# An Identity in Commutative Rings with Unity with Applications to Various Sums of Powers

Authors: Miomir Andjić; Romeo Meštrović
Journal: Discrete Dynamics in Nature and Society
 (2017)

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## Abstract

LetR=(R,+,·) be a commutative ring of characteristic m>0 (m may be equal to +∞) with unity e and zero 0. Given a positive integer n<m and the so-called n-symmetric set A=a1,a2,…,a2l-1,a2l such that al+i=ne-ai for each i=1,…,l, define the rth power sum Sr(A) as Sr(A)=∑i=12lair, for r=0,1,2,…. We prove that for each positive integer k there holds ∑i=02k-1(-1)i2k-1i22k-1-iniS2k-1-i(A)=0. As an application, we obtain two new Pascal-like identities for the sums of powers of the first n-1 positive integers.


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## Body

## 1. Introduction and Statements Results

### 1.1. Recurrence Formulas forn-Symmetric Sets of Commutative Rings with Unity

Here, as always in the sequel,R=(R,+,·) (briefly R) will denote acommutative ring ofcharacteristicm>0 (m may be equal to +∞) with unity e and zero 0. Throughout this paper N, Z, and Zn (n=2,3,…) will, respectively, denote the set of positive integers, the ring of integers, and the ring of residues modulo n.Definition 1.
Given a positive integern such that n<m, we say that a subset A of R is n-symmetric if it is satisfied: a∈R belongs to A if and only if ne-a also belongs to A.
Similarly,(a1,a2,…,an)⊂Rn is called n-symmetric tuple if it satisfies ne-ai=an+1-i for all i=1,2,…,n.It is easy to see that if2e is not invertible in R, then a finite subset A of R is n-symmetric if and only if (1)A=a1,…,al,ne-a1,…,ne-al,where {a1,…,al} is a finite subset of R such that ne-ai≠aj for all i and j with 1≤i<j≤n. Clearly, in this case B=(a1,…,al,ne-al,…,ne-a1) is a n-symmetric 2l-tuple of R.If2e is invertible element in R with the inverse (2e)-1, then a finite n-symmetric set A⊆R is of the above form or of the form (2)A=a1,…,al,ne·2e-1,ne-a1,…,ne-al,where {a1,…,al} is a finite subset of R such that ne-ai≠aj for all 1≤i<j≤n and ai≠(ne)·(2e)-1 for all 1≤i≤n (in particular, {(ne)·(2e)-1} is a finite n-symmetric set). Clearly, in this case, B=(a1,…,al,(ne)·(2e)-1,ne-al,…,ne-a1) is a n-symmetric (2l+1)-tuple of R.Our main result is as follows.Theorem 2.
LetR=(R,+,·) be a commutative ring of characteristic m>0 with unity e and zero 0. For a nonnegative integer r and a n-symmetric set A={a1,…,as}⊆R with 0<n<m, define the rth power sum Sr(A) as (3)SrA=∑i=1sair.Then for each positive integer k it holds (4)∑i=02k-1-1i2k-1i22k-1-iniS2k-1-iA=0.Using the obvious facts that(ie)r=ire, {e,2e,…,(n-1)e} is n-symmetric set of ring R and {0,e,2e,…,(n-1)e} is n-1-symmetric set of R (with S0(n)=ne) for each 1<n<m; Theorem 2 immediately yields the following result.Corollary 3.
LetR=(R,+,·) be a commutative ring of characteristic m>0 with unity e and zero 0. For positive integers r and n>1 and a set A={e,2e,…,(n-1)e}⊆R let (5)Srn=∑i=1n-1ire(we also define S0(n)=(n-1)e). Then for each positive integer k it holds (6)∑i=02k-1-1i2k-1i22k-1-iniS2k-1-in=0,∑i=02k-2-1i2k-1i22k-1-in-1iS2k-1-in=nn-12k-1e.


