ResearchArticle OnaSumInvolvingtheSum-of-DivisorsFunction
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Research Article On a Sum Involving the Sum-of-Divisors Function Feng Zhao 1 and Jie Wu 2 1 Department of Mathematics and Statistics, North China University of Water Resources and Electric Power, Jinshui E Road, Zhengzhou 450046, Henan, China 2 CNRS LAMA 8050, Laboratoire D’Analyse et de Math´ ematiques Appliqu´ ees, Universit´ e Paris-Est Cr´ eteil, Cr´ eteil Cedex 94010, France Correspondence should be addressed to Feng Zhao; [email protected] Received 14 February 2021; Accepted 12 March 2021; Published 5 April 2021 Academic Editor: Tianping Zhang Copyright © 2021 Feng Zhao and Jie Wu. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Let σ(n) be the sum of all divisors of n and let [t] be the integral part of t. In this paper, we shall prove that n≤x σ([x/n]) � (π 2 /6)x log x + O(x(log x) (2/3) (log 2 x) (4/3) ) for x ⟶∞, and that the error term of this asymptotic formula is Ω(x). 1. Introduction As usual, denote by φ(n) the Euler function and by [t] the integral part of real t, respectively. Recently, Bordell` es et al. [1] studied the asymptotic behaviour of the quantity S φ (x) ≔ n ≤ x φ x n ః ఄ ఁ ం, (1) for x ⟶∞. By exponential sum technique, they proved that 2629 4009 · 6 π 2 + o(1) x log x ≤ S φ (x) ≤ 2629 4009 · 6 π 2 + 1380 4009 + o(1) x log x, (2) and conjectured that S φ (x) ∼ 6 π 2 x log x, as x ⟶∞. (3) Very recently, Wu [2] improved (2) and Zhai [3] re- solved conjecture (3) by showing S φ (x)� 6 π 2 x log x + Ox(log x) (2/3) log 2 x ( (1/3) , (4) andalsoprovedthattheerrortermin(4)is Ω(x), where log 2 denotes the iterated logarithm. Some related works can be found in [4, 5]. Since the sum-of-divisors function σ (n) ≔ d|n d has similar properties as the Euler function φ(n) in many cases, it seems natural and interesting to consider its analogy of (3). Our result is as follows. Theorem 1 (i) For x ⟶∞, we have S σ (x) ≔ n ≤ x σ x n ః ఄ ఁ ం � π 2 6 x log x + Ox(log x) (2/3) log 2 x ( (4/3) . (5) (ii) Let E(x) betheerrortermin(5).en,for x ⟶∞, we have E(x)� Ω(x), i.e. limsup x⟶∞ |E(x)| x > 0. (6) Let μ(n) be the M¨ obius function and define id(n)� n and 1(n)� 1 for all integers n ≥ 1. en, φ � id ∗ μ and σ � id ∗ 1. In Zhai’s approach proving (4), the inequality Hindawi Journal of Mathematics Volume 2021, Article ID 5574465, 7 pages https://doi.org/10.1155/2021/5574465
Transcript of ResearchArticle OnaSumInvolvingtheSum-of-DivisorsFunction
Research ArticleOn a Sum Involving the Sum-of-Divisors Function
Feng Zhao 1 and Jie Wu2
1Department of Mathematics and Statistics North China University of Water Resources and Electric Power Jinshui E RoadZhengzhou 450046 Henan China2CNRS LAMA 8050 Laboratoire DrsquoAnalyse et de Mathematiques Appliquees Universite Paris-Est CreteilCreteil Cedex 94010 France
Correspondence should be addressed to Feng Zhao zhaofengncwueducn
Received 14 February 2021 Accepted 12 March 2021 Published 5 April 2021
Academic Editor Tianping Zhang
Copyright copy 2021 Feng Zhao and Jie Wu is is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited
Let σ(n) be the sum of all divisors of n and let [t] be the integral part of t In this paper we shall prove that 1113936nlexσ([xn])
(π26)x log x + O(x(log x)(23)(log2 x)(43)) for x⟶infin and that the error term of this asymptotic formula is Ω(x)
1 Introduction
As usual denote by φ(n) the Euler function and by [t] theintegral part of real t respectively Recently Bordelles et al[1] studied the asymptotic behaviour of the quantity
Sφ(x) ≔ 1113944nle x
φx
n1113876 11138771113874 1113875 (1)
for x⟶infin By exponential sum technique they provedthat
26294009
middot6π2 + o(1)1113888 1113889x log xle Sφ(x)
le26294009
middot6π2
+13804009
+ o(1)1113888 1113889x log x
(2)
and conjectured that
Sφ(x) sim6π2 x log x asx⟶infin (3)
Very recently Wu [2] improved (2) and Zhai [3] re-solved conjecture (3) by showing
Sφ(x) 6π2 x log x + O x(log x)
(23) log2 x( 1113857(13)
1113872 1113873 (4)
and also proved that the error term in (4) isΩ(x) where log2denotes the iterated logarithm Some related works can befound in [4 5] Since the sum-of-divisors functionσ(n) ≔ 1113936d|nd has similar properties as the Euler functionφ(n) in many cases it seems natural and interesting toconsider its analogy of (3)
Our result is as follows
Theorem 1(i) For x⟶infin we have
Sσ(x) ≔ 1113944nle x
σx
n1113876 11138771113874 1113875
π2
6x log x
+ O x(log x)(23) log2 x( 1113857
(43)1113872 1113873
(5)
(ii) Let E(x) be the error term in (5) en for x⟶infinwe have
E(x) Ω(x) ie lim supx⟶infin
|E(x)|
xgt 0 (6)
Let μ(n) be the Mobius function and define id(n) n
and 1(n) 1 for all integers nge 1 en φ idlowast μ andσ idlowast 1 In Zhairsquos approach proving (4) the inequality
HindawiJournal of MathematicsVolume 2021 Article ID 5574465 7 pageshttpsdoiorg10115520215574465
1113944nle x
μ(n)≪x exp minus c
log x
1113969
1113882 1113883 (xge 1) (7)
plays a key role where cgt 0 is a positive constant Clearlysuch a bound is not true for 1 By refining Zhairsquos approachwe shall prove our result
2 Preliminary Lemmas
As in [3] we need some bounds on exponential sums of thetype 1113936NlenltNprimee(Tn) where NltNprime le 2N For large values ofN Zhai used the theory of exponent pair and for smallerones the Vinogradov method Both estimates are containedin the following general theorem of Karatsuba [6 eorem1] which will be a key tool for proving eorem 1
Lemma 1 Let kge 2 and M and P be integers P beingpositive Let f isin Ck+1([M M + P]R) Suppose that thereexist positive absolute constants c0 c1 c2 c3 and c4 such thatc0 lt 1 c1 lt 1 and c2 + c4 lt c1 an integer r such thatc0kle rle k and distinct numbers sj ge 2(j 1 r) notexceeding k such that for Mle tleM + P the following in-equalities are satisfied
(i) |f(k+1)(t)(k + 1)|lePminus c1(k+1)(ii) Pminus c2sj le |f(sj)(t)sj|lePminus c3sj (j 1 r)
en for each positive integer P1 not exceeding P we have
1113944MlemleM+P1minus 1
e(f(m))
1113868111386811138681113868111386811138681113868111386811138681113868
1113868111386811138681113868111386811138681113868111386811138681113868leAP
1minus ck2( ) (8)
where e(t) ≔ e2πit and Agt 0 cgt 0 are absolute constants
e next two lemmas are essentially a special case of [7Lemmas 25 and 26] with a 1 e only difference is thatthe ranges of T and N here are slightly larger than those of[7 Lemmas 25 and 26] (TgeN2 in place of TgeN(32) andNlex(23) in place of Nlex(12) respectively) Although theproof is completely similar for the convenience of readerswe still reproduce a proof here
Lemma 2 Let e100 leNltNprime le 2N and TgeN(32) enthere exists an absolute positive constant c5 such that
1113944
Nle nltNprime
eT
n1113874 1113875≪N exp minus
c5log3
N
log2 T1113888 11138891113896 1113897 (9)
where the implied constant is absolute
Proof We apply Lemma 1 to f(t) ≔ (Tt) withM N P N P1 Nprime minus N For this we choose
c0 1100
c1 99100
c2 87100
c3 34
c4 1100
k 100log(Tn)
log N1113890 1113891
(10)
and take the sj to be all integers s such that
4log(Tn)
log Nle sle 5
log(Tn)
log Nmiddot (11)
Obviously the number r of sj is between c0k and k Nextwe shall verify that f(t) satisfies the conditions (i) and (ii) ofLemma 1 with the parameters chosen above
For Nle tle 2N we have
f(k+1)
(t)
(k + 1)
111386811138681113868111386811138681113868111386811138681113868
111386811138681113868111386811138681113868111386811138681113868 Tt
minus kminus 2 leTNminus kminus 2
Nminus η1 (12)
where
η1 ≔ k + 1 minuslog(Tn)
log Nge k + 1 minus
1100
kge99100
(k + 1) c1(k + 1)
(13)
Similarly for Nle tle 2N we find the inequality|f(sj)(t)sj|leNminus η3 where
η3 ≔ sj minuslog(Tn)
log Nge34sj c3sj (14)
For the lower bound of (ii) we have
fsj( 1113857
(t)
sj
11138681113868111386811138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868111386811138681113868 Tt
minus 1minus sj geT(2N)minus 1minus sj N
minus η2 (15)
where
η2 ≔ sj minuslog(Tn)
log N+log 2log N
sj + 11113872 1113873
le45sj +
log 2100
sj + 11113872 1113873le87100
sj c2sj
(16)
From Lemma 1 there exist two positive constants c andA such that
2 Journal of Mathematics
1113944
Nle nltNprime
eT
n1113874 1113875
1113868111386811138681113868111386811138681113868111386811138681113868
1113868111386811138681113868111386811138681113868111386811138681113868leAN
1minus ck2( ) leAN exp minusc5log
3N
log2(Tn)1113896 1113897
(17)
with c5 ≔ 10minus 4c is completes the proof of Lemma 2
Lemma 3 Define ψ(t) ≔ t minus [t] minus (12) Let c5 be theconstant defined by Lemma 2 and c6 ≔ (89)2c5
clowast ≔ ((35)c6)minus (13) en we have
1113944
Nle nltNprime
1nψ
x
n1113874 1113875≪ eminus c6(log N)3(log x)2(log N)
3
(log x)2 (18)
uniformly for xge 10 exp clowast(log x)(23)1113966 1113967leNlex(23) and
NltNprime le 2N
Proof By invoking a classical result on ψ(t) (see 8 page 39])we can write for any Hge 1
1113944
Nle nltNprime
ψx
n1113874 1113875≪NH
minus 1+ 1113944
1le hleH
hminus 1
1113944
Nle nltNprime
ehx
n1113888 1113889
1113868111386811138681113868111386811138681113868111386811138681113868
1113868111386811138681113868111386811138681113868111386811138681113868
(19)
An application of Lemma 2 with T hxgexgeN(32)
