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Research article Special Issues

The explicit formula and parity for some generalized Euler functions

  • Utilizing elementary methods and techniques, the explicit formula for the generalized Euler function φe(n)(e=8,12) has been developed. Additionally, a sufficient and necessary condition for φ8(n) or φ12(n) to be odd has been obtained, respectively.

    Citation: Shichun Yang, Qunying Liao, Shan Du, Huili Wang. The explicit formula and parity for some generalized Euler functions[J]. AIMS Mathematics, 2024, 9(5): 12458-12478. doi: 10.3934/math.2024609

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  • Utilizing elementary methods and techniques, the explicit formula for the generalized Euler function φe(n)(e=8,12) has been developed. Additionally, a sufficient and necessary condition for φ8(n) or φ12(n) to be odd has been obtained, respectively.



    Let Z and P denote the set of integers and primes, respectively. In order to generalize Lehmer's congruence (see [4] or [7]) for modulo prime squares to be modulo integer squares, Cai et al. [1] defined the following generalized Euler function for a positive integer n related to a given positive integer e:

    φe(n)=[ne]i=1,gcd(i,n)=11,

    where [x] is the greatest integer not more than x, i.e., φe(n) is the number of positive integers not greater than [ne] and prime to n. It is clear that φ1(n)=φ(n) is just the Euler function of n, φ2(n)=12φ(n), and

    φe(n)=d|nμ(nd)[de], (1.1)

    where μ(n) is the Möbius function. There are some good results for the generalized Euler function and its applications, especially those concerning φe(n)(e=2,3,4,6), which can be seen in [3].

    In 2013, Cai et al [2] gave the explicit formula for φ3(n) and obtained a criterion regarding the parity for φ2(n) or φ3(n), respectively. In [8], the authors derived the explicit formulae for φ4(n) and φ6(n), and then they obtained some sufficient and necessary conditions for the case that φe(n) or φe(n+1) is odd or even, respectively.

    Recently, Wang and Liao [9] gave the formula for φ5(n) in some special cases and then obtained some sufficient conditions for the case that φ5(n) is even. Liao and Luo [5] gave a computing formula for φe(n) (e=p,p2,pq), where p and q are distinct primes, and n satisfies some certain conditions. Liao [6] obtained the explicit formula for a special class of generalized Euler functions. However, the explicit formula for φe(n)(e3,4,6) was not obtained in the general case.

    In this paper, utilizing the methods and techniques given in [2,5,8], we study the explicit formula and the parity for φe(n)(e=8,12), obtain the corresponding computing formula, and then give a sufficient and necessary condition for the case that φe(n)(e=8,12) is odd or even, respectively.

    For convenience, throughout the paper, we denote Ω(n) and ω(n) to be the number of prime factors and distinct prime factors of a positive integer n, respectively. And for k primes p1,,pk, set Pk={p1,,pk},

    RPk={ripiri(mod8),0ri7,piPk,1ik},

    and

    RPk={ripiri(mod12),0ri11,piPk,1ik}.

    We have organized this paper as follows. In Section 2, we obtain the obvious formulas for [m8] and [m12] based on Jacobi symbol, and some important lemmas are given. In Sections 3 and 4, according to (1.1), and by using the property of the Möbius function μ(n), we derive the expressions for φe(8) and φe(12). In Section 5, we give the parities of φ8(n) and φ12(n), respectively. In the last section, we summarize the main advantage of the proposed method, and propose a further problem to be studied.

    In this section, we first present Lemmas 2.1 and 2.2, which are necessary for the derivations of both [m8] and [m12].

    Lemma 2.1. For any odd positive integer m, we have

    [m8]=18(m4+2(2m)+(1m)). (2.1)

    Furthermore, if gcd(m,6)=1, then we have

    [m12]=112(m6+3(1m)+2(3m)), (2.2)

    where (am) is the Jacobi symbol.

    Proof. For any odd positive integer m, by properties of the Jacobi symbol, we have

    (1m)={1,m1(mod4),1,m3(mod4),and(2m)={1,m1,3(mod8),1,m5,7(mod8).

    Thus from m1(mod8), we can get that 18(m4+2(2m)+(1m))=18(m1) and [m8]=18(m1), namely, (2.1) is true. Similarly, if m3,5,7(mod8), by direct computation, (2.1) holds.

    Furthermore, if gcd(m,6)=1, then by the properties for the Jacobi symbol and the quadratic reciprocity law, we have

    (3m)=(1m)(m3)(1)14(31)(m1)={1,m1,7(mod12),1,m5,11(mod12).

    Thus by m1(mod12), we have that 112(m6+3(1m)+2(3m))=112(m1)=[m12], i.e., (2.2) is true. Similarly, if m5,7,11(mod12), one can get (2.2) also.

    This completes the proof of Lemma 2.1.

    Now, we give a property for the Möbius function, which unifies the cases of Lemma 1.5 in [2] and Lemmas 1.4 and 1.5 in [8].

    Lemma 2.2. Let a be a nonzero integer, p1,,pk be distinct odd primes, and α1,,αk be positive integers. Suppose that n=ki=1pαii and gcd(pi,a)=1(1ik); then,

    d|nμ(nd)(ad)=ki=1((api)αi(api)αi1). (2.3)

    Proof. For a given integer x, set fx(m)=d|mμ(md)(xd).

    First, if m=pα, where p is an odd prime and α is a positive integer, then, by the definition of the Möbius function, we have

    fa(m)=μ(1)(apα)+μ(p)(apα1)=(ap)α(ap)α1.

    Second, if m=m1pα, where α is a positive integer, p is an odd prime with gcd(m1,p)=1, and m1 is an odd positive integer, then we have

    fa(m)=d|m1μ(m1d)(adpα)+d|m1μ(p)μ(m1d)(adpα1)=(ap)αd|m1μ(m1d)(ad)(ap)α1d|m1μ(m1d)(ad).=((ap)α(ap)α1)fa(m1).

    This means that fa(m) is a multiplicative function. Now denote pαn to be the case for both pαn and pα+1; then, we can get

    \begin{eqnarray} f_a(n) = \prod\limits_{p^{\alpha}\| n}\Big(\Big(\frac{a}{p}\Big)^{\alpha}-\Big(\frac{a}{p}\Big)^{\alpha-1}\Big) = \prod\limits_{i = 1}^{k}\Big(\Big(\frac{a}{p_{i}}\Big)^{\alpha_{i}}-\Big(\frac{a}{p_{i}}\Big)^{\alpha_{i}-1}\Big). \end{eqnarray}

    This completes the proof of Lemma 2.2.

    The following lemmas are necessary for proving our main results.

    Lemma 2.3. [2] Let p_1, \ldots, p_k be distinct primes and \alpha, \alpha_1, \ldots, \alpha_k be non-negative integers. If n = 3^{\alpha}\, \prod_{i = 1}^{k} p_{i}^{\alpha_{i}} > 3 and \gcd(p_{i}, 3) = 1 \, (1 \leq i \leq k) , then

    \begin{eqnarray} \label{eq:AB3-0} \varphi_{3}(n) = \begin{cases} \frac{1}{3}\, \varphi(n)+\frac{1}{3}\, (-1)^{\Omega(n)}2^{\omega(n)-\alpha-1}, & \mbox{if }~~ \alpha = 0 ~~ \mbox{or}~~ 1 , p_{i}\equiv2(\mathrm{mod}\, 3) , \\ \frac{1}{3}\, \varphi(n), & \mbox{otherwise}. \end{cases} \end{eqnarray}

    Lemma 2.4. [8] Let p_1, \ldots, p_k be distinct odd primes and \alpha, \alpha_1, \ldots, \alpha_k be non-negative integers. If n = 2^{\alpha}\, \prod_{i = 1}^{k} p_{i}^{\alpha_{i}} > 4 , then

    \begin{eqnarray} \varphi_{4}(n) = \begin{cases} \frac{1}{4}\, \varphi(n)+\frac{1}{4}\, (-1)^{\Omega(n)}2^{\omega(n)-\alpha}, & \mbox{if}~~ \alpha = 0 or 1 , p_{i}\equiv3(\mathrm{mod}\, 4) , \\ \frac{1}{4}\, \varphi(n), & \mbox{otherwise}. \end{cases} \end{eqnarray}

    Lemma 2.5. [8] Let p_1, \ldots, p_k be distinct primes and \alpha, \beta, \alpha_1, \ldots, \alpha_k be non-negative integers. If n = 2^{\alpha}\, 3^{\beta}\, \prod_{i = 1}^{k} p_{i}^{\alpha_{i}} > 6 and \gcd(p_{i}, 6) = 1 \, (1 \leq i \leq k) , then

    \begin{eqnarray} \varphi_{6}(n) = \begin{cases} \frac{1}{6}\, \varphi(n)+\frac{1}{6}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)+1-\beta}, & \mbox{if}~~ \alpha = 0 ~~\mbox{and}~~ \beta = 0 or 1 , p_{i}\equiv5(\mathrm{mod}\, 6) , \\ \frac{1}{6}\, \varphi(n)+\frac{1}{6}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)-1-\beta}, & \mbox{if}~~ \alpha = 1 ~~\mbox{and}~~ \beta = 0 or 1 , p_{i}\equiv5(\mathrm{mod}\, 6) , \\ \frac{1}{6}\, \varphi(n)-\frac{1}{6}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)-\beta}, & \mbox{if}~~ \alpha\geq2 ~~\mbox{and}~~ \beta = 0 or 1 , p_{i}\equiv5(\mathrm{mod}\, 6) , \\ \frac{1}{6}\, \varphi(n), & \mbox{otherwise}. \end{cases} \end{eqnarray}

    Lemma 2.6. [6] Let p_1, \ldots, p_k be distinct primes and \alpha_1, \ldots, \alpha_k be positive integers. If n = \prod_{i = 1}^{k} p_{i}^{\alpha_{i}} and e = \prod_{i = 1}^{k} p_{i}^{\beta_{i}} with 0\leq\beta_{i}\leq\alpha_{i}-1\, (1 \leq i \leq k) , then

    \begin{eqnarray} \varphi_{e}(n) = \frac{1}{e}\, \varphi(n). \end{eqnarray} (2.4)

    First, for a fixed positive integer \alpha and n = 2^{\alpha} , by Lemma 2.6 we can obtain the following:

    \begin{eqnarray} \varphi_{8}(2^{\alpha}) = \begin{cases} 0, & \mbox{if }~~ \alpha = 1, 2 , \\ 1, & \mbox{if}~~ \alpha = 3 , \\ 2^{\alpha-4}, & \mbox{if}~~ \alpha\geq 4 . \end{cases} \end{eqnarray} (3.1)

    Next, we consider the case that n = 2^{\alpha}n_{1} , where n_{1} > 1 is an odd integer. We have the following theorem.

    Theorem 3.1. Suppose that \alpha is a non-negative integer, p_{1}, \ldots, p_k are distinct odd primes, and n = 2^{\alpha}\, \prod_{i = 1}^{k} p_{i}^{\alpha_{i}} > 8. Then we have the following:

    \begin{eqnarray} \varphi_{8}(n) = \begin{cases} \frac{1}{8}\, \varphi(n)+\frac{1}{4}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)-\alpha}, \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \mbox{if}~~ \alpha = 0, 1, \mbox{and}\ R_{\mathbb{P}_{k}} = \{5, 7\}, \{5\} ;\\ \frac{1}{8}\, \varphi(n)+\frac{1}{8}\, (-1)^{\Omega(n)-[\frac{\alpha+1}{2}]}\, 2 \, ^{\omega(n)-\frac{1}{2}(1-(-1)^{\alpha})}, \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mbox{if }~~ \alpha = 0, 1, 2, \mbox{and}\ R_{\mathbb{P}_{k}} = \{3, 7\}, \{3\} ;\\ \frac{1}{8}\, \varphi(n)+\frac{1}{8}\, (-1)^{\Omega(n)-[\frac{\alpha}{2}]}\, 2 \, ^{\omega(n)-\frac{1}{2}(1-(-1)^{\alpha})}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\frac{1-[\frac{\alpha+1}{2}]}{4}\, (-1)^{\Omega(n)}\, 2^{\omega(n)}, \\ \;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \mbox{if}~~ \alpha = 0, 1, 2 , \mbox{and}\ R_{\mathbb{P}_{k}} = \{7\} ;\\ \frac{1}{8}\, \varphi(n), \;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\; \;\; \mbox{otherwise}. \end{cases} \end{eqnarray} (3.2)

    Proof. For n = 2^{\alpha}\, \prod_{i = 1}^{k} p_{i}^{\alpha_{i}} > 8 , set n_1 = \prod_{i = 1}^{k} p_{i}^{\alpha_{i}} ; then, \gcd(n_{1}, 2) = 1 . There are 4 cases as follows.

