In this paper, we prove some new Simpson's type inequalities for partial (p,q)-differentiable convex functions of two variables in the context of (p,q)-calculus. We also show that the findings in this paper are generalizations of comparable findings in the literature.
Citation: Xue-Xiao You, Muhammad Aamir Ali, Ghulam Murtaza, Saowaluck Chasreechai, Sotiris K. Ntouyas, Thanin Sitthiwirattham. Post-quantum Simpson's type inequalities for coordinated convex functions[J]. AIMS Mathematics, 2022, 7(2): 3097-3132. doi: 10.3934/math.2022172
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In this paper, we prove some new Simpson's type inequalities for partial (p,q)-differentiable convex functions of two variables in the context of (p,q)-calculus. We also show that the findings in this paper are generalizations of comparable findings in the literature.
We denote by (x,y) (resp. [x,y]) the greatest common divisor (resp. least common multiple) of any given integers x and y. Let a,b and n be positive integers and S={x1,...,xn} be a set of n distinct positive integers. Let f be an arithmetic function and we denote by (f(S)) (resp. (f[S])) the n×n matrix having f evaluated at (xi,xj) (resp. [xi,xj]) as its (i,j)-entry. Particularly, the n×n matrix (Sa)=((xi,xj)a), having the ath power (xi,xj)a as its (i,j) -entry, is called the ath power GCD matrix on S. The n×n matrix [Sa]=([xi,xj]a), having the ath power [xi,xj]a as its (i,j)-entry, is called the ath power LCM matrix on S. These are simply called the GCD matrix and LCM matrix respectively if a=1. The set S is said to be factor closed (FC) if it contains every divisor of x for any x∈S. The set S is said to be gcd closed (resp. lcm closed) if for all i and j, (xi,xj) (resp. [xi,xj]) is in S. Evidently, an FC set is gcd closed but not conversely. In 1875, Smith [33] published his famous theorem stating that the determinant of the GCD matrix (S) defined on the set S={1,...,n} is the product ∏nk=1φ(k), where φ is Euler's totient function. Since then many interesting generalizations of Smith's determinant and related results have been published (see, for example, [1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32] and [34,35,36,37,38,39,40,41,42]).
Divisibility is an important topic in the field of power GCD matrices and power LCM matrices. Bourque and Ligh [5] showed that if S is an FC set, then the GCD matrix (S) divides the LCM matrix [S] in the ring Mn(Z) of n×n matrices over the set Z of integers. That is, there exists an A∈Mn(Z) such that [S]=(S)A or [S]=A(S). Hong [19] showed that such factorization is no longer true in general if S is gcd closed. Bourque and Ligh [8] extended their own result showing that if S is factor closed, then the power GCD matrix (Sa) divides the power LCM matrix [Sa] in the ring Mn(Z). The set S is called a divisor chain if there exists a permutation σ on {1,...,n} such that xσ(1)|...|xσ(n). Obviously a divisor chain is gcd closed but the converse is not true. For x,y∈S, and x<y, if x|y and the conditions x|d|y and d∈S imply that d∈{x,y}, then we say that x is a greatest-type divisor of y in S. For x∈S, we denote by GS(x) the set of all greatest-type divisors of x in S. By [19], we know that there is gcd-closed set S with maxx∈S{|GS(x)|}=2 such that (S)−1[S]∉Mn(Z). In [26], Hong, Zhao and Yin showed that if S is gcd closed and maxx∈S{|GS(x)|}=1, then the GCD matrix (S) divides the LCM matrix [S] in Mn(Z). In [20], Hong showed that (f(S))|(f[S]) holds in the ring Mn(Z) when S is a divisor chain and f is an integer-valued multiplicative function satisfying that f(min(S))|f(x) for any x∈S.
