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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+1n; then, we can get

    fa(n)=pαn((ap)α(ap)α1)=ki=1((api)αi(api)αi1).

    This completes the proof of Lemma 2.2.

    The following lemmas are necessary for proving our main results.

    Lemma 2.3. [2] Let p1,,pk be distinct primes and α,α1,,αk be non-negative integers. If n=3αki=1pαii>3 and gcd(pi,3)=1(1ik), then

    φ3(n)={13φ(n)+13(1)Ω(n)2ω(n)α1,if   α=0  or  1,pi2(mod3),13φ(n),otherwise.

    Lemma 2.4. [8] Let p1,,pk be distinct odd primes and α,α1,,αk be non-negative integers. If n=2αki=1pαii>4, then

    φ4(n)={14φ(n)+14(1)Ω(n)2ω(n)α,if  α=0or1,pi3(mod4),14φ(n),otherwise.

    Lemma 2.5. [8] Let p1,,pk be distinct primes and α,β,α1,,αk be non-negative integers. If n=2α3βki=1pαii>6 and gcd(pi,6)=1(1ik), then

    φ6(n)={16φ(n)+16(1)Ω(n)2ω(n)+1β,if  α=0  and  β=0or1,pi5(mod6),16φ(n)+16(1)Ω(n)2ω(n)1β,if  α=1  and  β=0or1,pi5(mod6),16φ(n)16(1)Ω(n)2ω(n)β,if  α2  and  β=0or1,pi5(mod6),16φ(n),otherwise.

    Lemma 2.6. [6] Let p1,,pk be distinct primes and α1,,αk be positive integers. If n=ki=1pαii and e=ki=1pβii with 0βiαi1(1ik), then

    φe(n)=1eφ(n). (2.4)

    First, for a fixed positive integer α and n=2α, by Lemma 2.6 we can obtain the following:

    φ8(2α)={0,if   α=1,2,1,if  α=3,2α4,if  α4. (3.1)

    Next, we consider the case that n=2αn1, where n1>1 is an odd integer. We have the following theorem.

    Theorem 3.1. Suppose that α is a non-negative integer, p1,,pk are distinct odd primes, and n=2αki=1pαii>8. Then we have the following:

    φ8(n)={18φ(n)+14(1)Ω(n)2ω(n)α,if  α=0,1,and RPk={5,7},{5};18φ(n)+18(1)Ω(n)[α+12]2ω(n)12(1(1)α),if   α=0,1,2,and RPk={3,7},{3};18φ(n)+18(1)Ω(n)[α2]2ω(n)12(1(1)α)+1[α+12]4(1)Ω(n)2ω(n),if  α=0,1,2,and RPk={7};18φ(n),otherwise. (3.2)

    Proof. For n=2αki=1pαii>8, set n1=ki=1pαii; then, gcd(n1,2)=1. There are 4 cases as follows.

    Case 1. α=0, i.e., n=n1>8. By (1.1), (2.1) and Lemmas 2.1 and 2.2, we have

    φ8(n)=d|n1μ(n1d)[d8]=18d|n1μ(n1d)(d4+2(2d)+(1d))=18d|n1μ(n1d)d12d|n1μ(n1d)+14d|n1μ(n1d)(2d)+18d|n1μ(n1d)(1d)=18φ(n1)+14ki=1((2pi)αi(2pi)αi1)+18ki=1((1pi)αi(1pi)αi1). (3.3)

    If 1RPk, i.e., there exists an i(1ik) such that pi1(mod8), then (2pi)=(1pi)=1. Now by (3.3) we have

    φ8(n)=18φ(n1)=18φ(n). (3.4)

    If {3,5}RPk, i.e., there exist ij such that pi3(mod8) and pj5(mod8), which means that (2pi)=(1pj)=1, then, by (3.3) we also have

    φ8(n)=18φ(n1)=18φ(n).

    If RPk={5,7} or {5}, i.e., for any pPk, we have that p5,7(mod8) or p5(mod8), respectively. This means that there exists a prime p such that (2p)=1 and (1p)=1. Thus by (3.3) we can obtain

    φ8(n)=18φ(n1)+14ki=1(2(1)αi)=18φ(n)+14(1)Ω(n)2ω(n). (3.5)

    If RPk={3,7} or {3}, i.e., for any pPk,p3,7(mod8) or p3(mod8), respectively. This implies that for any pPk,(1p)=1, and there exists a prime pPk such that p3(mod8); then, (2p)=1. Thus by (3.3) we have

    φ8(n)=18φ(n1)+18ki=1(2(1)αi)=18φ(n)+18(1)Ω(n)2ω(n). (3.6)

    If RPk={7}, i.e., for any pPk,p7(mod8), then (2p)=(1p)=1. Thus by (3.3) we have

    φ8(n)=18φ(n1)+38ki=1(2(1)αi)=18φ(n)+38(1)Ω(n)2ω(n). (3.7)

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

    Case 2. α=1, i.e., n=2n1>8. Then from the definition we have

    φ8(n)=d|n1μ(2n1d)[d8]+d|n1μ(2n12d)[2d8]=φ8(n1)+φ4(n1).