    Case 1. \alpha = 0 , i.e., n = n_{1} > 8 . By (1.1), (2.1) and Lemmas 2.1 and 2.2, we have

    \begin{eqnarray} \varphi_{8}(n)& = &\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big)\Big[\frac{d}{8}\Big] = \frac{1}{8}\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big)\Big(d-4+2\Big(\frac{-2}{d}\Big)+\Big(\frac{-1}{d}\Big)\Big)\\ & = &\frac{1}{8}\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big)d-\frac{1}{2}\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big) +\frac{1}{4}\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big)\Big(\frac{-2}{d}\Big)\\ &&+\;\frac{1}{8}\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big)\Big(\frac{-1}{d}\Big)\\ & = &\frac{1}{8}\, \varphi(n_{1})+\frac{1}{4}\, \prod\limits_{i = 1}^{k}\Big(\Big(\frac{-2}{p_{i}}\Big)^{\alpha_{i}}-\Big(\frac{-2}{p_{i}}\Big)^{\alpha_{i}-1}\Big) \\ &&+\, \frac{1}{8}\, \prod\limits_{i = 1}^{k}\Big(\Big(\frac{-1}{p_{i}}\Big)^{\alpha_{i}}-\Big(\frac{-1}{p_{i}}\Big)^{\alpha_{i}-1}\Big). \end{eqnarray} (3.3)

    If 1\in R_{\mathbb{P}_{k}} , i.e., there exists an i\, (1\leq i\leq k) such that p_{i}\equiv1 \, (\mathrm{mod}\, 8) , then (\frac{-2}{p_{i}}) = (\frac{-1}{p_{i}}) = 1 . Now by (3.3) we have

    \begin{eqnarray} \varphi_{8}(n) = \frac{1}{8}\, \varphi(n_{1}) = \frac{1}{8}\, \varphi(n). \end{eqnarray} (3.4)

    If \{3, 5\}\subseteq R_{\mathbb{P}_{k}} , i.e., there exist i\neq j such that p_{i}\equiv3\, (\mathrm{mod}\, 8) and p_{j}\equiv5\, (\mathrm{mod}\, 8) , which means that (\frac{-2}{p_{i}}) = (\frac{-1}{p_{j}}) = 1 , then, by (3.3) we also have

    \begin{eqnarray} \label{eq:8n} \varphi_{8}(n) = \frac{1}{8}\, \varphi(n_{1}) = \frac{1}{8}\, \varphi(n). \end{eqnarray}

    If R_{\mathbb{P}_{k}} = \{5, 7\} or \{5\} , i.e., for any p\in\mathbb{P}_{k} , we have that p\equiv5, 7\, (\mathrm{mod}\, 8) or p\equiv5\, (\mathrm{mod}\, 8) , respectively. This means that there exists a prime p such that (\frac{-2}{p}) = -1 and (\frac{-1}{p}) = 1 . Thus by (3.3) we can obtain

    \begin{eqnarray} \varphi_{8}(n) = \frac{1}{8}\, \varphi(n_{1})+\frac{1}{4}\, \prod\limits_{i = 1}^{k}\Big(2\cdot(-1)^{\alpha_{i}}\Big) = \frac{1}{8}\, \varphi(n)+\frac{1}{4}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)}. \end{eqnarray} (3.5)

    If R_{\mathbb{P}_{k}} = \{3, 7\} or \{3\} , i.e., for any p\in\mathbb{P}_{k}, p\equiv3, 7\, (\mathrm{mod}\, 8) or p\equiv3\, (\mathrm{mod}\, 8) , respectively. This implies that for any p\in\mathbb{P}_{k}, (\frac{-1}{p}) = -1 , and there exists a prime p'\in\mathbb{P}_{k} such that p'\equiv3\, (\mathrm{mod}\, 8) ; then, (\frac{-2}{p' }) = 1 . Thus by (3.3) we have

    \begin{eqnarray} \varphi_{8}(n) = \frac{1}{8}\, \varphi(n_{1})+\frac{1}{8}\, \prod\limits_{i = 1}^{k}\Big(2\cdot(-1)^{\alpha_{i}}\Big) = \frac{1}{8}\, \varphi(n)+\frac{1}{8}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)}. \end{eqnarray} (3.6)

    If R_{\mathbb{P}_{k}} = \{7\} , i.e., for any p\in\mathbb{P}_{k}, p\equiv7 \, (\mathrm{mod}\, 8) , then (\frac{-2}{p}) = (\frac{-1}{p}) = -1 . Thus by (3.3) we have

    \begin{eqnarray} \varphi_{8}(n) = \frac{1}{8}\, \varphi(n_{1})+\frac{3}{8}\, \prod\limits_{i = 1}^{k}\Big(2\cdot(-1)^{\alpha_{i}}\Big) = \frac{1}{8}\, \varphi(n)+\frac{3}{8}\, (-1)^{\Omega(n)}\, 2\, ^{\omega(n)}. \end{eqnarray} (3.7)

    Now from (3.4)–(3.7) we know that Theorem 3.1 is true.

    Case 2. \alpha = 1 , i.e., n = 2n_{1} > 8 . Then from the definition we have

    \begin{eqnarray} \label{eq:2.2} \varphi_{8}(n) & = &\sum\limits_{d|n_{1}}\mu\Big(\frac{2n_{1}}{d}\Big)\Big[\frac{d}{8}\Big]+\sum\limits_{d|n_{1}}\mu\Big(\frac{2n_{1}}{2d}\Big)\Big[\frac{2d}{8}\Big]\\ & = &-\varphi_{8}(n_{1})+\varphi_{4}(n_{1}). \end{eqnarray}

    Now by Lemma 2.4 and the proof for Case 1, we can get the following:

    \begin{eqnarray} \varphi_{8}(n) = \begin{cases} \frac{1}{8}\, \varphi(n)+\frac{1}{4}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)-1}, & \mbox{if}~~ R_{\mathbb{P}_{k}} = \{5, 7\}, \{5\} ;\\ \frac{1}{8}\, \varphi(n)+\frac{1}{8}\, (-1)^{\Omega(n)-1}\, 2 \, ^{\omega(n)-1}, & \mbox{if}~~ R_{\mathbb{P}_{k}} = \{3, 7\}, \{3\} ;\\ \frac{1}{8}\, \varphi(n)+\frac{1}{8}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)-1}, & \mbox{if}~~ R_{\mathbb{P}_{k}} = \{7\} ;\\ \frac{1}{8}\, \varphi(n), & \mbox{otherwise}. \end{cases} \end{eqnarray} (3.8)

    This means that Theorem 3.1 is true in this case.

    Case 3. \alpha = 2 , i.e., n = 4n_{1} > 8 . Then from the definition we have

    \begin{eqnarray} \varphi_{8}(n) & = &\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{4n_{1}}{d}\Big)\Big[\frac{d}{8}\Big]+ \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{4n_{1}}{2d}\Big)\Big[\frac{2d}{8}\Big]+ \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{4n_{1}}{4d}\Big)\Big[\frac{4d}{8}\Big]\\ & = &\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{2n_{1}}{d}\Big)\Big[\frac{d}{4}\Big]+ \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big)\Big[\frac{d}{2}\Big]\\ & = &\varphi_{2}(n_{1})-\varphi_{4}(n_{1}) = \frac{1}{2}\varphi(n_{1})-\varphi_{4}(n_{1}). \end{eqnarray}

    Now from Lemma 2.4 and the proof for Case 1, we can also get the following:

    \begin{eqnarray} \varphi_{8}(n) = \begin{cases} \frac{1}{8}\, \varphi(n), & \mbox{if}~~ R_{\mathbb{P}_{k}} = \{5, 7\}, \{5\} , \\ \frac{1}{8}\, \varphi(n)+\frac{1}{8}\, (-1)^{\Omega(n)-1}\, 2 \, ^{\omega(n)}, & \mbox{if}~~ R_{\mathbb{P}_{k}} = \{3, 7\}, \{3\} ;\\ \frac{1}{8}\, \varphi(n)+\frac{1}{8}\, (-1)^{\Omega(n)-1}\, 2 \, ^{\omega(n)}, & \mbox{if }~~ R_{\mathbb{P}_{k}} = \{7\} ;\\ \frac{1}{8}\, \varphi(n), & \mbox{otherwise}. \end{cases} \end{eqnarray} (3.9)

    This means that Theorem 3.1 holds in this case.

    Case 4. \alpha\geq 3 . Note that \mu(2^{\gamma}) = 0 for any positive integer \gamma\geq 2 ; thus, by (1.1) and Lemma 2.4 we have

    \begin{eqnarray} \varphi_{8}(n) = \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{2n_{1}}{d}\Big)\Big[\frac{2^{\alpha-1}d}{8}\Big]+ \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big)\Big[\frac{2^{\alpha}d}{8}\Big]. \end{eqnarray} (3.10)

    If \alpha = 3 , then

    \varphi_{8}(n) = -\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big)\Big[\frac{d}{2}\Big] +\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big)d = -\frac{1}{2}\varphi(n_1)+\varphi(n_1) = \frac{1}{2}\varphi(n_1) = \frac{1}{8}\, \varphi(n)

    If \alpha\geq4 , then \varphi_{8}(n) = -2^{\alpha-4}\varphi(n_{1})+2^{\alpha-3}\varphi(n_{1}) = 2^{\alpha-4}\varphi(n_{1}) = \frac{1}{8}\, \varphi(n) , which means that Theorem 3.1 also holds.

    From the above, we have completed the proof of Theorem 3.1.

    In this section, we give the explicit formula for \varphi_{12}(n) . Obviously, \varphi_{12}(n) = 0 when n < 12 , and \varphi_{12}(n) = 1 when n = 12 or 24 ; then, we consider n > 12 and n\neq24 .

    Theorem 4.1. Let \alpha and \beta be non-negative integers. If n = 2^{\alpha}\, 3^{\beta} > 12 and n\neq 24 , then the following holds:

    \begin{eqnarray} \;\;\;\; \;\;\;\;\varphi_{12}(n) = \begin{cases} \frac{1}{2}\, (3^{\beta-2}-(-1)^{\alpha+\beta}), &\mbox{if}~~ \alpha = 0, \, or\, \alpha\geq1, \beta\geq2 ;\\ 2^{\alpha-2}\cdot 3^{\beta-2}, &\mbox{if }~~ \alpha\geq 2, \beta\geq 2 ;\\ \frac{1}{3}\, (2^{\alpha+\beta-3}+(-1)^{\alpha+\beta}), &\mbox{if}~~ \alpha\geq4, \beta = 0, 1 . \end{cases} \end{eqnarray} (4.1)

    Proof. (1) For the case that \alpha = 0 , i.e., n = 3^{\beta} > 12 , and \beta\geq3 , then we have

    \begin{eqnarray} \varphi_{12}(3^{\beta})& = &\sum\limits_{d\, |\, 3^{\beta}}\mu\Big(\frac{3^{\beta}}{d}\Big)\Big[\frac{d}{12}\Big] = \Big[\frac{3^{\beta}}{12}\Big]-\Big[\frac{3^{\beta-1}}{12}\Big] = \Big[\frac{3^{\beta-1}}{4}\Big]-\Big[\frac{3^{\beta-2}}{4}\Big]\\ & = & \frac{1}{4}\big(3^{\beta-1}-2+(-1)^{\beta-1}\big)-\frac{1}{4}\big(3^{\beta-2}-2+(-1)^{\beta-2}\big) \\ & = & \frac{1}{2}\big(3^{\beta-2}-(-1)^{\beta}\big). \end{eqnarray}

    (2) For the case that \alpha = 1 , i.e., n = 2\cdot3^{\beta} > 12 , and \beta\geq2 , by Lemma 2.5,

    \begin{eqnarray} \varphi_{12}(2\cdot3^{\beta})& = &\sum\limits_{d\, |\, 3^{\beta}}\mu\Big(\frac{2\cdot3^{\beta}}{d}\Big)\Big[\frac{d}{12}\Big] +\sum\limits_{d\, |\, 3^{\beta}}\mu\Big(\frac{2\cdot3^{\beta}}{2d}\Big)\Big[\frac{2d}{12}\Big]\\ & = &-\varphi_{12}(3^{\beta})+\varphi_{6}(3^{\beta}) = -\frac{1}{12}\varphi(3^{\beta})+\frac{1}{2}\, (-1)^{\beta}+\frac{1}{6}\varphi(3^{\beta})\\ & = &\frac{1}{2}\big(3^{\beta-2}-(-1)^{\beta+1}\big). \end{eqnarray}

