# On the Numerical Semigroup Generated by2k−m⋅2n+i−1|i∈ℕ

Authors: Ze Gu; Donglin Li; Yucong Huang
Journal: Journal of Mathematics
 (2022)

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## Abstract

Letk,n be two positive integers such that 2≤k≤2n and m be an odd number with 1≤m<2k−1. Denote by Tk,m,n a numerical semigroup generated by 2k−m⋅2n+i−1|i∈ℕ. In this paper, we give the formula for the embedding dimension of Tk,m,n. In particular, the formulas for computing the Frobenius number and the genus of Tk,2t+1,n with 2≤t<k−1 are given.


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## Body

## 1. Introduction and Preliminaries

Letℕ be the set of nonnegative integers. A numerical semigroup is a subset S of ℕ such that it is closed under addition, 0∈S, and ℕ\S is finite. The elements of GS=ℕ\S are the gaps of S, and its cardinality, denoted by gS, is called the genus of S. Another important invariant of S is the largest integer not belonging to S, known as the Frobenius number of S and denoted by FS.Given a nonempty subsetA of ℕ, we denote by A the submonoid of ℕ,+ generated by A, that is, A=λ1a1+⋯+λnan|n∈ℕ\0,a1,…,an∈A,λ1,…,λn∈ℕ. If M=A, then we say that A is a system of generators of M. In addition, if no proper subset of A generates M, then we say that A is a minimal system of generators of M. In [1], it is proved that every submonoid of ℕ,+ has a unique minimal system of generators and such a system is finite. It is well-known that A is a numerical semigroup if and only if gcdA=1. Let S be a numerical semigroup. The cardinality of the minimal system of generators of S is called the embedding dimension of S and is denoted by eS.The Frobenius problem [1, 2] is to find formulas for computing the Frobenius number and the genus of a numerical semigroup. The problem was solved by Sylvester in [3] for numerical semigroups with embedding dimension two. The formulas for some particular cases of numerical semigroups with embedding dimension three were given by Tripathi in [4]. Thus, this problem remains open for numerical semigroups with an embedding dimension greater than three.Arepunit is a number such as 11,111 or 1111 that contains only the digit 1. The set of repunits in base b is bn−1/b−1|n∈ℕ\0. In particular, repunits in binary is well-known Mersenne numbers. A positive integer x is a Thabit number if x=3⋅2n−1 for some n∈ℕ. The concept of Mersenne [5] (repunit [6] and Thabit [7]) numerical semigroups was introduced by Rosales et al. Moreover, they gave the formulas for the Frobenius number, the genus, and the embedding dimension of the three kinds of numerical semigroups. In [8], the first author and Tang generalized Thabit numerical semigroups to the class of numerical semigroups generated by 2k−1⋅2n+i−1|i∈ℕ, where k,n are positive integers such that 2≤k≤2n. In [9], the first author generalized Thabit numerical semigroups along another line to the class of numerical semigroups generated by bn+1+i+bn+i−1/b−1|i∈ℕ, where b,n are positive integers such that b≥2. In this paper, we further generalize the results of [8]. Let k,n be two positive integers such that 2≤k≤2n and m be an odd integer with 1≤m<2k−1. We define a numerical semigroup generated by 2k−m⋅2n+i−1|i∈ℕ and denote it by Tk,m,n. The main purpose of the rest of this paper is to give the formula for the embedding dimension of Tk,m,n, and the formulas for computing the Frobenius number and the genus of Tk,2t+1,n with 2≤t<k−1.