yields
1113944
Nle nltNprime
ψx
n1113874 1113875≪N H
minus 1+ e
minus c5(log N)3log2(Hx)log H1113874 1113875
(20)
Taking H exp (log N)3(log x)21113966 1113967le x(827) we easilydeduce that
1113944
Nle nltNprime
ψx
n1113874 1113875≪N e
minus (log N)3(log x)2+ e
minus c6(log N)3(log x)2(log N)3
(log x)21113888 1113889
(21)
e first term can be absorbed by the second since c5 canbe chosen small enough to ensure that c6 lt 1 and sinceexp clowast(log x)(23)
1113966 1113967leN implies (log N)3(log x)2 ge clowastHence
1113944
Nle nltNprime
ψx
n1113874 1113875≪Ne
minus c6(log N)3(log x)2(log N)3
(log x)2 (22)
and an Abel summation produces the required result
Lemma 4 Let 2le z1 lt z2 le x and Fx(t) ≔ (1t)ψ(xt)Denote by VFx
[z1 z2] the total variation of Fx on [z1 z2]en
VFxz1 z21113858 1113859≪
x
z21
+1z1
(23)
where the implied constant is absolute
Proof If z1 t0 lt t1 lt middot middot middot lt tn z2 is a partition of the in-terval [z1 z2] then
1113944
n
k1Fx tk( 1113857 minus Fx tkminus 1( 1113857
11138681113868111386811138681113868111386811138681113868 1113944
n
k1
1tk
ψx
tk
1113888 1113889 minus1
tkminus 1ψ
x
tkminus 11113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868
le 1113944n
k1
1tk
minus1
tkminus 11113888 1113889ψ
x
tk
1113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868+ 1113944
n
k1
1tkminus 1
ψx
tk
1113888 1113889 minus ψx
tkminus 11113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868
(24)
Since |ψ(t)|le 1 for all t we have
1113944n
k1
1tk
minus1
tkminus 11113888 1113889ψ
x
tk
1113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868le1z1
minus1z2lt1z1
middot (25)
On the other hand since ψ(u) is of period 1 we have
1113944
n
k1
1tkminus 1
ψx
tk
1113888 1113889 minus ψx
tkminus 11113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868
le1z1
x
z1+ 11113888 1113889Vψ[0 1]le 2
x
z21
+1z1
1113888 1113889
(26)
Inserting these two bounds into (24) we obtain therequired result
3 Proof of Theorem 1
31 A Formula on the Mean Value of σ(n)
Lemma 5
(i) For xge 2 and 1le zlex(13) we have
1113944nle x
σ(n) π2
12x2
minus x(z minus [z])
2+[z]
2z+ O
x
z1113874 1113875 minus Δ(x z)
(27)
where
Δ(x z) ≔ 1113944dle (xz)
x
dψ
x
d1113874 1113875 (28)
(ii) For x⟶infin we have
1113944nlex
σ(n) π2
12x2
+ O(x log x) (29)
Proof Using σ(n) 1113936dmnm the hyperbole principle ofDirichlet allows us to write
1113944nle x
σ(n) 1113944dmle x
m S1 + S2 minus S3 (30)
where
Journal of Mathematics 3
S1 ≔ 1113944dle(xz)
1113944mle(xd)
m
S2 ≔ 1113944mle z
1113944dle(xm)
m
S3 ≔ 1113944dle(xz)
1113944mlez
m
(31)
Firstly we have
S2 1113944mlez
mx
m1113876 1113877 x[z] + O z
21113872 1113873 (32)
S3 x
z1113876 1113877
[z]([z] + 1)
2 x
[z]([z] + 1)
2z+ O z
21113872 1113873 (33)
Secondly we can write
S1 12
1113944dle(xz)
x
dminus ψ
x
d1113874 1113875 minus
12
1113874 1113875x
dminus ψ
x
d1113874 1113875 +
12
1113874 1113875
12
1113944dle(xz)
x2
d2 minus 2
x
dψ
x
d1113874 1113875 + ψ
x
d1113874 1113875
2minus14
1113888 1113889
π2
12x2
minus12
xz minus Δ(x z) + O(xz)
(34)
where Δ(x z) is as in (28) Inserting (32) (33) and (34) into(30) and using z2 le (xz) we get (27)
Taking z 1 in (27) and noticing that
1113944dlex
1d2
π2
6+ O
1x
1113874 1113875
1113944dlex
x
dψ
x
d1113874 1113875
1113868111386811138681113868111386811138681113868111386811138681113868
1113868111386811138681113868111386811138681113868111386811138681113868≪ x log x
(35)
we obtain the required boundis completes the proof
32 Estimates of Error Terms
Lemma 6 Let N0 ≔ exp (6c6)(log x)(23)(log2 x)(13)1113966 1113967
where c6 is given as in Lemma 3 Let Δ(x z) be defined by(28) en for xge 10 and 2le zle
N0
1113968 we have
1113944N0ltnle
x
radicΔ
x
n z1113874 1113875
11138681113868111386811138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868111386811138681113868+ 1113944
N0ltnlex
radicΔ
x
nminus 1 z1113874 1113875
11138681113868111386811138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868111386811138681113868
≪ x1
(log x)3 +
log x
z1113888 1113889
(36)
Proof Denote by Δ1(x z) and Δ2(x z) two sums on theleft-hand side of (36) respectively By (28) of Lemma 5 wecan write
Δ1(x z) x 1113944N0ltnle
x
radic1113944
dle(x(nz))
1dn
ψx
dn1113874 1113875
x 1113944
dle x N0z( )( )
1d
1113944N0ltnlemin(
x
radicx(dz))
1nψ
x
dn1113874 1113875
xΔdagger1(x z) + xΔ♯1(x z)
(37)
where
Δdagger1(x z) ≔ 1113944
dle x N0z( )( )
1d
1113944
N0ltnle(xd)(23)
1nψ
x
dn1113874 1113875
Δ♯1(x z) ≔ 1113944
dle x N0z( )( )
1d
1113944
(xd)(23)lt nleminx
radic(xdz)
1nψ
x
dn1113874 1113875
(38)
For 0le kle (log((xd)(23)N0))log 2 let Nk ≔ 2kN0and define
Sk(d) ≔ 1113944Nklt nle 2Nk
1nψ
x
dn1113874 1113875 (39)
Noticing that N0 leNk le (xd)(23) we can apply Lemma3 to derive that
Sk(d)≪ eminus ϑ log Nk( )3(log(xd))2( 1113857
(40)
with ϑ(t) ≔ c6t minus log t It is clear that ϑ(t) is increasing on[c6infin) On the other hand for kge 0 and dge 1 we have
log Nk( 11138573(log(xd))
2 ge log N0( 11138573(log x)
2 6c6( 1113857log2 x
(41)
us
ϑlog Nk( 1113857
3
(log(xd))21113888 1113889ge ϑ
6c6
1113888 1113889log2 x1113888 1113889
6 log2 x minus log6c6
1113888 1113889log2 x1113888 1113889ge 5 log2 x
(42)
which implies that Sk(d)≪ (log x)minus 5 Inserting this intothe expression of Δdagger1(x z) we get
Δdagger1(x z)≪ 1113944
dle x N0z( )( )
1d
1113944
2kN0le(xd)(23)
Sk(d)1113868111386811138681113868
1113868111386811138681113868≪ (log x)minus 3
(43)
Next we bound Δ♯1(x z) Let F(t) be a function ofbounded variation on [n n + 1] for each integer n and letVF[n n + 1] be the total variation of F on [n n + 1] Inte-grating by parts we have
4 Journal of Mathematics
1113946n+1
nt minus n minus
12
1113874 1113875dF(t) 12(F(n + 1) + F(n)) minus 1113946
n+1
nF(t)dt
(44)
From this we can derive that
12(F(n + 1) + F(n)) 1113946
n+1
nF(t)dt + O VF[n n + 1]( 1113857
(45)
for nge 1 Summing over n we find that
1113944N1lt nleN2
F(n) 1113946N2
N1
F(t)dt
+12
F N1( 1113857 + F N2( 1113857( 1113857 + O VF N1 N21113858 1113859( 1113857
(46)
We apply this formula to
F(xd)(t) 1tψ
(xd)
t1113888 1113889
N1 (xd)(23)
1113960 1113961
N2 minx
radic
x
(dz)1113896 11138971113890 1113891
(47)
According to Lemma 4 we haveVF(xd)
[N1 N2]≪ (xd)minus (13) and thus by puttingu (xd)t we obtain with the notationxd1 max(
x
radicd tz) and xd2 (xd)(13)
1113944
(xd)(23)ltnleminx
radic(x(dz))
1nψ
x
dn1113874 1113875 1113946
xd2
xd1
ψ(u)
udu + O
x
d1113874 1113875
minus (13)
1113888 1113889
≪ zminus 1
+x
d1113874 1113875
minus (13)
≪ zminus 1
(48)
where we have used the fact that zleN0
1113968and
dle (x(N0z))rArrz le (xd)(13) and the bound
1113946xd2
xd1
ψ(u)
udu 1113946
xd2
xd1
1113946u
xd1
ψ(t)dt1113888 1113889du
u2 minus
1x
(23)d2
1113946xd2
xd1
ψ(t)dt
≪xminus 1d1 + xd21113872 1113873
minus (23)≪ z
minus 1+(xd)
minus (23)≪ zminus 1
(49)
Using (48) a simple partial integration allows us toderive that
Δ♯1(x z)≪ zminus 1
1113944
dlex N0z( )
dminus 1≪ z
minus 1log x(50)
Combining (43) and (50) it follows that
Δ1(x z)1113868111386811138681113868
1113868111386811138681113868≪ x(log x)minus 3
+ xzminus 1log x (51)
Similarly we can prove the same bound for |Δ2(x z)|is completes the proof
33 End of the Proof of eorem 1 Let c6 be the constantgiven as in Lemma 3 and N0 ≔ exp (6c6)1113864 (log x)(23)
(log2 x)(13) Let z isin [2N0
1113968] be a parameter to be chosen
laterPutting d [xn] we have (xn) minus 1lt dle (xn) and
x(d + 1)lt nle (xd) We have with the conventionσ(0) 0
Sσ(x) 1113944dlex
σ(d) 1113944(x(d+1))ltnle(xd)
1
1113944dnle x
σ(d) minus 1113944dnle xdge2
σ(d minus 1)
1113944dnlex
(σ(d) minus σ(d minus 1))
(52)
By the hyperbole principle of Dirichlet we can write
Sσ(x) S1(x σ) + S2(x σ) minus S3(x σ) (53)
where
S1(x σ) ≔ 1113944dle
x
radicdnle x
(σ(d) minus σ(d minus 1))
S2(x σ) ≔ 1113944nle
x
radicdnlex
(σ(d) minus σ(d minus 1))
S3(x σ) ≔ 1113944dle
x
radicnle
x
radic(σ(d) minus σ(d minus 1))
(54)
With the help of the bound σ(n)≪ n log2 n we canderive that
S3(x σ) [x
radic]σ([
x
radic])≪x log2 x (55)
For evaluating S1(x σ) we write
S1(x σ) 1113944dle
x
radic(σ(d) minus σ(d minus 1))
x
d1113876 1113877
x 1113944dle
x
radic
σ(d) minus σ(d minus 1)
d+ O 1113944
dlex
radic|σ(d) minus σ(d minus 1)|⎛⎝ ⎞⎠
(56)
With the help of Lemma 5 (ii) a simple partial inte-gration gives us
Journal of Mathematics 5
1113944dle
x
radic
σ(d) minus σ(d minus 1)
d 1113944
dlex
radic
σ(d)
d(d + 1)
1113944dle
x
radic
σ(d)
d2 minus 1113944
dlex
radic
σ(d)
d2(d + 1)
1113946
x
radic
1minustminus 2d
π2
12t2
+ O(t log t)1113888 1113889 + O(1)
π2
12log x + O(1)
1113944dle
x
radic|σ(d) minus σ(d minus 1)|
≪ 1113944dle
x
radicσ(d)≪x
(57)
Inserting these estimates into (56) we find that
S1(x σ) π2
12x log x + O(x) (58)
Finally we evaluate S2(x σ) For this we write
S2(x σ) Sdagger2(x σ) + S
♯2(x σ) (59)
where
Sdagger2(x σ) ≔ 1113944
nleN0dnle x
(σ(d) minus σ(d minus 1))
S♯2(x σ) ≔ 1113944
N0ltnlex
radicdnle x
(σ(d) minus σ(d minus 1))(60)
By the bound that σ(n)≪ n log2 n we have
Sdagger2(x σ) 1113944
nleN0
σx
n1113876 11138771113874 1113875≪ 1113944
nleN0
x
n1113874 1113875log2 x
≪ x(log x)(23) log2 x( 1113857
(43)
(61)
On the other hand (27) of Lemma 5 allows us to derivethat
1113944dlex
σ(d) minus 1113944dlexminus 1
σ(d) π2
12x2
minus (x minus 1)2