    (3) For the case that \alpha = 2 , i.e., n = 4\cdot 3^{\beta} > 12 , and so \beta\geq2 , then we have

    \begin{eqnarray} \varphi_{12}(4\cdot3^{\beta})& = & \sum\limits_{d\, |\, 4\cdot3^{\beta}}\mu\Big(\frac{4\cdot3^{\beta}}{d}\Big)\Big[\frac{d}{12}\Big]\\ & = &\mu(1)\Big[\frac{4\cdot3^{\beta}}{12}\Big]+ \mu(2)\Big[\frac{2\cdot3^{\beta}}{12}\Big] +\mu(3)\Big[\frac{4\cdot3^{\beta-1}}{12}\Big]+\mu(6)\Big[\frac{2\cdot3^{\beta-1}}{12}\Big]\\ & = &3^{\beta-1}-\Big[\frac{3^{\beta-1}}{2}\Big] -3^{\beta-2}+\Big[\frac{3^{\beta-2}}{2}\Big]\\ & = &3^{\beta-2}. \end{eqnarray}

    (4) For the case that \alpha = 3 , i.e., n = 8\cdot 3^{\beta} > 12 and n\neq24 , and \beta\geq2 , then

    \begin{eqnarray} \varphi_{12}(8\cdot3^{\beta})& = & \sum\limits_{d\, |\, 8\cdot3^{\beta}}\mu\Big(\frac{8\cdot3^{\beta}}{d}\Big)\Big[\frac{d}{12}\Big]\\ & = &\mu(1)\Big[\frac{8\cdot3^{\beta}}{12}\Big]+ \mu(2)\Big[\frac{4\cdot3^{\beta}}{12}\Big] +\mu(3)\Big[\frac{8\cdot3^{\beta-1}}{12}\Big]+\mu(6)\Big[\frac{4\cdot3^{\beta-1}}{12}\Big]\\ & = &2\cdot3^{\beta-1}-3^{\beta-1} -2\cdot3^{\beta-2}+3^{\beta-2}\\ & = &2\cdot3^{\beta-2}. \end{eqnarray}

    (5) For the case that \alpha\geq4 , i.e., n = 2^{\alpha}\cdot 3^{\beta} > 12 , and so \beta\geq0 , if \beta = 0 , i.e., n = 2^{\alpha}(\alpha\geq4) , then we have

    \begin{eqnarray} \varphi_{12}(2^{\alpha})& = &\sum\limits_{d\, |\, 2^{\alpha}}\mu\Big(\frac{2^{\alpha}}{d}\Big)\Big[\frac{d}{12}\Big] = \Big[\frac{2^{\alpha-2}}{3}\Big]-\Big[\frac{2^{\alpha-3}}{3}\Big]\\ & = & \frac{1}{3}\Big(2^{\alpha-2}-\frac{1}{2}\big(3-(-1)^{\alpha-2}\big)\Big) -\frac{1}{3}\Big(2^{\alpha-3}-\frac{1}{2}\big(3-(-1)^{\alpha-3}\big)\Big) \\ & = &\frac{1}{3}\Big(2^{\alpha-3}+(-1)^{\alpha}\Big) \end{eqnarray}

    If \beta = 1 , i.e., n = 3\cdot2^{\alpha} , then we have

    \begin{eqnarray} \varphi_{12}(3\cdot2^{\alpha})& = &\sum\limits_{d\, |\, 3\cdot2^{\alpha}}\mu\Big(\frac{3\cdot2^{\alpha}}{d}\Big)\Big[\frac{d}{12}\Big] = 2^{\alpha-2}-2^{\alpha-3}-\Big[\frac{2^{\alpha-2}}{3}\Big]+\Big[\frac{2^{\alpha-3}}{3}\Big]\\ & = &\frac{1}{3}\Big(2^{\alpha-2}+(-1)^{\alpha+1}\Big). \end{eqnarray}

    If \beta\geq2 , we have

    \begin{eqnarray} \varphi_{12}(2^{\alpha}\cdot3^{\beta})& = & \sum\limits_{d\, |\, 2^{\alpha}\cdot3^{\beta}}\mu\Big(\frac{2^{\alpha}\cdot3^{\beta}}{d}\Big)\Big[\frac{d}{12}\Big]\\ & = &\mu(1)\Big[\frac{2^{\alpha}\cdot3^{\beta}}{12}\Big]+ \mu(2)\Big[\frac{2^{\alpha-1}\cdot3^{\beta}}{12}\Big] +\mu(3)\Big[\frac{2^{\alpha}\cdot3^{\beta-1}}{12}\Big]+\mu(6)\Big[\frac{2^{\alpha-1}\cdot3^{\beta-1}}{12}\Big]\\ & = &2^{\alpha-2}\cdot3^{\beta-1}-\Big[\frac{2^{\alpha-2}\cdot3^{\beta-1}}{2}\Big] -2^{\alpha-2}\cdot3^{\beta-2}+\Big[\frac{2^{\alpha-2}\cdot3^{\beta-2}}{2}\Big]\\ & = &2^{\alpha-2}\cdot3^{\beta-2}. \end{eqnarray}

    This completes the proof of Theorem 4.1.

    Now consider the case that n = 2^{\alpha}\, 3^{\beta}n_{1} , where n_{1} > 1 and \gcd(n_{1}, 6) = 1 . We have the following theorem.

    Theorem 4.2. Let \alpha and \beta be non-negative integers, k, \alpha_{i}\, (1\leq i\leq k) be positive integers, and p_{1}, \ldots, p_k be distinct primes. Suppose that \gcd(p_{i}, 6) = 1 \, (1\leq i\leq k) and n = 2^{\alpha}\, 3^{\beta}\, \prod_{i = 1}^{k} p_{i}^{\alpha_{i}} > 12 ; then, we have the following:

    \begin{eqnarray} \;\;\;\;\varphi_{12}(n) = \begin{cases} \frac{1}{12}\, \varphi(n)+\frac{1}{4}\, (-1)^{\Omega(n)}\cdot 2 \, ^{\omega(n)-\alpha}, \\ \;\;\;\; \;\;\;\; \;\;\;\; \;\;\;\;\;\mbox{if}~~ \alpha = 0, 1 , \beta = 0 , ~~ \mbox{and}~~ R_{\mathbb{P}_{k}}' = \{7, 11\}, \{7\} ;\\ \frac{1}{12}\, \varphi(n)+\frac{1}{4}\, (-1)^{\Omega(n)+1}\cdot 2 \, ^{\omega(n)-\alpha}, \\ \;\;\;\; \;\;\;\; \;\;\;\; \;\;\;\;\;\mbox{if}~~ \alpha = 0, 1 , \beta\geq2 , ~~ \mbox{and}~~ R_{\mathbb{P}_{k}}' = \{7, 11\}, \{7\}, \{11\} ;\\ \frac{1}{12}\, \varphi(n)+\frac{1}{6}\, (-1)^{\Omega(n)+[\frac{\alpha+1}{2}]}\cdot 2 \, ^{\omega(n)-[\frac{\alpha+1}{2}]-\beta}, \\ \;\;\;\; \;\;\;\; \;\;\;\; \;\;\;\;\; \mbox{if }~~ \alpha = 0, 1, 2 , \beta = 0, 1 , ~~ \mbox{and}~~ R_{\mathbb{P}_{k}}' = \{5, 11\}, \{5\}, \\ \;\;\;\; \;\;\;\; \;\;\;\; \;\;\;\;\; \mbox{or }~~ \alpha = 0, 1 , \beta = 1 , ~~ \mbox{and}~~ R_{\mathbb{P}_{k}}' = \{11\} ;\\ \;\;\;\; \;\;\;\; \;\;\;\; \;\;\;\;\; \mbox{or} \alpha = 2 , \beta = 0, 1 , ~~ \mbox{and}~~ R_{\mathbb{P}_{k}}' = \{11\} ;\\ \frac{1}{12}\, \varphi(n)+\frac{1}{6}\, (-1)^{\Omega(n)}\cdot 2 \, ^{\omega(n)-\beta}, \\ \;\;\;\; \;\;\;\; \;\;\;\; \;\;\;\; \; \mbox{if}~~ \alpha\geq3 , \beta = 0, 1 , ~~ \mbox{and}~~ R_{\mathbb{P}_{k}}' = \{5, 11\}, \{5\}, \{11\} ;\\ \frac{1}{12}\, \varphi(n)+\frac{5}{12}\, (-1)^{\Omega(n)}\cdot2\, ^{\omega(n)}, \\ \;\;\;\; \;\;\;\; \;\;\;\; \;\;\;\;\; \mbox{if }~~ \alpha = 0 , \beta = 0 , ~~ \mbox{and}~~ R_{\mathbb{P}_{k}}' = \{11\} ;\\ \frac{1}{12}\, \varphi(n)+\frac{1}{12}\, (-1)^{\Omega(n)}\cdot2\, ^{\omega(n)-1}, \\ \;\;\;\; \;\;\;\; \;\;\;\; \;\;\;\;\; \mbox{if }~~ \alpha = 1 , \beta = 0 , ~~ \mbox{and}~~ R_{\mathbb{P}_{k}}' = \{11\} ;\\ \frac{1}{12}\, \varphi(n), \;\;\;\; \mbox{otherwise}. \end{cases} \end{eqnarray} (4.2)

    Proof. Set n_1 = \prod_{i = 1}^{k} p_{i}^{\alpha_{i}} ; then, \gcd(n_{1}, 6) = 1 and n = 2^{\alpha}\, 3^{\beta}\, n_1 .

    Case 1. \alpha = 0 .

    (A) If \beta = 0 , then n_{1} > 1 . Thus by (1.1), (2.2) and Lemmas 2.1 and 2.2, we have

    \begin{eqnarray} \varphi_{12}(n)& = &\varphi_{12}(n_{1}) = \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big)\Big[\frac{d}{12}\Big]\\ & = &\frac{1}{12}\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big)\Big(d-6+3\Big(\frac{-1}{d}\Big)+2\Big(\frac{-3}{d}\Big)\Big)\\ & = &\frac{1}{12}\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big)d-\frac{1}{2}\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big) +\frac{1}{4}\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big)\Big(\frac{-1}{d}\Big)\\ &&+\;\frac{1}{6}\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{n_{1}}{d}\Big)\Big(\frac{-3}{d}\Big)\\ & = &\frac{1}{12}\, \varphi(n_{1})+\frac{1}{4}\, \prod\limits_{i = 1}^{k}\Big(\Big(\frac{-1}{p_{i}}\Big)^{\alpha_{i}}- \Big(\frac{-1}{p_{i}}\Big)^{\alpha_{i}-1}\Big) \\ &&+\, \frac{1}{6}\, \prod\limits_{i = 1}^{k}\Big(\Big(\frac{-3}{p_{i}}\Big)^{\alpha_{i}}-\Big(\frac{-3}{p_{i}}\Big)^{\alpha_{i}-1}\Big). \end{eqnarray} (4.3)

    If 1\in R_{\mathbb{P}_{k}}' or \{5, 7\}\subseteq R_{\mathbb{P}_{k}}' , then there exists p_{i}\equiv1 \, (\mathrm{mod}\, 12) , or there exist p_j and p_l such that p_{j}\equiv \, 5(\mathrm{mod}\, 12) and p_{l}\equiv 7\, (\mathrm{mod}\, 12) ; then, (\frac{-1}{p_{i}}) = (\frac{-3}{p_{i}}) = 1 or (\frac{-1}{p_{j}}) = (\frac{-3}{p_{l}}) = 1 , respectively. Thus by (4.3) we can get

    \begin{eqnarray} \varphi_{12}(n) = \frac{1}{12}\, \varphi(n_{1}) = \frac{1}{12}\, \varphi(n). \end{eqnarray} (4.4)

    If R_{\mathbb{P}_{k}}' = \{7, 11\} or \{7\} , i.e., for any p\in\mathbb{P}_{k} , we have that p\equiv7, 11\, (\mathrm{mod}\, 12) or p\equiv7\, (\mathrm{mod}\, 12) , respectively. This means that (\frac{-1}{p}) = -1 and there exists a prime p'\equiv7\, (\mathrm{mod}\, 12) , i.e., (\frac{-3}{p'}) = 1 , in either of the two cases. Thus by (4.3) we can obtain

    \begin{eqnarray} \varphi_{12}(n) = \frac{1}{12}\, \varphi(n_{1})+\frac{1}{4}\, \prod\limits_{i = 1}^{k}\Big(2\, (-1)^{\alpha_{i}}\Big) = \frac{1}{12}\, \varphi(n)+\frac{1}{4}\, (-1)^{\Omega(n)}\, 2^{\omega(n)}. \end{eqnarray} (4.5)

    If R_{\mathbb{P}_{k}}' = \{5, 11\} or \{5\} , i.e., for any p\in\mathbb{P}_{k} , p\equiv5, 11\, (\mathrm{mod}\, 12) or p\equiv5\, (\mathrm{mod}\, 12) , respectively. Then (\frac{-3}{p}) = -1 , and there exists a prime p'\equiv5\, (\mathrm{mod}\, 12) , i.e., (\frac{-1}{p'}) = 1 in either case. Thus by (4.3) we can get

    \begin{eqnarray} \varphi_{12}(n) = \frac{1}{12}\, \varphi(n_{1})+\frac{1}{6}\, \prod\limits_{i = 1}^{k}\Big(2\, (-1)^{\alpha_{i}}\Big) = \frac{1}{12}\, \varphi(n)+\frac{1}{6}\, (-1)^{\Omega(n)}\, 2^{\omega(n)}. \end{eqnarray} (4.6)