Hong [22] initiated the investigation of divisibility among power GCD matrices and among power LCM matrices. In fact, Hong [22] proved that the power GCD matrix (Sa) divides the power GCD matrix (Sb) if a|b and S is a divisor chain. Hong also showed that the power LCM matrix [Sa] divides the power LCM matrix [Sb] if a|b and S is a divisor chain. But such factorizations are not true if a⧸|b and gcd(S)=1 as well |S|≥2, where by |S| and gcd(S) we denote the cardinality of the set S and the greatest common divisor of all the elements in S, respectively. We say that the set S consists of two coprime divisor chains if we can partition S as S=S1∪S2, where S1 and S2 are divisor chains and each element of S1 is coprime to each element of S2. Later on, Hong's results were extended by Tan et al. These results confirm partially Conjectures 4.2-4.4 of [22]. It was proved in [36] that if a|b, then (Sa)|(Sb), [Sa]|[Sb] and (Sa)|[Sb] hold in the ring Mn(Z) if and only if both xayb−1xaya−1 and xbya−1xaya−1 are integers, where S=S1∪S2 with S1 and S2 being divisor chains and x=min(S1) and y=min(S2). From this one can read that even though a|b and S consists of two coprime divisor chains, but if 1∉S, then the divisibility theorems among power GCD matrices and among power LCM matrices need not always hold. Meanwhile, Tan, Lin and Liu found surprisingly that the divisibility theorems among determinants of power GCD matrices and among determinants of power LCM matrices should always hold. That is, they showed that if a|b and S consists of two coprime divisor chains as well 1∉S, then det(Sa)|det(Sb), det[Sa]|det[Sb] and det(Sa)|det[Sb].
The main aim of this paper is to generalize this interesting result to the matrices of the forms det(fa(S)) and det(fa[S]), where the arithmetic function fa is defined for any positive integer x by fa(x)=(f(x))a. We will study the divisibility among det(fa(S)) and det(fb(S)) and among det(fa[S]) and det(fb[S]) when a|b. We also investigate the divisibility among det(f(Sa)) and det(f(Sb)) and among det(f[Sa]) and det(f[Sb]) when a|b, where Sa:={xa|x∈S} is the ath power set of S. In particular, we show that if S consists of two coprime divisor chains with 1∉S and f is an integer-valued multiplicative function (see, for instance, [2]), then for any positive integer a, we have det(f(Sa))|det(f[Sa]). But it is unclear whether or not the n×n matrix (f[Sa]) is divisible by the n×n matrix (f(Sa)) in the ring Mn(Z) when S consists of two coprime divisor chains with 1∉S and f is integer-valued and multiplicative. This problem remains open. We guess that the answer to this question is affirmative.
This paper is organized as follows. First of all, we recall in Section 2 Hong's formulas for det(f(S)) and det(f[S]) when S is gcd closed, and then use them to give formulae for the determinants of matrices associated with arithmetic functions on divisor chains. Consequently, in Section 3, we use these results to derive the formulae for the determinants of matrices associated with arithmetic functions on two coprime divisor chains. The final section is to present the main results and their proofs. Our results extend Hong's results[20,22] and the Tan-Lin-Liu results [36].
In the close future, we will explore the divisibility among the power matrices associated with integer-valued arithmetic functions.
In the present section, we provide formulas for the determinants of matrices associated with arithmetic functions on divisor chains. For this purpose, we need the concept of greatest-type divisor introduced by Hong in 1996 (see, for example, [16] and [17]). Notice that the concept of greatest-type divisor played central roles in Hong's solution [16,17] to the Bourque-Ligh conjecture [5], in Cao's partial answer [9] to Hong's conjecture [18] as well as in Li's partial answer [28] to Hong's conjecture [21]. We begin with the following formulas due to Hong.
Lemma 2.1. ([21]) Let f be an arithmetic function and S be a gcd-closed set. Then
det(f(S))=∏x∈S∑J⊂GS(x)(−1)|J|f(gcd(J∪{x}))) |
and if f is multiplicative, then
det(f[S])=∏x∈Sf(x)2∑J⊂GS(x)(−1)|J|f(gcd(J∪{x})). |
We can now use Hong's formulae to deduce the formulae for det(Sa) and det[Sa] when S is a divisor chain.