1113872 1113873 minus Δ(x z)
+ Δ(x minus 1 z) + Ox
z1113874 1113875
π2
6x minus Δ(x z) + Δ(x minus 1 z) + O
x
z1113874 1113875
(62)
where Δ(x z) is given by (28) us
S♯2(x σ) 1113944
N0ltnlex
radic
π2
6middotx
nminus Δ
x
n z1113874 1113875 + Δ
x
nminus 1 z1113874 1113875 + O
x
nz1113874 11138751113888 1113889
π2
12x log x + O x(log x)
(23) log2 x( 1113857(13)
+ xzminus 1log x1113872 1113873 minus Δ1(x z) + Δ2(x z)
(63)
where
Δ1(x z) ≔ 1113944N0ltnle
x
radicΔ
x
n z1113874 1113875≪ x(log x)
minus 3+ xz
minus 1log x
Δ2(x z) ≔ 1113944N0ltnle
x
radicΔ
x
nminus 1 z1113874 1113875≪x(log x)
minus 3+ xz
minus 1log x
(64)
thanks to Lemma 6 Inserting these estimates into (59) wefind that
S2(x σ) π2
12x log x + O x(log x)
(23) log2 x( 1113857(43)
+ xzminus 1log x1113872 1113873
(65)
Now (5) follows from (53) (55) (58) and (66) with thechoice of z (log x)(13)
34 Proof ofeorem1 (ii) For any odd prime p (52) allowsus to write
1113944d ∣p
(σ(d) minus σ(d minus 1)) Sσ(p) minus Sσ(p minus 1)
π2
6(log p minus log(p minus 1)) + E(p) minus E(p minus 1)
geE(p) minus E(p minus 1)ge minus 2Elowast(p)
(66)
where Elowast(p) ≔ max |E(p)| |E(p minus 1)|1113864 1113865 On the otherhand we have
1113944d|p
(σ(d) minus σ(d minus 1)) σ(p) minus σ(p minus 1) + 1
lep + 1 minus p minus 1 +12(p minus 1) + 2 + 11113874 1113875 + 1le minus
14p
(67)
us Elowast(p)ge (18)p for all odd primes
Data Availability
No data were used to support this study
6 Journal of Mathematics
Conflicts of Interest
e authors declare that they have no conflicts of interest
Acknowledgments
is study was supported in part by the National NaturalScience Foundation of China (grant nos 1187119311971370 and 12071375) Natural Science Foundation ofChongqing (grant no cstc2019jcyj-msxm1651) NationalNatural Science Foundation of Henan Province (grant no202300410274) and Research Projects for Overseas Re-searchers of Department of Human Resources and SocialSecurity of Henan Province 2020
References
[1] O Bordelles L Dai R Heyman H Pan and I E ShparlinskildquoOn a sum involving the Euler functionrdquo Journal of Numbereory vol 202 pp 278ndash297 2019
[2] J Wu ldquoOn a sum involving the Euler totient functionrdquoIndagationes Mathematicae vol 30 no 4 pp 536ndash541 2019
[3] W Zhai ldquoOn a sum involving the euler functionrdquo Journal ofNumber eory vol 211 pp 199ndash219 2020
[4] J Ma and J Wu ldquoOn a sum involving the Mangoldt functionrdquoPeriodica Mathematica Hungarica vol 80 2020
[5] J Wu ldquoNote on a paper by Bordelles Dai heyman Pan andShparlinskirdquo Periodica Mathematica Hungarica vol 80 no 1pp 95ndash102 2020
[6] A A Karatsuba ldquoEstimates for trigonometric sums byVinogradovrsquos method and some applicationsrdquo Proceedings ofthe Steklov Institute of Mathematics vol 112 pp 251ndash265 1971
[7] Y-F S Petermann and J Wu ldquoOn the sum of exponentialdivisors of an integerrdquo Acta Mathematica Hungarica vol 77no 12 pp 159ndash175 1997
[8] S W Graham and G Kolesnik Van der Corputrsquos Method ofExponential Sums Cambridge University Press CambridgeUK 1991
Journal of Mathematics 7
1113944nle x
μ(n)≪x exp minus c
log x
1113969
1113882 1113883 (xge 1) (7)
plays a key role where cgt 0 is a positive constant Clearlysuch a bound is not true for 1 By refining Zhairsquos approachwe shall prove our result
2 Preliminary Lemmas
As in [3] we need some bounds on exponential sums of thetype 1113936NlenltNprimee(Tn) where NltNprime le 2N For large values ofN Zhai used the theory of exponent pair and for smallerones the Vinogradov method Both estimates are containedin the following general theorem of Karatsuba [6 eorem1] which will be a key tool for proving eorem 1
Lemma 1 Let kge 2 and M and P be integers P beingpositive Let f isin Ck+1([M M + P]R) Suppose that thereexist positive absolute constants c0 c1 c2 c3 and c4 such thatc0 lt 1 c1 lt 1 and c2 + c4 lt c1 an integer r such thatc0kle rle k and distinct numbers sj ge 2(j 1 r) notexceeding k such that for Mle tleM + P the following in-equalities are satisfied
(i) |f(k+1)(t)(k + 1)|lePminus c1(k+1)(ii) Pminus c2sj le |f(sj)(t)sj|lePminus c3sj (j 1 r)
en for each positive integer P1 not exceeding P we have
1113944MlemleM+P1minus 1
e(f(m))
1113868111386811138681113868111386811138681113868111386811138681113868
1113868111386811138681113868111386811138681113868111386811138681113868leAP
1minus ck2( ) (8)
where e(t) ≔ e2πit and Agt 0 cgt 0 are absolute constants
e next two lemmas are essentially a special case of [7Lemmas 25 and 26] with a 1 e only difference is thatthe ranges of T and N here are slightly larger than those of[7 Lemmas 25 and 26] (TgeN2 in place of TgeN(32) andNlex(23) in place of Nlex(12) respectively) Although theproof is completely similar for the convenience of readerswe still reproduce a proof here
Lemma 2 Let e100 leNltNprime le 2N and TgeN(32) enthere exists an absolute positive constant c5 such that
1113944
Nle nltNprime
eT
n1113874 1113875≪N exp minus
c5log3
N
log2 T1113888 11138891113896 1113897 (9)
where the implied constant is absolute
Proof We apply Lemma 1 to f(t) ≔ (Tt) withM N P N P1 Nprime minus N For this we choose
c0 1100
c1 99100
c2 87100
c3 34
c4 1100
k 100log(Tn)
log N1113890 1113891
(10)
and take the sj to be all integers s such that
4log(Tn)
log Nle sle 5
log(Tn)
log Nmiddot (11)
Obviously the number r of sj is between c0k and k Nextwe shall verify that f(t) satisfies the conditions (i) and (ii) ofLemma 1 with the parameters chosen above
For Nle tle 2N we have
f(k+1)
(t)
(k + 1)
111386811138681113868111386811138681113868111386811138681113868
111386811138681113868111386811138681113868111386811138681113868 Tt
minus kminus 2 leTNminus kminus 2
Nminus η1 (12)
where
η1 ≔ k + 1 minuslog(Tn)
log Nge k + 1 minus
1100
kge99100
(k + 1) c1(k + 1)
(13)
Similarly for Nle tle 2N we find the inequality|f(sj)(t)sj|leNminus η3 where
η3 ≔ sj minuslog(Tn)
log Nge34sj c3sj (14)
For the lower bound of (ii) we have
fsj( 1113857
(t)
sj
11138681113868111386811138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868111386811138681113868 Tt
minus 1minus sj geT(2N)minus 1minus sj N
minus η2 (15)
where
η2 ≔ sj minuslog(Tn)
log N+log 2log N
sj + 11113872 1113873
le45sj +
log 2100
sj + 11113872 1113873le87100
sj c2sj
(16)
From Lemma 1 there exist two positive constants c andA such that
2 Journal of Mathematics
1113944
Nle nltNprime
eT
n1113874 1113875
1113868111386811138681113868111386811138681113868111386811138681113868
1113868111386811138681113868111386811138681113868111386811138681113868leAN
1minus ck2( ) leAN exp minusc5log
3N
log2(Tn)1113896 1113897
(17)
with c5 ≔ 10minus 4c is completes the proof of Lemma 2
Lemma 3 Define ψ(t) ≔ t minus [t] minus (12) Let c5 be theconstant defined by Lemma 2 and c6 ≔ (89)2c5
clowast ≔ ((35)c6)minus (13) en we have
1113944
Nle nltNprime
1nψ
x
n1113874 1113875≪ eminus c6(log N)3(log x)2(log N)
3
(log x)2 (18)
uniformly for xge 10 exp clowast(log x)(23)1113966 1113967leNlex(23) and
NltNprime le 2N
Proof By invoking a classical result on ψ(t) (see 8 page 39])we can write for any Hge 1
1113944
Nle nltNprime
ψx
n1113874 1113875≪NH
minus 1+ 1113944
1le hleH
hminus 1
1113944
Nle nltNprime
ehx
n1113888 1113889
1113868111386811138681113868111386811138681113868111386811138681113868
1113868111386811138681113868111386811138681113868111386811138681113868
(19)
An application of Lemma 2 with T hxgexgeN(32)
yields
1113944
Nle nltNprime
ψx
n1113874 1113875≪N H
minus 1+ e
minus c5(log N)3log2(Hx)log H1113874 1113875
(20)
Taking H exp (log N)3(log x)21113966 1113967le x(827) we easilydeduce that
1113944
Nle nltNprime
ψx
n1113874 1113875≪N e
minus (log N)3(log x)2+ e
minus c6(log N)3(log x)2(log N)3
(log x)21113888 1113889
(21)
e first term can be absorbed by the second since c5 canbe chosen small enough to ensure that c6 lt 1 and sinceexp clowast(log x)(23)
1113966 1113967leN implies (log N)3(log x)2 ge clowastHence
1113944
Nle nltNprime
ψx
n1113874 1113875≪Ne
minus c6(log N)3(log x)2(log N)3
(log x)2 (22)
and an Abel summation produces the required result
Lemma 4 Let 2le z1 lt z2 le x and Fx(t) ≔ (1t)ψ(xt)Denote by VFx
[z1 z2] the total variation of Fx on [z1 z2]en
VFxz1 z21113858 1113859≪
x
z21
+1z1
(23)
where the implied constant is absolute
Proof If z1 t0 lt t1 lt middot middot middot lt tn z2 is a partition of the in-terval [z1 z2] then
1113944
n
k1Fx tk( 1113857 minus Fx tkminus 1( 1113857
11138681113868111386811138681113868111386811138681113868 1113944
n
k1
1tk
ψx
tk
1113888 1113889 minus1
tkminus 1ψ
x
tkminus 11113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868
le 1113944n
k1
1tk
minus1
tkminus 11113888 1113889ψ
x
tk
1113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868+ 1113944
n
k1
1tkminus 1
ψx
tk
1113888 1113889 minus ψx
tkminus 11113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868
(24)
Since |ψ(t)|le 1 for all t we have
1113944n
k1
1tk
minus1
tkminus 11113888 1113889ψ
x
tk
1113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868le1z1
minus1z2lt1z1
middot (25)
On the other hand since ψ(u) is of period 1 we have
1113944
n
k1
1tkminus 1
ψx
tk
1113888 1113889 minus ψx
tkminus 11113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868
le1z1
x