    If R_{\mathbb{P}_{k}}' = \{11\} , i.e., for any p\in \mathbb{P}_{k} , p\equiv11 \, (\mathrm{mod}\, 12) ; then, (\frac{-1}{p}) = (\frac{-3}{p}) = -1 . Thus by (4.3) we have

    \begin{eqnarray} \;\;\;\varphi_{12}(n) = \frac{1}{12}\, \varphi(n_{1})+\frac{5}{12}\, \prod\limits_{i = 1}^{k}\Big(2\, (-1)^{\alpha_{i}}\Big) = \frac{1}{12}\, \varphi(n)+\frac{5}{12}\, (-1)^{\Omega(n)}\, 2\, ^{\omega(n)}. \end{eqnarray} (4.7)

    (B) If \beta\geq1 , then by (1.1) we have

    \begin{eqnarray} \label{eq:12n00} \varphi_{12}(n) & = &\varphi_{12}(3^{\beta}n_{1}) = \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{3^{\beta}n_{1}}{d}\Big)\Big[\frac{d}{12}\Big] +\sum\limits_{d\, |\, 3^{\beta-1}n_{1}}\mu\Big(\frac{3^{\beta}n_{1}}{3d}\Big)\Big[\frac{3d}{12}\Big]\\ & = &\mu(3^{\beta})\, \varphi_{12}(n_{1})+\varphi_{4}(3^{\beta-1}n_{1}). \end{eqnarray}

    Now from \beta = 1 , Lemma 2.4 and Case 1, we can get the following:

    \begin{eqnarray} \varphi_{12}(n)& = &-\varphi_{12}(n_{1})+\varphi_{4}(n_{1})\\ & = &\begin{cases} \frac{1}{12}\, \varphi(n), & \mbox{if}~~ R_{\mathbb{P}_{k}}' = \{7, 11\}, \{7\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{6}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)-1}, & \mbox{if }~~ R_{\mathbb{P}_{k}}' = \{5, 11\}, \{5\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{6}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)-1}, & \mbox{if }~~ R_{\mathbb{P}_{k}}' = \{11\} , \\ \frac{1}{12}\, \varphi(n), & \mbox{otherwise}. \end{cases} \end{eqnarray} (4.8)

    For the case that \beta\geq2 , note that \mu(3^{\gamma}) = 0 with \gamma\geq2 ; thus, by Lemma 2.4 we have the following:

    \begin{eqnarray} \varphi_{12}(n)& = &\varphi_{4}(3^{\beta-1} n_{1})\\ & = &\begin{cases} \frac{1}{12}\, \varphi(n)+\frac{1}{4}\, (-1)^{\Omega(n)+1}\, 2 \, ^{\omega(n)}, & \mbox{if }~~ R_{\mathbb{P}_{k}}' = \{7, 11\}, \{7\} , \\ \frac{1}{12}\, \varphi(n), & \mbox{if} ~~R_{\mathbb{P}_{k}}' = \{5, 11\}, \{5\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{4}\, (-1)^{\Omega(n)+1}\, 2 \, ^{\omega(n)}, & \mbox{if }~~ R_{\mathbb{P}_{k}}' = \{11\} , \\ \frac{1}{12}\, \varphi(n), & \mbox{otherwise}. \end{cases} \end{eqnarray} (4.9)

    From the above (4.3)–(4.9), Theorem 4.2 is proved in this case.

    Case 2. \alpha = 1 .

    (A) If \beta = 0 , i.e., n = 2n_{1} , then by (1.1), Case 1 and Lemma 2.4, we have

    \begin{eqnarray} \varphi_{12}(n) & = &\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{2n_{1}}{d}\Big)\Big[\frac{d}{12}\Big]+ \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{2n_{1}}{2d}\Big)\Big[\frac{2d}{12}\Big]\\ & = &-\varphi_{12}(n_{1})+\varphi_{6}(n_{1})\\ & = &\begin{cases} \frac{1}{12}\, \varphi(n)+\frac{1}{4}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)-1}, & \mbox{if }~~ R_{\mathbb{P}_{k}}' = \{7, 11\}, \{7\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{12}\, (-1)^{\Omega(n)+1}\, 2 \, ^{\omega(n)}, & \mbox{if}~~ R_{\mathbb{P}_{k}}' = \{5, 11\}, \{5\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{12}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)-1}, & \mbox{if }~~ R_{\mathbb{P}_{k}}' = \{11\} , \\ \frac{1}{12}\, \varphi(n), & \mbox{otherwise}. \end{cases} \end{eqnarray} (4.10)

    (B) If \beta = 1 , i.e., n = 6n_{1} , then from (1.1) we can get

    \begin{eqnarray} \label{eq:12n11} \varphi_{12}(n) & = &\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{6n_{1}}{d}\Big)\Big[\frac{d}{12}\Big]+ \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{6n_{1}}{2d}\Big)\Big[\frac{2d}{12}\Big]+ \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{6n_{1}}{3d}\Big)\Big[\frac{3d}{12}\Big]\\ && + \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{6n_{1}}{6d}\Big)\Big[\frac{6d}{12}\Big]\\ & = &\varphi_{12}(n_{1})-\varphi_{6}(n_{1})-\varphi_{4}(n_{1})+\varphi_{2}(n_{1}). \end{eqnarray}

    Now by Lemmas 2.4 and 2.5 and Case 1, we have the following:

    \begin{eqnarray} \varphi_{12}(n) = \begin{cases} \frac{1}{12}\, \varphi(n), & \mbox{if}~~ R_{\mathbb{P}_{k}}' = \{7, 11\}, \{7\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{12}\, (-1)^{\Omega(n)+1}\, 2 \, ^{\omega(n)-1}, & \mbox{if }~~ R_{\mathbb{P}_{k}}' = \{5, 11\}, \{5\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{12}\, (-1)^{\Omega(n)+1}\, 2 \, ^{\omega(n)-1}, & \mbox{if }~~ R_{\mathbb{P}_{k}}' = \{11\} , \\ \frac{1}{12}\, \varphi(n), & \mbox{otherwise}. \end{cases} \end{eqnarray} (4.11)

    (C) If \beta\geq 2 , then by (1.1) one can easily see that

    \begin{eqnarray} \label{eq:12n12} \varphi_{12}(n) & = &\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{2\cdot3^{\beta}n_{1}}{d}\Big)\Big[\frac{d}{12}\Big]+ \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{2\cdot3^{\beta}n_{1}}{2d}\Big)\Big[\frac{2d}{12}\Big]\\ &&+\, \sum\limits_{d\, |\, 3^{\beta-1}n_{1}}\mu\Big(\frac{2\cdot3^{\beta} n_{1}}{3d}\Big)\Big[\frac{3d}{12}\Big] + \sum\limits_{d\, |\, 3^{\beta-1}n_{1}}\mu\Big(\frac{2\cdot3^{\beta}n_{1}}{6d}\Big)\Big[\frac{6d}{12}\Big]\\ & = &-\varphi_{4}(3^{\beta-1}n_{1})+\varphi_{2}(3^{\beta-1}n_{1}). \end{eqnarray}

    Now by Lemma 2.4 we can get the following:

    \begin{eqnarray} \varphi_{12}(n) = \begin{cases} \frac{1}{12}\, \varphi(n)+\frac{1}{4}\, (-1)^{\Omega(n)+1}\, 2 \, ^{\omega(n)-1}, & \mbox{if}~~ R_{\mathbb{P}_{k}}' = \{7, 11\}, \{7\} , \\ \frac{1}{12}\, \varphi(n), & \mbox{if}~~ R_{\mathbb{P}_{k}}' = \{5, 11\}, \{5\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{4}\, (-1)^{\Omega(n)+1}\, 2 \, ^{\omega(n)-1}, & \mbox{if}~~ R_{\mathbb{P}_{k}}' = \{11\} , \\ \frac{1}{12}\, \varphi(n), & \mbox{otherwise}. \end{cases} \end{eqnarray} (4.12)

    From the above (4.10) and (4.12), Theorem 4.2 is true in this case.

    Case 3. \alpha = 2 .

    (A) If \beta = 0 , i.e., n = 4n_{1} , then from Lemmas 2.3 and 2.5, we can obtain

    \begin{eqnarray} \varphi_{12}(n)& = &\sum\limits_{d\, |\, 4n_{1}}\mu\Big(\frac{4n_{1}}{d}\Big)\Big[\frac{d}{12}\Big]\\ & = &\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{4n_{1}}{d}\Big)\Big[\frac{d}{12}\Big]+ \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{4n_{1}}{2d}\Big)\Big[\frac{2d}{12}\Big]+ \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{4n_{1}}{4d}\Big)\Big[\frac{4d}{12}\Big]\\ & = &-\varphi_{6}(n_{1})+\varphi_{3}(n_{1})\\ & = &\begin{cases} \frac{1}{12}\, \varphi(n), & \mbox{if}~~ R_{\mathbb{P}_{k}}' = \{7, 11\}, \{7\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{12}\, (-1)^{\Omega(n)+1}\, 2 \, ^{\omega(n)}, & \mbox{if}~~ R_{\mathbb{P}_{k}}' = \{5, 11\}, \{5\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{12}\, (-1)^{\Omega(n)+1}\, 2 \, ^{\omega(n)}, & \mbox{if}~~ R_{\mathbb{P}_{k}}' = \{11\} , \\ \frac{1}{12}\, \varphi(n), & \mbox{otherwise}. \end{cases} \end{eqnarray} (4.13)

    (B) If \beta = 1 , i.e., n = 12\, n_{1} , then by the definition we have

    \begin{eqnarray} \label{eq:12n21} \varphi_{12}(n)& = &\sum\limits_{d|12n_{1}}\mu\Big(\frac{12n_{1}}{d}\Big)\Big[\frac{d}{12}\Big]\\ & = &\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{12 n_{1}}{d}\Big)\Big[\frac{d}{12}\Big]+ \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{12 n_{1}}{2d}\Big)\Big[\frac{2d}{12}\Big]+ \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{12 n_{1}}{4d}\Big)\Big[\frac{4d}{12}\Big]\\ && + \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{12 n_{1}}{3d}\Big)\Big[\frac{3d}{12}\Big]+ \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{12 n_{1}}{6d}\Big)\Big[\frac{6d}{12}\Big]+ \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{12 n_{1}}{12d}\Big)\Big[\frac{12 d}{12}\Big]\\ & = &\varphi_{6}(n_{1})-\varphi_{3}(n_{1})-\varphi_{2}(n_{1})+\varphi(n_{1}). \end{eqnarray}

    Now by Lemmas 2.3 and 2.4 and Case 1, we can get the following:

    \begin{eqnarray} \varphi_{12}(n) = \begin{cases} \frac{1}{12}\, \varphi(n), & \mbox{if }~~ R_{\mathbb{P}_{k}}' = \{7, 11\}, \{7\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{12}\, (-1)^{\Omega(n)+1}\, 2 \, ^{\omega(n)-1}, & \mbox{if}~~ R_{\mathbb{P}_{k}}' = \{5, 11\}, \{5\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{12}\, (-1)^{\Omega(n)+1}\, 2 \, ^{\omega(n)-1}, & \mbox{if }~~ R_{\mathbb{P}_{k}}' = \{11\}, \\ \frac{1}{12}\, \varphi(n), & \mbox{otherwise}. \end{cases} \end{eqnarray} (4.14)

    (C) If \beta\geq 2 , then from n = 4\cdot3^{\beta}\, n_{1} and the definition, we know that

    \begin{eqnarray} \varphi_{12}(n) & = &\sum\limits_{d\, |\, 4\cdot3^{\beta}n_{1}}\mu\Big(\frac{4\cdot3^{\beta}n_{1}}{d}\Big)\Big[\frac{d}{12}\Big] = \sum\limits_{d\, |\, 2\cdot3^{\beta-1}n_{1}}\mu\Big(\frac{4\cdot3^{\beta}n_{1}}{6d}\Big)\Big[\frac{6d}{12}\Big]\\ & = &\sum\limits_{d\, |\, 3^{\beta-1}n_{1}}\mu\Big(\frac{2\cdot3^{\beta}n_{1}}{2d}\Big)\Big[\frac{2d}{2}\Big]+ \sum\limits_{d\, |\, 3^{\beta-1}n_{1}}\mu\Big(\frac{2\cdot3^{\beta}n_{1}}{d}\Big)\Big[\frac{d}{2}\Big]\\ & = &\varphi(3^{\beta-1}n_{1})-\varphi_{2}(3^{\beta-1}n_{1}) = \frac{1}{2}\varphi(3^{\beta-1}n_{1})\\ & = &\frac{1}{12}\varphi(4\cdot3^{\beta}n_{1}) = \frac{1}{12}\varphi(n). \end{eqnarray} (4.15)

    From the above (4.13)–(4.15), Theorem 4.2 is proved in this case.