Theorem 2.2. Let f be an arithmetic function and S={x1,...,xn} be a divisor chain such that x1|...|xn and n≥2. Then
det(f(S))=f(x1)n∏i=2(f(xi)−f(xi−1)) |
and if f is multiplicative, then
det(f[S])=(−1)n−1f(xn)n∏i=2(f(xi)−f(xi−1)). |
Proof. Since x1|x2|...|xn, we have GS(x1)=ϕ and GS(xi)={xi−1} for 2≤i≤n. Then Theorem 2.2 follows immediately from Lemma 2.1.
This completes the proof of Theorem 2.2.
In this section, we give the formulae calculating the determinants of matrices associated with arithmetic functions on two coprime divisor chains.
Theorem 3.1. Let f be an arithmetic function and S={x1,...,xn,y1,...,ym}, where x1|...|xn, y1|...|ym and gcd(xn,ym)=1. Then
det(f(S))=(f(x1)f(y1)−f(1)2)(n−1∏i=1(f(xi+1)−f(xi)))(m−1∏j=1(f(yj+1)−f(yj))) |
and if f is multiplicative, then
det(f[S])=(−1)m+n−1f(xn)f(ym)(f(x1)f(y1)−1)(n−1∏i=1(f(xi+1)−f(xi)))(m−1∏j=1(f(yj+1)−f(yj))). |
Proof. Write Si:={x1,...,xi} and Tj:={y1,...,yj} for all integers i and j with 1≤i≤n and 1≤j≤m. Then S=Sn∪Tm.
First let n=1. Then
det(f(S))=det(f(S1∪Tm))=det(f(x1)f(1)f(1)⋯f(1)f(1)f(y1)f(y1)⋯f(y1)f(1)f(y1)f(y2)⋯f(y2)⋯⋯⋯⋯⋯f(1)f(y1)f(y2)⋯f(ym)). |
Let f(y1)=0. If m=1, then it is clear that
det(f(S))=f(x1)f(y1)−f(1)2 |
as expected. If m≥2, then we can calculate that
det(f(S))=−f(1)2det(f(˜Tm−1)), |
where ˜Tm−1:=Tm∖{y1}. If m=2, then det(f(S))=−f(1)2f(y2) since det(f(˜T1))=f(y2). If m≥3, then it follows from Theorem 2.2 that
det(f(S))=−f(1)2f(y2)m−1∏j=2(f(yj+1)−f(yj)) |
as desired.
Now let f(y1)≠0. Then replacing the first row by the sum of itself and −f(1)f(y1) multiple of the second row and using Theorem 2.2, one arrives at
det(f(S))=det(f(x1)−f(1)2f(y1)00⋯0f(1)f(y1)f(y1)⋯f(y1)f(1)f(y1)f(y2)⋯f(y2)⋯⋯⋯⋯⋯f(1)f(y1)f(y2)⋯f(ym))=(f(x1)−f(1)2f(y1))det(f(Tm))=(f(x1)−f(1)2f(y1))f(y1)m−1∏j=1(f(yj+1)−f(yj))=(f(x1)f(y1)−f(1)2)m−1∏j=1(f(yj+1)−f(yj)) |
as required. Thus the first formula of Theorem 3.1 is true when n=1.
