The Higher Power Moments of the Coefficients of the Dedekind Zeta Function over a Polynomial in Six Variables

#A101 INTEGERS 25 (2025)
THE HIGHER POWER MOMENTS OF THE COEFFICIENTS OF
THE DEDEKIND ZETA FUNCTION OVER A POLYNOMIAL IN
SIX VARIABLES
Na een K. Goda a
Depa men o Ma hema ics, Indian Ins i u e o Technology Mad as, Chennai,
Tamil Nadu, India
[email p o ec ed]
P ashan Tiwa i
Depa men o Ma hema ics, Indian Ins i u e o Technology Ropa , Rupnaga ,
Punjab, India
[email p o ec ed]
Recei ed: 4/10/25, Accep ed: 10/26/25, Published: 11/5/25
Abs ac
Le Kbe a non-no mal ield o deg ee 3 o e Q. Le ℓ≥2 be an in ege . In
his pape , we in es iga e he ℓ h powe momen o he coe icien s a ached o
he Dedekind ze a unc ion ζK(s) o e a sequence. In pa icula , we conside he
ollowing:
Sℓ(x) := X
n=x2
1+x2
2+x2
3+x2
4+x2
5+x2
6≤x
(x1,x2,x3,x4,x5,x6)∈Z6
aℓ
K(n)
and es ablish an asymp o ic esul , whe e ζK(s) = P∞
n=1
aK(n)
nsis he Dedekind ze a
unc ion.
1. In oduc ion
Le Kbe a numbe ield o deg ee [K:Q] = dand OKbe i s ing o in ege s. Then
he Dedekind ze a unc ion a ached o Kis de ined as
ζK(s) := Y
p⊆OK
p=0 1−1
(Np)s−1
=X
u⊆OK
u=0
1
(Nu)s(1)
o Re(s)>1, whe e he p oduc is o e non-ze o p ime ideals in OKand Nu
deno es he absolu e no m o u. Fo K=Q, he Dedekind ze a unc ion ζK(s) is
DOI: 10.5281/zenodo.17535302
INTEGERS: 25 (2025) 2
he Riemann ze a unc ion. The unc ion ζK(s) ex ends analy ically o he en i e
complex plane excep o a simple pole a s= 1, and he esidue a s= 1 is gi en
by he analy ic class numbe o mula
lim
s→1+(s−1)ζK(s) = 2 1(2π) 2hR
ωp|DK|,
whe e 1is he numbe o eal embeddings o K, 2 2is he numbe o complex
embeddings o K, h deno es he class numbe , Ris he egula o , ωis he numbe o
oo s o uni y in K, and DKis he disc iminan o K. The Dedekind ze a unc ion
sa is ies a unc ional equa ion simila o he Riemann ze a unc ion,
ξK(s) = ξK(s−1),
whe e
ξK(s) := p|DK|
2 2πn/2s
Γs
2 1
Γ(s) 2ζK(s),
which is analy ic in he whole complex plane excep o he simple poles a s= 0
and s= 1.
We ew i e Equa ion (1) as a Di ichle se ies
ζK(s) =
∞
X
n=1
aK(n)
ns,Re(s)>1,(2)
whe e aK(n) deno es he numbe o in eg al ideals in Kwi h no m n. The coe icien s
aK(n) o he Dedekind ze a unc ion ζK(s) in Equa ion (2) sa is y he ollowing
p ope ies: o all n≥1, we ha e aK(n)≥0; o all cop ime in ege s mand n,
aK(mn) = aK(m)aK(n); and o any ϵ > 0,
aK(n)≤d(n)[K:Q]≪nϵ,
whe e d(n) deno es he di iso unc ion. Since he coe icien s aK(n) a e mul iplica-
i e, he Dedekind ze a unc ion has he Eule p oduc
ζK(s) =
∞
X
n=1
aK(n)
ns=Y
p1 + aK(p)
ps+aK(p2)
p2s+· · · +aK(pk)
pks +· · · (3)
o Re(s)>1.
In 1949, Landau [13] in es iga ed he i s momen o aK(n) o a numbe ield
Kwi h deg ee d≥2 and p o ed ha
X
n≤x
aK(n) = cKx+O(x1−2
d+1 +ϵ)
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o some cons an cKdepending on K. La e , Chand asekha an and Na asimhan
[4] s udied he second momen o aK(n) and p o ed ha
X
n≤x
a2
K(n)≪x(log x)d−1.