z1+ 11113888 1113889Vψ[0 1]le 2
x
z21
+1z1
1113888 1113889
(26)
Inserting these two bounds into (24) we obtain therequired result
3 Proof of Theorem 1
31 A Formula on the Mean Value of σ(n)
Lemma 5
(i) For xge 2 and 1le zlex(13) we have
1113944nle x
σ(n) π2
12x2
minus x(z minus [z])
2+[z]
2z+ O
x
z1113874 1113875 minus Δ(x z)
(27)
where
Δ(x z) ≔ 1113944dle (xz)
x
dψ
x
d1113874 1113875 (28)
(ii) For x⟶infin we have
1113944nlex
σ(n) π2
12x2
+ O(x log x) (29)
Proof Using σ(n) 1113936dmnm the hyperbole principle ofDirichlet allows us to write
1113944nle x
σ(n) 1113944dmle x
m S1 + S2 minus S3 (30)
where
Journal of Mathematics 3
S1 ≔ 1113944dle(xz)
1113944mle(xd)
m
S2 ≔ 1113944mle z
1113944dle(xm)
m
S3 ≔ 1113944dle(xz)
1113944mlez
m
(31)
Firstly we have
S2 1113944mlez
mx
m1113876 1113877 x[z] + O z
21113872 1113873 (32)
S3 x
z1113876 1113877
[z]([z] + 1)
2 x
[z]([z] + 1)
2z+ O z
21113872 1113873 (33)
Secondly we can write
S1 12
1113944dle(xz)
x
dminus ψ
x
d1113874 1113875 minus
12
1113874 1113875x
dminus ψ
x
d1113874 1113875 +
12
1113874 1113875
12
1113944dle(xz)
x2
d2 minus 2
x
dψ
x
d1113874 1113875 + ψ
x
d1113874 1113875
2minus14
1113888 1113889
π2
12x2
minus12
xz minus Δ(x z) + O(xz)
(34)
where Δ(x z) is as in (28) Inserting (32) (33) and (34) into(30) and using z2 le (xz) we get (27)
Taking z 1 in (27) and noticing that
1113944dlex
1d2
π2
6+ O
1x
1113874 1113875
1113944dlex
x
dψ
x
d1113874 1113875
1113868111386811138681113868111386811138681113868111386811138681113868
1113868111386811138681113868111386811138681113868111386811138681113868≪ x log x
(35)
we obtain the required boundis completes the proof
32 Estimates of Error Terms
Lemma 6 Let N0 ≔ exp (6c6)(log x)(23)(log2 x)(13)1113966 1113967
where c6 is given as in Lemma 3 Let Δ(x z) be defined by(28) en for xge 10 and 2le zle
N0
1113968 we have
1113944N0ltnle
x
radicΔ
x
n z1113874 1113875
11138681113868111386811138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868111386811138681113868+ 1113944
N0ltnlex
radicΔ
x
nminus 1 z1113874 1113875
11138681113868111386811138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868111386811138681113868
≪ x1
(log x)3 +
log x
z1113888 1113889
(36)
Proof Denote by Δ1(x z) and Δ2(x z) two sums on theleft-hand side of (36) respectively By (28) of Lemma 5 wecan write
Δ1(x z) x 1113944N0ltnle
x
radic1113944
dle(x(nz))
1dn
ψx
dn1113874 1113875
x 1113944
dle x N0z( )( )
1d
1113944N0ltnlemin(
x
radicx(dz))
1nψ
x
dn1113874 1113875
xΔdagger1(x z) + xΔ♯1(x z)
(37)
where
Δdagger1(x z) ≔ 1113944
dle x N0z( )( )
1d
1113944
N0ltnle(xd)(23)
1nψ
x
dn1113874 1113875
Δ♯1(x z) ≔ 1113944
dle x N0z( )( )
1d
1113944
(xd)(23)lt nleminx
radic(xdz)
1nψ
x
dn1113874 1113875
(38)
For 0le kle (log((xd)(23)N0))log 2 let Nk ≔ 2kN0and define
Sk(d) ≔ 1113944Nklt nle 2Nk
1nψ
x
dn1113874 1113875 (39)
Noticing that N0 leNk le (xd)(23) we can apply Lemma3 to derive that
Sk(d)≪ eminus ϑ log Nk( )3(log(xd))2( 1113857
(40)
with ϑ(t) ≔ c6t minus log t It is clear that ϑ(t) is increasing on[c6infin) On the other hand for kge 0 and dge 1 we have
log Nk( 11138573(log(xd))
2 ge log N0( 11138573(log x)
2 6c6( 1113857log2 x
(41)
us
ϑlog Nk( 1113857
3
(log(xd))21113888 1113889ge ϑ
6c6
1113888 1113889log2 x1113888 1113889
6 log2 x minus log6c6
1113888 1113889log2 x1113888 1113889ge 5 log2 x
(42)
which implies that Sk(d)≪ (log x)minus 5 Inserting this intothe expression of Δdagger1(x z) we get
Δdagger1(x z)≪ 1113944
dle x N0z( )( )
1d
1113944
2kN0le(xd)(23)
Sk(d)1113868111386811138681113868
1113868111386811138681113868≪ (log x)minus 3
(43)
Next we bound Δ♯1(x z) Let F(t) be a function ofbounded variation on [n n + 1] for each integer n and letVF[n n + 1] be the total variation of F on [n n + 1] Inte-grating by parts we have
4 Journal of Mathematics
1113946n+1
nt minus n minus
12
1113874 1113875dF(t) 12(F(n + 1) + F(n)) minus 1113946
n+1
nF(t)dt
(44)
From this we can derive that
12(F(n + 1) + F(n)) 1113946
n+1
nF(t)dt + O VF[n n + 1]( 1113857
(45)
for nge 1 Summing over n we find that
1113944N1lt nleN2
F(n) 1113946N2
N1
F(t)dt
+12
F N1( 1113857 + F N2( 1113857( 1113857 + O VF N1 N21113858 1113859( 1113857
(46)
We apply this formula to
F(xd)(t) 1tψ
(xd)
t1113888 1113889
N1 (xd)(23)
1113960 1113961
N2 minx
radic
x
(dz)1113896 11138971113890 1113891
(47)
According to Lemma 4 we haveVF(xd)
[N1 N2]≪ (xd)minus (13) and thus by puttingu (xd)t we obtain with the notationxd1 max(
x
radicd tz) and xd2 (xd)(13)
1113944
(xd)(23)ltnleminx
radic(x(dz))
1nψ
x
dn1113874 1113875 1113946
xd2
xd1
ψ(u)
udu + O
x
d1113874 1113875
minus (13)
1113888 1113889
≪ zminus 1
+x
d1113874 1113875
minus (13)
≪ zminus 1
(48)
where we have used the fact that zleN0
1113968and
dle (x(N0z))rArrz le (xd)(13) and the bound
1113946xd2
xd1
ψ(u)
udu 1113946
xd2
xd1
1113946u
xd1
ψ(t)dt1113888 1113889du
u2 minus
1x
(23)d2
1113946xd2
xd1
ψ(t)dt
≪xminus 1d1 + xd21113872 1113873
minus (23)≪ z
minus 1+(xd)
minus (23)≪ zminus 1
(49)
Using (48) a simple partial integration allows us toderive that
Δ♯1(x z)≪ zminus 1
1113944
dlex N0z( )
dminus 1≪ z
minus 1log x(50)
Combining (43) and (50) it follows that
Δ1(x z)1113868111386811138681113868
1113868111386811138681113868≪ x(log x)minus 3
+ xzminus 1log x (51)
Similarly we can prove the same bound for |Δ2(x z)|is completes the proof
33 End of the Proof of eorem 1 Let c6 be the constantgiven as in Lemma 3 and N0 ≔ exp (6c6)1113864 (log x)(23)
(log2 x)(13) Let z isin [2N0
1113968] be a parameter to be chosen
laterPutting d [xn] we have (xn) minus 1lt dle (xn) and
x(d + 1)lt nle (xd) We have with the conventionσ(0) 0
Sσ(x) 1113944dlex
σ(d) 1113944(x(d+1))ltnle(xd)
1
1113944dnle x
σ(d) minus 1113944dnle xdge2
σ(d minus 1)
1113944dnlex
(σ(d) minus σ(d minus 1))
(52)
By the hyperbole principle of Dirichlet we can write
Sσ(x) S1(x σ) + S2(x σ) minus S3(x σ) (53)
where
S1(x σ) ≔ 1113944dle
x
radicdnle x
(σ(d) minus σ(d minus 1))
S2(x σ) ≔ 1113944nle
x
radicdnlex
(σ(d) minus σ(d minus 1))
S3(x σ) ≔ 1113944dle
x
radicnle
x
radic(σ(d) minus σ(d minus 1))
(54)
With the help of the bound σ(n)≪ n log2 n we canderive that
S3(x σ) [x
radic]σ([
x
radic])≪x log2 x (55)
For evaluating S1(x σ) we write
S1(x σ) 1113944dle
x
radic(σ(d) minus σ(d minus 1))
x
d1113876 1113877
x 1113944dle
x
radic
σ(d) minus σ(d minus 1)
d+ O 1113944
dlex
radic|σ(d) minus σ(d minus 1)|⎛⎝ ⎞⎠
(56)
With the help of Lemma 5 (ii) a simple partial inte-gration gives us
Journal of Mathematics 5
1113944dle
x
radic
σ(d) minus σ(d minus 1)
d 1113944
dlex
radic
σ(d)
d(d + 1)
1113944dle
x
radic
σ(d)
d2 minus 1113944
dlex
radic
σ(d)
d2(d + 1)
1113946
x
radic
1minustminus 2d
π2
12t2
+ O(t log t)1113888 1113889 + O(1)
π2
12log x + O(1)
1113944dle
x
radic|σ(d) minus σ(d minus 1)|
≪ 1113944dle
x
radicσ(d)≪x
(57)
Inserting these estimates into (56) we find that
S1(x σ) π2
12x log x + O(x) (58)
Finally we evaluate S2(x σ) For this we write
S2(x σ) Sdagger2(x σ) + S
♯2(x σ) (59)
where
Sdagger2(x σ) ≔ 1113944
nleN0dnle x
(σ(d) minus σ(d minus 1))
S♯2(x σ) ≔ 1113944
N0ltnlex
radicdnle x
(σ(d) minus σ(d minus 1))(60)
By the bound that σ(n)≪ n log2 n we have
Sdagger2(x σ) 1113944
nleN0
σx
n1113876 11138771113874 1113875≪ 1113944
nleN0
x
n1113874 1113875log2 x
≪ x(log x)(23) log2 x( 1113857
(43)
(61)
On the other hand (27) of Lemma 5 allows us to derivethat
1113944dlex
σ(d) minus 1113944dlexminus 1
σ(d) π2
12x2
minus (x minus 1)2
1113872 1113873 minus Δ(x z)
+ Δ(x minus 1 z) + Ox
z1113874 1113875
π2
6x minus Δ(x z) + Δ(x minus 1 z) + O
x
z1113874 1113875
(62)
where Δ(x z) is given by (28) us
S♯2(x σ) 1113944
N0ltnlex
radic
π2
6middotx
nminus Δ
x
n z1113874 1113875 + Δ
x
nminus 1 z1113874 1113875 + O
x
nz1113874 11138751113888 1113889
π2
12x log x + O x(log x)
(23) log2 x( 1113857(13)
+ xzminus 1log x1113872 1113873 minus Δ1(x z) + Δ2(x z)
(63)
where
Δ1(x z) ≔ 1113944N0ltnle
x
radicΔ
x
n z1113874 1113875≪ x(log x)
minus 3+ xz
minus 1log x
Δ2(x z) ≔ 1113944N0ltnle
x
radicΔ
x
nminus 1 z1113874 1113875≪x(log x)
minus 3+ xz
minus 1log x
(64)
thanks to Lemma 6 Inserting these estimates into (59) wefind that
S2(x σ) π2
12x log x + O x(log x)
(23) log2 x( 1113857(43)
+ xzminus 1log x1113872 1113873
(65)
Now (5) follows from (53) (55) (58) and (66) with thechoice of z (log x)(13)
34 Proof ofeorem1 (ii) For any odd prime p (52) allowsus to write
1113944d ∣p
(σ(d) minus σ(d minus 1)) Sσ(p) minus Sσ(p minus 1)
π2
6(log p minus log(p minus 1)) + E(p) minus E(p minus 1)
geE(p) minus E(p minus 1)ge minus 2Elowast(p)
(66)
where Elowast(p) ≔ max |E(p)| |E(p minus 1)|1113864 1113865 On the otherhand we have
1113944d|p
(σ(d) minus σ(d minus 1)) σ(p) minus σ(p minus 1) + 1
lep + 1 minus p minus 1 +12(p minus 1) + 2 + 11113874 1113875 + 1le minus
14p
(67)
us Elowast(p)ge (18)p for all odd primes
Data Availability
No data were used to support this study
6 Journal of Mathematics
Conflicts of Interest
e authors declare that they have no conflicts of interest
Acknowledgments
is study was supported in part by the National NaturalScience Foundation of China (grant nos 1187119311971370 and 12071375) Natural Science Foundation ofChongqing (grant no cstc2019jcyj-msxm1651) NationalNatural Science Foundation of Henan Province (grant no202300410274) and Research Projects for Overseas Re-searchers of Department of Human Resources and SocialSecurity of Henan Province 2020