Consequently, let n>1. Then we deduce that
det(f(S))=det(f(Sn∪Tm))=det(f(x1)f(x1)⋯f(x1)f(x1)f(1)f(1)⋯f(1)f(x1)f(x2)⋯f(x2)f(x2)f(1)f(1)⋯f(1)⋮⋮⋱⋮⋮⋮⋮⋱⋮f(x1)f(x2)⋯f(xn−1)f(xn−1)f(1)f(1)⋯f(1)f(x1)f(x2)⋯f(xn−1)f(xn)f(1)f(1)⋯f(1)f(1)f(1)⋯f(1)f(1)f(y1)f(y1)⋯f(y1)f(1)f(1)⋯f(1)f(1)f(y1)f(y2)⋯f(y2)⋮⋮⋱⋮⋮⋮⋮⋱⋮f(1)f(1)⋯f(1)f(1)f(y1)f(y2)⋯f(ym)). |
Replacing nth row by the sum of itself and (−1) multiple of (n−1)th row gives us that
det(f(S))=det(f(x1)f(x1)⋯f(x1)f(x1)f(1)f(1)⋯f(1)f(x1)f(x2)⋯f(x2)f(x2)f(1)f(1)⋯f(1)⋮⋮⋱⋮⋮⋮⋮⋱⋮f(x1)f(x2)⋯f(xn−1)f(xn−1)f(1)f(1)⋯f(1)00⋯0f(xn)−f(xn−1)00⋯0f(1)f(1)⋯f(1)f(1)f(y1)f(y1)⋯f(y1)f(1)f(1)⋯f(1)f(1)f(y1)f(y2)⋯f(y2)⋮⋮⋱⋮⋮⋮⋮⋱⋮f(1)f(1)⋯f(1)f(1)f(y1)f(y2)⋯f(ym))=(f(xn)−f(xn−1))det(f(Sn−1∪Tm))=(f(xn)−f(xn−1))(f(xn−1)−f(xn−2))…(f(x2)−f(x1))det(f(S1∪Tm))=(f(x1)f(y1)−f(1)2)(n−1∏i=1(f(xi+1)−f(xi)))(m−1∏j=1(f(yj+1)−f(yj))) |
as desired. This concludes the proof of the first part of Theorem 3.1.
We are now in the position to show the second part of Theorem 3.1. Since f is multiplicative, one has f(1)=1 and
f(gcd(xi,xj))f(lcm(xi,xj))=f(xi)f(xj). |
It then follows that
(f[S])=Λ⋅(1f(S))⋅Λ, |
where
Λ:=diag(f(x1),...,f(xn),f(y1),...,f(ym)) |
is the (n+m)×(n+m) diagonal matrix with f(x1),...,f(xn),f(y1),...,f(ym) as its diagonal elements. Therefore
det(f[S])=(n∏i=1f2(xi))(m∏j=1f2(yj))det(1f(S)). |
By the first part of Theorem 3.1, one derives that
det(1f(S))=(1f(x1)f(y1)−1f2(1))⋅n−1∏i=1(1f(xi+1)−1f(xi))⋅m−1∏j=1(1f(yj+1)−1f(yj))=1−f(x1)f(y1)f(x1)f(y1)⋅n−1∏i=1(f(xi)−f(xi+1))f(x1)f2(x2)⋯f2(xn−1)f(xn)⋅m−1∏j=1(f(yj)−f(yj+1))f(y1)f2(y2)⋯f2(ym−1)f(ym). |
So we obtain that
det(f[S])=(n∏i=1f2(xi))(m∏j=1f2(yj))×1−f(x1)f(y1)f(x1)f(y1)⋅n−1∏i=1(f(xi)−f(xi+1))f(x1)f2(x2)⋯f2(xn−1)f(xn)⋅m−1∏j=1(f(yj)−f(yj+1))f(yj)f2(y2)⋯f2(ym−1)f(ym)=(−1)m+n−1f(xn)f(ym)(f(x1)f(y1)−1)(n−1∏i=1(f(xi+1)−f(xi)))(m−1∏j=1(f(yj+1)−f(yj))) |
as desired.
This ends the proof of Theorem 3.1.
In this last section, we first study the divisibility among determinants of power matrices associated with integer-valued arithmetic functions. We begin with the following result that is the first main result of this section.
Theorem 4.1. Let f be an integer-valued arithmetic function and let a and b be positive integers such that a|b. Let S consist of two coprime divisor chains with 1∉S. Then det(fa(S))|det(fb(S)). Furthermore, if f is multiplicative, then det(fa[S])|det(fb[S]) and det(fa(S))|det(fb[S]).