Fo a Galois ex ension Ko e Qo deg ee d > 1, Chand asekha an and Good [3]
p o ed ha X
n≤x
aℓ
K(n) = xPℓ(log x) + O(x1−2d−ℓ+ϵ),
o e e y ϵ > 0 and in ege ℓ≥2, whe e Pℓ(log x) deno es a sui able polynomial o
deg ee dℓ−1−1. Fo a non-no mal ield ex ension Ko e Qo deg ee 3 gi en by
an i educible polynomial x3+Ax2+Bx +Co disc iminan D < 0, Fomenko [6]
in es iga ed he i s and second momen s and p o ed
X
n≤x
a2
K(n) = c1xlog x+c2x+O(x9/11+ϵ)
and
X
n≤x
a3
K(n) = xP3(log x) + O(x73/79+ϵ),
whe e P3(log x) is a polynomial in log xo deg ee 4. L¨u [17] imp o ed he e o
e ms, and Liu [16] u he imp o ed he abo e-s a ed esul o Fomenko. Fo a
mo e ecen de elopmen , see [7].
Le Kbe a non-no mal ield o e Qo deg ee 3 gi en by an i educible polynomial
x3+Ax2+Bx +Co disc iminan D < 0. Fo a gi en na u al numbe ℓ≥2, we
conside he ℓ h powe momen o he Dedekind ze a unc ion associa ed wi h K
gi en by
Sℓ(x) := X
n=x2
1+x2
2+x2
3+x2
4+x2
5+x2
6≤x
(x1,x2,x3,x4,x5,x6)∈Z6
aℓ
K(n).(4)
In his pape , we p o e he ollowing esul s.
Theo em 1. Le Kbe a non-no mal ield o e Qo deg ee 3gi en by an i educible
polynomial x3+Ax2+Bx +Co disc iminan D < 0. Then o any ϵ > 0, we ha e
Sℓ(x) = (c1x3+c2x3log x+O(x25
9+ϵ)i ℓ= 2,
x3P3(log x) + O(x79
27 +ϵ)i ℓ= 3,
whe e c1, c2a e some sui able cons an s and P3(log x)is a polynomial in log xo
deg ee 4.
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Theo em 2. Le Kbe a non-no mal ield o e Qo deg ee 3gi en by an i educible
polynomial x3+Ax2+Bx +Co disc iminan D < 0, and le ℓ≥4be an in ege .
Then o any ϵ > 0, we ha e
Sℓ(x) = (x3Pℓ(log x) + O(x3−2
3ℓ+ϵ) o e en ℓ≥4,
x3Qℓ(log x) + O(x3−2
3ℓ+ϵ) o odd ℓ≥5,
whe e Pℓ(log x), Qℓ(log x)a e polynomials in log xo deg ee a0,ℓ +a3,ℓ −1. He e,
a0,ℓ = 1 +
⌊ℓ
2⌋
X
i=1 ℓ
2i(2i)!
i!(i+ 1)!
and
a3,ℓ =
⌈ℓ
2−1⌉
X
i=1 ℓ
2i+ 14(2i+ 1)!
(i−1)!(i+ 3)!.
2. P elimina ies
Th oughou he pape , ϵdeno es an a bi a ily small posi i e cons an , bu no
necessa ily he same one as o he s, and any implied cons an may depend on ϵ.
Le Hk(SL(2,Z)) be he se o no malized Hecke eigen o ms o e en in eg al
weigh k o he ull modula g oup SL(2,Z). Le {λ (n)}n∈Nbe he no malized
Fou ie coe icien s o he cusp o m ∈Hk(SL(2,Z)) a in ini y, i.e.,
(z) =
∞
X
n=1
λ (n)nk−1
2e2πιnz
o all z∈H, whe e His he Poinca ´e uppe hal -plane. The Fou ie coe icien s
λ (n) a e he Hecke eigen alues o , and hese a e eal numbe s. Also, λ (n)
sa is ies he ollowing ela ion:
λ (m)λ (n) = X
d|gcd (m,n)
λ mn
d2.(5)
As a consequence o Deligne’s [5] p oo o Weil’s conjec u es, we ha e
|λ (n)| ≤ d(n)≪ϵnϵ
o any ϵ > 0, whe e d(n) is he di iso unc ion.