### 1.2. The Application of Theorem2 for the Sums of Powers on a Finite Arithmetic Progression in Z

Using the obvious fact that for1<n<m and a,d∈Z with d≠0, {a,a+d,…,a+(n-1)d} is a (2a+(n-1)d)-symmetric set of the ring Z of integers, as a consequence of Theorem 2 we immediately obtain the following recurrence formula for the sums of powers on a finite arithmetic progression in Z.Corollary 4.
Leta, d≠0, r≥1, and n≥2 be integers, and let (7)Srn;a,d=∑j=0n-1a+jdr(we also define S0(n;a,d)=n). Then for each positive integer k it holds that(8)∑i=02k-1-1i2k-1i22k-1-i2a+n-1diS2k-1-in;a,d=0.Ifa and d≠0 are arbitrary real numbers, then, expanding via binomial formula every term (a+jd)r (j=0,…,n-1) of the power sum Sr(n;a,d)=∑j=0n-1(a+jd)r (with a fixed r∈{1,2,…}), it follows by Bernoulli’s formula given in Remark 8 that Sr(n;a,d) can be expressed as a polynomial in variables n,a, and d. Hence, this is also true for the sum on the left hand side of (8) which vanishes for all integers a and d≠0. This yields the following extension of Corollary 4.Corollary 5.
For complex numbersa and d≠0, and positive integers r≥1 and n≥2 set (9)Srn;a,d=∑j=0n-1a+jdr.Then formula (8) of Corollary 4 is satisfied for each positive integer k.Remark 6.
A calculation of sumsSr(n;a,d) can be traced back to Faulhaber [1] in 1631 and Bernoulli [2] in 1713. The recurrence formula (8) is in fact a summation formula which expresses S2k-1(n;a,d) as a sum involving power sums S2k-1-i(n;a,d)(i=1,2,…,2k-1) with the corresponding “coefficients” (-1)i-12k-1in/2i.In particular, since the set{1,2,…,n-1} is n-symmetric (the case when a=d=1 with n-1 instead of n in Corollary 4) and the set {0,1,2,…,n-1} is n-1-symmetric (the case a=0, d=1 in Corollary 4), the formula (8) immediately yields the following two Pascal-like identities for the sums of powers of the first n-1 positive integers (cf. (6) with R=Z and e=1).Corollary 7.
For nonnegative integersr and n≥2 define (10)Srn≔Srn;0,1=∑j=1n-1jr.Then for each positive integer k it holds that(11)∑i=02k-1-1i2k-1i22k-1-iniS2k-1-in=0,(12)∑i=02k-2-1i2k-1i22k-1-in-1iS2k-1-in=nn-12k-1.Remark 8.
Finding formulas for sums of powersSk(n) has interested mathematicians for more than 300 years since the early 17th-century mathematical publications of Faulhaber (1580–1635) [1]. Bernoulli (1654–1705) had given a comprehensive account of these sums in his famous workArs Conjectandi [2], published posthumously in 1713. The second section ofArs Conjectandi (also see [3, pp. 269-270]) contains the fundamental Bernoulli’s formula which expresses the sum Sr(n)=∑i=1n-1ir(r=0,1,2,…) as a (r+1)th-degree polynomial function on n whose coefficients involve Bernoulli numbers. Namely, the celebratedBernoulli’s formula (sometimes calledFaulhaber’s formula) gives the sum Sk(n)(k≥1) explicitly given as a (k+1)th-degree polynomial of n (see, e.g., [3, Section  6.5, pp. 269-270], where it is given an induction proof on k). (13)Skn=1k+1∑i=0kk+1ink+1-iBi,where Bi(i=0,1,2,…) areBernoulli numbers defined by the generating function (see, e.g., [4–6]) (14)∑i=0∞Bixii!