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

    φ8(n)={18φ(n)+14(1)Ω(n)2ω(n)1,if  RPk={5,7},{5};18φ(n)+18(1)Ω(n)12ω(n)1,if  RPk={3,7},{3};18φ(n)+18(1)Ω(n)2ω(n)1,if  RPk={7};18φ(n),otherwise. (3.8)

    This means that Theorem 3.1 is true in this case.

    Case 3. α=2, i.e., n=4n1>8. Then from the definition we have

    φ8(n)=d|n1μ(4n1d)[d8]+d|n1μ(4n12d)[2d8]+d|n1μ(4n14d)[4d8]=d|n1μ(2n1d)[d4]+d|n1μ(n1d)[d2]=φ2(n1)φ4(n1)=12φ(n1)φ4(n1).

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

    φ8(n)={18φ(n),if  RPk={5,7},{5},18φ(n)+18(1)Ω(n)12ω(n),if  RPk={3,7},{3};18φ(n)+18(1)Ω(n)12ω(n),if   RPk={7};18φ(n),otherwise. (3.9)

    This means that Theorem 3.1 holds in this case.

    Case 4. α3. Note that μ(2γ)=0 for any positive integer γ2; thus, by (1.1) and Lemma 2.4 we have

    φ8(n)=d|n1μ(2n1d)[2α1d8]+d|n1μ(n1d)[2αd8]. (3.10)

    If α=3, then

    φ8(n)=d|n1μ(n1d)[d2]+d|n1μ(n1d)d=12φ(n1)+φ(n1)=12φ(n1)=18φ(n)

    If α4, then φ8(n)=2α4φ(n1)+2α3φ(n1)=2α4φ(n1)=18φ(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 φ12(n). Obviously, φ12(n)=0 when n<12, and φ12(n)=1 when n=12 or 24; then, we consider n>12 and n24.

    Theorem 4.1. Let α and β be non-negative integers. If n=2α3β>12 and n24, then the following holds:

    φ12(n)={12(3β2(1)α+β),if  α=0,orα1,β2;2α23β2,if   α2,β2;13(2α+β3+(1)α+β),if  α4,β=0,1. (4.1)

    Proof. (1) For the case that α=0, i.e., n=3β>12, and β3, then we have

    φ12(3β)=d|3βμ(3βd)[d12]=[3β12][3β112]=[3β14][3β24]=14(3β12+(1)β1)14(3β22+(1)β2)=12(3β2(1)β).

    (2) For the case that α=1, i.e., n=23β>12, and β2, by Lemma 2.5,

    φ12(23β)=d|3βμ(23βd)[d12]+d|3βμ(23β2d)[2d12]=φ12(3β)+φ6(3β)=112φ(3β)+12(1)β+16φ(3β)=12(3β2(1)β+1).

    (3) For the case that α=2, i.e., n=43β>12, and so β2, then we have

    φ12(43β)=d|43βμ(43βd)[d12]=μ(1)[43β12]+μ(2)[23β12]+μ(3)[43β112]+μ(6)[23β112]=3β1[3β12]3β2+[3β22]=3β2.

    (4) For the case that α=3, i.e., n=83β>12 and n24, and β2, then

    φ12(83β)=d|83βμ(83βd)[d12]=μ(1)[83β12]+μ(2)[43β12]+μ(3)[83β112]+μ(6)[43β112]=23β13β123β2+3β2=23β2.

    (5) For the case that α4, i.e., n=2α3β>12, and so β0, if β=0, i.e., n=2α(α4), then we have

    φ12(2α)=d|2αμ(2αd)[d12]=[2α23][2α33]=13(2α212(3(1)α2))13(2α312(3(1)α3))=13(2α3+(1)α)

    If β=1, i.e., n=32α, then we have

    φ12(32α)=d|32αμ(32αd)[d12]=2α22α3[2α23]+[2α33]=13(2α2+(1)α+1).

    If β2, we have

    φ12(2α3β)=d|2α3βμ(2α3βd)[d12]=μ(1)[2α3β12]+μ(2)[2α13β12]+μ(3)[2α3β112]+μ(6)[2α13β112]=2α23β1[2α23β12]2α23β2+[2α23β22]=2α23β2.

    This completes the proof of Theorem 4.1.

    Now consider the case that n=2α3βn1, where n1>1 and gcd(n1,6)=1. We have the following theorem.

    Theorem 4.2. Let α and β be non-negative integers, k,αi(1ik) be positive integers, and p1,,pk be distinct primes. Suppose that gcd(pi,6)=1\, (1ik) and n=2α3βki=1pαii>12; then, we have the following:

    φ12(n)={112φ(n)+14(1)Ω(n)2ω(n)α,if  α=0,1,β=0,  and  RPk={7,11},{7};112φ(n)+14(1)Ω(n)+12ω(n)α,if  α=0,1,β2,  and  RPk={7,11},{7},{11};112φ(n)+16(1)Ω(n)+[α+12]2ω(n)[α+12]β,if   α=0,1,2,β=0,1,  and  RPk={5,11},{5},or   α=0,1,β=1,  and  RPk={11};orα=2,β=0,1,  and  RPk={11};112φ(n)+16(1)Ω(n)2ω(n)β,if  α3,β=0,1,  and  RPk={5,11},{5},{11};112φ(n)+512(1)Ω(n)2ω(n),if   α=0,β=0,  and  RPk={11};112φ(n)+112(1)Ω(n)2ω(n)1,if   α=1,β=0,  and  RPk={11};112φ(n),otherwise. (4.2)

    Proof. Set n1=ki=1pαii; then, gcd(n1,6)=1 and n=2α3βn1.