## 2. The Embedding Dimension

Letk,n be two positive integers such that 2≤k≤2n and m be an odd number with 1≤m<2k−1. Then, Tk,m,n is a submonoid of ℕ,+. We denote by si the element 2k−m⋅2n+i−1 for all i∈ℕ. It is clear that si+1=2si+1 for all i∈ℕ. Hence, gcdsi,si+1=1 and Tk,m,n is a numerical semigroup. In the following, we give the formula for the embedding dimension of Tk,m,n. First, we need a preliminary result.Lemma 1 (Lemma 2 [6]).
Letp,q be integers and let A be a nonempty set of positive integers such that S=A. Then, the following conditions are equivalent:(1)
pa+q∈M for all a∈A(2)
ps+q∈M for all s∈S\0Theorem 1.
The sets0,s1,…,sn+k−1 is the minimal system of generators of Tk,m,n and eTk,m,n=n+k.Proof.
First, we show thats0,s1,…,sn+k−1 is a system of generators of Tk,m,n. Let S=s0,s1,…,sn+k−1. It is easy to get S⊆Tk,m,n. Next, we prove another inclusion. If i∈0,1,…,n+k−2, then 2si+1=si+1∈S. For i=n+k−1, we have 2sn+k−1+1=sn+k=2k−m⋅22n+k−1=2k−1−m⋅2n+1+12k−m⋅2n−1+m⋅2n2k−m⋅2n+1−1=2k−1−m⋅2n+1+1⋅s0+m⋅2n⋅s1∈S.
From Lemma1, we obtain that 2s+1∈S for all s∈S\0. By induction, we can easily show that sn+i∈S for all i≥k.
Suppose that it is not a minimal system of generators ofTk,m,n. Then, there exists h∈1,2,…,n+k−1 such that sh∈s0,s1,…,sh−1. Denote by T the numerical semigroup s0,s1,…,sh−1. If i∈0,1,…,h−2, then 2si+1=si+1∈T. Besides, 2sh−1+1=sh∈T. Therefore, it follows from Lemma 1 that 2x+1∈T for all x∈T\0. Hence, by induction we can obtain that si∈T for all i≥h, which is a contradiction with sn+k−1∉T. Indeed, if sn+k−1∈s0,s1,…,sn+k−2. Then, sn+k−1=∑i=0n+k−2aisi=∑i=0n+k−2ai2k−m⋅2n+i−∑i=0n+k−2ai for some a0,a1,…,an+k−2∈ℕ. Hence, ∑i=0n+k−2ai≡1mod2k−m⋅2n and ∑i=0n+k−2ai=1+l2k−m⋅2n for some l∈ℕ. Moreover, it is obvious that l≠0 and thus ∑i=0n+k−2ai≥1+2k−m⋅2n. Consequently, sn+k−1=∑i=0n+k−2aisi≥∑i=0n+k−2ais0≥2k−m⋅2n+12k−m⋅2n−1=2k−m2⋅22n−1>2k−m⋅2k−1⋅22n−1=2k−m⋅22n+k−1−1=sn+k−1.
This is impossible.