The L- unc ion associa ed wi h he no malized Hecke eigen o m (z) =
∞
X
n=1
λ (n)nk−1
2e2πιnz
is de ined as
L( , s) =
∞
X
n=1
λ (n)
ns(6)
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o Re(s)>1. Since λ (mn) = λ (m)λ (n) o all posi i e in ege s mand nsuch
ha gcd(m, n) = 1, he L- unc ion has he Eule p oduc
L( , s) = Y
p1−λ (p)
ps+1
p2s−1
=Y
p1−αp
ps−11−βp
ps−1
,
whe e αp+βp=λ (p) and αpβp=|αp|=|βp|= 1.
Fo a gi en Di ichle cha ac e χo modulus N, he wis ed L- unc ion is de ined
as
L( ⊗χ, s) :=
∞
X
n=1
λ (n)χ(n)
ns
o Re(s)>1.No e ha bo h he L- unc ions L( , s) and L( ⊗χ, s) ha e analy ic
con inua ions o he whole complex plane (see, [11, Sec ion 7.2]).
Fo j≥2, he j h symme ic powe L- unc ion o deg ee j+ 1 is de ined as
L(symj , s) := Y
p
j
Y
i=0 1−αpj−iβpip−s−1=
∞
X
n=1
λsymj (n)
ns o Re(s)>1,(7)
whe e λsymj (n) is a eal- alued mul iplica i e unc ion. F om Deligne’s bound [5],
we ha e
|λsymj (n)| ≤ dj+1(n)≪ϵnϵ
o any ϵ > 0, whe e dj(n) deno es he numbe o j ac o s o a posi i e in ege n.
Fo j≥2, we de ine he wis ed j h symme ic powe L- unc ion as
L(symj ⊗χ, s) :=
∞
X
n=1
λsymj (n)χ(n)
ns o Re(s)>1.(8)
Also, bo h he L- unc ions L(symj , s) and L(symj ⊗χ, s) ha e analy ic con in-
ua ions o he whole complex plane and sa is y nice unc ional equa ions ( o mo e
de ails, see [18, 19]).
Fo a gi en Di ichle cha ac e χo modulus N, he Di ichle L- unc ion is de ined
as
L(s, χ) =
∞
X
n=1
χ(n)
ns o Re(s)>1.
Deno e
L(symj , s) := (ζ(s) i j= 0,
L( , s) i j= 1
and
L(symj ⊗χ, s) := (L(s, χ) i j= 0,
L( ⊗χ, s) i j= 1.

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Le =
∞
X
n=1
λ (n)nk−1
2qn∈Hk(SL(2,Z)). The j h symme ic powe L- unc ion
can be w i en as
L(symj , s) = Y
p1 + λsymj (p)
ps+λsymj (p2)
p2s+· · · +λsymj (pk)
pks +· · · (9)
o Re(s)>1. The coe icien s λsymj (n) o he Di ichle se ies in Equa ion (7) and
he Fou ie coe icien s λ (n) sa is y
λ (pj) = λsymj (p) = αj+1
p−βj+1
p
αp−βp
.(10)
Le 6(n) deno e he numbe o ep esen a ions o a posi i e in ege nby a
polynomial x2
1+x2
2+x2
3+x2
4+x2
5+x2
6∈Q[x1, x2, x3, x4, x5, x6], i.e.,
6(n) := #{x:= (x1, x2, x3, x4, x5, x6)∈Z6|n=x2
1+x2
2+x2
3+x2
4+x2
5+x2
6}.