    Case 1. α=0.

    (A) If β=0, then n1>1. Thus by (1.1), (2.2) and Lemmas 2.1 and 2.2, we have

    φ12(n)=φ12(n1)=d|n1μ(n1d)[d12]=112d|n1μ(n1d)(d6+3(1d)+2(3d))=112d|n1μ(n1d)d12d|n1μ(n1d)+14d|n1μ(n1d)(1d)+16d|n1μ(n1d)(3d)=112φ(n1)+14ki=1((1pi)αi(1pi)αi1)+16ki=1((3pi)αi(3pi)αi1). (4.3)

    If 1RPk or {5,7}RPk, then there exists pi1(mod12), or there exist pj and pl such that pj5(mod12) and pl7(mod12); then, (1pi)=(3pi)=1 or (1pj)=(3pl)=1, respectively. Thus by (4.3) we can get

    φ12(n)=112φ(n1)=112φ(n). (4.4)

    If RPk={7,11} or {7}, i.e., for any pPk, we have that p7,11(mod12) or p7(mod12), respectively. This means that (1p)=1 and there exists a prime p7(mod12), i.e., (3p)=1, in either of the two cases. Thus by (4.3) we can obtain

    φ12(n)=112φ(n1)+14ki=1(2(1)αi)=112φ(n)+14(1)Ω(n)2ω(n). (4.5)

    If RPk={5,11} or {5}, i.e., for any pPk, p5,11(mod12) or p5(mod12), respectively. Then (3p)=1, and there exists a prime p5(mod12), i.e., (1p)=1 in either case. Thus by (4.3) we can get

    φ12(n)=112φ(n1)+16ki=1(2(1)αi)=112φ(n)+16(1)Ω(n)2ω(n). (4.6)

    If RPk={11}, i.e., for any pPk, p11(mod12); then, (1p)=(3p)=1. Thus by (4.3) we have

    φ12(n)=112φ(n1)+512ki=1(2(1)αi)=112φ(n)+512(1)Ω(n)2ω(n). (4.7)

    (B) If β1, then by (1.1) we have

    φ12(n)=φ12(3βn1)=d|n1μ(3βn1d)[d12]+d|3β1n1μ(3βn13d)[3d12]=μ(3β)φ12(n1)+φ4(3β1n1).

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

    φ12(n)=φ12(n1)+φ4(n1)={112φ(n),if  RPk={7,11},{7},112φ(n)+16(1)Ω(n)2ω(n)1,if   RPk={5,11},{5},112φ(n)+16(1)Ω(n)2ω(n)1,if   RPk={11},112φ(n),otherwise. (4.8)

    For the case that β2, note that μ(3γ)=0 with γ2; thus, by Lemma 2.4 we have the following:

    φ12(n)=φ4(3β1n1)={112φ(n)+14(1)Ω(n)+12ω(n),if   RPk={7,11},{7},112φ(n),if  RPk={5,11},{5},112φ(n)+14(1)Ω(n)+12ω(n),if   RPk={11},112φ(n),otherwise. (4.9)

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

    Case 2. α=1.

    (A) If β=0, i.e., n=2n1, then by (1.1), Case 1 and Lemma 2.4, we have

    φ12(n)=d|n1μ(2n1d)[d12]+d|n1μ(2n12d)[2d12]=φ12(n1)+φ6(n1)={112φ(n)+14(1)Ω(n)2ω(n)1,if   RPk={7,11},{7},112φ(n)+112(1)Ω(n)+12ω(n),if  RPk={5,11},{5},112φ(n)+112(1)Ω(n)2ω(n)1,if   RPk={11},112φ(n),otherwise. (4.10)

    (B) If β=1, i.e., n=6n1, then from (1.1) we can get

    φ12(n)=d|n1μ(6n1d)[d12]+d|n1μ(6n12d)[2d12]+d|n1μ(6n13d)[3d12]+d|n1μ(6n16d)[6d12]=φ12(n1)φ6(n1)φ4(n1)+φ2(n1).

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

    φ12(n)={112φ(n),if  RPk={7,11},{7},112φ(n)+112(1)Ω(n)+12ω(n)1,if   RPk={5,11},{5},112φ(n)+112(1)Ω(n)+12ω(n)1,if   RPk={11},112φ(n),otherwise. (4.11)

    (C) If β2, then by (1.1) one can easily see that

    φ12(n)=d|n1μ(23βn1d)[d12]+d|n1μ(23βn12d)[2d12]+d|3β1n1μ(23βn13d)[3d12]+d|3β1n1μ(23βn16d)[6d12]=φ4(3β1n1)+φ2(3β1n1).