=xex-1.It is easy to find the values B0=1, B1=-1/2, B2=1/6, B4=-1/30, and Bi=0 for odd i≥3. Furthermore, (-1)i-1B2i>0 for all i≥1. It is well known that Bn=Bn(0), where Bn(x) is the classical Bernoulli polynomial (see, e.g., [4, 7–10]). These and many other properties can be found, for instance, in [11, Section  5.3. pp. 525–538] or [12]. In particular (see, e.g., [13], where a similar formula was established), the usual recurrence is well known(15)Bk=-1k+1∑i=0k-1k+1iBi,B0=1,k=1,2,….Notice that Bernoulli numbers and polynomials from a more general point of view were studied by many authors (e.g., in [14, 15] the method of generating function was applied to introduce new forms of Bernoulli numbers and polynomials; also see [16]). Finding formulas for Sk(n) has interested mathematicians for more than 300 years since the time of Bernoulli (see, e.g., [17, 18]). Recall that, by the well-knownPascal’s identity proven by Pascal (1623–1662) in 1654 [19], (16)∑i=0kk+1iSin=nk+1-1.Recently, q-analogues of the sums of powers of consecutive integers have been investigated extensively (see, e.g., [20]).
For given positive integerk, the recurrence (11) presents a formula for expressing S2k-1(n) as a sum involving power sums S2k-1-i(n)(i=1,2,…,2k-1) with the corresponding “coefficients” (-1)i-12k-1in/2i. For example, taking k=1,2,3,4 into (11), we, respectively, obtain (17)S1n=n2S0n,(18)S3n=3n2S2n-3n24S1n+n38S0n,(19)S5n=5n2S4n-5n22S3n+5n34S2n-5n416S1n+n532S0n,(20)S7n=7n2S6n-21n24S5n+35n38S4n-35n416S3n+21n532S2n-7n664S1n+n7128S0n,where S0(n)=n-1.
Substituting (17) into (18), we get (21)S3n=3n2S2n-n34S0n.Substituting the above formula and (17) into (19), we obtain (22)S5n=5n2S4n-5n32S2n+n52S0n.Substituting the above two formulas and (17) into (20), we find that (23)S7n=7n2S6n-35n34S4n+21n52S2n-17n78S0n.More generally, for any fixed k≥1, iterating the formula (11) we can express the sum S2k-1(n) as a linear combination of the polynomials n2k-1-2iS2i(n) in n with i=0,1,…,k-1. This is given by the following result.Corollary 9.
Let(akl), k=1,2,…, l=1,2,…,k, be real numbers recursively defined as (24)ak+1,l+1=122k+12k+12l22l-∑j=1k-l2k+12j22k+1-2jak+1-j,l+1with a11=1/2, k=1,2,…, and 1≤l≤k. Then for each k≥1 it holds that(25)S2k-1n=∑i=0k-1ak,i+1n2k-1-2iS2i.Furthermore, for each k≥1 (25) is a unique representation of the power sum S2k-1(n) as a linear combination of the functions n2k-1S0(n),…,nS2k-2(n).Remark 10.
Unfortunately, the analogous formula to (11) with 2k instead of 2k-1 does not exist. Namely, suppose that for some even m≥2 there exist real numbers a1,a2,…,am such that for every n(26)Smn=∑i=1mainiSm-in.Considering the sum Sm(n) as a polynomial in n, then it is well known that Sm(n) is divisible by n for each m≥1, and Sm(n) is divisible by n2 if and only if m is odd (see, e.g., [17]; this also immediately follows by Bernoulli’s formula). However, the first of these facts and the equality (26) show that S2k(n) is divisible by n2 for each k≥1, a contradiction.
Although formula (11) cannot be directly used for recursive determination of the expressions for Sk(n), it can be useful for establishing various congruence involving these sums [21].