## 3. The Frobenius number

LetS be a numerical semigroup and a∈S\0. The Apéry set [10] of a in S is ApS,a=s∈S|s−a∉S. It is well-known from [1] that ApS,a=w0,w1,…,wm−1, where wi is the least element of S congruent to i modulo a for all i∈0,1,…,a−1.Lemma 2.
LetApTk,m,n,s0=w0,w1,…,ws0−1. Then, w0<w1<⋯<ws0−1.Proof.
We show thatwi<wi+1 for all i∈0,1,…,s0−2. Since wi+1∈Tk,m,n, there exist a0,a1,…,an+k−1 such that wi+1=∑j=0n+k−1ajsj. Moreover, wi+1≢0mods0 and so there exists j∈1,2,…,n+k−1 such that aj≠0. Therefore,(1)wi+1−1=a0s0+⋯+aj−1sj+⋯+an+k−1sn+k−1+2k−m⋅2n+j−2=a0s0+⋯+aj−1sj+⋯+an+k−1sn+k−1+22k−m⋅2n+j−1−1=a0s0+⋯+aj−1+2sj−1+aj−1sj+⋯+an+k−1sn+k−1∈Tk,m,n.
Aswi+1−1≡imods0, we have wi≤wi+1−1.Next, our aim purpose is to give an explicit description of the elements in the setApTk,2t+1,n,s0 with 2≤t≤k−1. Given a positive integer r, we define Ar=a1,a2,…,ar∈0,1,2r|if1≤i<j≤randaj=2,thenai=0. It is well-known from [8] that the relation ≤r on Ar defined by(2)a1,a2,…,ar≤rb1,b2,…,br⟺∀i∈1,…,rai=bior∃j=maxi∈1,…,r|ai≠biaj<bj,is a colexicographic order and a total order. Moreover, the minimum and the maximum element in Ar are 0,…,0 and 0,…,0,2, respectively. The following result is direct.Lemma 3.
a1,a2,…,ar≤rb1,b2,…,br is equivalent to ∑i=1rai2i−1≤∑i=1rbi2i−1.Remark 1.
(1) From Lemma3, we know that ∑i=1rai2i−1|a1,a2,…,ar∈Ar=0,1,…,2r+1−2 and the cardinality of Ar is ♯Ar=2r+1−1.
(2) Letk,t,n be positive integers such that 3≤t+1<k≤2n. Then, 1≤k−3≤2n−3. Hence, from (1), there exists k1,k2,…,kn−1∈An−1 such that k=∑i=1n−1ki2i−1+3.Lemma 4.
maxApTk,2t+1,n,s0≤∑i=1n−1kisi+∑j=nn+k−1sj−sn+t.Proof.
Obviously,si=2is0+2i−1 for all i∈ℕ. Therefore, we can obtain(3)∑i=1n−1kisi+∑j=nn+k−1sj−sn+t=∑i=1n−1ki2is0+2i−1+∑j=nn+k−12js0+2j−1−2n+ts0+2n+t−1=∑i=1n−1ki2i+∑j=nn+k−12j−2n+ts0+∑i=1n−1ki2i−1+∑j=nn+k−12j−1−2n+t−1=∑i=1n−1ki2i+∑j=nn+k−12j−2n+ts0+k−3+2k−2t+1⋅2n−k+1=∑i=1n−1ki2i+∑j=nn+k−12j−2n+ts0+2k−2t+1⋅2n−2=∑i=1n−1ki2i+∑j=nn+k−12j−2n+ts0+s0−1.
Hence,∑i=1n−1kisi+∑j=nn+k−1sj−sn+t≡s0−1mods0. Consequently, ws0−1≤∑i=1n−1kisi+∑j=nn+k−1sj−sn+t and by Lemma 2 we complete our proof.LetX be a set and P be a condition. We denote X|P the subset of X satisfying the condition P. Let k=∑i=1n−1ki2i−1+3 with k1,k2,…,kn−1∈An−1 and m=2t+1 with 2≤t<k−1. For convenience, we define the condition C by(4)a1,…,an−1,an,…,an+t−1,an+t,an+t+1,…,an+k−1≤n+k−1k1,…,kn−1,1,…,1,0,1,…,1.Moreover, we denote(5)Rn+k−1=An+k−1|C,R==Rn+k−1|an,…,an+t−1,an+t,an+t+1,…,an+k−1=1,…,1,0,1,…,1,R<=Rn+k−1|an,…,an+t−1,an+t,an+t+1,…,an+k−1<k1,…,1,0,1,…,1.Then,Rn+k−1=R=∪R<.Lemma 5.
♯Rn+k−1=2k−2t+1⋅2n−1.Proof.