    Now by Lemma 2.4 we can get the following:

    φ12(n)={112φ(n)+14(1)Ω(n)+12ω(n)1,if  RPk={7,11},{7},112φ(n),if  RPk={5,11},{5},112φ(n)+14(1)Ω(n)+12ω(n)1,if  RPk={11},112φ(n),otherwise. (4.12)

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

    Case 3. α=2.

    (A) If β=0, i.e., n=4n1, then from Lemmas 2.3 and 2.5, we can obtain

    φ12(n)=d|4n1μ(4n1d)[d12]=d|n1μ(4n1d)[d12]+d|n1μ(4n12d)[2d12]+d|n1μ(4n14d)[4d12]=φ6(n1)+φ3(n1)={112φ(n),if  RPk={7,11},{7},112φ(n)+112(1)Ω(n)+12ω(n),if  RPk={5,11},{5},112φ(n)+112(1)Ω(n)+12ω(n),if  RPk={11},112φ(n),otherwise. (4.13)

    (B) If β=1, i.e., n=12n1, then by the definition we have

    φ12(n)=d|12n1μ(12n1d)[d12]=d|n1μ(12n1d)[d12]+d|n1μ(12n12d)[2d12]+d|n1μ(12n14d)[4d12]+d|n1μ(12n13d)[3d12]+d|n1μ(12n16d)[6d12]+d|n1μ(12n112d)[12d12]=φ6(n1)φ3(n1)φ2(n1)+φ(n1).

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

    φ12(n)={112φ(n),if   RPk={7,11},{7},112φ(n)+112(1)Ω(n)+12ω(n)1,if  RPk={5,11},{5},112φ(n)+112(1)Ω(n)+12ω(n)1,if   RPk={11},112φ(n),otherwise. (4.14)

    (C) If β2, then from n=43βn1 and the definition, we know that

    φ12(n)=d|43βn1μ(43βn1d)[d12]=d|23β1n1μ(43βn16d)[6d12]=d|3β1n1μ(23βn12d)[2d2]+d|3β1n1μ(23βn1d)[d2]=φ(3β1n1)φ2(3β1n1)=12φ(3β1n1)=112φ(43βn1)=112φ(n). (4.15)

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

    Case 4. α3.

    (A) If β=0, i.e., n=2αn1, then by Lemma 2.5 we have

    φ12(n)=d|n1μ(2αn1d)[d12]+d|2α1n1μ(2αn12d)[2d12]=φ6(2α1n1)={112φ(n),if   RPk={7,11},{7},112φ(n)+16(1)Ω(n)2ω(n),if  RPk={5,11},{5},112φ(n)+16(1)Ω(n)2ω(n),if   RPk={11},112φ(n),otherwise. (4.16)

    (B) If β=1, i.e., n=32αn1, then by the definition we have

    φ12(n)=d|n1μ(32αn1d)[d12]+d|2α1n1μ(32αn12d)[2d12]+d|n1μ(32αn13d)[3d112]+d|2α1n1μ(32αn16d)[6d12]=φ6(2α1n1)+φ2(2α1n1)={112φ(n),if  RPk={7,11},{7},112φ(n)+112(1)Ω(n)2ω(n),if  RPk={5,11},{5},112φ(n)+112(1)Ω(n)2ω(n),if  RPk={11},112φ(n),otherwise. (4.17)

    (C) If β2, then by Lemma 2.6 we can get

    φ12(n)=112φ(2α3βn1)=112φ(n). (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 φ8(n) and φ12(n), respectively.

    Theorem 5.1. If n is a positive integer, then φ8(n) is odd if and only if n=8,16 or n is given by Table 1. In the above table, p,p1,p2 are odd primes with p1p2, and α,α1,α2 are positive integers.

    Table 1.  the conditions of φ8(n) is odd.
    n Conditions
    pα p9,15(mod16); p3,5(mod16), 2|α; p11,13(mod16), 2α;
    2pα p7,9(mod16); p3,13(mod16), 2|α; p5,11(mod16), 2α;
    4pα p3,5(mod8);
    8pα p3,7(mod8);
    pα11pα22 p1p23(mod8); p1p25(mod8); p13(mod8),p25(mod8);
    2pα11pα22 p1p23(mod8); p1p25(mod8); p13(mod8),p25(mod8).

     | Show Table
    DownLoad: CSV

    Proof. For n=2α, by (3.1) we know that φ8(n) is odd if and only if n=8,16.

    Now suppose that n=2αki=1pαii, where α0, α1,,αk are positive integers, and p1,,pk are distinct odd primes. Set n1=ki=1pαii; then, n1>1 is odd. By Theorem 3.1, we have the following four cases.

    Case 1. RPk={5,7} or {5}.

    (A) If α=0, i.e., n=n1 is odd, then, by (3.2) we have that φ8(n)=18φ(n)+14(1)Ω(n)2ω(n). Note that there exists a prime factor p of n such that p5(mod8); thus, we must have that ω(n)2 if φ8(n) is odd. For ω(n)=2, i.e., n=pα11pα22, by (3.2) we have

    φ8(n)=18pα111(p11)pα212(p21)+(1)α1+α2.