### 1.3. The Application of Theorem2 to the Sums φk(n)

TheEuler totient functionφ(n) is defined to be equal to the number of positive integers less than n which are relatively prime to n. Each of these φ(n) integers is called atotative (or “totitive”) of n (see [11, Section  3.4, p. 242], where this notion is attributed to J. J. Sylvester). Let t(n) denote the set of all totatives of n; that is, t(n)={j∈N:1≤j<n,gcd⁡(j,n)=1}. Given any fixed nonnegative integer k, in 1850 A. Thacker (see [11, p. 242]) introduced the function φk(n) defined as (27)φkn=∑t∈tntk,where the summation ranges over all totatives t of n (in addition, we define φk(1)=0 for all k). Notice that φ0(n)=φ(n) and there holds φk(n)=Sk(n) if and only if n=1 or n is a prime number.Using the obvious fact thatt(n) is a n-symmetric set in the ring Z, Theorem 2 immediately yields the following recurrence relation involving the functions φk(n).Corollary 11.
Letk≥1 and n≥2 be positive integers. Then (28)∑i=02k-1-1i2k-1i22k-1-iniφ2k-1-in=0.Remark 12.
The following recurrence relation for the functionsφk(n) was established in 1857 by J. Liouville (cf. [11, p. 243]): (29)∑d∣nndkφkd=Skn+1≔1k+2k+⋯+nkwhich for k=0 reduces toGauss’ formula∑d∣nφ(d)=n. Furthermore, in 1985 Bruckman and Lossers [22] established an explicit Bernoulli’s-like formula for theDirichlet series of φk(n) defined as fk(s)=∑k=1∞φk(n)/ns (there φk(n) is called generalized Euler function).For given integersa, d≠0, k≥1, and n≥2 define the function φk(n;a,d) as (30)φkn;a,d=∑t∈tna+tdk.Notice that φk(n;0,1)=φk(n) with the function φk(n) defined above. From the obvious fact that t∈t(n) if and only if n-t∈t(n), it follows immediately that the set {a+td:t∈t(n)} is a (2a+nd)-symmetric set in the ring Z. Therefore, by Theorem 2 we immediately obtain the following recurrence formula for the functions φk(n;a,d).Corollary 13.
Letk be a positive integer. Then (31)∑i=02k-1-1i2k-1i22k-1-i2a+ndiφ2k-1-in;a,d=0.


## 1.1. Recurrence Formulas forn-Symmetric Sets of Commutative Rings with Unity

Here, as always in the sequel,R=(R,+,·) (briefly R) will denote acommutative ring ofcharacteristicm>0 (m may be equal to +∞) with unity e and zero 0. Throughout this paper N, Z, and Zn (n=2,3,…) will, respectively, denote the set of positive integers, the ring of integers, and the ring of residues modulo n.Definition 1.
Given a positive integern such that n<m, we say that a subset A of R is n-symmetric if it is satisfied: a∈R belongs to A if and only if ne-a also belongs to A.
Similarly,(a1,a2,…,an)⊂Rn is called n-symmetric tuple if it satisfies ne-ai=an+1-i for all i=1,2,…,n.It is easy to see that if2e is not invertible in R, then a finite subset A of R is n-symmetric if and only if (1)A=a1,…,al,ne-a1,…,ne-al,where {a1,…,al} is a finite subset of R such that ne-ai≠aj for all i and j with 1≤i<j≤n. Clearly, in this case B=(a1,…,al,ne-al,…,ne-a1) is a n-symmetric 2l-tuple of R.If2e is invertible element in R with the inverse (2e)-1, then a finite n-symmetric set A⊆R is of the above form or of the form (2)A=a1,…,al,ne·2e-1,ne-a1,…,ne-al,where {a1,…,al} is a finite subset of R such that ne-ai≠aj for all 1≤i<j≤n and ai≠(ne)·(2e)-1 for all 1≤i≤n (in particular, {(ne)·(2e)-1} is a finite n-symmetric set). Clearly, in this case, B=(a1,…,al,(ne)·(2e)-1,ne-al,…,ne-a1) is a n-symmetric (2l+1)-tuple of R.Our main result is as follows.Theorem 2.
LetR=(R,+,·) be a commutative ring of characteristic m>0 with unity e and zero 0. For a nonnegative integer r and a n-symmetric set A={a1,…,as}⊆R with 0<n<m, define the rth power sum Sr(A) as (3)SrA=∑i=1sair.Then for each positive integer k it holds (4)∑i=02k-1-1i2k-1i22k-1-iniS2k-1-iA=0.Using the obvious facts that(ie)r=ire, {e,2e,…,(n-1)e} is n-symmetric set of ring R and {0,e,2e,…,(n-1)e} is n-1-symmetric set of R (with S0(n)=ne) for each 1<n<m; Theorem 2 immediately yields the following result.Corollary 3.
LetR=(R,+,·) be a commutative ring of characteristic m>0 with unity e and zero 0. For positive integers r and n>1 and a set A={e,2e,…,(n-1)e}⊆R let (5)Srn=∑i=1n-1ire(we also define S0(n)=(n-1)e). Then for each positive integer k it holds (6)∑i=02k-1-1i2k-1i22k-1-iniS2k-1-in=0,∑i=02k-2-1i2k-1i22k-1-in-1iS2k-1-in=nn-12k-1e.