Leta1,…,an+k−1∈An+k−1.(1)
Ifan,…,an+t−1,an+t,an+t+1,…,an+k−1=1,…,1,0,1,…,1, then a1,…,an+k−1∈R= is equivalent to a1,…,an−1≤n−1k1,…,kn−1. By Remark 1, we have ♯An−1|a1,…,an−1≤n−1k1,…,kn−1=∑i=1n−1ki2i−1+1=k−2. Hence, ♯R==k−2.(2)
Letan,…,an+t−1,an+t,an+t+1,…,an+k−1<k1,…,1,0,1,…,1. From Remark 1, we know that ♯Ak|an,…,an+t−1,an+t,an+t+1,…,an+k−1<k1,…,1,0,1,…,1=∑i=1k2i−1−2t−1=2k+1−2t−k−1. We distinguish two cases.(i)
If2∉an,…,an+k−1, then ♯R<|2∉an,…,an+k−1=♯an,…,an+k−1∈0,1k|an,…,an+t−1,an+t,an+t+1,…,an+k−1<k1,…,1,0,1,…,1⋅♯An−1=2k−2t+1⋅2n−1.(ii)
If2∈an,…,an+k−1, then a1=⋯=an−1=0. Hence, we have ♯R<|2∈an,…,an+k−1=2k+1−2t−k−1−(2k−2t+1=2k−2t−k.
Therefore,♯Rn+k−1=k−2+(2k−2t+1⋅2n−1+2k−2t−k=2k−2t+1⋅2n−1.Theorem 2.
ApTk,2t+1,n,s0=∑i=1n+k−1aisi|a1,…,an+k−1∈Rn+k−1.Proof.
DenoteH=∑i=1n+k−1aisi|a1,…,an+k−1∈Rn+k−1. By Lemma 3 and 5, we obtain ♯H=♯Rn+k−1=2k−2t+1⋅2n−1=s0=♯ApTk,2t+1,n,s0. Next, we need only to show that ApTk,2t+1,n,s0⊆H. Let x∈ApTk,2t+1,n,s0. First, we need to show that x=∑i=1n+k−1aisi for some a1,…,an+k−1∈An+k−1. By induction over x, if x=0 then the result holds obviously. Suppose that x>0 and j=mini∈0,1,…,n+k−1|x−si∈Tk,2t+1,n. Since x∈ApTk,2t+1,n,s0, we have j≠0 and x−sj∈ApTk,2t+1,n,s0. By induction hypothesis, there exists b1,b2,…,bn+k−1∈An+k−1 such that x−sj=∑i=1n+k−1bisi. Hence, x=b1s1+⋯+bj+1sj+⋯+bn+k−1sn+k−1. Next, we check that b1,…,bj+1,…,bn+k−1∈An+k−1.(1)
To proveb1,…,bj+1,…,bn+k−1∈0,1,2n+k−1, it is suffices to show that bj+1≠3. If bj+1=3, then bj+1sj=3sj=2sj−1+sj+1. Thus x−sj−1∈Tk,2t+1,n which contradicts the minimality of j.(2)
From the minimality ofj, we obtain that bi=0 for all 1≤i<j. Moreover, bl≠2 for all j<l≤n+k−1. In fact, if there exists j<l≤n+k−1 such that bl=2, then sj+blsl=sj+2sl=2sj−1+sl+1. Hence, x−sj−1∈Tk,2t+1,n which is impossible.
Moreover, we can deduce from Lemma4 that b1,…,bj+1,…,bn+k−1∈Rn+k−1 and so x∈H. Therefore, we obtain our conclusion.LetS be a numerical semigroup and m∈S\0. It is well known from [1] that FS=maxApS,m−m. Hence, we obtain the following formula for the Frobenius number of Tk,2t+1,n.Corollary 1.
FTk,2t+1,n=k−2+∑i=1n−1kis0+s02−1.Proof.
From Lemma4 and Theorem 2, we have that maxApTk,2t+1,n,s0=∑i=1n−1kisi+∑j=nn+k−1sj−sn+t. Moreover, si=2is0+2i−1. Thus(6)FTk,2t+1,n=∑i=1n−1kisi+∑j=nn+k−1sj−sn+t−s0=∑i=1n−1ki2is0+2i−1+∑j=nn+k−12js0+2j−1−2n+ts0+2n+t−1−s0=∑i=1n−1ki2i+∑j=nn+k−12j−2n+ts0+∑i=1n−1ki2i−1+∑j=nn+k−12j−k−2n+t−1−s0=k−3+∑i=1n−1ki+2n+k−2n−2n+ts0+k−3+2n+k−2n−k−2n+t+1−s0=k−2+∑i=1n−1ki+2k−2t−1⋅2n−1s0+2k−2t+1⋅2n−2−s0=k−2+∑i=1n−1ki+s0s0+s0−1−s0=k−2+∑i=1n−1kis0+s02−1.
Hence, we complete this proof.