References
[1] O Bordelles L Dai R Heyman H Pan and I E ShparlinskildquoOn a sum involving the Euler functionrdquo Journal of Numbereory vol 202 pp 278ndash297 2019
[2] J Wu ldquoOn a sum involving the Euler totient functionrdquoIndagationes Mathematicae vol 30 no 4 pp 536ndash541 2019
[3] W Zhai ldquoOn a sum involving the euler functionrdquo Journal ofNumber eory vol 211 pp 199ndash219 2020
[4] J Ma and J Wu ldquoOn a sum involving the Mangoldt functionrdquoPeriodica Mathematica Hungarica vol 80 2020
[5] J Wu ldquoNote on a paper by Bordelles Dai heyman Pan andShparlinskirdquo Periodica Mathematica Hungarica vol 80 no 1pp 95ndash102 2020
[6] A A Karatsuba ldquoEstimates for trigonometric sums byVinogradovrsquos method and some applicationsrdquo Proceedings ofthe Steklov Institute of Mathematics vol 112 pp 251ndash265 1971
[7] Y-F S Petermann and J Wu ldquoOn the sum of exponentialdivisors of an integerrdquo Acta Mathematica Hungarica vol 77no 12 pp 159ndash175 1997
[8] S W Graham and G Kolesnik Van der Corputrsquos Method ofExponential Sums Cambridge University Press CambridgeUK 1991
Journal of Mathematics 7
1113944
Nle nltNprime
eT
n1113874 1113875
1113868111386811138681113868111386811138681113868111386811138681113868
1113868111386811138681113868111386811138681113868111386811138681113868leAN
1minus ck2( ) leAN exp minusc5log
3N
log2(Tn)1113896 1113897
(17)
with c5 ≔ 10minus 4c is completes the proof of Lemma 2
Lemma 3 Define ψ(t) ≔ t minus [t] minus (12) Let c5 be theconstant defined by Lemma 2 and c6 ≔ (89)2c5
clowast ≔ ((35)c6)minus (13) en we have
1113944
Nle nltNprime
1nψ
x
n1113874 1113875≪ eminus c6(log N)3(log x)2(log N)
3
(log x)2 (18)
uniformly for xge 10 exp clowast(log x)(23)1113966 1113967leNlex(23) and
NltNprime le 2N
Proof By invoking a classical result on ψ(t) (see 8 page 39])we can write for any Hge 1
1113944
Nle nltNprime
ψx
n1113874 1113875≪NH
minus 1+ 1113944
1le hleH
hminus 1
1113944
Nle nltNprime
ehx
n1113888 1113889
1113868111386811138681113868111386811138681113868111386811138681113868
1113868111386811138681113868111386811138681113868111386811138681113868
(19)
An application of Lemma 2 with T hxgexgeN(32)
yields
1113944
Nle nltNprime
ψx
n1113874 1113875≪N H
minus 1+ e
minus c5(log N)3log2(Hx)log H1113874 1113875
(20)
Taking H exp (log N)3(log x)21113966 1113967le x(827) we easilydeduce that
1113944
Nle nltNprime
ψx
n1113874 1113875≪N e
minus (log N)3(log x)2+ e
minus c6(log N)3(log x)2(log N)3
(log x)21113888 1113889
(21)
e first term can be absorbed by the second since c5 canbe chosen small enough to ensure that c6 lt 1 and sinceexp clowast(log x)(23)
1113966 1113967leN implies (log N)3(log x)2 ge clowastHence
1113944
Nle nltNprime
ψx
n1113874 1113875≪Ne
minus c6(log N)3(log x)2(log N)3
(log x)2 (22)
and an Abel summation produces the required result
Lemma 4 Let 2le z1 lt z2 le x and Fx(t) ≔ (1t)ψ(xt)Denote by VFx
[z1 z2] the total variation of Fx on [z1 z2]en
VFxz1 z21113858 1113859≪
x
z21
+1z1
(23)
where the implied constant is absolute
Proof If z1 t0 lt t1 lt middot middot middot lt tn z2 is a partition of the in-terval [z1 z2] then
1113944
n
k1Fx tk( 1113857 minus Fx tkminus 1( 1113857
11138681113868111386811138681113868111386811138681113868 1113944
n
k1
1tk
ψx
tk
1113888 1113889 minus1
tkminus 1ψ
x
tkminus 11113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868
le 1113944n
k1
1tk
minus1
tkminus 11113888 1113889ψ
x
tk
1113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868+ 1113944
n
k1
1tkminus 1
ψx
tk
1113888 1113889 minus ψx
tkminus 11113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868
(24)
Since |ψ(t)|le 1 for all t we have
1113944n
k1
1tk
minus1
tkminus 11113888 1113889ψ
x
tk
1113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868le1z1
minus1z2lt1z1
middot (25)
On the other hand since ψ(u) is of period 1 we have
1113944
n
k1
1tkminus 1
ψx
tk
1113888 1113889 minus ψx
tkminus 11113888 1113889
11138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868
le1z1
x
z1+ 11113888 1113889Vψ[0 1]le 2
x
z21
+1z1
1113888 1113889
(26)
Inserting these two bounds into (24) we obtain therequired result
3 Proof of Theorem 1
31 A Formula on the Mean Value of σ(n)
Lemma 5
(i) For xge 2 and 1le zlex(13) we have
1113944nle x
σ(n) π2
12x2
minus x(z minus [z])
2+[z]
2z+ O
x
z1113874 1113875 minus Δ(x z)
(27)
where
Δ(x z) ≔ 1113944dle (xz)
x
dψ
x
d1113874 1113875 (28)
(ii) For x⟶infin we have
1113944nlex
σ(n) π2
12x2
+ O(x log x) (29)
Proof Using σ(n) 1113936dmnm the hyperbole principle ofDirichlet allows us to write
1113944nle x
σ(n) 1113944dmle x
m S1 + S2 minus S3 (30)
where
Journal of Mathematics 3
S1 ≔ 1113944dle(xz)
1113944mle(xd)
m
S2 ≔ 1113944mle z
1113944dle(xm)
m
S3 ≔ 1113944dle(xz)
1113944mlez
m
(31)
Firstly we have
S2 1113944mlez
mx
m1113876 1113877 x[z] + O z
21113872 1113873 (32)
S3 x
z1113876 1113877
[z]([z] + 1)
2 x
[z]([z] + 1)
2z+ O z
21113872 1113873 (33)
Secondly we can write
S1 12
1113944dle(xz)
x
dminus ψ
x
d1113874 1113875 minus
12
1113874 1113875x
dminus ψ
x
d1113874 1113875 +
12
1113874 1113875
12
1113944dle(xz)
x2
d2 minus 2
x
dψ
x
d1113874 1113875 + ψ
x
d1113874 1113875
2minus14
1113888 1113889
π2
12x2
minus12
xz minus Δ(x z) + O(xz)
(34)
where Δ(x z) is as in (28) Inserting (32) (33) and (34) into(30) and using z2 le (xz) we get (27)
Taking z 1 in (27) and noticing that
1113944dlex
1d2
π2
6+ O
1x
1113874 1113875
1113944dlex
x
dψ
x
d1113874 1113875
1113868111386811138681113868111386811138681113868111386811138681113868
1113868111386811138681113868111386811138681113868111386811138681113868≪ x log x
(35)
we obtain the required boundis completes the proof
32 Estimates of Error Terms
Lemma 6 Let N0 ≔ exp (6c6)(log x)(23)(log2 x)(13)1113966 1113967
where c6 is given as in Lemma 3 Let Δ(x z) be defined by(28) en for xge 10 and 2le zle
N0
1113968 we have
1113944N0ltnle
x
radicΔ
x
n z1113874 1113875
11138681113868111386811138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868111386811138681113868+ 1113944
N0ltnlex
radicΔ
x
nminus 1 z1113874 1113875
11138681113868111386811138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868111386811138681113868
≪ x1
(log x)3 +
log x
z1113888 1113889
(36)
Proof Denote by Δ1(x z) and Δ2(x z) two sums on theleft-hand side of (36) respectively By (28) of Lemma 5 wecan write
Δ1(x z) x 1113944N0ltnle
x
radic1113944
dle(x(nz))
1dn
ψx
dn1113874 1113875
x 1113944
dle x N0z( )( )
1d
1113944N0ltnlemin(
x
radicx(dz))
1nψ
x
dn1113874 1113875
xΔdagger1(x z) + xΔ♯1(x z)
(37)
where
Δdagger1(x z) ≔ 1113944
dle x N0z( )( )
1d
1113944
N0ltnle(xd)(23)
1nψ
x
dn1113874 1113875
Δ♯1(x z) ≔ 1113944
dle x N0z( )( )
1d
1113944
(xd)(23)lt nleminx
radic(xdz)
1nψ
x
dn1113874 1113875
(38)
For 0le kle (log((xd)(23)N0))log 2 let Nk ≔ 2kN0and define
Sk(d) ≔ 1113944Nklt nle 2Nk
1nψ
x
dn1113874 1113875 (39)
Noticing that N0 leNk le (xd)(23) we can apply Lemma3 to derive that
Sk(d)≪ eminus ϑ log Nk( )3(log(xd))2( 1113857
(40)
with ϑ(t) ≔ c6t minus log t It is clear that ϑ(t) is increasing on[c6infin) On the other hand for kge 0 and dge 1 we have
log Nk( 11138573(log(xd))
2 ge log N0( 11138573(log x)
2 6c6( 1113857log2 x
(41)
us
ϑlog Nk( 1113857
3
(log(xd))21113888 1113889ge ϑ
6c6
1113888 1113889log2 x1113888 1113889
6 log2 x minus log6c6
1113888 1113889log2 x1113888 1113889ge 5 log2 x
(42)
which implies that Sk(d)≪ (log x)minus 5 Inserting this intothe expression of Δdagger1(x z) we get
Δdagger1(x z)≪ 1113944
dle x N0z( )( )
1d
1113944
2kN0le(xd)(23)
Sk(d)1113868111386811138681113868
1113868111386811138681113868≪ (log x)minus 3
(43)
Next we bound Δ♯1(x z) Let F(t) be a function ofbounded variation on [n n + 1] for each integer n and letVF[n n + 1] be the total variation of F on [n n + 1] Inte-grating by parts we have
4 Journal of Mathematics
1113946n+1
nt minus n minus
12
1113874 1113875dF(t) 12(F(n + 1) + F(n)) minus 1113946
n+1
nF(t)dt
(44)
From this we can derive that
12(F(n + 1) + F(n)) 1113946
n+1
nF(t)dt + O VF[n n + 1]( 1113857
(45)
for nge 1 Summing over n we find that
1113944N1lt nleN2
F(n) 1113946N2
N1
F(t)dt
+12
F N1( 1113857 + F N2( 1113857( 1113857 + O VF N1 N21113858 1113859( 1113857
(46)
We apply this formula to
F(xd)(t) 1tψ
(xd)
t1113888 1113889
N1 (xd)(23)
1113960 1113961
N2 minx
radic
x
(dz)1113896 11138971113890 1113891
(47)
According to Lemma 4 we haveVF(xd)
[N1 N2]≪ (xd)minus (13) and thus by puttingu (xd)t we obtain with the notationxd1 max(
x
radicd tz) and xd2 (xd)(13)
1113944
(xd)(23)ltnleminx
radic(x(dz))
1nψ
x
dn1113874 1113875 1113946
xd2
xd1
ψ(u)
udu + O
x
d1113874 1113875
minus (13)
1113888 1113889
≪ zminus 1
+x
d1113874 1113875
minus (13)
≪ zminus 1
(48)
where we have used the fact that zleN0
1113968and
dle (x(N0z))rArrz le (xd)(13) and the bound
1113946xd2
xd1
ψ(u)
udu 1113946
xd2
xd1
1113946u
xd1
ψ(t)dt1113888 1113889du
u2 minus
1x
(23)d2
1113946xd2
xd1
ψ(t)dt
≪xminus 1d1 + xd21113872 1113873
minus (23)≪ z
minus 1+(xd)
minus (23)≪ zminus 1
(49)
Using (48) a simple partial integration allows us toderive that
Δ♯1(x z)≪ zminus 1
1113944
dlex N0z( )
dminus 1≪ z
minus 1log x(50)