## 1.2. The Application of Theorem2 for the Sums of Powers on a Finite Arithmetic Progression in Z

Using the obvious fact that for1<n<m and a,d∈Z with d≠0, {a,a+d,…,a+(n-1)d} is a (2a+(n-1)d)-symmetric set of the ring Z of integers, as a consequence of Theorem 2 we immediately obtain the following recurrence formula for the sums of powers on a finite arithmetic progression in Z.Corollary 4.
Leta, d≠0, r≥1, and n≥2 be integers, and let (7)Srn;a,d=∑j=0n-1a+jdr(we also define S0(n;a,d)=n). Then for each positive integer k it holds that(8)∑i=02k-1-1i2k-1i22k-1-i2a+n-1diS2k-1-in;a,d=0.Ifa and d≠0 are arbitrary real numbers, then, expanding via binomial formula every term (a+jd)r (j=0,…,n-1) of the power sum Sr(n;a,d)=∑j=0n-1(a+jd)r (with a fixed r∈{1,2,…}), it follows by Bernoulli’s formula given in Remark 8 that Sr(n;a,d) can be expressed as a polynomial in variables n,a, and d. Hence, this is also true for the sum on the left hand side of (8) which vanishes for all integers a and d≠0. This yields the following extension of Corollary 4.Corollary 5.
For complex numbersa and d≠0, and positive integers r≥1 and n≥2 set (9)Srn;a,d=∑j=0n-1a+jdr.Then formula (8) of Corollary 4 is satisfied for each positive integer k.Remark 6.
A calculation of sumsSr(n;a,d) can be traced back to Faulhaber [1] in 1631 and Bernoulli [2] in 1713. The recurrence formula (8) is in fact a summation formula which expresses S2k-1(n;a,d) as a sum involving power sums S2k-1-i(n;a,d)(i=1,2,…,2k-1) with the corresponding “coefficients” (-1)i-12k-1in/2i.In particular, since the set{1,2,…,n-1} is n-symmetric (the case when a=d=1 with n-1 instead of n in Corollary 4) and the set {0,1,2,…,n-1} is n-1-symmetric (the case a=0, d=1 in Corollary 4), the formula (8) immediately yields the following two Pascal-like identities for the sums of powers of the first n-1 positive integers (cf. (6) with R=Z and e=1).Corollary 7.
For nonnegative integersr and n≥2 define (10)Srn≔Srn;0,1=∑j=1n-1jr.Then for each positive integer k it holds that(11)∑i=02k-1-1i2k-1i22k-1-iniS2k-1-in=0,(12)∑i=02k-2-1i2k-1i22k-1-in-1iS2k-1-in=nn-12k-1.Remark 8.
Finding formulas for sums of powersSk(n) has interested mathematicians for more than 300 years since the early 17th-century mathematical publications of Faulhaber (1580–1635) [1]. Bernoulli (1654–1705) had given a comprehensive account of these sums in his famous workArs Conjectandi [2], published posthumously in 1713. The second section ofArs Conjectandi (also see [3, pp. 269-270]) contains the fundamental Bernoulli’s formula which expresses the sum Sr(n)=∑i=1n-1ir(r=0,1,2,…) as a (r+1)th-degree polynomial function on n whose coefficients involve Bernoulli numbers. Namely, the celebratedBernoulli’s formula (sometimes calledFaulhaber’s formula) gives the sum Sk(n)(k≥1) explicitly given as a (k+1)th-degree polynomial of n (see, e.g., [3, Section  6.5, pp. 269-270], where it is given an induction proof on k). (13)Skn=1k+1∑i=0kk+1ink+1-iBi,where Bi(i=0,1,2,…) areBernoulli numbers defined by the generating function (see, e.g., [4–6]) (14)∑i=0∞Bixii!