No e ha
6(n) = 16 X
m|n
χn
mm2−4X
m|n
χ(m)m2
:= 16 (n)−4 (n),(11)
whe e χis a non-p incipal Di ichle cha ac e o modulus 4. No e ha bo h (n)
and (n) a e mul iplica i e unc ions. Using Equa ion (11), he sum Sℓ(x) de ined
in Equa ion (4) can be exp essed as
Sℓ(x) = X
n=x2
1+x2
2+x2
3+x2
4+x2
5+x2
6≤x
aℓ
K(n) = X
n≤x
aℓ
K(n) 6(n)
= 16 X
n≤x
aℓ
K(n) (n)−4X
n≤x
aℓ
K(n) (n).(12)
Also, no ice ha
(p) = p2+χ(p), (p) = 1 + p2χ(p).(13)
F om [6], we lea n ha
ζK(s) = ζ(s)L( , s),
whe e is a holomo phic cusp o m o weigh 1 o he cong uence subg oup Γ0(|D|).
This implies
aK(n) = X
m|n
λ (m)
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and consequen ly, we ge
aK(p) = 1 + λ (p).(14)
Recen yea s ha e wi nessed an inc easing ocus on he a e age beha iou o
a i hme ical unc ions; e e o [2, 21] and he associa ed e e ences o u he
de ails.
3. Auxilia y Resul s
Fo a eal numbe m, we w i e ⌊m⌋and ⌈m⌉ o he loo and ceiling o m, espec-
i ely.
Since aℓ
K(p) = (1 + λ (p))ℓ,we ha e
aℓ
K(p) =
ℓ
X
i=0 ℓ
iλi
(p).(15)
Following Equa ion (10) and [14], we w i e
aℓ
K(p) = a0,ℓ +a1,ℓλ (p) +
ℓ
X
i=2
ai,ℓλsymi (p).(16)
Le ℓ= 2m o some m≥1. F om [14, Lemma 7.1, Equa ion (38)], we ha e
λℓ
(p) = λ2m
(p) = (2m)!
m!(m+ 1)! +
m−1
X
=1 (2m)!(2 + 1)!
(m− )!(m+ + 1)!!λsym2 (p) + λsym2m (p).
(17)
Le ℓ= 2m+ 1 be an odd in ege o some m≥1. Then we ha e
λ2m+1
(p) = 2(2m+ 1)!
m!(m+ 2)! +
m−1
X
=1 (2m+ 1)!(2 + 2)
(m− )!(m+ + 2)!!λsym2 +1 (p) + λsym2m+1 (p).
(18)
F om Equa ions (15), (16), (17), and (18), we ge
a0,ℓ =















1 +
ℓ
2
X
i=1 ℓ
2i(2i)!
i!(i+ 1)! o e en ℓ,
1 +
ℓ−1
2
X
i=1 ℓ
2i(2i)!
i!(i+ 1)! o odd ℓ
(19)
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and
a3,ℓ =















ℓ
2−1
X
i=1 ℓ
2i+ 14(2i+ 1)!
(i−1)!(i+ 3)! o e en ℓ,
ℓ−1
2
X
i=1 ℓ
2i+ 14(2i+ 1)!
(i−1)!(i+ 3)! o odd ℓ.
(20)
Nex , we conside he Di ichle se ies associa ed wi h aℓ
K(n) (n) and aℓ
K(n) (n)
gi en by
Rℓ(s) =
∞
X
n=1
aℓ
K(n) (n)
ns(21)
and
Tℓ(s) =
∞
X
n=1
aℓ
K(n) (n)
ns(22)
o Re(s)>3.
In his sec ion, we ind he L-decomposi ions o Rℓ(s) and Tℓ(s) in ol ing known
au omo phic L- unc ions. Th oughou , s∈Cdeno es he complex numbe s=
γ+ι , whe e Re(s) = γand Im(s) = .
Lemma 1. We ha e
Rℓ(s) = (ζ(s−2)a0,ℓ L(sym3 , s −2)a3,ℓ Wℓ(s)Hℓ(s) o e en ℓ≥4,
ζ(s−2)a0,ℓ L(sym3 , s −2)a3,ℓ W′
ℓ(s)H′
ℓ(s) o odd ℓ≥5,
whe e
Wℓ(s) = Y
1≤ 1≤ℓ
1=3
L(sym 1 , s −2)a 1,ℓ Y
0≤ 2≤ℓ
L(sym 2 ⊗χ, s)a 2,ℓ
and
W′
ℓ(s) = Y
1≤ ′
1≤ℓ
′
1=3
L(sym ′
1 , s −2)a ′
1,ℓ Y
0≤ ′
2≤ℓ
L(sym ′
2 ⊗χ, s)a ′
2,ℓ
o some sui able cons an s a 1,ℓ , a 2,ℓ , a ′
1,ℓ, a ′
2,ℓ. He e, Hℓ(s),H′
ℓ(s)con e ge ab-
solu ely and uni o mly in he igh hal -plane Re(s)>5
2wi h Hℓ(s),H′
ℓ(s)non-ze o
o Re(s)=3, whe e a0,ℓ and a3,ℓ a e gi en in Equa ions (19) and (20).