Combining (43) and (50) it follows that
Δ1(x z)1113868111386811138681113868
1113868111386811138681113868≪ x(log x)minus 3
+ xzminus 1log x (51)
Similarly we can prove the same bound for |Δ2(x z)|is completes the proof
33 End of the Proof of eorem 1 Let c6 be the constantgiven as in Lemma 3 and N0 ≔ exp (6c6)1113864 (log x)(23)
(log2 x)(13) Let z isin [2N0
1113968] be a parameter to be chosen
laterPutting d [xn] we have (xn) minus 1lt dle (xn) and
x(d + 1)lt nle (xd) We have with the conventionσ(0) 0
Sσ(x) 1113944dlex
σ(d) 1113944(x(d+1))ltnle(xd)
1
1113944dnle x
σ(d) minus 1113944dnle xdge2
σ(d minus 1)
1113944dnlex
(σ(d) minus σ(d minus 1))
(52)
By the hyperbole principle of Dirichlet we can write
Sσ(x) S1(x σ) + S2(x σ) minus S3(x σ) (53)
where
S1(x σ) ≔ 1113944dle
x
radicdnle x
(σ(d) minus σ(d minus 1))
S2(x σ) ≔ 1113944nle
x
radicdnlex
(σ(d) minus σ(d minus 1))
S3(x σ) ≔ 1113944dle
x
radicnle
x
radic(σ(d) minus σ(d minus 1))
(54)
With the help of the bound σ(n)≪ n log2 n we canderive that
S3(x σ) [x
radic]σ([
x
radic])≪x log2 x (55)
For evaluating S1(x σ) we write
S1(x σ) 1113944dle
x
radic(σ(d) minus σ(d minus 1))
x
d1113876 1113877
x 1113944dle
x
radic
σ(d) minus σ(d minus 1)
d+ O 1113944
dlex
radic|σ(d) minus σ(d minus 1)|⎛⎝ ⎞⎠
(56)
With the help of Lemma 5 (ii) a simple partial inte-gration gives us
Journal of Mathematics 5
1113944dle
x
radic
σ(d) minus σ(d minus 1)
d 1113944
dlex
radic
σ(d)
d(d + 1)
1113944dle
x
radic
σ(d)
d2 minus 1113944
dlex
radic
σ(d)
d2(d + 1)
1113946
x
radic
1minustminus 2d
π2
12t2
+ O(t log t)1113888 1113889 + O(1)
π2
12log x + O(1)
1113944dle
x
radic|σ(d) minus σ(d minus 1)|
≪ 1113944dle
x
radicσ(d)≪x
(57)
Inserting these estimates into (56) we find that
S1(x σ) π2
12x log x + O(x) (58)
Finally we evaluate S2(x σ) For this we write
S2(x σ) Sdagger2(x σ) + S
♯2(x σ) (59)
where
Sdagger2(x σ) ≔ 1113944
nleN0dnle x
(σ(d) minus σ(d minus 1))
S♯2(x σ) ≔ 1113944
N0ltnlex
radicdnle x
(σ(d) minus σ(d minus 1))(60)
By the bound that σ(n)≪ n log2 n we have
Sdagger2(x σ) 1113944
nleN0
σx
n1113876 11138771113874 1113875≪ 1113944
nleN0
x
n1113874 1113875log2 x
≪ x(log x)(23) log2 x( 1113857
(43)
(61)
On the other hand (27) of Lemma 5 allows us to derivethat
1113944dlex
σ(d) minus 1113944dlexminus 1
σ(d) π2
12x2
minus (x minus 1)2
1113872 1113873 minus Δ(x z)
+ Δ(x minus 1 z) + Ox
z1113874 1113875
π2
6x minus Δ(x z) + Δ(x minus 1 z) + O
x
z1113874 1113875
(62)
where Δ(x z) is given by (28) us
S♯2(x σ) 1113944
N0ltnlex
radic
π2
6middotx
nminus Δ
x
n z1113874 1113875 + Δ
x
nminus 1 z1113874 1113875 + O
x
nz1113874 11138751113888 1113889
π2
12x log x + O x(log x)
(23) log2 x( 1113857(13)
+ xzminus 1log x1113872 1113873 minus Δ1(x z) + Δ2(x z)
(63)
where
Δ1(x z) ≔ 1113944N0ltnle
x
radicΔ
x
n z1113874 1113875≪ x(log x)
minus 3+ xz
minus 1log x
Δ2(x z) ≔ 1113944N0ltnle
x
radicΔ
x
nminus 1 z1113874 1113875≪x(log x)
minus 3+ xz
minus 1log x
(64)
thanks to Lemma 6 Inserting these estimates into (59) wefind that
S2(x σ) π2
12x log x + O x(log x)
(23) log2 x( 1113857(43)
+ xzminus 1log x1113872 1113873
(65)
Now (5) follows from (53) (55) (58) and (66) with thechoice of z (log x)(13)
34 Proof ofeorem1 (ii) For any odd prime p (52) allowsus to write
1113944d ∣p
(σ(d) minus σ(d minus 1)) Sσ(p) minus Sσ(p minus 1)
π2
6(log p minus log(p minus 1)) + E(p) minus E(p minus 1)
geE(p) minus E(p minus 1)ge minus 2Elowast(p)
(66)
where Elowast(p) ≔ max |E(p)| |E(p minus 1)|1113864 1113865 On the otherhand we have
1113944d|p
(σ(d) minus σ(d minus 1)) σ(p) minus σ(p minus 1) + 1
lep + 1 minus p minus 1 +12(p minus 1) + 2 + 11113874 1113875 + 1le minus
14p
(67)
us Elowast(p)ge (18)p for all odd primes
Data Availability
No data were used to support this study
6 Journal of Mathematics
Conflicts of Interest
e authors declare that they have no conflicts of interest
Acknowledgments
is study was supported in part by the National NaturalScience Foundation of China (grant nos 1187119311971370 and 12071375) Natural Science Foundation ofChongqing (grant no cstc2019jcyj-msxm1651) NationalNatural Science Foundation of Henan Province (grant no202300410274) and Research Projects for Overseas Re-searchers of Department of Human Resources and SocialSecurity of Henan Province 2020
References
[1] O Bordelles L Dai R Heyman H Pan and I E ShparlinskildquoOn a sum involving the Euler functionrdquo Journal of Numbereory vol 202 pp 278ndash297 2019
[2] J Wu ldquoOn a sum involving the Euler totient functionrdquoIndagationes Mathematicae vol 30 no 4 pp 536ndash541 2019
[3] W Zhai ldquoOn a sum involving the euler functionrdquo Journal ofNumber eory vol 211 pp 199ndash219 2020
[4] J Ma and J Wu ldquoOn a sum involving the Mangoldt functionrdquoPeriodica Mathematica Hungarica vol 80 2020
[5] J Wu ldquoNote on a paper by Bordelles Dai heyman Pan andShparlinskirdquo Periodica Mathematica Hungarica vol 80 no 1pp 95ndash102 2020
[6] A A Karatsuba ldquoEstimates for trigonometric sums byVinogradovrsquos method and some applicationsrdquo Proceedings ofthe Steklov Institute of Mathematics vol 112 pp 251ndash265 1971
[7] Y-F S Petermann and J Wu ldquoOn the sum of exponentialdivisors of an integerrdquo Acta Mathematica Hungarica vol 77no 12 pp 159ndash175 1997
[8] S W Graham and G Kolesnik Van der Corputrsquos Method ofExponential Sums Cambridge University Press CambridgeUK 1991
Journal of Mathematics 7
S1 ≔ 1113944dle(xz)
1113944mle(xd)
m
S2 ≔ 1113944mle z
1113944dle(xm)
m
S3 ≔ 1113944dle(xz)
1113944mlez
m
(31)
Firstly we have
S2 1113944mlez
mx
m1113876 1113877 x[z] + O z
21113872 1113873 (32)
S3 x
z1113876 1113877
[z]([z] + 1)
2 x
[z]([z] + 1)
2z+ O z
21113872 1113873 (33)
Secondly we can write
S1 12
1113944dle(xz)
x
dminus ψ
x
d1113874 1113875 minus
12
1113874 1113875x
dminus ψ
x
d1113874 1113875 +
12
1113874 1113875
12
1113944dle(xz)
x2
d2 minus 2
x
dψ
x
d1113874 1113875 + ψ
x
d1113874 1113875
2minus14
1113888 1113889
π2
12x2
minus12
xz minus Δ(x z) + O(xz)
(34)
where Δ(x z) is as in (28) Inserting (32) (33) and (34) into(30) and using z2 le (xz) we get (27)
Taking z 1 in (27) and noticing that
1113944dlex
1d2
π2
6+ O
1x
1113874 1113875
1113944dlex
x
dψ
x
d1113874 1113875
1113868111386811138681113868111386811138681113868111386811138681113868
1113868111386811138681113868111386811138681113868111386811138681113868≪ x log x
(35)
we obtain the required boundis completes the proof
32 Estimates of Error Terms
Lemma 6 Let N0 ≔ exp (6c6)(log x)(23)(log2 x)(13)1113966 1113967
where c6 is given as in Lemma 3 Let Δ(x z) be defined by(28) en for xge 10 and 2le zle
N0
1113968 we have
1113944N0ltnle
x
radicΔ
x
n z1113874 1113875
11138681113868111386811138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868111386811138681113868+ 1113944
N0ltnlex
radicΔ
x
nminus 1 z1113874 1113875
11138681113868111386811138681113868111386811138681113868111386811138681113868
11138681113868111386811138681113868111386811138681113868111386811138681113868
≪ x1
(log x)3 +
log x
z1113888 1113889
(36)
Proof Denote by Δ1(x z) and Δ2(x z) two sums on theleft-hand side of (36) respectively By (28) of Lemma 5 wecan write
Δ1(x z) x 1113944N0ltnle
x
radic1113944
dle(x(nz))
1dn
ψx
dn1113874 1113875
x 1113944
dle x N0z( )( )
1d
1113944N0ltnlemin(
x
radicx(dz))
1nψ
x
dn1113874 1113875
xΔdagger1(x z) + xΔ♯1(x z)
(37)
where
Δdagger1(x z) ≔ 1113944
dle x N0z( )( )
1d
1113944
N0ltnle(xd)(23)
1nψ
x
dn1113874 1113875
Δ♯1(x z) ≔ 1113944
dle x N0z( )( )
1d
1113944
(xd)(23)lt nleminx
radic(xdz)
1nψ
x
dn1113874 1113875
(38)
For 0le kle (log((xd)(23)N0))log 2 let Nk ≔ 2kN0and define
Sk(d) ≔ 1113944Nklt nle 2Nk
1nψ
x
dn1113874 1113875 (39)
Noticing that N0 leNk le (xd)(23) we can apply Lemma3 to derive that
Sk(d)≪ eminus ϑ log Nk( )3(log(xd))2( 1113857
(40)
with ϑ(t) ≔ c6t minus log t It is clear that ϑ(t) is increasing on[c6infin) On the other hand for kge 0 and dge 1 we have
log Nk( 11138573(log(xd))
2 ge log N0( 11138573(log x)
2 6c6( 1113857log2 x
(41)
us
ϑlog Nk( 1113857
3
(log(xd))21113888 1113889ge ϑ
6c6
1113888 1113889log2 x1113888 1113889
6 log2 x minus log6c6
1113888 1113889log2 x1113888 1113889ge 5 log2 x
(42)
which implies that Sk(d)≪ (log x)minus 5 Inserting this intothe expression of Δdagger1(x z) we get
Δdagger1(x z)≪ 1113944
dle x N0z( )( )
1d
1113944
2kN0le(xd)(23)
Sk(d)1113868111386811138681113868
1113868111386811138681113868≪ (log x)minus 3
(43)
Next we bound Δ♯1(x z) Let F(t) be a function ofbounded variation on [n n + 1] for each integer n and letVF[n n + 1] be the total variation of F on [n n + 1] Inte-grating by parts we have
4 Journal of Mathematics
1113946n+1
nt minus n minus
12
1113874 1113875dF(t) 12(F(n + 1) + F(n)) minus 1113946
n+1
nF(t)dt
(44)
From this we can derive that
12(F(n + 1) + F(n)) 1113946
n+1
nF(t)dt + O VF[n n + 1]( 1113857
(45)
for nge 1 Summing over n we find that
1113944N1lt nleN2
F(n) 1113946N2
N1
F(t)dt
+12
F N1( 1113857 + F N2( 1113857( 1113857 + O VF N1 N21113858 1113859( 1113857
(46)
We apply this formula to
F(xd)(t) 1tψ
(xd)
t1113888 1113889
N1 (xd)(23)
1113960 1113961
N2 minx
radic
x
(dz)1113896 11138971113890 1113891
(47)
According to Lemma 4 we haveVF(xd)
[N1 N2]≪ (xd)minus (13) and thus by puttingu (xd)t we obtain with the notationxd1 max(