    Case 4. \alpha\geq 3 .

    (A) If \beta = 0 , i.e., n = 2^{\alpha}\, n_{1} , then by Lemma 2.5 we have

    \begin{eqnarray} \varphi_{12}(n) & = &\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{2^{\alpha}n_{1}}{d}\Big)\Big[\frac{d}{12}\Big]+ \sum\limits_{d\, |\, 2^{\alpha-1}n_{1}}\mu\Big(\frac{2^{\alpha}n_{1}}{2d}\Big)\Big[\frac{2d}{12}\Big]\\ & = &\varphi_{6}(2^{\alpha-1}n_{1})\\ & = &\begin{cases} \frac{1}{12}\, \varphi(n), & \mbox{if }~~ R_{\mathbb{P}_{k}}' = \{7, 11\}, \{7\}, \\ \frac{1}{12}\, \varphi(n)+\frac{1}{6}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)}, & \mbox{if}~~ R_{\mathbb{P}_{k}}' = \{5, 11\}, \{5\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{6}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)}, & \mbox{if }~~ R_{\mathbb{P}_{k}}' = \{11\} , \\ \frac{1}{12}\, \varphi(n), & \mbox{otherwise}. \end{cases} \end{eqnarray} (4.16)

    (B) If \beta = 1 , i.e., n = 3\cdot2^{\alpha}\, n_{1} , then by the definition we have

    \begin{eqnarray} \varphi_{12}(n) & = &\sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{3\cdot2^{\alpha} n_{1}}{d}\Big)\Big[\frac{d}{12}\Big]+ \sum\limits_{d\, |\, 2^{\alpha-1}n_{1}}\mu\Big(\frac{3\cdot2^{\alpha} n_{1}}{2d}\Big)\Big[\frac{2d}{12}\Big]\\ && + \sum\limits_{d\, |\, n_{1}}\mu\Big(\frac{3\cdot2^{\alpha} n_{1}}{3d}\Big)\Big[\frac{3d_{1}}{12}\Big]+ \sum\limits_{d\, |\, 2^{\alpha-1}n_{1}}\mu\Big(\frac{3\cdot2^{\alpha} n_{1}}{6d}\Big)\Big[\frac{6d}{12}\Big]\\ & = &-\varphi_{6}(2^{\alpha-1}n_{1})+\varphi_{2}(2^{\alpha-1}n_{1})\\ & = &\begin{cases} \frac{1}{12}\, \varphi(n), & \mbox{if}~~ R_{\mathbb{P}_{k}}' = \{7, 11\}, \{7\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{12}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)}, & \mbox{if}~~ R_{\mathbb{P}_{k}}' = \{5, 11\}, \{5\} , \\ \frac{1}{12}\, \varphi(n)+\frac{1}{12}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)}, & \mbox{if}~~ R_{\mathbb{P}_{k}}' = \{11\} , \\ \frac{1}{12}\, \varphi(n), & \mbox{otherwise}. \end{cases} \end{eqnarray} (4.17)

    (C) If \beta\geq 2 , then by Lemma 2.6 we can get

    \begin{eqnarray} \varphi_{12}(n) = \frac{1}{12}\varphi(2^{\alpha}\cdot3^{\beta}n_{1}) = \frac{1}{12}\varphi(n). \end{eqnarray} (4.18)

    Now from (4.16)–(4.18), Theorem 4.2 is proved in this case.

    From the above, we complete the proof for Theorem 4.2.

    Based on Theorems 3.1, 4.1 and 4.2, this section gives the parity of \varphi_{8}(n) and \varphi_{12}(n) , respectively.

    Theorem 5.1. If n is a positive integer, then \varphi_{8}(n) is odd if and only if n = 8, 16 or n is given by Table 1. In the above table, p, p_{1}, p_{2} are odd primes with p_1\neq p_2 , and \alpha, \alpha_{1}, \alpha_{2} are positive integers.

    Table 1.  the conditions of \varphi_{8}(n) is odd.
    n Conditions
    p^{\alpha} p\equiv9, 15\, (\mathrm{mod}\, 16) ; p\equiv3, 5\, (\mathrm{mod}\, 16) , 2\, |\alpha ; p\equiv11, 13\, (\mathrm{mod}\, 16) , 2\nmid\alpha ;
    2p^{\alpha} p\equiv7, 9\, (\mathrm{mod}\, 16) ; p\equiv3, 13\, (\mathrm{mod}\, 16) , 2\, |\alpha ; p\equiv5, 11\, (\mathrm{mod}\, 16) , 2\nmid\alpha ;
    4p^{\alpha} p\equiv3, 5\, (\mathrm{mod}\, 8) ;
    8p^{\alpha} p\equiv3, 7\, (\mathrm{mod}\, 8) ;
    p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} p_{1}\equiv p_{2}\equiv3\, (\mathrm{mod}\, 8) ; p_{1}\equiv p_{2}\equiv5\, (\mathrm{mod}\, 8) ; p_{1}\equiv3\, (\mathrm{mod}\, 8), p_{2}\equiv5\, (\mathrm{mod}\, 8) ;
    2p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} p_{1}\equiv p_{2}\equiv3\, (\mathrm{mod}\, 8) ; p_{1}\equiv p_{2}\equiv5\, (\mathrm{mod}\, 8) ; p_{1}\equiv3\, (\mathrm{mod}\, 8), p_{2}\equiv5\, (\mathrm{mod}\, 8) .

     | Show Table
    DownLoad: CSV

    Proof. For n = 2^{\alpha} , by (3.1) we know that \varphi_{8}(n) is odd if and only if n = 8, 16 .

    Now suppose that n = 2^{\alpha}\, \prod_{i = 1}^{k} p_{i}^{\alpha_{i}} , where \alpha \geq0 , \alpha_{1}, \ldots, \alpha_k are positive integers, and p_{1}, \ldots, p_k are distinct odd primes. Set n_{1} = \prod_{i = 1}^{k} p_{i}^{\alpha_{i}} ; then, n_{1} > 1 is odd. By Theorem 3.1, we have the following four cases.

    Case 1. R_{\mathbb{P}_{k}} = \{5, 7\} or \{5\} .

    (A) If \alpha = 0 , i.e., n = n_1 is odd, then, by (3.2) we have that \varphi_{8}(n) = \frac{1}{8}\varphi(n)+\frac{1}{4}(-1)^{\Omega(n)}2 ^{\omega(n)} . Note that there exists a prime factor p of n such that p\equiv5 \, (\mathrm{mod}\, 8) ; thus, we must have that \omega(n)\leq2 if \varphi_{8}(n) is odd. For \omega(n) = 2 , i.e., n = p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} , by (3.2) we have

    \begin{eqnarray} \varphi_{8}(n) = \frac{1}{8}\, p_{1}^{\alpha_{1}-1}(p_{1}-1)\, p_{2}^{\alpha_{2}-1}(p_{2}-1)+(-1)^{\alpha_{1}+\alpha_{2}}. \end{eqnarray}

    Therefore \varphi_{8}(n) is odd if and only if p_{1}\equiv p_{2}\equiv5\, (\mathrm{mod}\, 8) , which is true. Now for \omega(n) = 1 , i.e., n = p_1^{\alpha_1} with p_1\equiv 5\, (\mathrm{mod}\, 8) , similarly, by (3.2) we have

    \begin{eqnarray} \varphi_{8}(n) = \frac{1}{8}\, p_1^{\alpha_{1}-1}(p_{1}-1)+\frac{1}{2}(-1)^{\alpha_{1}} = \frac{1}{8}\, \big(p_{1}^{\alpha_{1}-1}(p_{1}-1)+4\cdot(-1)^{\alpha_{1}}\big). \end{eqnarray}

    From p_{1}\equiv5\, (\mathrm{mod}\, 8) , we have that p_{1}\equiv5, 13\, (\mathrm{mod}\, 16) . If p_{1}\equiv5\, (\mathrm{mod}\, 16) , then

    p_{1}^{\alpha_{1}-1}(p_{1}-1)+4(-1)^{\alpha_{1}}\equiv4\cdot5^{\alpha_{1}-1}+4(-1)^{\alpha_{1}}(\mathrm{mod}\, 16).

    Thus, \varphi_{8}(n) is odd if and only if 2\mid\alpha_{1} . If p_{1}\equiv13\, (\mathrm{mod}\, 16) , then

    p_{1}^{\alpha_{1}-1}(p_{1}-1)+4(-1)^{\alpha_{1}}\equiv12\cdot(-3)^{\alpha_{1}-1}+4(-1)^{\alpha_{1}}(\mathrm{mod}\, 16).

    Thus, \varphi_{8}(n) is odd if and only if \alpha_{1} is odd.

    (B) If \alpha = 1 , i.e., \omega(n)\geq 2 , by (3.2) we have that \varphi_{8}(n) = \frac{1}{8}\varphi(n)+\frac{1}{4}(-1)^{\Omega(n)}2 ^{\omega(n)-1} . Then we must have that \omega(n)\leq3 if \varphi_{8}(n) is odd. For \omega(n) = 3 , namely, n = 2p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} , using the same method as (A), \varphi_{8}(n) is odd if and only if p_{1}\equiv p_{2}\equiv5\, (\mathrm{mod}\, 8) . Now for \omega(n) = 2 , i.e., n = 2p_{1}^{\alpha_{1}} with p_{1}\equiv5\, (\mathrm{mod}\, 8) , similar to (A), \varphi_{8}(n) is odd if and only if p_{1}\equiv5\, (\mathrm{mod}\, 16) and \alpha_{1} is odd, or if p_{1}\equiv13\, (\mathrm{mod}\, 16) and 2\mid \alpha_{1} .

    (C) If \alpha = 2 , i.e., \omega(n)\geq2 , then by (3.2), we have that \varphi_{8}(n) = \frac{1}{8}\varphi(n) = \frac{1}{4}\prod _{i = 1}^{k} p_{i}^{\alpha_{i}-1}(p_{i}-1) . Thus from the assumption that p_{i}\equiv 5, 7\, (\mathrm{mod}\, 8) or p_{i}\equiv 5\, (\mathrm{mod}\, 8) , we know that \omega(n) = 2 if \varphi_{8}(n) is odd. In this case, n = 4p_{1}^{\alpha_{1}} with p_{1}\equiv5\, (\mathrm{mod}\, 8) ; then, p_{1}^{\alpha_{1}-1}(p_{1}-1)\equiv4\, (\mathrm{mod}\, 8) , namely, \varphi_{8}(n) is odd.

    (D) If \alpha\geq3 , then by (3.2), \varphi_{8}(n) = \frac{1}{8}\varphi(n) = 2^{\alpha-4}\prod _{i = 1} ^{k} p_{i}^{\alpha_{i}-1}(p_{i}-1) . Thus we must have that \alpha = 3 and k = 1 if \varphi_{8}(n) is odd, namely, n = 8p_1^{\alpha_1} with p_{1}\equiv 5\, (\mathrm{mod}\, 8) . In this case, \varphi_{8}(n) = \frac{1}{8}\varphi(n) = \frac{1}{2}p_{1}^{\alpha_{1}-1}(p_{1}-1) is always even.

    Case 2. R_{\mathbb{P}_{k}} = \{3, 7\} or \{3\} .

    (A) If \alpha = 0 , by (3.2) we have that \varphi_{8}(n) = \frac{1}{8}\varphi(n)+\frac{1}{8}(-1)^{\Omega(n)}2 ^{\omega(n)} . Thus we must have that \omega(n)\leq 3 if \varphi_{8}(n) is odd. For the case that \omega(n) = 3 , i.e, n = p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3} , where p_i\equiv 3\, (\mathrm{mod}\, 4) \, (i = 1, 2, 3) , it is easy to see that \varphi_{8}(n) is always even in this case. Therefore we must have that \omega(n) = 1, 2 . Consider that \omega(n) = 2 , i.e., n = p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} . Note that R_{\mathbb{P}_{k}} = \{3, 7\} or \{3\} ; then, by (3.2), \varphi_{8}(n) = \frac{1}{8}\big(p_{1}^{\alpha_{1}-1}(p_{1}-1)\, p_{2}^{\alpha_{2}-1}(p_{2}-1)+4\cdot(-1)^{\alpha_{1}+\alpha_{2}}\big) is odd if and only if p_1\equiv p_2\equiv3\, (\mathrm{mod}\, 8) . Now, for \omega(n) = 1 , i.e., n = p_{1}^{\alpha_{1}} with p_{1}\equiv3\, (\mathrm{mod}\, 8) , then by (3.2) we have that \varphi_{8}(n) = \frac{1}{8}\big(p_{1}^{\alpha_{1}-1}(p_{1}-1)+2(-1)^{\alpha_{1}}\big). Thus, \varphi_{8}(n) is odd if and only if p_{1}\equiv 3\, (\mathrm{mod}\, 16) and 2\mid\alpha_{1} , or if p_{1}\equiv 11\, (\mathrm{mod}\, 16) and \alpha_{1} is odd.