    Therefore φ8(n) is odd if and only if p1p25(mod8), which is true. Now for ω(n)=1, i.e., n=pα11 with p15(mod8), similarly, by (3.2) we have

    φ8(n)=18pα111(p11)+12(1)α1=18(pα111(p11)+4(1)α1).

    From p15(mod8), we have that p15,13(mod16). If p15(mod16), then

    pα111(p11)+4(1)α145α11+4(1)α1(mod16).

    Thus, φ8(n) is odd if and only if 2α1. If p113(mod16), then

    pα111(p11)+4(1)α112(3)α11+4(1)α1(mod16).

    Thus, φ8(n) is odd if and only if α1 is odd.

    (B) If α=1, i.e., ω(n)2, by (3.2) we have that φ8(n)=18φ(n)+14(1)Ω(n)2ω(n)1. Then we must have that ω(n)3 if φ8(n) is odd. For ω(n)=3, namely, n=2pα11pα22, using the same method as (A), φ8(n) is odd if and only if p1p25(mod8). Now for ω(n)=2, i.e., n=2pα11 with p15(mod8), similar to (A), φ8(n) is odd if and only if p15(mod16) and α1 is odd, or if p113(mod16) and 2α1.

    (C) If α=2, i.e., ω(n)2, then by (3.2), we have that φ8(n)=18φ(n)=14ki=1pαi1i(pi1). Thus from the assumption that pi5,7(mod8) or pi5(mod8), we know that ω(n)=2 if φ8(n) is odd. In this case, n=4pα11 with p15(mod8); then, pα111(p11)4(mod8), namely, φ8(n) is odd.

    (D) If α3, then by (3.2), φ8(n)=18φ(n)=2α4ki=1pαi1i(pi1). Thus we must have that α=3 and k=1 if φ8(n) is odd, namely, n=8pα11 with p15(mod8). In this case, φ8(n)=18φ(n)=12pα111(p11) is always even.

    Case 2. RPk={3,7} or {3}.

    (A) If α=0, by (3.2) we have that φ8(n)=18φ(n)+18(1)Ω(n)2ω(n). Thus we must have that ω(n)3 if φ8(n) is odd. For the case that ω(n)=3, i.e, n=pα11pα22pα33, where pi3(mod4)(i=1,2,3), it is easy to see that φ8(n) is always even in this case. Therefore we must have that ω(n)=1,2. Consider that ω(n)=2, i.e., n=pα11pα22. Note that RPk={3,7} or {3}; then, by (3.2), φ8(n)=18(pα111(p11)pα212(p21)+4(1)α1+α2) is odd if and only if p1p23(mod8). Now, for ω(n)=1, i.e., n=pα11 with p13(mod8), then by (3.2) we have that φ8(n)=18(pα111(p11)+2(1)α1). Thus, φ8(n) is odd if and only if p13(mod16) and 2α1, or if p111(mod16) and α1 is odd.

    (B) If α=1, i.e., ω2, by (3.2) we have that φ8(n)=18φ(n)+18(1)Ω(n)12ω(n)1. Thus we must have that ω(n)3 if φ8(n) is odd. Using the same method as (A) in case 1, we can get that φ8(n) is odd if and only if n=2pα11pα22 with p1p23(mod8), or if n=2pα11 with p13(mod16) and 2α1, or if p111(mod16) and α1 is odd.

    (C) If α=2, i.e., ω(n)2, by (3.2) we have that φ8(n)=18φ(n)+18(1)Ω(n)12ω(n). Therefore we must have that ω(n)3 if φ8(n) is odd. For the case that ω(n)=3, i.e., n=4pα11pα22, we know that

    φ8(n)=14pα111(p11)pα212(p21)+(1)α1+α2+1,

    which is always even. Now for the case that ω(n)=2, i.e., n=4pα11 with p13(mod8), by (3.2) we have that φ8(n)=14(pα111(p11)+2(1)α1+1). Since

    pα111(p11)+2(1)α1+123α11+2(1)α1+14(mod8),

    it follows that φ8(n) is odd.

    (D) If α3, by (3.2) we have that φ8(n)=18φ(n)=2α4ki=1pαi1i(pi1). From RPk={3,7} or {3}, we must have that α=3 and k=1 if φ8(n) is odd, namely, n=8pα11 with p13(mod8). Obviously, φ8(n)=12pα111(p11) is odd in this case.

    Case 3. RPk={7}.

    (A) If α=0, by (3.2), φ8(n)=18φ(n)+38(1)Ω(n)2ω(n). Then we must have that ω(n)2 if φ8(n) is odd. For ω(n)=2, i.e., n=pα11pα22, it follows that

    φ8(n)=12(pα111pα212p112p212+3(1)α1+α2).