P oo . Since aℓ
K(n) (n) is a mul iplica i e unc ion, Rℓ(s) has he Eule p oduc
Rℓ(s) = Y
p1 + aℓ
K(p) (p)
ps+
∞
X
k=2
aℓ
K(pk) (pk)
pks (23)
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o Re(s)>3.
Le ℓ≥4 be an e en in ege . No e ha
aℓ
K(p) (p)=(a0,ℓ +a1,ℓλ (p) +
ℓ
X
i=2
ai,ℓλsymi (p))(p2+χ(p)) := c(p),
whe e a0,ℓ and a3,ℓ a e gi en in Equa ions (19) and (20). F om he s uc u e o
c(p), we de ine he Di ichle se ies associa ed wi h he coe icien s c(n) as
∞
X
n=1
c(n)
ns
=ζ(s−2)a0,ℓ L(sym3 , s −2)a3,ℓ Y
1≤ 1≤ℓ
1=3
L(sym 1 , s −2)a 1,ℓ Y
0≤ 2≤ℓ
L(sym 2 ⊗χ, s)a 2,ℓ ,
which is absolu ely con e gen in Re(s)>3, whe e a 1,ℓ , a 2,ℓ , a ′
1,ℓ, a ′
2,ℓ a e some
sui able cons an s. We also no e ha
Y
p1 + c(p)
ps+· · · +c(pm)
pms +· · · 
=ζ(s−2)a0,ℓ L(sym3 , s −2)a3,ℓ Y
1≤ 1≤ℓ
1=3
L(sym 1 , s −2)a 1,ℓ
×Y
0≤ 2≤ℓ
L(sym 2 ⊗χ, s)a 2,ℓ
=: Gℓ(s)
o Re(s)>3. Obse e ha c(n)≪ϵn2+ϵ o any ϵ > 0 and

c(p)
ps+c(p2)
p2s+· · · +c(pm)
pms +· · · ≪
∞
X
m=1
p(2+ϵ)m
pmγ <1
o Re(s)≥3 + ϵ.
W i e
P=aℓ
K(p) (p)
ps+· · · +aℓ
K(pm) (pm)
pms +· · ·
and
Q=c(p)
ps+· · · +c(pm)
pms +· · · .
F om he abo e calcula ions, we obse e ha |Q|<1 in Re(s)≥3 +ϵ. No ice ha ,
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and
Z5
2+ϵ+ιT
5
2+ϵ−ιT
T2(s)xs
sds≪x(5
2+ϵ)T5
4+ϵ.
The e o e,
4X
n≤x
a2
K(n) (n)≪x3+ϵ
T+x5
2+ϵT5
4+ϵ.
F om Equa ion (12), we ob ain
S2(x) = x3P2(log x) + Ox3+ϵ
T+Ox5
2+ϵT5
4+ϵ.
To balance he e o e ms, we choose T=x2
9, which implies
S2(x) = x3P2(log x) + O(x25
9+ϵ).
Fo ℓ= 3: Conside he sum Pn≤xa3
K(n) (n). F om Lemma 3 and Pe on’s
o mula, we ha e
X
n≤x
a3
K(n) (n) = 1
2πι Z3+ϵ+ιT
3+ϵ−ιT
R3(s)xs
sds +Ox3+ϵ
T,
whe e 1 ≤T≤x, o some T o be chosen la e . Using Cauchy’s esidue heo em,
we ge
X
n≤x
a3
K(n) (n) = Ress=3R3(s)xs
s
+1
2πι(Z5
2+ϵ+ιT
5
2+ϵ−ιT
+Z5
2+ϵ−ιT
3+ϵ−ιT
+Z3+ϵ+ιT
5
2+ϵ+ιT )R3(s)xs
sds
+Ox3+ϵ
T.