    (B) If \alpha = 1 , i.e., \omega\geq2 , by (3.2) we have that \varphi_{8}(n) = \frac{1}{8}\varphi(n)+\frac{1}{8}(-1)^{\Omega(n)-1}2 ^{\omega(n)-1} . Thus we must have that \omega(n)\leq 3 if \varphi_{8}(n) is odd. Using the same method as (A) in case 1, we can get that \varphi_{8}(n) is odd if and only if n = 2p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} with p_{1}\equiv p_{2}\equiv3\, (\mathrm{mod}\, 8) , or if n = 2p_{1}^{\alpha_{1}} with p_{1}\equiv3\, (\mathrm{mod}\, 16) and 2\mid \alpha_{1} , or if p_{1}\equiv11\, (\mathrm{mod}\, 16) and \alpha_{1} is odd.

    (C) If \alpha = 2 , i.e., \omega(n)\geq2 , by (3.2) we have that \varphi_{8}(n) = \frac{1}{8}\varphi(n)+\frac{1}{8}\, (-1)^{\Omega(n)-1}\, 2 \, ^{\omega(n)} . Therefore we must have that \omega(n)\leq3 if \varphi_{8}(n) is odd. For the case that \omega(n) = 3 , i.e., n = 4p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} , we know that

    \varphi_{8}(n) = \frac{1}{4}p_{1}^{\alpha_{1}-1}(p_1-1)p_{2}^{\alpha_{2}-1}(p_2-1)+(-1)^{\alpha_{1}+\alpha_{2}+1},

    which is always even. Now for the case that \omega(n) = 2 , i.e., n = 4p_{1}^{\alpha_{1}} with p_{1}\equiv 3\, (\mathrm{mod}\, 8) , by (3.2) we have that \varphi_{8}(n) = \frac{1}{4}(p_{1}^{\alpha_{1}-1}(p_1-1)+2(-1)^{\alpha_{1}+1}) . Since

    p_{1}^{\alpha_{1}-1}(p_1-1)+2(-1)^{\alpha_{1}+1}\equiv2\cdot3^{\alpha_{1}-1}+2(-1)^{\alpha_{1}+1}\equiv 4\, (\mathrm{mod}\, 8),

    it follows that \varphi_{8}(n) is odd.

    (D) If \alpha\geq3 , by (3.2) we have that \varphi_{8}(n) = \frac{1}{8}\varphi(n) = 2^{\alpha-4}\prod _{i = 1} ^{k} p_{i}^{\alpha_{i}-1}(p_{i}-1) . From R_{\mathbb{P}_{k}} = \{3, 7\} or \{3\} , we must have that \alpha = 3 and k = 1 if \varphi_{8}(n) is odd, namely, n = 8p_{1}^{\alpha_{1}} with p_{1}\equiv3\, (\mathrm{mod}\, 8) . Obviously, \varphi_{8}(n) = \frac{1}{2}\, p_{1}^{\alpha_{1}-1}(p_{1}-1) is odd in this case.

    Case 3. R_{\mathbb{P}_{k}} = \{7\} .

    (A) If \alpha = 0 , by (3.2), \varphi_{8}(n) = \frac{1}{8}\varphi(n)+\frac{3}{8}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)} . Then we must have that \omega(n)\leq2 if \varphi_{8}(n) is odd. For \omega(n) = 2 , i.e., n = p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} , it follows that

    \varphi_{8}(n) = \frac{1}{2}\Big(p_{1}^{\alpha_{1}-1}p_{2}^{\alpha_{2}-1}\cdot\frac{p_{1}-1}{2}\cdot\frac{p_{2}-1}{2} +3\cdot(-1)^{\alpha_{1}+\alpha_{2}}\Big).

    Since p_{1}\equiv p_{2}\equiv7\, (\mathrm{mod}\, 8) , we have that \frac{p_{1}-1}{2}\cdot\frac{p_{2}-1}{2}\equiv1\, (\mathrm{mod}\, 4) and

    p_{1}^{\alpha_{1}-1}p_{2}^{\alpha_{2}-1}\cdot\frac{p_{1}-1}{2}\cdot\frac{p_{2}-1}{2} +3\cdot(-1)^{\alpha_{1}+\alpha_{2}} \equiv (-1)^{\alpha_{1}+\alpha_{2}-2}+3\cdot(-1)^{\alpha_{1}+\alpha_{2}} \equiv 0\, (\mathrm{mod}\, 4),

    which means that \varphi_{8}(n) is even. Now for \omega(n) = 1 , i.e., n = p_{1}^{\alpha_{1}} , by (3.2) we have

    \varphi_{8}(n) = \frac{1}{4}\Big(p_{1}^{\alpha_{1}-1}\cdot\frac{p_{1}-1}{2}+3\cdot(-1)^{\alpha_{1}}\Big).

    Now from p_{1}\equiv7\, (\mathrm{mod}\, 8) , we have that p_{1}\equiv7, 15\, (\mathrm{mod}\, 16) . If p_{1}\equiv7\, (\mathrm{mod}\, 16) , then

    p_{1}^{\alpha_{1}-1}\cdot\frac{p_{1}-1}{2}+3\cdot(-1)^{\alpha_{1}}\equiv3\cdot(-1)^{\alpha_{1}-1}+3\cdot(-1)^{\alpha_{1}}\equiv0\, (\mathrm{mod}\, 8),

    namely, \varphi_{8}(n) is even. Thus, p_{1}\equiv15(\mathrm{mod}\, 16) , then

    p_{1}^{\alpha_{1}-1}\cdot\frac{p_{1}-1}{2}+3\cdot(-1)^{\alpha_{1}}\equiv7\cdot(-1)^{\alpha_{1}-1}+3\cdot(-1)^{\alpha_{1}}\equiv4\, (\mathrm{mod}\, 8),

    namely, \varphi_{8}(n) is odd.

    (B) If \alpha = 1 , by (3.2), \varphi_{8}(n) = \frac{1}{8}\varphi(n)+\frac{1}{8}(-1)^{\Omega(n)}2\, ^{\omega(n)-1} . Using a similar proof as that for (A) in case 1, \varphi_{8}(n) is odd if and only if n = 2p_{1}^{\alpha_{1}} and p_{1}\equiv7\, (\mathrm{mod}\, 16) .

    (C) If \alpha = 2 , i.e., \omega(n)\geq2 , by (3.2), \varphi_{8}(n) = \frac{1}{8}\varphi(n)+\frac{1}{8}\, (-1)^{\Omega(n)-1}\, 2 \, ^{\omega(n)} . Then we must have that \omega(n)\leq3 if \varphi_{8}(n) is odd. For \omega(n) = 3 , i.e., n = 4p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} with p_{1}\equiv p_{2}\equiv7\, (\mathrm{mod}\, 8) , we know that

    \varphi_{8}(n) = p_{1}^{\alpha_{1}-1}p_{2}^{\alpha_{2}-1}\cdot\frac{p_{1}-1}{2}\cdot\frac{p_{2}-1}{2} +(-1)^{\alpha_{1}+\alpha_{2}-1}

    is always even. Now for \omega(n) = 2 , i.e., n = 4p_{1}^{\alpha_{1}} with p_{1}\equiv7\, (\mathrm{mod}\, 8) , we can verify that

    \varphi_{8}(n) = \frac{1}{2}\Big(p_{1}^{\alpha_{1}-1}\cdot\frac{p_{1}-1}{2}+(-1)^{\alpha_{1}-1}\Big)

    is also even.

    (D) If \alpha\geq3 , by (3.2), \varphi_{8}(n) = \frac{1}{8}\varphi(n) = 2^{\alpha-4}\prod _{i = 1} ^{k} p_{i}^{\alpha_{i}-1}(p_{i}-1) . Hence, by R_{\mathbb{P}_{k}} = \{7\} we know that \varphi_{8}(n) is odd if and only if \alpha = 3 and k = 1 , i.e., n = 8p_{1}^{\alpha_{1}} with p_{1}\equiv7\, (\mathrm{mod}\, 8) .

    Case 4. \{3, 5\}\subseteq R_{\mathbb{P}_{k}} or 1\in R_{\mathbb{P}_{k}} .

    (A) If \{3, 5\}\subseteq R_{\mathbb{P}_{k}} , i.e., k\geq 2 , then by (3.2) we have that \varphi_{8}(n) = \frac{1}{8}\varphi(n) = \frac{1}{8}\varphi(2^{\alpha})\prod _{i = 1} ^{k}p_{i}^{\alpha_{i}-1} (p_{i}-1) . Thus we must have that k = 2 and \alpha\leq1 if \varphi_{8}(n) is odd, namely, n = p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} or 2p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} , where p_{1}\equiv3\, (\mathrm{mod}\, 8) and p_{2}\equiv5\, (\mathrm{mod}\, 8) . Obviously, \varphi_{8}(n) = \frac{1}{8}p_{1}^{\alpha_{1}-1}(p_{1}-1)p_{2}^{\alpha_{2}-1}(p_{2}-1) is always odd in this case.

    (B) If 1\in R_{\mathbb{P}_{k}} , by (3.2), \varphi_{8}(n) = \frac{1}{8}\varphi(n) = \frac{1}{8}\varphi(2^{\alpha})\prod _{i = 1} ^{k} p_{i}^{\alpha_{i}-1}(p_{i}-1) . Thus, we must have that \alpha\leq1 and k = 1 if \varphi_{8}(n) is odd. Namely, n = p_{1}^{\alpha_{1}}, 2p_{1}^{\alpha_{1}} with p_{1}\equiv1\, (\mathrm{mod}\, 8) ; then, \varphi_{8}(n) = \frac{1}{8}p^{\alpha_{1}-1}(p_{1}-1) . Obviously, \varphi_{8}(n) is odd if and only if p_{1}\equiv9\, (\mathrm{mod}\, 16) .

    From the above, we have completed the proof of Theorem 5.1.

    Theorem 5.2. If n is a positive integer, then \varphi_{12}(n) is odd if and only if n = 2^{\alpha}\, (\alpha\geq4) , 3\cdot2^{\alpha}\, (\alpha\geq2) , 2\cdot3^{\beta}\, (\beta\geq2) , 4\cdot3^{\beta}\, (\beta\geq2) , or if it satisfies the conditions given in Table 2. Here, p > 3 is an odd prime and \alpha\geq 1 .

    Table 2.  the conditions of \varphi_{12}(n) is odd.
    n Conditions
    p^{\alpha} p\equiv13, 17, 19, 23\, (\mathrm{mod}\, 24) ;
    2p^{\alpha} p\equiv7, 11, 13, 17\, (\mathrm{mod}\, 24) ;
    3p^{\alpha} p\equiv5, 7\, (\mathrm{mod}\, 12) ;
    4p^{\alpha} p\equiv5, 7\, (\mathrm{mod}\, 12) ;
    6p^{\alpha} p\equiv5, 7\, (\mathrm{mod}\, 12) ;
    12p^{\alpha} p\equiv5, 11\, (\mathrm{mod}\, 12) .

     | Show Table
    DownLoad: CSV

    Proof. Obviously, by the definition of \varphi_{12}(n) we can get that \varphi_{12}(n) = 0 for n < 12 and \varphi_{12}(n) = 1 for n = 12, 24 ; then, we consider that n > 12 and n\neq 24 . First, we consider the case that n = 2^{\alpha}\cdot3^{\beta} .

    If \alpha = 0 , we have that \beta\geq3 ; then, by (4.1), \varphi_{12}(n) = \frac{1}{2}\big(3^{\beta-2}-(-1)^{\beta}\big) is even.

    If \alpha = 1 , we have that \beta\geq2 ; then, by (4.1), \varphi_{12}(n) = \frac{1}{2}\big(3^{\beta-2}-(-1)^{\beta+1}\big) is odd.

    If \alpha = 2 , we have that \beta\geq2 ; then, by (4.1), \varphi_{12}(n) = 3^{\beta-2} is odd.

    If \alpha = 3 , we have that \beta\geq2 ; then, by (4.1), \varphi_{12}(8\cdot3^{\beta}) = 2\cdot3^{\beta-2} is even.

    If \alpha\geq4 , then that \beta\geq0 . For \beta = 0 , by (4.1), \varphi_{12}(n) = \frac{1}{3}(2^{\alpha-3}+(-1)^{\alpha}) is odd. For \beta = 1 , by (4.1), \varphi_{12}(n) = \frac{1}{3}\big(2^{\alpha-2}+(-1)^{\alpha+1}\big) is odd. For \beta\geq2 , by (4.1), \varphi_{12}(2^{\alpha}\cdot3^{\beta}) = 2^{\alpha-2}\cdot3^{\beta-2}, which is always even.