x
radicd tz) and xd2 (xd)(13)
1113944
(xd)(23)ltnleminx
radic(x(dz))
1nψ
x
dn1113874 1113875 1113946
xd2
xd1
ψ(u)
udu + O
x
d1113874 1113875
minus (13)
1113888 1113889
≪ zminus 1
+x
d1113874 1113875
minus (13)
≪ zminus 1
(48)
where we have used the fact that zleN0
1113968and
dle (x(N0z))rArrz le (xd)(13) and the bound
1113946xd2
xd1
ψ(u)
udu 1113946
xd2
xd1
1113946u
xd1
ψ(t)dt1113888 1113889du
u2 minus
1x
(23)d2
1113946xd2
xd1
ψ(t)dt
≪xminus 1d1 + xd21113872 1113873
minus (23)≪ z
minus 1+(xd)
minus (23)≪ zminus 1
(49)
Using (48) a simple partial integration allows us toderive that
Δ♯1(x z)≪ zminus 1
1113944
dlex N0z( )
dminus 1≪ z
minus 1log x(50)
Combining (43) and (50) it follows that
Δ1(x z)1113868111386811138681113868
1113868111386811138681113868≪ x(log x)minus 3
+ xzminus 1log x (51)
Similarly we can prove the same bound for |Δ2(x z)|is completes the proof
33 End of the Proof of eorem 1 Let c6 be the constantgiven as in Lemma 3 and N0 ≔ exp (6c6)1113864 (log x)(23)
(log2 x)(13) Let z isin [2N0
1113968] be a parameter to be chosen
laterPutting d [xn] we have (xn) minus 1lt dle (xn) and
x(d + 1)lt nle (xd) We have with the conventionσ(0) 0
Sσ(x) 1113944dlex
σ(d) 1113944(x(d+1))ltnle(xd)
1
1113944dnle x
σ(d) minus 1113944dnle xdge2
σ(d minus 1)
1113944dnlex
(σ(d) minus σ(d minus 1))
(52)
By the hyperbole principle of Dirichlet we can write
Sσ(x) S1(x σ) + S2(x σ) minus S3(x σ) (53)
where
S1(x σ) ≔ 1113944dle
x
radicdnle x
(σ(d) minus σ(d minus 1))
S2(x σ) ≔ 1113944nle
x
radicdnlex
(σ(d) minus σ(d minus 1))
S3(x σ) ≔ 1113944dle
x
radicnle
x
radic(σ(d) minus σ(d minus 1))
(54)
With the help of the bound σ(n)≪ n log2 n we canderive that
S3(x σ) [x
radic]σ([
x
radic])≪x log2 x (55)
For evaluating S1(x σ) we write
S1(x σ) 1113944dle
x
radic(σ(d) minus σ(d minus 1))
x
d1113876 1113877
x 1113944dle
x
radic
σ(d) minus σ(d minus 1)
d+ O 1113944
dlex
radic|σ(d) minus σ(d minus 1)|⎛⎝ ⎞⎠
(56)
With the help of Lemma 5 (ii) a simple partial inte-gration gives us
Journal of Mathematics 5
1113944dle
x
radic
σ(d) minus σ(d minus 1)
d 1113944
dlex
radic
σ(d)
d(d + 1)
1113944dle
x
radic
σ(d)
d2 minus 1113944
dlex
radic
σ(d)
d2(d + 1)
1113946
x
radic
1minustminus 2d
π2
12t2
+ O(t log t)1113888 1113889 + O(1)
π2
12log x + O(1)
1113944dle
x
radic|σ(d) minus σ(d minus 1)|
≪ 1113944dle
x
radicσ(d)≪x
(57)
Inserting these estimates into (56) we find that
S1(x σ) π2
12x log x + O(x) (58)
Finally we evaluate S2(x σ) For this we write
S2(x σ) Sdagger2(x σ) + S
♯2(x σ) (59)
where
Sdagger2(x σ) ≔ 1113944
nleN0dnle x
(σ(d) minus σ(d minus 1))
S♯2(x σ) ≔ 1113944
N0ltnlex
radicdnle x
(σ(d) minus σ(d minus 1))(60)
By the bound that σ(n)≪ n log2 n we have
Sdagger2(x σ) 1113944
nleN0
σx
n1113876 11138771113874 1113875≪ 1113944
nleN0
x
n1113874 1113875log2 x
≪ x(log x)(23) log2 x( 1113857
(43)
(61)
On the other hand (27) of Lemma 5 allows us to derivethat
1113944dlex
σ(d) minus 1113944dlexminus 1
σ(d) π2
12x2
minus (x minus 1)2
1113872 1113873 minus Δ(x z)
+ Δ(x minus 1 z) + Ox
z1113874 1113875
π2
6x minus Δ(x z) + Δ(x minus 1 z) + O
x
z1113874 1113875
(62)
where Δ(x z) is given by (28) us
S♯2(x σ) 1113944
N0ltnlex
radic
π2
6middotx
nminus Δ
x
n z1113874 1113875 + Δ
x
nminus 1 z1113874 1113875 + O
x
nz1113874 11138751113888 1113889
π2
12x log x + O x(log x)
(23) log2 x( 1113857(13)
+ xzminus 1log x1113872 1113873 minus Δ1(x z) + Δ2(x z)
(63)
where
Δ1(x z) ≔ 1113944N0ltnle
x
radicΔ
x
n z1113874 1113875≪ x(log x)
minus 3+ xz
minus 1log x
Δ2(x z) ≔ 1113944N0ltnle
x
radicΔ
x
nminus 1 z1113874 1113875≪x(log x)
minus 3+ xz
minus 1log x
(64)
thanks to Lemma 6 Inserting these estimates into (59) wefind that
S2(x σ) π2
12x log x + O x(log x)
(23) log2 x( 1113857(43)
+ xzminus 1log x1113872 1113873
(65)
Now (5) follows from (53) (55) (58) and (66) with thechoice of z (log x)(13)
34 Proof ofeorem1 (ii) For any odd prime p (52) allowsus to write
1113944d ∣p
(σ(d) minus σ(d minus 1)) Sσ(p) minus Sσ(p minus 1)
π2
6(log p minus log(p minus 1)) + E(p) minus E(p minus 1)
geE(p) minus E(p minus 1)ge minus 2Elowast(p)
(66)
where Elowast(p) ≔ max |E(p)| |E(p minus 1)|1113864 1113865 On the otherhand we have
1113944d|p
(σ(d) minus σ(d minus 1)) σ(p) minus σ(p minus 1) + 1
lep + 1 minus p minus 1 +12(p minus 1) + 2 + 11113874 1113875 + 1le minus
14p
(67)
us Elowast(p)ge (18)p for all odd primes
Data Availability
No data were used to support this study
6 Journal of Mathematics
Conflicts of Interest
e authors declare that they have no conflicts of interest
Acknowledgments
is study was supported in part by the National NaturalScience Foundation of China (grant nos 1187119311971370 and 12071375) Natural Science Foundation ofChongqing (grant no cstc2019jcyj-msxm1651) NationalNatural Science Foundation of Henan Province (grant no202300410274) and Research Projects for Overseas Re-searchers of Department of Human Resources and SocialSecurity of Henan Province 2020
References
[1] O Bordelles L Dai R Heyman H Pan and I E ShparlinskildquoOn a sum involving the Euler functionrdquo Journal of Numbereory vol 202 pp 278ndash297 2019
[2] J Wu ldquoOn a sum involving the Euler totient functionrdquoIndagationes Mathematicae vol 30 no 4 pp 536ndash541 2019
[3] W Zhai ldquoOn a sum involving the euler functionrdquo Journal ofNumber eory vol 211 pp 199ndash219 2020
[4] J Ma and J Wu ldquoOn a sum involving the Mangoldt functionrdquoPeriodica Mathematica Hungarica vol 80 2020
[5] J Wu ldquoNote on a paper by Bordelles Dai heyman Pan andShparlinskirdquo Periodica Mathematica Hungarica vol 80 no 1pp 95ndash102 2020
[6] A A Karatsuba ldquoEstimates for trigonometric sums byVinogradovrsquos method and some applicationsrdquo Proceedings ofthe Steklov Institute of Mathematics vol 112 pp 251ndash265 1971
[7] Y-F S Petermann and J Wu ldquoOn the sum of exponentialdivisors of an integerrdquo Acta Mathematica Hungarica vol 77no 12 pp 159ndash175 1997
[8] S W Graham and G Kolesnik Van der Corputrsquos Method ofExponential Sums Cambridge University Press CambridgeUK 1991
Journal of Mathematics 7
1113946n+1
nt minus n minus
12
1113874 1113875dF(t) 12(F(n + 1) + F(n)) minus 1113946
n+1
nF(t)dt
(44)
From this we can derive that
12(F(n + 1) + F(n)) 1113946
n+1
nF(t)dt + O VF[n n + 1]( 1113857
(45)
for nge 1 Summing over n we find that
1113944N1lt nleN2
F(n) 1113946N2
N1
F(t)dt
+12
F N1( 1113857 + F N2( 1113857( 1113857 + O VF N1 N21113858 1113859( 1113857
(46)
We apply this formula to
F(xd)(t) 1tψ
(xd)
t1113888 1113889
N1 (xd)(23)
1113960 1113961
N2 minx
radic
x
(dz)1113896 11138971113890 1113891
(47)
According to Lemma 4 we haveVF(xd)
[N1 N2]≪ (xd)minus (13) and thus by puttingu (xd)t we obtain with the notationxd1 max(
x
radicd tz) and xd2 (xd)(13)
1113944
(xd)(23)ltnleminx
radic(x(dz))
1nψ
x
dn1113874 1113875 1113946
xd2
xd1
ψ(u)
udu + O
x
d1113874 1113875
minus (13)
1113888 1113889
≪ zminus 1
+x
d1113874 1113875
minus (13)
≪ zminus 1
(48)
where we have used the fact that zleN0
1113968and
dle (x(N0z))rArrz le (xd)(13) and the bound
1113946xd2
xd1
ψ(u)
udu 1113946
xd2
xd1
1113946u
xd1
ψ(t)dt1113888 1113889du
u2 minus
1x
(23)d2
1113946xd2
xd1
ψ(t)dt
≪xminus 1d1 + xd21113872 1113873
minus (23)≪ z
minus 1+(xd)
minus (23)≪ zminus 1
(49)
Using (48) a simple partial integration allows us toderive that
Δ♯1(x z)≪ zminus 1
1113944
dlex N0z( )
dminus 1≪ z
minus 1log x(50)
Combining (43) and (50) it follows that
Δ1(x z)1113868111386811138681113868
1113868111386811138681113868≪ x(log x)minus 3
+ xzminus 1log x (51)
Similarly we can prove the same bound for |Δ2(x z)|is completes the proof
33 End of the Proof of eorem 1 Let c6 be the constantgiven as in Lemma 3 and N0 ≔ exp (6c6)1113864 (log x)(23)
(log2 x)(13) Let z isin [2N0
1113968] be a parameter to be chosen
laterPutting d [xn] we have (xn) minus 1lt dle (xn) and
x(d + 1)lt nle (xd) We have with the conventionσ(0) 0
Sσ(x) 1113944dlex
σ(d) 1113944(x(d+1))ltnle(xd)
1
1113944dnle x
σ(d) minus 1113944dnle xdge2
σ(d minus 1)
1113944dnlex
(σ(d) minus σ(d minus 1))
(52)
By the hyperbole principle of Dirichlet we can write
Sσ(x) S1(x σ) + S2(x σ) minus S3(x σ) (53)
where
S1(x σ) ≔ 1113944dle
x
radicdnle x
(σ(d) minus σ(d minus 1))
S2(x σ) ≔ 1113944nle
x
radicdnlex
(σ(d) minus σ(d minus 1))
S3(x σ) ≔ 1113944dle
x
radicnle
x
radic(σ(d) minus σ(d minus 1))
(54)
With the help of the bound σ(n)≪ n log2 n we canderive that
S3(x σ) [x
radic]σ([
x
radic])≪x log2 x (55)
For evaluating S1(x σ) we write
S1(x σ) 1113944dle
x
radic(σ(d) minus σ(d minus 1))
x
d1113876 1113877
x 1113944dle
x
radic
σ(d) minus σ(d minus 1)
d+ O 1113944
dlex
radic|σ(d) minus σ(d minus 1)|⎛⎝ ⎞⎠
(56)
With the help of Lemma 5 (ii) a simple partial inte-gration gives us
Journal of Mathematics 5
1113944dle
x
radic
σ(d) minus σ(d minus 1)
d 1113944
dlex
radic
σ(d)
d(d + 1)
1113944dle
x
radic
σ(d)
d2 minus 1113944
dlex
radic
σ(d)
d2(d + 1)
1113946
x
radic
1minustminus 2d
π2
12t2
+ O(t log t)1113888 1113889 + O(1)
π2
12log x + O(1)
1113944dle
x
radic|σ(d) minus σ(d minus 1)|
≪ 1113944dle
x
radicσ(d)≪x
(57)
Inserting these estimates into (56) we find that
S1(x σ) π2
12x log x + O(x) (58)
Finally we evaluate S2(x σ) For this we write
S2(x σ) Sdagger2(x σ) + S
♯2(x σ) (59)