Proof. Since S consists of two coprime divisor chains and 1∉S, without loss of any generality, we may let S={x1,...,xn,y1,...,ym}, where x1|...|xn, y1|...|ym and gcd(xn,ym)=1. Then with f replaced by fa and fb, Theorem 3.1 tells us that
det(fa(S))=(fa(x1)fa(y1)−f(1)2a)(n−1∏i=1(fa(xi+1)−fa(xi)))(m−1∏j=1(fa(yj+1)−fa(yj))), |
det(fb(S))=(fb(x1)fb(y1)−f(1)2b)(n−1∏i=1(fb(xi+1)−fb(xi)))(m−1∏j=1(fb(yj+1)−fb(yj))), |
det(fa[S])=(−1)m+n−1fa(xn)fa(ym)(fa(x1)fa(y1)−1)×(n−1∏i=1(fa(xi+1)−fa(xi)))(m−1∏j=1(fa(yj+1)−fa(yj))) |
and
det(fb[S])=(−1)m+n−1fb(xn)fb(ym)(fb(x1)fb(y1)−1)(n−1∏i=1(fb(xi+1)−fb(xi)))(m−1∏j=1(fb(yj+1)−fb(yj))). |
Now let det(fa(S))=0. Then fa(x1)fa(y1)−f(1)2a=0, or fa(xi+1)−fa(xi)=0 for some integer i with 1≤i≤n−1, or fa(yj+1)−fa(yj))=0 for some integer j with 1≤j≤m−1. Since a|b, one then deduces that fb(x1)fb(y1)−f(1)2b=0, or fb(xi+1)−fb(xi)=0 for some integer i with 1≤i≤n−1, or fb(yj+1)−fb(yj))=0 for some integer j with 1≤j≤m−1. Thus det(fb(S))=det(fb[S])=0 which infers that det(fa(S))|det(fb(S)), det(fa[S])|det(fb[S]) and det(fa(S))|det(fb[S]) as desired. Likewise, if det(fa[S])=0, then we can deduce that det(fb[S])=0. Hence det(fa[S])|det(fb[S]) as expected. So Theorem 4.1 holds in this case.
In what follows, we let det(fa(S))≠0 and det(fa[S])≠0. Since a|b, we may let b=ka for some integer k. Therefore
det(fb(S))det(fa(S))=(fb(x1)fb(y1)−f(1)2b)(n−1∏i=1(fb(xi+1)−fb(xi)))(m−1∏j=1(fb(yj+1)−fb(yj)))(fa(x1)fa(y1)−f(1)2a)(n−1∏i=1(fa(xi+1)−fa(xi)))(m−1∏j=1(fa(yj+1)−fa(yj)))=(fka(x1)fka(y1)−f(1)2ka)(n−1∏i=1(fka(xi+1)−fka(xi)))(m−1∏j=1(fka(yj+1)−fka(yj)))(fa(x1)fa(y1)−f(1)2a)(n−1∏i=1(fa(xi+1)−fa(xi)))(m−1∏j=1(fa(yj+1)−fa(yj)))=(k∑t=1(f(x1)f(y1))(k−t)af2(t−1)a(1))(n−1∏i=1k∑t=1(f(xi+1))(k−t)af(t−1)a(xi))×(m−1∏j=1k∑t=1(f(yj+1))(k−t)af(t−1)a(yj))∈Z. |
This implies that det(fa(S))|det(fb(S)).
Similarly, if f is multiplicative and integer-valued, then one deduces that f(1)=1,
det(fb[S])det(fa[S])=fb(xn)fb(ym)(fb(x1)fb(y1)−1)(n−1∏i=1(fb(xi+1)−fb(xi)))(m−1∏j=1(fb(yj+1)−fb(yj)))fa(xn)fa(ym)(fa(x1)fa(y1)−1)(n−1∏i=1(fa(xi+1)−fa(xi)))(m−1∏j=1(fa(yj+1)−fa(yj)))=(f(xn)f(ym))(k−1)a(k∑t=1(f(x1)f(y1))(k−t)a)(n−1∏i=1k∑t=1(f(xi+1))(k−t)af(t−1)a(xi))×(m−1∏j=1k∑t=1(f(yj+1))(k−t)af(t−1)a(yj))∈Z |
and
det(fb[S])det(fa(S))=(−1)m+n−1×fb(xn)fb(ym)(fb(x1)fb(y1)−1)(n−1∏i=1(fb(xi+1)−fb(xi)))(m−1∏j=1(fb(yj+1)−fb(yj)))(fa(x1)fa(y1)−1)(n−1∏i=1(fa(xi+1)−fa(xi)))(m−1∏j=1(fa(yj+1)−fa(yj)))=(−1)m+n−1fb(xn)fb(ym)(k∑t=1(f(x1)f(y1))(k−t)a)×(n−1∏i=1k∑t=1(f(xi+1))(k−t)af(t−1)a(xi))(m−1∏j=1k∑t=1(f(yj+1))(k−t)af(t−1)a(yj))∈Z |
as one requires. Thus Theorem 4.1 is true if det(fa(S))≠0 and det(fa[S])≠0.