    Next, we consider the case that n = 2^{\alpha}\, 3^{\beta}n_{1} , where \alpha \geq0 , \beta \geq0 , n_{1} > 1 and \gcd(n_{1}, 6) = 1 . For convenience, we set n = 2^{\alpha}\, 3^{\beta}\prod_{i = 1}^{k} p_{i}^{\alpha_{i}} , where \alpha_{i} \geq1 , p_{i} is an odd prime and p_{i} > 3\, (1\leq i\leq k) . By Theorem 4.2 we have the following four cases.

    Case 1. R_{\mathbb{P}_{k}}' = \{7, 11\} or \{7\} .

    (A) \alpha = 0 . If \beta = 0 , i.e., \omega(n)\geq 1 , from (4.2) we have that \varphi_{12}(n) = \frac{1}{12}\varphi(n)+\frac{1}{4}\, (-1)^{\Omega(n)}\, 2 \, ^{\omega(n)} . Thus, from the assumption that R_{\mathbb{P}_{k}}' = \{7, 11\}\ \text{or}\ \{7\} , we must have that \omega(n)\leq 2 if \varphi_{12}(n) is odd. For \omega(n) = 2 , i.e., n = p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} , note that R_{\mathbb{P}_{k}}' = \{7, 11\}\ \text{or}\ \{7\} ; then,

    \varphi_{12}(n) = \frac{1}{3}\, p_{1}^{\alpha_{1}-1}\, p_{2}^{\alpha_{2}-1}\cdot\frac{p_{1}-1}{2}\cdot\frac{p_{2}-1}{2}+(-1)^{\alpha_{1} +\alpha_{2}}

    is always even in this case. Thus, \omega(n) = 1 , i.e., n = p_{1}^{\alpha_{1}} ; then, by (4.5),

    \varphi_{12}(n) = \frac{1}{12}\big(p_{1}^{\alpha_{1}-1}(p_{1}-1) +6\cdot(-1)^{\alpha_{1}}\big).

    Note that p_{1}\equiv7\, (\mathrm{mod}\, 12) , i.e., p_{1}\equiv7, 19\, (\mathrm{mod}\, 24) . If p_{1}\equiv7\, (\mathrm{mod}\, 24) , then p_{1}^{\alpha_{1}-1}(p-1)+6\cdot(-1)^{\alpha_{1}}\equiv0\, (\mathrm{mod}\, 24) , which means that \varphi_{12}(n) is even. Thus, p_{1}\equiv19\, (\mathrm{mod}\, 24) ; then, p_{1}^{\alpha_{1}-1}(p_{1}-1)+6\cdot(-1)^{\alpha_{1}}\equiv12\, (\mathrm{mod}\, 24) , namely, \varphi_{12}(n) is odd.

    If \beta = 1 , i.e., \omega(n)\geq 2 , by (4.2), \varphi_{12}(n) = \frac{1}{12}\varphi(n) = \frac{1}{6}\prod _{i = 1} ^{k} p_{i}^{\alpha_{i}-1}(p_{i}-1) . Similarly, we must have that \omega(n) = 2 if \varphi_{12}(n) is odd. In this case, n = 3p_{1}^{\alpha} with p_{1}\equiv7\, (\mathrm{mod}\, 12) ; then, \varphi_{12}(n) = \frac{1}{6}p_{1}^{\alpha-1}(p_{1}-1) , easy to see that \varphi_{12}(n) is always even. If \beta\geq2 , i.e., \omega(n)\geq 2 , by (4.2), \varphi_{12}(n) = \frac{1}{12}\varphi(n)+\frac{1}{4}\, (-1)^{\Omega(n)+1}\, 2 \, ^{\omega(n)} . Similarly, we must have that \omega(n) = 2 , i.e., n = 3^{\beta}p_{1}^{\alpha_{1}} if \varphi_{12}(n) is odd. Since p_{1}\equiv7\, (\mathrm{mod}\, 12) , it follows that \varphi_{12}(n) = 3^{\beta-2}p_{1}^{\alpha_{1}-1}\cdot\frac{p_{1}-1}{2}+(-1)^{\beta+\alpha_{1}+1} is always even.

    (B) \alpha = 1 . If \beta = 0 , i.e., \omega(n)\geq2 , by (4.10), \varphi_{12}(n) = \frac{1}{12}\varphi(n)+\frac{1}{4}(-1)^{\Omega(n)}2^{\omega(n)-1} . Similarly, we must have that \omega(n)\leq3 if \varphi_{12}(n) is odd. For \omega(n) = 3 , i.e., n = 2p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} , note that R_{\mathbb{P}_{k}}' = \{7, 11\}\ \text{or}\ \{7\} ; then, it is easy to see that \varphi_{12}(n) is always even. Thus, \omega(n) = 2 , i.e., n = 2p_{1}^{\alpha_{1}} ; note that p_{1}\equiv7\, (\mathrm{mod}\, 12) , namely, p_{1}\equiv7, 19\, (\mathrm{mod}\, 24) . In this case, \varphi_{12}(n) is odd if and only if p_{1}\equiv7\, (\mathrm{mod}\, 24) .

    If \beta = 1 , i.e., \omega(n)\geq3 , by (4.2), \varphi_{12}(n) = \frac{1}{12}\varphi(n) = \frac{1}{6}\prod _{i = 1} ^{k} p_{i}^{\alpha_{i}-1}(p_{i}-1) . Similarly, from R_{\mathbb{P}_{k}}' = \{7, 11\}\ \text{or}\ \{7\} , we can get that \varphi_{12}(n) is odd if and only if \omega(n) = 3 , i.e., n = 6p_{1}^{\alpha} with p_{1}\equiv7\, (\mathrm{mod}\, 12) .

    If \beta\geq2 , i.e., \omega(n)\geq3 , by(4.2), \varphi_{12}(n) = \frac{1}{12}\varphi(n)+\frac{1}{4}(-1)^{\Omega(n)+1}2 ^{\omega(n)-1} . Then we must have that \omega(n) = 3 if \varphi_{12}(n) is odd, namely, n = 2\cdot3^{\beta}p_{1}^{\alpha_{1}} with p_{1}\equiv7\, (\mathrm{mod}\, 12) . Obviously, \varphi_{12}(n) = 3^{\beta-2}p^{\alpha-1}\cdot\frac{p-1}{2}+(-1)^{2+\beta+\alpha} is always even in this case.

    (C) \alpha = 2 . If \beta = 0 , i.e., \omega(n)\geq 2 , by (4.2), \varphi_{12}(n) = \frac{1}{12}\varphi(n) = \frac{1}{6}\prod _{i = 1} ^{k} p_{i}^{\alpha_{i}-1}(p_{i}-1) . Thus, we must have that \omega(n) = 2 if \varphi_{12}(n) is odd. In this case, n = 4p_{1}^{\alpha_{1}} with p_{1}\equiv7\, (\mathrm{mod}\, 12) ; then, p_1^{\alpha_1}(p_1-1)\equiv 6\, (\mathrm{mod}\, 12) , which means that \varphi_{12}(n) is always odd in this case.

    If \beta\geq1 , i.e., \omega(n)\geq 3 , by (4.2), \varphi_{12}(n) = \frac{1}{12}\varphi(n) = 3^{\beta-2}\prod _{i = 1} ^{k} p_{i}^{\alpha_{i}-1}(p_{i}-1) . Note that R_{\mathbb{P}_{k}}' = \{7, 11\}\ \text{or}\ \{7\} ; then, \varphi_{12}(n) is always even.

    (D) \alpha\geq3 . By (4.2) and R_{\mathbb{P}_{k}}' = \{7, 11\}\ \text{or}\ \{7\} , \varphi_{12}(n) = \frac{1}{12}\varphi(n) = \frac{1}{3}\cdot2^{\alpha-3}\varphi(3^{\beta}n_{1}) is always even in this case.

    Case 2. R_{\mathbb{P}_{k}}' = \{5, 11\} or \{5\} .

    (A) \alpha = 0 . If \beta = 0 , i.e., \omega(n)\geq1 , from (4.6), we can get that \varphi_{12}(n) = \frac{1}{12}\, \varphi(n)+\frac{1}{6}\, (-1)^{\Omega(n)}\, 2^{\omega(n)} . Thus, we must have that \omega(n) = 1 if \varphi_{12}(n) is odd, namely, n = p_{1}^{\alpha_{1}} with p_{1}\equiv5\, (\mathrm{mod}\, 12) . Hence

    \varphi_{12}(p_{1}^{\alpha_{1}}) = \frac{1}{12}\, p_{1}^{\alpha_{1}-1}(p_{1}-1)+\frac{1}{3}\, (-1)^{\alpha_{1}} = \frac{1}{3}\Big(p_{1}^{\alpha_{1}-1}\cdot\frac{p_{1}-1}{4}+(-1)^{\alpha_{1}}\Big).

    Note that p_{1}\equiv5\, (\mathrm{mod}\, 12) , i.e., p_{1}\equiv5, 17\, (\mathrm{mod}\, 24) . If p_{1}\equiv5(\mathrm{mod}\, 24) , then p_{1}^{\alpha-1}\cdot\frac{p_{1}-1}{4}+(-1)^{\alpha_{1}}\equiv0(\mathrm{mod}\, 6) , which means that \varphi_{12}(p_{1}^{\alpha}) is always even. Thus, p_{1}\equiv17\, (\mathrm{mod}\, 24) , in this case p_{1}^{\alpha_{1}-1}\cdot\frac{p_{1}-1}{4}+(-1)^{\alpha_{1}}\equiv3\, (\mathrm{mod}\, 6) , namely, \varphi_{12}(p^{\alpha_{1}}) is odd.

    If \beta = 1 , i.e., \omega(n)\geq2 , from (4.8) we have that \varphi_{12}(n) = \frac{1}{12}\, \varphi(n)+\frac{1}{6}\, (-1)^{\Omega(n)}\, 2^{\omega(n)-1} . Thus, we must have that \omega(n) = 2 if \varphi_{12}(n) is odd, namely, n = 3p_{1}^{\alpha_{1}} with p_{1}\equiv5\, (\mathrm{mod}\, 12) ; in this case

    \varphi_{12}(3p_{1}^{\alpha_{1}}) = \frac{1}{3}\Big(2\, p_{1}^{\alpha_{1}-1}\cdot\frac{p_{1}-1}{4}+(-1)^{\alpha_{1}}\Big)

    is always odd.

    If \beta\geq2 , i.e., \omega(n)\geq2 , from (4.9) we can get that \varphi_{12}(n) = \frac{1}{12}\, \varphi(n) . We must have that \omega(n) = 2 , i.e., n = 3^{\beta}p^{\alpha}(\beta\geq2) , if \varphi_{12}(n) is odd. From the assumption p_{1}\equiv5\, (\mathrm{mod}\, 12) , \varphi_{12}(3^{\beta}\, p_{1}^{\alpha_{1}}) = \frac{1}{12}\, \varphi(3^{\beta}\, p_{1}^{\alpha_{1}}) = 2\cdot 3^{\beta-1}\, p_{1}^{\alpha_{1}-1}\cdot\frac{p_{1}-1}{4} is always even.

    (B) \alpha = 1 , i.e., \omega(n)\geq 2 . By (4.10)–(4.12), we must have \omega(n)\leq3 if \varphi_{12}(n) is odd. Namely, n = 2p_{1}^{\alpha_{1}}, 2p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}, 6p_{1}^{\alpha_{1}} , or 2\cdot3^{\beta}p_{1}^{\alpha_{1}}\, (\beta\geq2) . Similar to the proof of (A) in case 1, \varphi_{12}(n) is odd if and only if n = 2p_{1}^{\alpha_{1}} with p_{1}\equiv17\, (\mathrm{mod}\, 24) , or if n = 6p_{1}^{\alpha_{1}} with p_{1}\equiv5\, (\mathrm{mod}\, 12) .

    (C) \alpha = 2 . If \beta = 0 , i.e., \omega(n)\geq2 , by (4.13), \varphi_{12}(n) = \frac{1}{12}\varphi(n)+\frac{1}{12}(-1)^{\Omega(n)+1}2^{\omega(n)} ; then, we must have that \omega(n) = 2 if \varphi_{12}(n) is odd. In this case, n = 4p_{1}^{\alpha_{1}} with p_{1}\equiv5\, (\mathrm{mod}\, 12) . Hence, \varphi_{12}(n) = \frac{1}{6}\, p_{1}^{\alpha_{1}-1}(p_{1}-1)+\frac{1}{3}\, (-1)^{\alpha_{1}+3} = \frac{1}{3}\Big(p_{1}^{\alpha_{1}-1}\frac{p_{1}-1}{2}+(-1)^{\alpha_{1}+3}\Big) is always odd.