Since R3(s) has a pole a s= 3 o o de 5 coming ou om ζ(s−2)4and L(sym3 , s−
2) (see Rema k 1), we ind
16 Ress=3R3(s)xs
s=x3P3(log x),
whe e P3(log x) is a polynomial o deg ee 4 in log x. Again, by using Lemmas 5-7,

INTEGERS: 25 (2025) 17
we ge
Z5
2+ϵ−ιT
3+ϵ−ιT
+Z3+ϵ+ιT
5
2+ϵ+ιT )R3(s)xs
sds≪Z1+ϵ
1
2+ϵ
|R3(γ+ιT)|xγ+2
Tdγ
≪max
1
2+ϵ≤γ≤1+ϵ
xγ+2T(26
21 +10
3+18
5+2)(1−γ)−1+ϵ
≪x3+ϵ
T+x5
2+ϵT753
210 −1+ϵ
and
Z5
2+ϵ+ιT
5
2+ϵ−ιT
R3(s)xs
sds≪x(5
2+ϵ)T534
105 −13
14 +ϵ.
The e o e,
16 X
n≤x
a3
K(n) (n) = x3P3(log x) + Ox3+ϵ
T+Ox5
2+ϵT534
105 −13
14 +ϵ.
Now, conside Pn≤xa3
K3(n) (n). Since T3(s) is analy ic in he ob ained egion,
we apply Pe on’s o mula and Cauchy’s esidue heo em o ob ain
X
n≤x
a3
K(n) (n) = 1
2πι(Z5
2+ϵ+ιT
5
2+ϵ−ιT
+Z5
2+ϵ−ιT
3+ϵ−ιT
+Z3+ϵ+ιT
5
2+ϵ+ιT )T3(s)xs
sds +Ox3+ϵ
T.
On using Lemmas 5-7, we deduce
Z5
2+ϵ−ιT
3+ϵ−ιT
+Z3+ϵ+ιT
5
2+ϵ+ιT )T3(s)xs
sds≪x3+ϵ
T+x5
2+ϵT77
12 −1+ϵ
and applying he Cauchy-Schwa z inequali y, we ge
Z5
2+ϵ+ιT
5
2+ϵ−ιT
T3(s)xs
sds≪x(5
2+ϵ)T23
4+ϵ.
Hence,
4X
n≤x
a3
K(n) (n)≪x3+ϵ
T+x5
2+ϵT23
4+ϵ.
F om Equa ion (12), we ha e
S3(x) = x3P3(log x) + Ox3+ϵ
T+Ox5
2+ϵT23
4+ϵ.
INTEGERS: 25 (2025) 18
We balance he e o e ms by choosing T=x2
27 , which u he p o ides he desi ed
asymp o ic o mula
S3(x) = x3P3(log x) + O(x79
27 +ϵ).
This comple es he p oo .
P oo o Theo em 2. We only gi e he p oo when ℓ≥4 is an e en in ege , and
he o he case ollows simila ly. Conside he sum Pn≤xaℓ
K(n) (n). F om Lemma
1, we ha e
Rℓ(s) = ζ(s−2)a0,ℓ L(sym3 , s −2)a3,ℓ Y
1≤ 1≤ℓ
1=3
L(sym 1 , s −2)a 1,ℓ
×Y
0≤ 2≤ℓ
L(sym 2 ⊗χ, s)a 2,ℓ Hℓ(s).
F om [14, Lemma 2.4], we lea n ha he deg ee o
L(sym3 , s −2)a3,ℓ Y
1≤ 1≤ℓ
1=3
L(sym 1 , s −2)a 1,ℓ
is (3ℓ−a0,ℓ).F om Lemma 1 and Pe on’s o mula, we ha e
X
n≤x
aℓ
K(n) (n) = 1
2πι Z3+ϵ+ιT
3+ϵ−ιT
Rℓ(s)xs
sds +Ox3+ϵ
T.