=xex-1.It is easy to find the values B0=1, B1=-1/2, B2=1/6, B4=-1/30, and Bi=0 for odd i≥3. Furthermore, (-1)i-1B2i>0 for all i≥1. It is well known that Bn=Bn(0), where Bn(x) is the classical Bernoulli polynomial (see, e.g., [4, 7–10]). These and many other properties can be found, for instance, in [11, Section  5.3. pp. 525–538] or [12]. In particular (see, e.g., [13], where a similar formula was established), the usual recurrence is well known(15)Bk=-1k+1∑i=0k-1k+1iBi,B0=1,k=1,2,….Notice that Bernoulli numbers and polynomials from a more general point of view were studied by many authors (e.g., in [14, 15] the method of generating function was applied to introduce new forms of Bernoulli numbers and polynomials; also see [16]). Finding formulas for Sk(n) has interested mathematicians for more than 300 years since the time of Bernoulli (see, e.g., [17, 18]). Recall that, by the well-knownPascal’s identity proven by Pascal (1623–1662) in 1654 [19], (16)∑i=0kk+1iSin=nk+1-1.Recently, q-analogues of the sums of powers of consecutive integers have been investigated extensively (see, e.g., [20]).
For given positive integerk, the recurrence (11) presents a formula for expressing S2k-1(n) as a sum involving power sums S2k-1-i(n)(i=1,2,…,2k-1) with the corresponding “coefficients” (-1)i-12k-1in/2i. For example, taking k=1,2,3,4 into (11), we, respectively, obtain (17)S1n=n2S0n,(18)S3n=3n2S2n-3n24S1n+n38S0n,(19)S5n=5n2S4n-5n22S3n+5n34S2n-5n416S1n+n532S0n,(20)S7n=7n2S6n-21n24S5n+35n38S4n-35n416S3n+21n532S2n-7n664S1n+n7128S0n,where S0(n)=n-1.
Substituting (17) into (18), we get (21)S3n=3n2S2n-n34S0n.Substituting the above formula and (17) into (19), we obtain (22)S5n=5n2S4n-5n32S2n+n52S0n.Substituting the above two formulas and (17) into (20), we find that (23)S7n=7n2S6n-35n34S4n+21n52S2n-17n78S0n.More generally, for any fixed k≥1, iterating the formula (11) we can express the sum S2k-1(n) as a linear combination of the polynomials n2k-1-2iS2i(n) in n with i=0,1,…,k-1. This is given by the following result.Corollary 9.
Let(akl), k=1,2,…, l=1,2,…,k, be real numbers recursively defined as (24)ak+1,l+1=122k+12k+12l22l-∑j=1k-l2k+12j22k+1-2jak+1-j,l+1with a11=1/2, k=1,2,…, and 1≤l≤k. Then for each k≥1 it holds that(25)S2k-1n=∑i=0k-1ak,i+1n2k-1-2iS2i.Furthermore, for each k≥1 (25) is a unique representation of the power sum S2k-1(n) as a linear combination of the functions n2k-1S0(n),…,nS2k-2(n).Remark 10.
Unfortunately, the analogous formula to (11) with 2k instead of 2k-1 does not exist. Namely, suppose that for some even m≥2 there exist real numbers a1,a2,…,am such that for every n(26)Smn=∑i=1mainiSm-in.Considering the sum Sm(n) as a polynomial in n, then it is well known that Sm(n) is divisible by n for each m≥1, and Sm(n) is divisible by n2 if and only if m is odd (see, e.g., [17]; this also immediately follows by Bernoulli’s formula). However, the first of these facts and the equality (26) show that S2k(n) is divisible by n2 for each k≥1, a contradiction.
Although formula (11) cannot be directly used for recursive determination of the expressions for Sk(n), it can be useful for establishing various congruence involving these sums [21].