    Since p1p27(mod8), we have that p112p2121(mod4) and

    pα111pα212p112p212+3(1)α1+α2(1)α1+α22+3(1)α1+α20(mod4),

    which means that φ8(n) is even. Now for ω(n)=1, i.e., n=pα11, by (3.2) we have

    φ8(n)=14(pα111p112+3(1)α1).

    Now from p17(mod8), we have that p17,15(mod16). If p17(mod16), then

    pα111p112+3(1)α13(1)α11+3(1)α10(mod8),

    namely, φ8(n) is even. Thus, p115(mod16), then

    pα111p112+3(1)α17(1)α11+3(1)α14(mod8),

    namely, φ8(n) is odd.

    (B) If α=1, by (3.2), φ8(n)=18φ(n)+18(1)Ω(n)2ω(n)1. Using a similar proof as that for (A) in case 1, φ8(n) is odd if and only if n=2pα11 and p17(mod16).

    (C) If α=2, i.e., ω(n)2, by (3.2), φ8(n)=18φ(n)+18(1)Ω(n)12ω(n). Then we must have that ω(n)3 if φ8(n) is odd. For ω(n)=3, i.e., n=4pα11pα22 with p1p27(mod8), we know that

    φ8(n)=pα111pα212p112p212+(1)α1+α21

    is always even. Now for ω(n)=2, i.e., n=4pα11 with p17(mod8), we can verify that

    φ8(n)=12(pα111p112+(1)α11)

    is also even.

    (D) If α3, by (3.2), φ8(n)=18φ(n)=2α4ki=1pαi1i(pi1). Hence, by RPk={7} we know that φ8(n) is odd if and only if α=3 and k=1, i.e., n=8pα11 with p17(mod8).

    Case 4. {3,5}RPk or 1RPk.

    (A) If {3,5}RPk, i.e., k2, then by (3.2) we have that φ8(n)=18φ(n)=18φ(2α)ki=1pαi1i (pi1). Thus we must have that k=2 and α1 if φ8(n) is odd, namely, n=pα11pα22 or 2pα11pα22, where p13(mod8) and p25(mod8). Obviously, φ8(n)=18pα111(p11)pα212(p21) is always odd in this case.

    (B) If 1RPk, by (3.2), φ8(n)=18φ(n)=18φ(2α)ki=1pαi1i(pi1). Thus, we must have that α1 and k=1 if φ8(n) is odd. Namely, n=pα11,2pα11 with p11(mod8); then, φ8(n)=18pα11(p11). Obviously, φ8(n) is odd if and only if p19(mod16).

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

    Theorem 5.2. If n is a positive integer, then φ12(n) is odd if and only if n=2α(α4), 32α(α2), 23β(β2), 43β(β2), or if it satisfies the conditions given in Table 2. Here, p>3 is an odd prime and α1.

    Table 2.  the conditions of φ12(n) is odd.
    n Conditions
    pα p13,17,19,23(mod24);
    2pα p7,11,13,17(mod24);
    3pα p5,7(mod12);
    4pα p5,7(mod12);
    6pα p5,7(mod12);
    12pα p5,11(mod12).

     | Show Table
    DownLoad: CSV

    Proof. Obviously, by the definition of φ12(n) we can get that φ12(n)=0 for n<12 and φ12(n)=1 for n=12,24; then, we consider that n>12 and n24. First, we consider the case that n=2α3β.

    If α=0, we have that β3; then, by (4.1), φ12(n)=12(3β2(1)β) is even.

    If α=1, we have that β2; then, by (4.1), φ12(n)=12(3β2(1)β+1) is odd.

    If α=2, we have that β2; then, by (4.1), φ12(n)=3β2 is odd.

    If α=3, we have that β2; then, by (4.1), φ12(83β)=23β2 is even.

    If α4, then that β0. For β=0, by (4.1), φ12(n)=13(2α3+(1)α) is odd. For β=1, by (4.1), φ12(n)=13(2α2+(1)α+1) is odd. For β2, by (4.1), φ12(2α3β)=2α23β2, which is always even.

    Next, we consider the case that n=2α3βn1, where α0, β0, n1>1 and gcd(n1,6)=1. For convenience, we set n=2α3βki=1pαii, where αi1, pi is an odd prime and pi>3(1ik). By Theorem 4.2 we have the following four cases.

    Case 1. RPk={7,11} or {7}.

    (A) α=0. If β=0, i.e., ω(n)1, from (4.2) we have that φ12(n)=112φ(n)+14(1)Ω(n)2ω(n). Thus, from the assumption that RPk={7,11} or {7}, we must have that ω(n)2 if φ12(n) is odd. For ω(n)=2, i.e., n=pα11pα22, note that RPk={7,11} or {7}; then,

    φ12(n)=13pα111pα212p112p212+(1)α1+α2

    is always even in this case. Thus, ω(n)=1, i.e., n=pα11; then, by (4.5),

    φ12(n)=112(pα111(p11)+6(1)α1).

    Note that p17(mod12), i.e., p17,19(mod24). If p17(mod24), then pα111(p1)+6(1)α10(mod24), which means that φ12(n) is even. Thus, p119(mod24); then, pα111(p11)+6(1)α112(mod24), namely, φ12(n) is odd.

