ − − −
7= (∆ + ∆ + 2) n − p (5 △ +3) + 3 △ − △ − < (∆ + ∆ + 2) n − p (∆ + ∆ + 2) + p + 9 p + 28 p + 26 . since − p (5 △ +3) + 3 △ − △ − < − p (∆ + ∆ + 2) + p + 9 p + 28 p + 26 . Subcase 2.3 p + 3 ≤ △ − n + 2 n + . . . + n △− = ( △ −
2) + ( p + 3) . Therefore( n , n , . . . , n p +4 , · · · , n △− , n ∆ ) = ((∆ − k + p − , , . . . , , , · · · , , , k − p ≤ △ − △ ≥
5, then we have F ( B ) ≤ F max ( B n, ∆ ) = n + 2 n + . . . + ∆ n ∆ = (∆ − k + p − p + 4) + ∆ − △ +3 △ − ( k − △ + ∆ − k + p + ( p + 4) − △ +3 △ −
2= (∆ + ∆ + 2) n − p (∆ + ∆ + 2) + p + 12 p + 49 p − + 3∆ + 62 < (∆ + ∆ + 2) n − p (∆ + ∆ + 2) + p + 9 p + 28 p + 26 , since for ∆ ≥ p + 12 p + 49 p − + 3∆ + 62 < p + 9 p + 28 p + 26. Case 3. ≤ r < ∆ − p − n + 2 n + . . . + (∆ − n ∆ − = (∆ − r + ( p + r + 2). Since r < ∆ − p − n , n , . . . , n p + r +3 , . . . , n ∆ − , n ∆ − , n ∆ ) = ((∆ − k + p − , , . . . , , , , . . . , , r, k − r )14hich is the optimal solution. On the other hand, we deduce from p ≤ ∆ − r < ∆ − p − △ ≥
6. Thus F ( B ) ≤ n + 2 n + . . . + (∆ − n ∆ − + ∆ n ∆ = (∆ − k + p − p + r + 3) + (∆ − r + ∆ ( k − r )= (∆ + ∆ − k + p − p + ( r + 3) + 3 p ( r + 3) + 3 p ( r + 3) ) − r + 3∆ r − r = (∆ + ∆ − k + p + 9 p + 28 p − r + 3) + 3 pr ( p + r + 2) − r + 3∆ r − r< (∆ + ∆ − k + p + 9 p + 28 p + 26 + r + 9 r + 27 r + 3 pr (∆ + 1) − r + 3∆ r − r = (∆ + ∆ − k + p + 9 p + 28 p + 26 + r ( r ( r + 9) + 27 + 3 p (∆ + 1) − + 3∆ − < (∆ + ∆ − k + p + 9 p + 28 p + 26 + r ((∆ − p − − p + 8) + 26 + 3 p (∆ + 1) − + 3∆)= (∆ + ∆ − k + p + 9 p + 28 p + 26 + r ( − + 10∆ + p ∆ + p ( p −
4) + 18) < (∆ + ∆ − k + p + 9 p + 28 p + 26 + r ( − + 10∆ + ∆(∆ − − −
7) + 18)= (∆ + ∆ + 2)( n − p ) + p + 9 p + 28 p + 26 + r ( −
3∆ + 39) < (∆ + ∆ + 2)( n − p ) + p + 9 p + 28 p + 26 . Case 4. ∆ − p − ≤ r ≤ k − . Let p + r = t (∆ −
2) + s. By substituting in (4), we have n + 2 n + . . . + (∆ − n ∆ − = (∆ − r + t ) + s. If s = 0 then by Corollary 1,( n , n , . . . , n ∆ − , n ∆ − , n ∆ ) = ((∆ − k + p − ( t + 1) , , , , . . . , , r + t, k − r )which is the optimal solution. Since ∆ − p ≤ r + 1 and p < ∆ −
3, and clearly △ − △ +10 △ − < p + 9 p + 26 p + 1 < ∆ − ∆ − . Thus F ( B ) ≤ n + 2 n + . . . + (∆ − n ∆ − + ∆ n ∆ = (∆ − k + p − ( t + 1) + (3) + (∆ − ( t + r ) + ∆ ( k − r )= (∆ + ∆ − k + p + 26 + ∆ t − t + 3∆ t − t − r + 3∆ r − r
15 (∆ + ∆ − k + p + 26 − t (∆ + 3∆ −
3∆ + 2) − r (3∆ −
3∆ + 1) < (∆ + ∆ − k + p + 26 − + 3∆ −
3∆ + 2) + ( p − ∆ + 1)(3∆ −
3∆ + 1)= (∆ + ∆ − k + p + 26 + p (3∆ −
3∆ + 1) + 2∆ − − ∆ + 25)= (∆ + ∆ + 2) n − p ( − + 4∆ − − + 3∆ − ∆ + 25 < (∆ + ∆ + 2) n − p (∆ + ∆ + 2) + p + 9 p + 28 p + 26 . Now let 0 < s . Since s < ∆ −
2, it follows from Corollary 1 that( n , n , . . . , n s , n s +1 , n s +2 , . . . , n ∆ − , n ∆ − , n ∆ ) = ((∆ − k + p − ( t +1) , , . . . , , , , . . . , , , r + t, k − r )which is the optimal solution. Since 2 ≤ p < ∆ − < s ≤ ∆ − − p ( − + 4∆ − − ∆ + 3∆ − ∆ − < △ − △ +10 △ − < p + 9 p + 28 p + 26 . Thus F ( B ) ≤ n + 2 n + . . . + (∆ − n ∆ − + ∆ n ∆ = (∆ − k + p − ( t + 1) + ( s + 3) + (∆ − ( t + r ) + ∆ ( k − r )= (∆ + ∆ − k + p − s + 3) − t ( − ∆ + 3∆ −
3∆ + 2) − r (3∆ −
3∆ + 1) < (∆ + ∆ − k + p − − − ∆ + 3∆ −
3∆ + 2) + ( p + 1 − ∆)(3∆ −
3∆ + 1)= (∆ + ∆ + 2) n − p ( − ∆ + ∆ + 2) + p + p (3∆ + 3∆ + 1) − ∆ + 3∆ − ∆ −
2= (∆ + ∆ + 2) n − p ( − + 4∆ − − ∆ + 3∆ − ∆ − < (∆ + ∆ + 2) n − p (∆ + ∆ + 2) + p + 9 p + 28 p + 26 . This completes the proof.
Funding Information:
J. Rodr´ıguez was supported by MINEDUC-UA project, codeANT-1899 and Funded by the Initiation Program in Research - Universidad de Antofa-gasta, INI-19-06.
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