## 4. The Genus

For a numerical semigroupS and m∈S\0, from [1], we know that gS=1/m∑w∈ApS,mw−m−1/2. Hence, to give the formula for the genus of Tk,2t+1,n, we need only to compute ∑w∈ApTk,2t+1,n,s0w. From Theorem 2, we obtain(7)∑w∈ApTk,2t+1,n,s0w=∑a1,…,an+k−1∈Rn+k−1∑i=1n+k−1aisi=∑a1,…,an+k−1∈R=∑i=1n+k−1aisi+∑a1,…,an+k−1∈R<∑i=1n+k−1aisi=∑a1,…,an−1∈R=n−1∑i=1n−1aisi+k−2∑i=nn+t−1si+k−2∑i=n+t+1n+k−1si+∑a1,…,an+k−1∈R<∑i=1n+k−1aisi,where R=n−1=a1,…,an−1|a1,…,an+k−1∈R=. Furthermore,(8)∑a1,…,an+k−1∈R<∑i=1n+k−1aisi=∑i=1n+k−1∑a1,…,an+k−1∈R<|ai=1si+∑i=1n+k−1∑a1,…,an+k−1∈R<|ai=22si.To our purpose, we need two preliminary results.Lemma 6.
♯R<|ai=2=2n−1−i⋅2k−2t−11≤i≤n−1;2n−1−i⋅2k−2t−1n≤i≤n+t−1;2n+k−1−i−1n+t≤i≤n+k−1.Proof.
(1)
Ifai=2 for some i∈1,…,n−1, then a1=⋯=ai−1=0, ai+1,…,an+k−1∈0,1 and an,…,an+t−1,an+t,an+t+1,…,an+k−1<k1,…,1,0,1,…,1.Thus, we have(9)♯R<|ai=2=2n−1−i⋅♯Ak|an,…,an+t−1,an+t,an+t+1,…,an+k−1<k1,…,1,0,1,…,1=2n−1−i⋅2k−2t−1.(2)
Ifai=2 for some i∈n,…,n+t−1, then a1=⋯=ai−1=0, ai+1,…,an+k−1∈0,1 and ai+1,…,an+t−1,an+t,an+t+1,…,an+k−1<n+k−1−i1,…,1,0,1,…,1.Hence, we obtain(10)♯R<|ai=2=♯An+k−1−i|ai+1,…,an+t,…,an+k−1<n+k−1−i1,…,0,…,1=2n+k−1−i−2n+t−1−i−1=2n−1−i⋅2k−2t−1.(3)
Ifai=2 for some i∈n+t,…,n+k−1, then a1=⋯=ai−1=0, ai+1,…,an+k−1∈0,1 and ai+1,…,an+k−1<n+k−1−i1,…,1. Therefore, ♯R<|ai=2=♯An+k−1−i|ai+1,…,an+k−1<n+k−1−i1,…,1=2n+k−1−i−1.Lemma 7.
♯R<|ai=1=2n−1−2n−1−i⋅2k−2t−11≤i≤n−1;2n−1−2n−1−i⋅2k−2t−2n−i+n+1n≤i≤n+t−1;2k−t−1−1⋅2n+t−1i=n+t;2n⋅2k−1−2t−1−2n+k−1−i−i+n+2n+t+1≤i≤n+k−1.Proof.
Leta1,…,an+k−1∈R<.(1)
Letai=1 with i∈1,…,n−1. We distinguish two cases.(i)
If2∉a1,…,ai−1, then a1,…,ai−1,ai+1,…,an+k−1∈0,1 and an,…,an+t−1,an+t,an+t+1,…,an+k−1<k1,…,1,0,1,…,1. Thus ♯R<|ai=1,2∉a1,…,ai−1=2n−2⋅2k−2t−1.(ii)
If2∈a1,…,ai−1, then there exists a unique j∈1,…,i−1 such that aj=2. Hence, a1=⋯=aj−1=0,aj+1,…,ai−1,ai+1,…,an+k−1∈0,1 and an,…,an+t−1,an+t,an+t+1,…,an+k−1<k1,…,1,0,1,…,1. Therefore, ♯R<|ai=1,aj=2=2n−2−j⋅2k−2t−1.Consequently,♯R<|ai=1=2n−2⋅2k−2t−1+∑j=1i−12n−2−j⋅2k−2t−1=2n−1−2n−1−i⋅2k−2t−1.(2)
Letai=1 for some i∈n,…,n+t−1. We distinguish two cases.(iii)
If2∉a1,…,ai−1, then a1,…,ai−1,ai+1,…,an+k−1∈0,1 and an,…,an+t−1,an+t,an+t+1,…,an+k−1<k1,…,1,0,1,…,1. Thus ♯R<|ai=1,2∉a1,…,ai−1=2n−1⋅2k−1−2t−1−1.(iv)