Using Cauchy’s esidue heo em, we ge
X
n≤x
aℓ
K(n) (n) = Ress=3Rℓ(s)xs
s
+1
2πι(Z5
2+ϵ+ιT
5
2+ϵ−ιT
+Z5
2+ϵ−ιT
3+ϵ−ιT
+Z3+ϵ+ιT
5
2+ϵ+ιT )Rℓ(s)xs
sds
+Ox3+ϵ
T.
F om Rema k 1, he o de o he pole a s= 3 in Rℓ(s) is (a0,ℓ +a3,ℓ); he e o e,
we ha e
16Ress=3Rℓ(s)xs
s=x3Pℓ(log x),
whe e Pℓ(log x) is a polynomial in log xo deg ee (a0,ℓ +a3,ℓ)−1. We use esul s
om Subsec ion 3.1 o es ima e he in eg als appea ing in he p oo o his heo em,
INTEGERS: 25 (2025) 19
om which one can easily ind
Z5
2+ϵ−ιT
3+ϵ−ιT
+Z3+ϵ+ιT
5
2+ϵ+ιT )Rℓ(s)xs
sds≪Z1+ϵ
1
2+ϵ
|Rℓ(γ+ιT)|xγ+2
Tdγ
≪max
1
2+ϵ≤γ≤1+ϵ
xγ+2T13
42 (a0,ℓ)+ 3ℓ
2−a0,ℓ
2(1−γ)−1+ϵ
≪x3+ϵ
T+x5
2+ϵT13a0,ℓ+21(3ℓ−a0,ℓ)
84 −1+ϵ
and
Z5
2+ϵ+ιT
5
2+ϵ−ιT
Rℓ(s)xs
sds≪x(5
2+ϵ)T
−13
14 +13a0,ℓ+21(3ℓ−a0,ℓ)
84 +ϵ
.
The e o e,
16 X
n≤x
aℓ
K(n) (n) = x3Pℓ(log x)+Ox3+ϵ
T+Ox5
2+ϵT
−13
14 +13a0,ℓ+21(3ℓ−a0,ℓ)
84 +ϵ.
Now, conside he sum Pn≤xaℓ
K(n) (n). Since Tℓ(s) is analy ic (as in Lemma 2)
in he ob ained egion, we apply Pe on’s o mula and Cauchy’s esidue heo em o
ge
X
n≤x
aℓ
K(n) (n) = 1
2πι(Z5
2+ϵ+ιT
5
2+ϵ−ιT
+Z5
2+ϵ−ιT
3+ϵ−ιT
+Z3+ϵ+ιT
5
2+ϵ+ιT )Tℓ(s)xs
sds +Ox3+ϵ
T.
No e ha he deg ee o
Y
1≤ 2≤ℓ
L(sym 2 ⊗χ, s −2)a 2,ℓ
is (3ℓ−a0,ℓ).The e o e, we ob ain
Z5
2+ϵ−ιT
3+ϵ−ιT
+Z3+ϵ+ιT
5
2+ϵ+ιT )Tℓ(s)xs
sds≪x3+ϵ
T+x5
2+ϵT4a0,ℓ+6(3ℓ−a0,ℓ)
24 −1+ϵ
and applying he Cauchy-Schwa z inequali y, we ge
Z5
2+ϵ+ιT
5
2+ϵ−ιT
Tℓ(s)xs
sds≪x(5
2+ϵ)T3ℓ
4−1+ϵ.
INTEGERS: 25 (2025) 20
Hence,
4X
n≤x
aℓ
K(n) (n)≪x3+ϵ
T+x5
2+ϵT3ℓ
4−1+ϵ.
F om Equa ion (12), we ha e
Sℓ(x) = x3Pℓ(log x) + Ox3+ϵ
T+Ox5
2+ϵT3ℓ
4−1+ϵ.
We choose T=x2
3ℓ o ge
Sℓ(x) = x3Pℓ(log x) + Ox3−2
3ℓ+ϵ,
whe e Pℓ(log x) is a polynomial in log xo deg ee a0,ℓ +a3,ℓ −1. This comple es he
p oo .
Acknowledgemen . The au ho s hank hei a ilia ed ins i u ions o suppo
and he anonymous e e ees and managing edi o B uce Landman o hei aluable
commen s and sugges ions, which imp o ed he cla i y o he manusc ip . The
second au ho also acknowledges suppo om he NBHM pos doc o al ellowship.
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