This finishes the proof of Theorem 4.1.
We point out that the condition a|b in Theorem 4.1 is not necessary as the following example shows.
Example 4.1. (ⅰ). Let f(x)=x+1, a=2,b=5 and S={2,4,3}. Then a⧸|b. Clearly, one has
(f2(S))=(99492544416) and (f5(S))=(2432433224331253232321024). |
So we can compute and get that
det(f2(S))=2048 and det(f5(S))=714182656. |
Hence
det(f5(S))det(f2(S))=348722∈Z. |
That is, one has det(f2(S))|det(f5(S)).
(ⅱ). Let f(x)=φ(x), a=2,b=3 and S={2,4,7}. Then a⧸|b and
(φ2(S))=(1111411136),(φ2[S])=(1436441443614436) |
and
(φ3[S])=(182168817282161728216). |
One can easily calculate and obtain that
det(φ2(S))=105,det(φ2[S])=15120 and det(φ3[S])=2600640. |
Thus
det(φ3[S])det(φ2(S))=24768∈Z and det(φ3[S])det(φ2[S])=172∈Z. |
In other words, we have det(φ2[S])|det(φ3[S]) and det(φ2(S))|det(φ3[S]).
It is also remarked that the condition that f is multiplicative in Theorem 4.1 is necessary as the following example shows.
Example 4.2. Letting f(x):=x+1, a:=1,b:=3 and S:={2,4,3} gives us that
(f(S))=(332352224),(f[S])=(35755137134) |
and
(f3(S))=(272782712588864),(f3[S])=(271253431251252197343219764). |
So we obtain that det(f(S))=16,det(f[S])=118,det(f3(S))=163072 and det(f3[S])=42578782. Thus
det(f3(S))det(f(S))=10192∈Z,det(f3[S])det(f(S))=212893918∉Z and det(f3[S])det(f[S])=2128939159∉Z. |
So det(f(S))|det(f3(S)), det(f(S))⧸|det(f3[S]) and det(f[S])⧸|det(f3[S]).
Subsequently, we explore the divisibility of determinants of the matrices associated to the integer-valued multiplicative function on the power set Sa. We present the second main result of this section as follows.
Theorem 4.2. Let f be an integer-valued arithmetic function and let a and b be positive integers such that a|b. Let S consist of two coprime divisor chains with 1∉S. Then each of the following is true:
(ⅰ). If f is multiplicative, then det(f(Sa))|det(f[Sa]).
(ⅱ). If f is completely multiplicative, then we have det(f(Sa))|det(f(Sb)), det(f[Sa])|det(f[Sb]) and det(f(Sa))|det(f[Sb]).
Moreover, there exist multiplicative functions f, positive integers a and b with a|b and b>a, and a set S consisting of two coprime divisor chains with 1∉S, such that det(f(Sa))∤det(f(Sb)), det(f[Sa])∤det(f[Sb]) and det(f(Sa))∤det(f[Sb]).
Proof. We begin with the proof of the first part of Theorem 4.2.