    If β=1, i.e., ω(n)2, by (4.2), φ12(n)=112φ(n)=16ki=1pαi1i(pi1). Similarly, we must have that ω(n)=2 if φ12(n) is odd. In this case, n=3pα1 with p17(mod12); then, φ12(n)=16pα11(p11), easy to see that φ12(n) is always even. If β2, i.e., ω(n)2, by (4.2), φ12(n)=112φ(n)+14(1)Ω(n)+12ω(n). Similarly, we must have that ω(n)=2, i.e., n=3βpα11 if φ12(n) is odd. Since p17(mod12), it follows that φ12(n)=3β2pα111p112+(1)β+α1+1 is always even.

    (B) α=1. If β=0, i.e., ω(n)2, by (4.10), φ12(n)=112φ(n)+14(1)Ω(n)2ω(n)1. Similarly, we must have that ω(n)3 if φ12(n) is odd. For ω(n)=3, i.e., n=2pα11pα22, note that RPk={7,11} or {7}; then, it is easy to see that φ12(n) is always even. Thus, ω(n)=2, i.e., n=2pα11; note that p17(mod12), namely, p17,19(mod24). In this case, φ12(n) is odd if and only if p17(mod24).

    If β=1, i.e., ω(n)3, by (4.2), φ12(n)=112φ(n)=16ki=1pαi1i(pi1). Similarly, from RPk={7,11} or {7}, we can get that φ12(n) is odd if and only if ω(n)=3, i.e., n=6pα1 with p17(mod12).

    If β2, i.e., ω(n)3, by(4.2), φ12(n)=112φ(n)+14(1)Ω(n)+12ω(n)1. Then we must have that ω(n)=3 if φ12(n) is odd, namely, n=23βpα11 with p17(mod12). Obviously, φ12(n)=3β2pα1p12+(1)2+β+α is always even in this case.

    (C) α=2. If β=0, i.e., ω(n)2, by (4.2), φ12(n)=112φ(n)=16ki=1pαi1i(pi1). Thus, we must have that ω(n)=2 if φ12(n) is odd. In this case, n=4pα11 with p17(mod12); then, pα11(p11)6(mod12), which means that φ12(n) is always odd in this case.

    If β1, i.e., ω(n)3, by (4.2), φ12(n)=112φ(n)=3β2ki=1pαi1i(pi1). Note that RPk={7,11} or {7}; then, φ12(n) is always even.

    (D) α3. By (4.2) and RPk={7,11} or {7}, φ12(n)=112φ(n)=132α3φ(3βn1) is always even in this case.

    Case 2. RPk={5,11} or {5}.

    (A) α=0. If β=0, i.e., ω(n)1, from (4.6), we can get that φ12(n)=112φ(n)+16(1)Ω(n)2ω(n). Thus, we must have that ω(n)=1 if φ12(n) is odd, namely, n=pα11 with p15(mod12). Hence

    φ12(pα11)=112pα111(p11)+13(1)α1=13(pα111p114+(1)α1).

    Note that p15(mod12), i.e., p15,17(mod24). If p15(mod24), then pα11p114+(1)α10(mod6), which means that φ12(pα1) is always even. Thus, p117(mod24), in this case pα111p114+(1)α13(mod6), namely, φ12(pα1) is odd.

    If β=1, i.e., ω(n)2, from (4.8) we have that φ12(n)=112φ(n)+16(1)Ω(n)2ω(n)1. Thus, we must have that ω(n)=2 if φ12(n) is odd, namely, n=3pα11 with p15(mod12); in this case

    φ12(3pα11)=13(2pα111p114+(1)α1)

    is always odd.

    If β2, i.e., ω(n)2, from (4.9) we can get that φ12(n)=112φ(n). We must have that ω(n)=2, i.e., n=3βpα(β2), if φ12(n) is odd. From the assumption p15(mod12), φ12(3βpα11)=112φ(3βpα11)=23β1pα111p114 is always even.

    (B) α=1, i.e., ω(n)2. By (4.10)–(4.12), we must have ω(n)3 if φ12(n) is odd. Namely, n=2pα11,2pα11pα22,6pα11, or 23βpα11(β2). Similar to the proof of (A) in case 1, φ12(n) is odd if and only if n=2pα11 with p117(mod24), or if n=6pα11 with p15(mod12).

    (C) α=2. If β=0, i.e., ω(n)2, by (4.13), φ12(n)=112φ(n)+112(1)Ω(n)+12ω(n); then, we must have that ω(n)=2 if φ12(n) is odd. In this case, n=4pα11 with p15(mod12). Hence, φ12(n)=16pα111(p11)+13(1)α1+3=13(pα111p112+(1)α1+3) is always odd.

    If β=1, i.e., ω(n)3, by (4.14), φ12(n)=112φ(n)+112(1)Ω(n)+12ω(n)1; we must have that ω(n)=3 if φ12(n) is odd. In this case, n=12pα11 with p15(mod12); then, φ12(n)=13pα111(p11)+13(1)α1+4=13(pα111(p11)+(1)α1+4) is odd.