If2∈a1,…,ai−1, then there exists a unique j∈1,…,i−1 such that aj=2. Thus a1=⋯=aj−1=0 and aj+1,…,ai−1,ai+1,…,an+k−1∈0,1. If 1≤j≤n−1, then ♯R<|ai=1,aj=2=2n−1−j⋅2k−1−2t−1−1. On the other hand, aj+1,…,an+t−1,an+t,an+t+1,…,an+k−1<n+k−1−j1,…,1,0,1,…,1 for n≤j≤i−1. Therefore, ♯R<|ai=1,aj=2=2n+k−2−j−2n+t−2−j−1.Consequently, we have♯R<|ai=1=2n−1⋅2k−1−2t−1−1+∑j=1n−12n−1−j⋅2k−1−2t−1−1+∑j=ni−12n+k−2−j−2n+t−2−j−1=2n−1−2n−1−i⋅2k−2t−2n−i+n+1.(3)
Letan+t=1. We distinguish two cases.(v)
If2∉a1,…,an+t−1, then a1,…,an+t−1,an+t+1,…,an+k−1∈0,1 and an+t+1,…,an+k−1≠1,…,1. Hence, ♯R<|an+t=1,2∉a1,…,an+t−1=2n+t−1⋅2k−t−1−1.(vi)
If2∈a1,…,an+t−1, then aj=2 for a unique j∈1,…,n+t−1. Thus a1=⋯=aj−1=0,aj+1,…,an+t−1,an+t+1,…,an+k−1∈0,1 and an+t+1,…,an+k−1≠1,…,1. Hence, ♯R<|an+t=1,aj=2=2n+t−1−j⋅2k−t−1−1.Conseqently,♯R<|ai=1=2n+t−1⋅2k−t−1−1+∑j=1n+t−12n+t−1−j⋅2k−t−1−1=2k−t−1−1⋅2n+t−1.(4)
Letai=1 for some i∈n+t+1,…,n+k−1. We distinguish two cases.(vii)
If2∉a1,…,ai−1, then a1,…,ai−1,ai+1,…,an+k−1∈0,1 and an,…,an+t−1,an+t,an+t+1,…,an+k−1<k1,…,1,0,1,…,1. Thus ♯R<|ai=1,2∉a1,…,ai−1=2n−1⋅2k−1−2t−1.(viii)
If2∈a1,…,ai−1, then there exists a unique j∈1,…,i−1 such that aj=2. Thus a1=⋯=aj−1=0 and aj+1,…,ai−1,ai+1,…,an+k−1∈0,1. If 1≤j≤n−1, then ♯R<|ai=1,aj=2=2n−1−j⋅2k−1−2t−1. If n≤j≤n+t−1, then aj+1,…,an+k−1<n+k−1−j1,…,1,0,1,…,1. Hence, ♯R<|ai=1,aj=2=2n+k−2−j−2n+t−1−j−1. If n+t≤j≤i−1, then aj+1,…,an+k−1≠1,…,1. Therefore, R<|ai=1,aj=2=2n+k−2−j−1.
Consequently,(11)♯R<|ai=1=2n−1⋅2k−1−2t−1+∑j=1n−12n−1−j⋅2k−1−2t−1+∑j=nn+t−12n+k−2−j−2n+t−1−j−1+∑j=n+ti−12n+k−2−j−1=2n⋅2k−1−2t−1−2n+k−1−i−i+n+2.As the main result of this section, we give the following formula for the genus ofTk,2t+1,n.Theorem 3.
gTk,2t+1,n=1/s0∑w∈ApSk,n,s0w−s0−1/2, where(12)∑w∈ApTk,2t+1,n,s0w=∑a1,…,an−1∈R=n−1∑i=1n−1aisi+2k−2t−13⋅23n−1+2k−2t−12kn+k−3−2tn+2k−t−3−n−2k+5⋅22n−1−2kn+k−2tn+2k−t−n−2k+2⋅2n−1−k−1k−42.Proof.
From Lemma6 and 7, we have(13)∑i=1n+k−1∑a1,…,an+k−1∈R<|ai=1si=∑i=1n−12n−1−2n−1−i⋅2k−2t−1si+∑i=nn+t−12n−1−2n−1−i⋅2k−2t−2n−i+n+1si+2k−t−1−1⋅2n+t−1sn+t+∑i=n+t+1n+k−12n⋅2k−1−2t−1−2n+k−1−i−i+n+2si,∑i=2n+k−1∑a1,…,an+k−1∈R<|ai=12si=∑i=1n−12n−1−i⋅2k−2t−1⋅2si+∑i=nn+t−12n−1−i⋅2k−2t−1⋅2si+∑i=n+tn+k−12n+k−1−i−1⋅2si=∑i=1n−12n−i⋅2k−2t−1si+∑j=nn+t−12n−i⋅2k−2t−2si+∑i=n+tn+k−12n+k−i−2si.