(ⅰ). Since S consists of two coprime divisor chains with 1∉S, the power set Sa consists of two coprime divisor chains with gcd(Sa)=1∉Sa. Furthermore, since f is multiplicative, one has either f(1)=0 or f(1)=1. If f(1)=0, then f is the zero function and so one has det(f(Sa))=det(f[Sa])=0. Thus det(f(Sa))|det(f[Sa]) as desired. Now let f(1)=1. Then by Lemma 3.1, we have
(−1)m+n−1f(xan)f(yam)det(f(Sa))=(−1)m+n−1f(xan)f(yam)(f(xa1)f(ya1)−1)(n−1∏i=1(f(xai+1)−f(xai)))(m−1∏j=1(f(yaj+1)−f(yaj)))=det(f[Sa]) |
However, since f is integer valued, one has f(xan)f(yam)∈Z. Therefore the desired result det(f(Sa))|det(f[Sa]) follows. Part (ⅰ) is proved.
(ⅱ). If f is complete multiplicative, then it is clear that f(xa)=fa(x) for any positive integers a and x. So one has
(f(Sa))=(fa(S)),(f(Sb))=(fb(S)),(f[Sa])=(fa[S]) and (f[Sb])=(fb[S]). |
Since a|b and S consists of two coprime divisor chains with 1∉S, it then follows from Theorem 4.1 that det(fa(S))|det(fb(S)), det(fa[S])|det(fb[S]) and det(fa(S))|det(fb[S]). Thus the desired results det(f(Sa))|det(f(Sb)), det(f[Sa])|det(f[Sb]) and det(f(Sa))|det(f[Sb]) follow immediately. Part (ⅱ) is proved.
Finally, we turn our attention to the proof of the second part of Theorem 4.2. Letting S:={2,4,3} and a:=2,b:=4 gives us that
(S)=(221241113),(S2)=(4414161119),(S4)=(161611625611181) |
and
[S]=(24644126123),[S2]=(416361616144361449),[S4]=(1625612962562562073612962073681). |
Therefore picking f=φ to be the Euler's totient function tells us that
(f(S2))=(φ(S2))=(221281116),(f(S4))=(φ(S4))=(881812811154) |
and
(f[S2])=(φ[S2])=(2812884812486),(f[S4])=(φ[S4])=(81284321281286912432691254). |
So one deduces that
det(f(Sb))det(f(Sa))=det(φ(S4))det(φ(S2))=5172066=862011∉Z, |
det(f[Sb])det(f[Sa])=det(φ[S4])det(φ[S2])=3574886403168=124128011∉Z |
and
det(f[Sb])det(f(Sa))=det(φ[S4])det(φ(S2))=35748864066=5958144011∉Z. |
So det(f(Sa))∤det(f(Sb)), det(f[Sa])∤det(f[Sb]) and det(f(Sa))∤det(f[Sb]) as desired.
This concludes the proof of Theorem 4.2.
Remark 4.3. (ⅰ). If S consists of at least three coprime divisor chains, then the divisibility result in Theorem 4.2 (i) may be false. For instance, letting S:={2,4,3,5} and a:=2 gives us that
(φ(S2))=(22112811116111120) and (φ[S2])=(2812408848160124861204016012020). |
Hence
det(φ[S2])det(φ(S2))=−148320107∉Z. |
That is, det(φ(S2))⧸|det(φ[S2]).
On the other hand, the condition that f is multiplicative in Theorem 4.2 (i) is necessary. For example, let f(x):=x+1 and S:={2,4,3},a:=1. Then f is not multiplicative and
det(f[S])det(f(S))=598∉Z. |
Hence det(f(S))⧸|det(f[S]).
(ⅱ). We remark that the condition a|b is not necessary for the divisibility result in Theorem 4.2 (ii). For example, letting f(x):=x,S:={2,6,5},a:=2 and b:=5 gives us that
(S2)=(44143611125),(S5)=(323213277761113125) |
and
[S2]=(436100363690010090025),[S5]=(3277761000007776777624300000100000243000003125). |
Then we compute and get that
det((S5))det((S2))=244442∈Z,det([S5])det([S2])=659993400∈Z and det([S5])det((S2))=5939940600000∈Z. |
It follows immediately that det((S2))|det((S5)),det([S2])|det([S5]) and det((S2))|det([S5]).
The authors would like to thank the anonymous referees for careful readings of the manuscript and helpful comments. S.F. Hong was supported partially by National Science Foundation of China Grant # 11771304.
We declare that we have no conflict of interest.
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