## 1.3. The Application of Theorem2 to the Sums φk(n)

TheEuler totient functionφ(n) is defined to be equal to the number of positive integers less than n which are relatively prime to n. Each of these φ(n) integers is called atotative (or “totitive”) of n (see [11, Section  3.4, p. 242], where this notion is attributed to J. J. Sylvester). Let t(n) denote the set of all totatives of n; that is, t(n)={j∈N:1≤j<n,gcd⁡(j,n)=1}. Given any fixed nonnegative integer k, in 1850 A. Thacker (see [11, p. 242]) introduced the function φk(n) defined as (27)φkn=∑t∈tntk,where the summation ranges over all totatives t of n (in addition, we define φk(1)=0 for all k). Notice that φ0(n)=φ(n) and there holds φk(n)=Sk(n) if and only if n=1 or n is a prime number.Using the obvious fact thatt(n) is a n-symmetric set in the ring Z, Theorem 2 immediately yields the following recurrence relation involving the functions φk(n).Corollary 11.
Letk≥1 and n≥2 be positive integers. Then (28)∑i=02k-1-1i2k-1i22k-1-iniφ2k-1-in=0.Remark 12.
The following recurrence relation for the functionsφk(n) was established in 1857 by J. Liouville (cf. [11, p. 243]): (29)∑d∣nndkφkd=Skn+1≔1k+2k+⋯+nkwhich for k=0 reduces toGauss’ formula∑d∣nφ(d)=n. Furthermore, in 1985 Bruckman and Lossers [22] established an explicit Bernoulli’s-like formula for theDirichlet series of φk(n) defined as fk(s)=∑k=1∞φk(n)/ns (there φk(n) is called generalized Euler function).For given integersa, d≠0, k≥1, and n≥2 define the function φk(n;a,d) as (30)φkn;a,d=∑t∈tna+tdk.Notice that φk(n;0,1)=φk(n) with the function φk(n) defined above. From the obvious fact that t∈t(n) if and only if n-t∈t(n), it follows immediately that the set {a+td:t∈t(n)} is a (2a+nd)-symmetric set in the ring Z. Therefore, by Theorem 2 we immediately obtain the following recurrence formula for the functions φk(n;a,d).Corollary 13.
Letk be a positive integer. Then (31)∑i=02k-1-1i2k-1i22k-1-i2a+ndiφ2k-1-in;a,d=0.