where
Sdagger2(x σ) ≔ 1113944
nleN0dnle x
(σ(d) minus σ(d minus 1))
S♯2(x σ) ≔ 1113944
N0ltnlex
radicdnle x
(σ(d) minus σ(d minus 1))(60)
By the bound that σ(n)≪ n log2 n we have
Sdagger2(x σ) 1113944
nleN0
σx
n1113876 11138771113874 1113875≪ 1113944
nleN0
x
n1113874 1113875log2 x
≪ x(log x)(23) log2 x( 1113857
(43)
(61)
On the other hand (27) of Lemma 5 allows us to derivethat
1113944dlex
σ(d) minus 1113944dlexminus 1
σ(d) π2
12x2
minus (x minus 1)2
1113872 1113873 minus Δ(x z)
+ Δ(x minus 1 z) + Ox
z1113874 1113875
π2
6x minus Δ(x z) + Δ(x minus 1 z) + O
x
z1113874 1113875
(62)
where Δ(x z) is given by (28) us
S♯2(x σ) 1113944
N0ltnlex
radic
π2
6middotx
nminus Δ
x
n z1113874 1113875 + Δ
x
nminus 1 z1113874 1113875 + O
x
nz1113874 11138751113888 1113889
π2
12x log x + O x(log x)
(23) log2 x( 1113857(13)
+ xzminus 1log x1113872 1113873 minus Δ1(x z) + Δ2(x z)
(63)
where
Δ1(x z) ≔ 1113944N0ltnle
x
radicΔ
x
n z1113874 1113875≪ x(log x)
minus 3+ xz
minus 1log x
Δ2(x z) ≔ 1113944N0ltnle
x
radicΔ
x
nminus 1 z1113874 1113875≪x(log x)
minus 3+ xz
minus 1log x
(64)
thanks to Lemma 6 Inserting these estimates into (59) wefind that
S2(x σ) π2
12x log x + O x(log x)
(23) log2 x( 1113857(43)
+ xzminus 1log x1113872 1113873
(65)
Now (5) follows from (53) (55) (58) and (66) with thechoice of z (log x)(13)
34 Proof ofeorem1 (ii) For any odd prime p (52) allowsus to write
1113944d ∣p
(σ(d) minus σ(d minus 1)) Sσ(p) minus Sσ(p minus 1)
π2
6(log p minus log(p minus 1)) + E(p) minus E(p minus 1)
geE(p) minus E(p minus 1)ge minus 2Elowast(p)
(66)
where Elowast(p) ≔ max |E(p)| |E(p minus 1)|1113864 1113865 On the otherhand we have
1113944d|p
(σ(d) minus σ(d minus 1)) σ(p) minus σ(p minus 1) + 1
lep + 1 minus p minus 1 +12(p minus 1) + 2 + 11113874 1113875 + 1le minus
14p
(67)
us Elowast(p)ge (18)p for all odd primes
Data Availability
No data were used to support this study
6 Journal of Mathematics
Conflicts of Interest
e authors declare that they have no conflicts of interest
Acknowledgments
is study was supported in part by the National NaturalScience Foundation of China (grant nos 1187119311971370 and 12071375) Natural Science Foundation ofChongqing (grant no cstc2019jcyj-msxm1651) NationalNatural Science Foundation of Henan Province (grant no202300410274) and Research Projects for Overseas Re-searchers of Department of Human Resources and SocialSecurity of Henan Province 2020
References
[1] O Bordelles L Dai R Heyman H Pan and I E ShparlinskildquoOn a sum involving the Euler functionrdquo Journal of Numbereory vol 202 pp 278ndash297 2019
[2] J Wu ldquoOn a sum involving the Euler totient functionrdquoIndagationes Mathematicae vol 30 no 4 pp 536ndash541 2019
[3] W Zhai ldquoOn a sum involving the euler functionrdquo Journal ofNumber eory vol 211 pp 199ndash219 2020
[4] J Ma and J Wu ldquoOn a sum involving the Mangoldt functionrdquoPeriodica Mathematica Hungarica vol 80 2020
[5] J Wu ldquoNote on a paper by Bordelles Dai heyman Pan andShparlinskirdquo Periodica Mathematica Hungarica vol 80 no 1pp 95ndash102 2020
[6] A A Karatsuba ldquoEstimates for trigonometric sums byVinogradovrsquos method and some applicationsrdquo Proceedings ofthe Steklov Institute of Mathematics vol 112 pp 251ndash265 1971
[7] Y-F S Petermann and J Wu ldquoOn the sum of exponentialdivisors of an integerrdquo Acta Mathematica Hungarica vol 77no 12 pp 159ndash175 1997
[8] S W Graham and G Kolesnik Van der Corputrsquos Method ofExponential Sums Cambridge University Press CambridgeUK 1991
Journal of Mathematics 7
1113944dle
x
radic
σ(d) minus σ(d minus 1)
d 1113944
dlex
radic
σ(d)
d(d + 1)
1113944dle
x
radic
σ(d)
d2 minus 1113944
dlex
radic
σ(d)
d2(d + 1)
1113946
x
radic
1minustminus 2d
π2
12t2
+ O(t log t)1113888 1113889 + O(1)
π2
12log x + O(1)
1113944dle
x
radic|σ(d) minus σ(d minus 1)|
≪ 1113944dle
x
radicσ(d)≪x
(57)
Inserting these estimates into (56) we find that
S1(x σ) π2
12x log x + O(x) (58)
Finally we evaluate S2(x σ) For this we write
S2(x σ) Sdagger2(x σ) + S
♯2(x σ) (59)
where
Sdagger2(x σ) ≔ 1113944
nleN0dnle x
(σ(d) minus σ(d minus 1))
S♯2(x σ) ≔ 1113944
N0ltnlex
radicdnle x
(σ(d) minus σ(d minus 1))(60)
By the bound that σ(n)≪ n log2 n we have
Sdagger2(x σ) 1113944
nleN0
σx
n1113876 11138771113874 1113875≪ 1113944
nleN0
x
n1113874 1113875log2 x
≪ x(log x)(23) log2 x( 1113857
(43)
(61)
On the other hand (27) of Lemma 5 allows us to derivethat
1113944dlex
σ(d) minus 1113944dlexminus 1
σ(d) π2
12x2
minus (x minus 1)2
1113872 1113873 minus Δ(x z)
+ Δ(x minus 1 z) + Ox
z1113874 1113875
π2
6x minus Δ(x z) + Δ(x minus 1 z) + O
x
z1113874 1113875
(62)
where Δ(x z) is given by (28) us
S♯2(x σ) 1113944
N0ltnlex
radic
π2
6middotx
nminus Δ
x
n z1113874 1113875 + Δ
x
nminus 1 z1113874 1113875 + O
x
nz1113874 11138751113888 1113889
π2
12x log x + O x(log x)
(23) log2 x( 1113857(13)
+ xzminus 1log x1113872 1113873 minus Δ1(x z) + Δ2(x z)
(63)
where
Δ1(x z) ≔ 1113944N0ltnle
x
radicΔ
x
n z1113874 1113875≪ x(log x)
minus 3+ xz
minus 1log x
Δ2(x z) ≔ 1113944N0ltnle
x
radicΔ
x
nminus 1 z1113874 1113875≪x(log x)
minus 3+ xz
minus 1log x
(64)
thanks to Lemma 6 Inserting these estimates into (59) wefind that
S2(x σ) π2
12x log x + O x(log x)
(23) log2 x( 1113857(43)
+ xzminus 1log x1113872 1113873
(65)
Now (5) follows from (53) (55) (58) and (66) with thechoice of z (log x)(13)
34 Proof ofeorem1 (ii) For any odd prime p (52) allowsus to write
1113944d ∣p
(σ(d) minus σ(d minus 1)) Sσ(p) minus Sσ(p minus 1)
π2
6(log p minus log(p minus 1)) + E(p) minus E(p minus 1)
geE(p) minus E(p minus 1)ge minus 2Elowast(p)
(66)
where Elowast(p) ≔ max |E(p)| |E(p minus 1)|1113864 1113865 On the otherhand we have
1113944d|p
(σ(d) minus σ(d minus 1)) σ(p) minus σ(p minus 1) + 1
lep + 1 minus p minus 1 +12(p minus 1) + 2 + 11113874 1113875 + 1le minus
14p
(67)
us Elowast(p)ge (18)p for all odd primes
Data Availability
No data were used to support this study
6 Journal of Mathematics
Conflicts of Interest
e authors declare that they have no conflicts of interest
Acknowledgments
is study was supported in part by the National NaturalScience Foundation of China (grant nos 1187119311971370 and 12071375) Natural Science Foundation ofChongqing (grant no cstc2019jcyj-msxm1651) NationalNatural Science Foundation of Henan Province (grant no202300410274) and Research Projects for Overseas Re-searchers of Department of Human Resources and SocialSecurity of Henan Province 2020
References
[1] O Bordelles L Dai R Heyman H Pan and I E ShparlinskildquoOn a sum involving the Euler functionrdquo Journal of Numbereory vol 202 pp 278ndash297 2019
[2] J Wu ldquoOn a sum involving the Euler totient functionrdquoIndagationes Mathematicae vol 30 no 4 pp 536ndash541 2019
[3] W Zhai ldquoOn a sum involving the euler functionrdquo Journal ofNumber eory vol 211 pp 199ndash219 2020
[4] J Ma and J Wu ldquoOn a sum involving the Mangoldt functionrdquoPeriodica Mathematica Hungarica vol 80 2020
[5] J Wu ldquoNote on a paper by Bordelles Dai heyman Pan andShparlinskirdquo Periodica Mathematica Hungarica vol 80 no 1pp 95ndash102 2020
[6] A A Karatsuba ldquoEstimates for trigonometric sums byVinogradovrsquos method and some applicationsrdquo Proceedings ofthe Steklov Institute of Mathematics vol 112 pp 251ndash265 1971
[7] Y-F S Petermann and J Wu ldquoOn the sum of exponentialdivisors of an integerrdquo Acta Mathematica Hungarica vol 77no 12 pp 159ndash175 1997
[8] S W Graham and G Kolesnik Van der Corputrsquos Method ofExponential Sums Cambridge University Press CambridgeUK 1991
Journal of Mathematics 7
Conflicts of Interest
e authors declare that they have no conflicts of interest
Acknowledgments
is study was supported in part by the National NaturalScience Foundation of China (grant nos 1187119311971370 and 12071375) Natural Science Foundation ofChongqing (grant no cstc2019jcyj-msxm1651) NationalNatural Science Foundation of Henan Province (grant no202300410274) and Research Projects for Overseas Re-searchers of Department of Human Resources and SocialSecurity of Henan Province 2020
References
[1] O Bordelles L Dai R Heyman H Pan and I E ShparlinskildquoOn a sum involving the Euler functionrdquo Journal of Numbereory vol 202 pp 278ndash297 2019
[2] J Wu ldquoOn a sum involving the Euler totient functionrdquoIndagationes Mathematicae vol 30 no 4 pp 536ndash541 2019
[3] W Zhai ldquoOn a sum involving the euler functionrdquo Journal ofNumber eory vol 211 pp 199ndash219 2020
[4] J Ma and J Wu ldquoOn a sum involving the Mangoldt functionrdquoPeriodica Mathematica Hungarica vol 80 2020
[5] J Wu ldquoNote on a paper by Bordelles Dai heyman Pan andShparlinskirdquo Periodica Mathematica Hungarica vol 80 no 1pp 95ndash102 2020
[6] A A Karatsuba ldquoEstimates for trigonometric sums byVinogradovrsquos method and some applicationsrdquo Proceedings ofthe Steklov Institute of Mathematics vol 112 pp 251ndash265 1971
[7] Y-F S Petermann and J Wu ldquoOn the sum of exponentialdivisors of an integerrdquo Acta Mathematica Hungarica vol 77no 12 pp 159ndash175 1997
[8] S W Graham and G Kolesnik Van der Corputrsquos Method ofExponential Sums Cambridge University Press CambridgeUK 1991
Journal of Mathematics 7