    If \beta = 1 , i.e., \omega(n)\geq3 , by (4.14), \varphi_{12}(n) = \frac{1}{12}\varphi(n)+\frac{1}{12}(-1)^{\Omega(n)+1}2^{\omega(n)-1} ; we must have that \omega(n) = 3 if \varphi_{12}(n) is odd. In this case, n = 12 \, p_{1}^{\alpha_{1}} with p_{1}\equiv5\, (\mathrm{mod}\, 12) ; then, \varphi_{12}(n) = \frac{1}{3}\, p_{1}^{\alpha_{1}-1}(p_{1}-1)+\frac{1}{3}\, (-1)^{\alpha_{1}+4} = \frac{1}{3}\big(p_{1}^{\alpha_{1}-1}(p_{1}-1)+(-1)^{\alpha_{1}+4}\big) is odd.

    If \beta\geq 2 , i.e., \omega(n)\geq3 , by (4.15), \varphi_{12}(n) = \frac{1}{12}\varphi(n) ; we must have that \omega(n) = 3 if \varphi_{12}(n) is odd. Namely, n = 4\cdot3^{\beta} p_{1}^{\alpha_{1}} with p_{1}\equiv5\, (\mathrm{mod}\, 12) ; then,

    \varphi_{12}(n) = \frac{1}{12}\, \varphi(n) = 3^{\beta-2}p_{1}^{\alpha_{1}-1}(p_{1}-1)

    is always even.

    (D) \alpha\geq3 , i.e., \omega(n)\geq 2 . If \beta = 0 , then by (4.16) and R_{\mathbb{P}_{k}}' = \{5, 11\}\ \text{or}\ \{5\} , we konw that \varphi_{12}(n) = \frac{1}{12}\varphi(n)+\frac{1}{6}(-1)^{\Omega(n)}\, 2 \, ^{\omega(n)} is always even in this case.

    If \beta = 1 , by (4.17) and R_{\mathbb{P}_{k}}' = \{5, 11\}\ \text{or}\ \{5\} , we know that \varphi_{12}(n) = \frac{1}{12}\varphi(n)+\frac{1}{12}(-1)^{\Omega(n)}\, 2 \, ^{\omega(n)} is always even in this case.

    If \beta\geq2 , by (4.18) and R_{\mathbb{P}_{k}}' = \{5, 11\}\ \text{or}\ \{5\} , \varphi_{12}(n) = \frac{1}{12}\varphi(n) is always even in this case.

    Case 3. R_{\mathbb{P}_{k}}' = \{11\} .

    (A) \alpha = 0 , i.e., \omega(n)\geq 1 . From (4.7)–(4.9), we must have that \omega(n)\leq2 if \varphi_{12}(n) is odd. Consider that \omega(n) = 2 , i.e., n = p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} , or 3^{\beta} p_{1}^{\alpha_{1}}\, (\beta\geq1) with p_{1}\equiv p_{2}\equiv11\, (\mathrm{mod}\, 12) . Thus, by (4.7)–(4.9), \varphi_{12}(n) is always even. Hence, \omega(n) = 1 , i.e., n = p_{1}^{\alpha_{1}} with p\equiv11\, (\mathrm{mod}\, 12) ; then, (4.7) we can get

    \varphi_{12}(p_{1}^{\alpha_{1}}) = \frac{1}{12}\, p_{1}^{\alpha_{1}-1}(p_{1}-1)+\frac{5}{6}\, (-1)^{\alpha_{1}} = \frac{1}{6}\Big(p_{1}^{\alpha_{1}-1}\cdot\frac{p_{1}-1}{2}+5(-1)^{\alpha_{1}}\Big).

    Note that p_{1}\equiv11(\mathrm{mod}\, 12) , i.e., p_{1}\equiv11, 23\, (\mathrm{mod}\, 24) . If p_{1}\equiv11\, (\mathrm{mod}\, 24) , then

    p_{1}^{\alpha_{1}-1}\cdot\frac{p_{1}-1}{2}+5(-1)^{\alpha_{1}} \equiv5(-1)^{\alpha_{1}-1}+5(-1)^{\alpha_{1}}\equiv0\, (\mathrm{mod}\, 12),

    namely, \varphi_{12}(n) is even. If p_{1}\equiv23\, (\mathrm{mod}\, 24) , then

    p_{1}^{\alpha_{1}-1}\cdot\frac{p_{1}-1}{2}+5(-1)^{\alpha_{1}} \equiv11(-1)^{\alpha_{1}-1}+5(-1)^{\alpha_{1}}\equiv6\, (\mathrm{mod}\, 12),

    namely, \varphi_{12}(n) is odd.

    (B) \alpha = 1 , i.e., \omega(n)\geq 2 . From (4.10)–(4.12), we must have that \omega(n)\leq3 if \varphi_{12}(n) is odd. Namely, n = 2p_{1}^{\alpha_{1}} , 2p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} , 6p_{1}^{\alpha_{1}} , or 2\cdot3^{\beta} p_{1}^{\alpha_{1}}\, (\beta\geq2) with p_{1}\equiv p_{2}\equiv11\, (\mathrm{mod}\, 12) . Using the same method as for (A) in case 1, \varphi_{12}(n) is odd if and only if n = 2p_{1}^{\alpha_{1}} with p_{1}\equiv11\, (\mathrm{mod}\, 24) .

    (C) \alpha = 2 , i.e., \omega(n)\geq 2 . If \beta = 0 , by (4.13), we must have that \omega(n) = 2 if \varphi_{12}(n) is odd, namely, n = 4p_{1}^{\alpha_{1}} with p_{1}\equiv11\, (\mathrm{mod}\, 12) . Then by (4.13),

    \varphi_{12}(4p_{1}^{\alpha_{1}}) = \frac{1}{3}\Big(p_{1}^{\alpha_{1}-1}\frac{p_{1}-1}{2}+(-1)^{\alpha_{1}+3}\Big)

    is always even.

    If \beta\geq1 , i.e., \omega(n)\geq3 , by (4.14)–(4.15), we must have that \omega(n) = 3 if \varphi_{12}(n) is odd. Namely, n = 4\cdot3^{\beta} p_{1}^{\alpha_{1}}\, (\beta\geq 1) with p_{1}\equiv11\, (\mathrm{mod}\, 12) . If \beta\geq2 , then by (4.15),

    \varphi_{12}(4\cdot3^{\beta} p_{1}^{\alpha_{1}}) = \frac{1}{12}\varphi(4\cdot3^{\beta} p_{1}^{\alpha_{1}}) = 3^{\beta-2}\, p_{1}^{\alpha_{1}-1}(p_{1}-1)

    is always even. Thus, \beta = 1 ; by (4.14), \varphi_{12}(12\, p_{1}^{\alpha_{1}}) = \frac{1}{3}\big(p_{1}^{\alpha_{1}-1}(p_{1}-1)+(-1)^{\alpha_{1}+3}\big) is odd.

    (D) \alpha\geq3 . If \beta = 0 , i.e., \omega(n)\geq 2 , then by (4.16) and R_{\mathbb{P}_{k}}' = \{11\} , we know that \varphi_{12}(n) = \frac{1}{12}\varphi(n)+\frac{1}{6}(-1)^{\Omega(n)}\, 2 \, ^{\omega(n)} is always even in this case.

    If \beta = 1 , i.e., \omega(n)\geq 3 , then by (4.17) and R_{\mathbb{P}_{k}}' = \{11\} , we know that \varphi_{12}(n) = \frac{1}{12}\varphi(n)+\frac{1}{12}(-1)^{\Omega(n)}\, 2 \, ^{\omega(n)} is always even in this case.

    If \beta\geq2 , i.e., \omega(n)\geq 3 , then by (4.18) and R_{\mathbb{P}_{k}}' = \{11\} , we know that \varphi_{12}(n) = \frac{1}{12}\varphi(n) = 2^{\alpha-2}\cdot3^{\beta-1}\prod_{i = 1}^{k} p_{i}^{\alpha_{i}-1}(p_{i}-1) is always even.

    Case 4. \{5, 7\}\subseteq R_{\mathbb{P}_{k}}' or 1\in R_{\mathbb{P}_{k}}' .

    (A) If \{5, 7\}\subseteq R_{\mathbb{P}_{k}}' , by (4.2) we have that \varphi_{12}(n) = \frac{1}{12}\varphi(n) is always even.

    (B) If 1\in R_{\mathbb{P}_{k}}' , then by (4.2), \varphi_{12}(n) = \frac{1}{12}\varphi(n) ; thus, we must have that k = 1 , \alpha\leq1 , and \beta = 0 if \varphi_{12}(n) is odd. Namely, n = p_{1}^{\alpha_{1}} or 2p_{1}^{\alpha_{1}} with p_{1}\equiv1\, (\mathrm{mod}\, 12) . In this case, \varphi_{12}(n) = \frac{1}{12}\, p_{1}^{\alpha_{1}-1}(p_{1}-1) is odd if and only if p_{1}\equiv13\, (\mathrm{mod}\, 24) .

    From the above, we have completed the proof of Theorem 5.2.

    In [2,8], Cai, et al. gave the explicit formulae for the generalized Euler functions denoted by \varphi_{e}(n) for e = 3, 4, 6 . The key point is that the derivation of [\frac{n}{e}] can be obtained by utilizing the corresponding Jacobi symbol for e = 3, 4, 6 . In the present paper, by applying Lemmas 2.1 and 2.2, the exact formulae for \varphi_8(n) and \varphi_{12}(n) have been given and the parity has been determined. Therefore, the obvious expression for [\frac{n}{e}] depends on the Jacobi symbol, seems to be the key to finding the exact formulae for \varphi_e(n) .

    We propose the following conjecture.

    Conjecture 6.1. Let e > 1 be a given integer. For any integer d > 2 with \gcd(d, e) = 1 , there exist u\in\mathbb{Q} , a_{1}, a_{2}, a_{3} , b_{i}\, (1\leq j\leq r) \in \mathbb{Z} , and q_{j}\, (1\leq j\leq r)\in \mathbb{P} , such that

    \begin{eqnarray} \;\;\;\;\;\;\Big[\frac{d}{e}\Big] = u\Big(a_{1}d+a_{2}+a_{3}\Big(\frac{-1}{d}\Big)+\sum\limits_{j = 1}^{r}b_{j}\Big(\frac{\varepsilon_{j}\, q_{j}}{d}\Big)\Big)\;\;{ ( 2\nmid d ) }, \end{eqnarray} (6.1)

    or

    \begin{eqnarray} \Big[\frac{d}{e}\Big] = u\Big(a_{1}d+a_{2}+\sum\limits_{j = 1}^{r}b_{j}\Big(\frac{\varepsilon_{j}\, d}{q_{j}}\Big)\Big) \;\;{ ( 2\, |\, d ) }, \end{eqnarray} (6.2)

    where r\geq 1 and \varepsilon_{j}\in\{1, -1\} .

    It is easy to see that Conjecture 6.1 is true for e = 2, 3, 4, 6, 8 and 12 . (see [2,8] and (2.1), (2.2)). If the formulas for (6.1) and (6.2) in the above conjecture can be obtained, then, by (1.1), using the properties of Möbius functions, we can find the exact formulae for \varphi_e(n) .

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    This work was supported by Natural Science Foundation of China under grant numbers 12361001, 12161001 and 12071321, and Research Projects of ABa Teachers University (AS-XJPT2023-02, AS-KCTD2023-02).

    The authors sincerely thanks Professor Cai Tianxin for his guidance and help, as he visited ABa Teachers University in October 2023.

    We would like to thank the referee for his/her detailed comments.

    The authors declare that there are no conflicts of interest regarding the publication of this paper.



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    [2] T. X. Cai, Z. Y. Shen, M. J. Hu, On the parity of the generalized Euler function (I), Adv. Math., 42 (2013), 505–510.
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    [5] Q. Y. Liao, W. L. Luo, The computing formula for two classes of generalized Euler functions, J. Math., 39 (2019), 97–110.
    [6] Q. Y. Liao, The explicit formula for a special class of generalized Euler functions (Chinese), Journal of Sichuan Normal University (Natural Science Edition), 42 (2019), 354–357. 10.3969/j.issn.1001-8395.2019.03.010 doi: 10.3969/j.issn.1001-8395.2019.03.010
    [7] P. Ribenboim, 13 Lectures on Fermat's last theorem, New York: Springer, 1979. https://doi.org/10.1007/978-1-4684-9342-9
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    [9] R. Wang, Q. Y. Liao, On the generalized Euler function \varphi_{5}(n) (Chinese), Journal of Sichuan Normal University (Natural Science Edition), 42 (2018), 445-449. 10.3969/j.issn.1001-8395.2018.04.003 doi: 10.3969/j.issn.1001-8395.2018.04.003
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