By the computation of sum of subsections, we obtain(14)∑i=1n−12n−1−2n−1−i⋅2k−2t−1si+∑i=1n−12n−i⋅2k−2t−1si=∑i=1n−12n−1−2n−1−i⋅2k−2t−12k−2t−12n+i−1=∑i=1n−12k−12n−1−2n−1−i2k−1⋅2n+i−1=2k−2t−12∑i=1n−122n+i−1+2k−2t−12∑i=1n−122n−1−2k−2t−1∑i=1n−12n−1−2k−2t−1∑i=1n−12n−i−1=2k−2t−1223n−1−22n+2k−2t−12n−122n−1−2k−2t−1n−12n−1−2k−2t−12n−1−1=2k−2t−12⋅23n−1+2k−2t−12n−3⋅22n−1−n2k−2t−1⋅2n−1+2k−2t−1,k−2∑i=nn+t−1si+∑i=nn+t−12n−1−2n−1−i⋅2k−2t−2n−i+n+1si+∑i=nn+t−12n−i⋅2k−2t−2si=∑i=nn+t−12k−2t2n−1+2n−1−i−2n+n+k−i−3⋅2k−2t−12n+i−1=2k−2t−12k−2t2n−1−2n+n+k−3∑i=nn+t−12n+i+∑i=nn+t−12k−2t−12k−2t22n−1−2k−2t2n−1+2n−2k−2t−1∑i=nn+t−1i2n+i−∑i=nn+t−12k−2t2n−i−1−∑i=nn+t−1n+k−i−3=2k−2t−12k−2t2n−1−2n+n+k−322n+t−22n+t2k−2t−12k−2t22n−1−2k−2t2n−1+2n−2k−2t−1n+t−222n+t+22n+1−n22n−2k−2k−t−2t+1−t2k−t−52=2k−2t−12k−2t−22k−1⋅23n−1+2k−2t−1t2k+2k−3t−22t−2k+2⋅22n−1−t2k−2t−2⋅2n−1−2k+2k−t+2t−1−t2k−t−52,k−2∑i=n+t+1n+k−1si+2k−t−1−1⋅2n+t−1sn+t+∑i=n+t+1n+k−12n⋅2k−1−2t−1−2n+k−1−i−i+n+2si+∑i=n+tn+k−12n+k−i−2si=2k−t−1−12n+t+1sn+t+∑i=n+t+1n+k−12n2k−1−2t−1+2n+k−1−i+n+k−2−isi=2k−t−1−12n+t+12k−2t−122n+t−1+∑i=n+t+1n+k−12n2k−1−2t−1+2n+k−1−i+n+k−2−i⋅2k−2t−12n+i−1=2k−t−1−12n+t+12k−2t−122n+t−1+2k−2t−1∑i=n+t+1n+k−12n2k−1−2t−1+n+k−22n+i+∑i=n+t+1n+k−12k−2t−122n+k−1−2n2k−1−2t−1−2k−2t−1∑i=n+t+1n+k−1i2n+i−∑i=n+t+1n+k−12n+k−1−i−∑i=n+t+1n+k−1n+k−2−i=2t2k−2t+12k−2t−123n−1+2k−2t+12k−2t−122n−1−2k−2t+12n−1−2k−t−1−1+2k+1−2t+22k−2t−12k−1−2t−123n−1+n+k−222n−1+k−t−12k2k−2t−122n−1−2k−2t+1−22n−1−2k−2t−1n+t−22k+1+2t+3−n+t+12t+222n−1−2k−t−1−1−k−t−1k−t−42=2k−2t+12k−2t−12k−2t−2⋅23n−1+2k−2t−12kk−t−2t+12k−2t−1⋅22n−1−k−t2k−2t+1−2k−t−1⋅2n−1−2k−t+2−k−t−1k−t−42.
Hence,(15)∑w∈ApTk,2t+1,n,s0w=∑a1,…,an−1∈R=n−1∑i=1n−1aisi+2k−2t−13⋅23n−1+2k−2t−12kn+k−3−2tn+2k−t−3−n−2k+5⋅22n−1−2kn+k−2tn+2k−t−n−2k+2⋅2n−1−k−1k−42.We end this paper with an example by illustrating the previous results.Example 1.
Letk=4,n=2 and t=2. Then k1=1 and R=1=0,1. By applying Theorem 1, Corollary 1, and Theorem 3, we obtain that eT4,5,2=6, FS4,5,2=1977 and gT4,5,2=1016.



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