## 2. Proofs of Theorem2 and Corollary 9

Proof of Theorem2.
Suppose first that2e is not invertible element in R. Then as noticed above, the set A has the form(32)A=a1,a2,…,as=b1,…,bl,ne-b1,…,ne-bl,where {b1,…,bl} (s=2l) is a finite subset of R such that ne-bi≠bj for all i,j with 1≤i<j≤n.
Since the binomial formula holds in any commutative ring with unity, we have(33)2aj-ne2k-1=∑i=02k-1-1i2k-1i22k-1-iniaj2k-1-ifor all positive integers n and j with 1≤j≤l. After summation in (33) over j=1,…,2l we obtain (34)∑j=1s2aje-ne2k-1=∑j=1s∑i=02k-1-1i2k-1i22k-1-iniaj2k-1-i=∑i=02k-1-1i2k-1i22k-1-ini∑j=1saj2k-1-i=∑i=02k-1-1i2k-1i22k-1-iniS2k-1-iA.On the other hand, from (32) we see that s=2l and we can assume that ai=bi for each i=1,…,l and ai=ne-bi for each i=l+1,…,2l. Using this and observing that -(2bje-ne)=2(ne-bje)-ne for each j=1,…,l, we find that(35)∑j=1s2aje-ne2k-1=∑j=1l2bj-ne2k-1+∑j=l+12l2ne-bj-ne2k-1=∑j=1l2bje-ne2k-1+2ne-bj-ne2k-1=∑j=1l2bj-ne2k-1+-12k-12bj-ne2k-1=0.Comparing (34) and (35) immediately gives the desired identity (4).
Similarly, if2e is invertible element in R, then as noticed in the previous section, the set A may be of the form (32) or of the form (36)A=a1,…,al,ne·2e-1,ne-a1,…,ne-al.Then since the central element (ne)·(2e)-1 of A satisfies the equality 2(ne)·(2e)-1-ne=0, in the same manner as in the first case, we arrive to (32). This completes the proof.Proof of Corollary9 is based on Corollary 7 and the following lemma.Lemma 14.
LetS~k(x), k=0,1,2,… be a sequence of polynomials of the real variable x defined as S~0(x)=x-1 and (37)S~kx=1k+1∑i=0kk+1ixk+1-iBi,k=1,2,…,where Bi(i=0,1,2,…) are Bernoulli numbers. Then for each nonnegative integer k the polynomials x2k+1S~0(x),x2k-1S~2(x),…,x3S~2k-2(x),xS~2k(x) are linearly independent over the field of real numbers R.Proof.
The proof easily follows by induction onk≥0. The base of induction (k=0) is satisfied because S~0(x)≢0. Suppose that the polynomials x2k+1S~0(x),x2k-1S~2(x),…,x3S~2k-2(x),xS~2k(x) are linearly independent for some k≥0. Now consider the k+1 polynomials (38)x2k+3S~0x,x2k+1S~2x,…,x3S~2kx,xS~2k+2xand assume that (39)∑i=0k+1aix2k+3-2iS~2ix≡0,where a0,a1,…,ak+1 are real numbers. Note that the equality (39) can be written as (40)ak+1xS~2k+2x+x2∑i=0kaix2k+1-2iS~2ix≡0.From (37) we see that the linear term of the polynomial S~2k+2(x) is B2k+2x. This together with (40) yields that the linear term of the polynomial on the left hand side of (40) is ak+1B2k+2x. Accordingly, in view of the fact that B2k+2≠0 for each k≥0, the identical equality (40) yields ak+1=0, which substituting in (40) gives (41)∑i=0kaix2k+1-2iS~2ix≡0.It follows from (41) that a0=a1=⋯=ak=0 because the polynomials x2k+1S~0(x),x2k-1S~2(x),…,xS~2k(x) are linearly independent by induction hypothesis. Therefore, (39) yields a0=a1=⋯=ak=ak+1=0. This completes the induction proof.Proof of Corollary9.
By using the identity (11) of Corollary 7 with 2k+1 instead of 2k-1, we find that(42)22k+1S2k+1n=∑i=12k+1-1i-12k+1i22k+1-iniS2k+1-in=∑iodd·+∑ieven·=∑j=0k2k+12j+122k-2jn2j+1S2k-2jn-∑j=1k2k+12j22k+1-2jn2jS2k+1-2jn.As noticed above, iterating the formula (11), for any k≥1 we can express the power sum S2k-1(n) as a linear combination of the polynomials n2k-1-2iS2i(n) of n with i=0,1,…,k-1. This means that for each k≥1 there exist real numbers bk1,bk2,…,bkk such that (43)S2k-1n=∑i=0k-1bk,i+1n2k-1-2iS2i.Replacing k by k-j+1 into formula (43) yields (44)S2k+1-2jn=∑i=0k-jbk-j+1,i+1n2k+1-2j-2iS2i.Substituting (44) into (42) we obtain(45)22k+1S2k+1n=∑j=0k2k+12j+122k-2jn2j+1S2k-2jn-∑j=1k2k+12j22k+1-2jn2j∑i=0k-jbk+1-j,i+1n2k+1-2j-2iS2i=∑i=0k2k+12k+1-2i22in2k+1-2iS2in-∑j=1k2k+12j22k+1-2j∑i=0k-jn2k+1-2iS2inbk+1-j,i+1=∑i=0k2k+12i22in2k+1-2iS2in-∑i=0k-1∑j=1k-i2k+12j22k+1-2jn2k+1-2iS2inbk+1-j,i+1.Notice that, for each k≥1 by Bernoulli’s formula (see the first formula of Remark 8), (46)Skn=1k+1∑i=0kk+1ink+1-iBi.From (37) and (46) we see that S~k(n)=Sk(n) for each n=1,2,…, and thus the polynomial identity (45) with respect to the positive integer variable n can be extended to the real variable x instead of n; that is, (45) is identically satisfied (on R) if we replace Sl(n) by S~l(x) (with l=2k+1 and l=2i). Accordingly, in view of Lemma 14, by comparing in (46) the coefficients with n2k+1-2iS2i(n), for i=0,1,…,k, we immediately obtain (47)bk+1,l+1=122k+12k+12l22l-∑j=1k-l2k+12j22k+1-2jbk+1-j,l+1.Finally, since a11=1/2=S1(n)/(nS0(n))=b11, from the recurrence relations (24) and (47) we conclude that akl=bkl for all k=1,2,… and l=1,2,…,k.
Notice also that, by Lemma14, (25) is a unique representation of the power sum S2k-1(n) as a linear combination of the functions n2k-1S0(n),…,nS2k-2(n). This completes proof of Corollary 9.



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*Source: 9092515-2017-02-27.xml*