    If β2, i.e., ω(n)3, by (4.15), φ12(n)=112φ(n); we must have that ω(n)=3 if φ12(n) is odd. Namely, n=43βpα11 with p15(mod12); then,

    φ12(n)=112φ(n)=3β2pα111(p11)

    is always even.

    (D) α3, i.e., ω(n)2. If β=0, then by (4.16) and RPk={5,11} or {5}, we konw that φ12(n)=112φ(n)+16(1)Ω(n)2ω(n) is always even in this case.

    If β=1, by (4.17) and RPk={5,11} or {5}, we know that φ12(n)=112φ(n)+112(1)Ω(n)2ω(n) is always even in this case.

    If β2, by (4.18) and RPk={5,11} or {5}, φ12(n)=112φ(n) is always even in this case.

    Case 3. RPk={11}.

    (A) α=0, i.e., ω(n)1. From (4.7)–(4.9), we must have that ω(n)2 if φ12(n) is odd. Consider that ω(n)=2, i.e., n=pα11pα22, or 3βpα11(β1) with p1p211(mod12). Thus, by (4.7)–(4.9), φ12(n) is always even. Hence, ω(n)=1, i.e., n=pα11 with p11(mod12); then, (4.7) we can get

    φ12(pα11)=112pα111(p11)+56(1)α1=16(pα111p112+5(1)α1).

    Note that p111(mod12), i.e., p111,23(mod24). If p111(mod24), then

    pα111p112+5(1)α15(1)α11+5(1)α10(mod12),

    namely, φ12(n) is even. If p123(mod24), then

    pα111p112+5(1)α111(1)α11+5(1)α16(mod12),

    namely, φ12(n) is odd.

    (B) α=1, i.e., ω(n)2. From (4.10)–(4.12), we must have that ω(n)3 if φ12(n) is odd. Namely, n=2pα11, 2pα11pα22, 6pα11, or 23βpα11(β2) with p1p211(mod12). Using the same method as for (A) in case 1, φ12(n) is odd if and only if n=2pα11 with p111(mod24).

    (C) α=2, i.e., ω(n)2. If β=0, by (4.13), we must have that ω(n)=2 if φ12(n) is odd, namely, n=4pα11 with p111(mod12). Then by (4.13),

    φ12(4pα11)=13(pα111p112+(1)α1+3)

    is always even.

    If β1, i.e., ω(n)3, by (4.14)–(4.15), we must have that ω(n)=3 if φ12(n) is odd. Namely, n=43βpα11(β1) with p111(mod12). If β2, then by (4.15),

    φ12(43βpα11)=112φ(43βpα11)=3β2pα111(p11)

    is always even. Thus, β=1; by (4.14), φ12(12pα11)=13(pα111(p11)+(1)α1+3) is odd.

    (D) α3. If β=0, i.e., ω(n)2, then by (4.16) and RPk={11}, we know that φ12(n)=112φ(n)+16(1)Ω(n)2ω(n) is always even in this case.

    If β=1, i.e., ω(n)3, then by (4.17) and RPk={11}, we know that φ12(n)=112φ(n)+112(1)Ω(n)2ω(n) is always even in this case.

    If β2, i.e., ω(n)3, then by (4.18) and RPk={11}, we know that φ12(n)=112φ(n)=2α23β1ki=1pαi1i(pi1) is always even.

    Case 4. {5,7}RPk or 1RPk.

    (A) If {5,7}RPk, by (4.2) we have that φ12(n)=112φ(n) is always even.

    (B) If 1RPk, then by (4.2), φ12(n)=112φ(n); thus, we must have that k=1, α1, and β=0 if φ12(n) is odd. Namely, n=pα11 or 2pα11 with p11(mod12). In this case, φ12(n)=112pα111(p11) is odd if and only if p113(mod24).

    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 φe(n) for e=3,4,6. The key point is that the derivation of [ne] 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 φ8(n) and φ12(n) have been given and the parity has been determined. Therefore, the obvious expression for [ne] depends on the Jacobi symbol, seems to be the key to finding the exact formulae for φe(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 uQ, a1,a2,a3, bi(1jr)Z, and qj(1jr)P, such that

    [de]=u(a1d+a2+a3(1d)+rj=1bj(εjqjd))(2d), (6.1)

    or

    [de]=u(a1d+a2+rj=1bj(εjdqj))(2|d), (6.2)

    where r1 and εj{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 φe(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.
    [3] T. X. Cai, H. Zhong, S. Chern, A congruence involving the quotients of Euler and its applications (III), Acta Math. Sinica. Chin. Ser., 62 (2019), 529–540.
    [4] E. Lehmer, On congruences involving Bernoulli numbers and the quotients of Fermat and Wilson, Ann. Math., 39 (1938), 350–360. https://doi.org/10.2307/1968791 doi: 10.2307/1968791
    [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
    [8] Z. Y. Shen, T. X. Cai, M. J. Hu, On the parity of the generalized Euler function (II), Adv. Math., 45 (2016), 509–519.
    [9] R. Wang, Q. Y. Liao, On the generalized Euler function φ5(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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