dekoninck7. 256KB Jun 04 2011 12:08:35 AM
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Journal of Integer Sequences, Vol. 13 (2010),
Article 10.1.2
23 11
Some Remarks on a Paper of L. Toth
Jean-Marie De Koninck
D´ep. de math´ematiques
Universit´e Laval
Qu´ebec, PQ G1V 0A6
Canada
jmdk@mat.ulaval.ca
Imre K´atai
Computer Algebra Department
E¨otv¨os Lor´and University
P´azm´any P´eter S´et´any I/C
1117 Budapest
Hungary
katai@compalg.inf.elte.hu
Abstract
Pn
Consider the functions P (n) :=
k=1 gcd(k, n) (studied by Pillai in 1933) and
Q
e
P (n) := n p|n (2 − 1/p) (studied by Toth in 2009). From their results, one can
P
P
obtain asymptotic expansions for n≤x P (n)/n and n≤x Pe(n)/n. We consider two
wide classes of functions R and U of arithmetical functions which include P (n)/n and
Pe(n)/n
given R ∈ R and U ∈PU, we obtain asymptotic expansions
P For any P
P respectively.
for n≤x R(n), n≤x U (n), p≤x R(p − 1) and p≤x U (p − 1).
1
Introduction
In 1933, Pillai [6] introduced the function
P (n) :=
n
X
k=1
1
gcd(k, n)
and showed that
P (n) =
X
dϕ(n/d)
d|n
and
X
P (d) = nτ (n) =
d|n
X
σ(d)ϕ(n/d),
d|n
where ϕ stands for the Euler function and where τ (n) and σ(n) stand for the number of
divisors of n and the sum of the divisors of n respectively.
It is easily shown that
Y
a/(a + 1)
P (n) = nτ (n)
1−
.
p
a
p kn
In 1985, Chidambaraswamy and Sitaramachandrarao [2] showed that, given an arbitrary
ε > 0,
X
(1)
P (n) = e1 x2 log x + e2 x2 + O x1+θ+ε ,
n≤x
1
1 ζ ′ (2)
1
where e1 =
2γ − −
, and where θ is the constant appearing
and e2 =
2ζ(2)
2ζ(2)
2
ζ(2)
below in Lemma 5, ζ stands for the Riemann Zeta Function and γ stands for Euler’s constant.
Using partial summation, one easily deduces from (1) that
X P (n)
n≤x
n
= e1 x log x + (2e2 − e1 )x + O xθ+ε .
In [10], Toth introduced the function
Y
Y
1
1
ω(n)
e
1−
2−
P (n) = n
=n·2
,
p
2p
p|n
(2)
p|n
where ω(n) stands for the number of distinct prime factors of n. Toth obtained an estiP
mate for n≤x Pe(n), analogous to (1) and from which one can easily derive an asymptotic
P
expansion for n≤x Pe(n)/n.
In this paper, we consider two wide classes of arithmetical functions R and U, the first
of which includes the function P (n)/n, and theP
second of which includes Pe(n)/n. Given
R ∈ R, we obtain an asymptotic
expansion P
for n≤x R(n); similarly for U ∈ U. We then
P
examine the behavior of p≤x R(p − 1) and p≤x U (p − 1).
More precisely, the class R is made
Q of the following functions R. First, let γ(n) stand for
the kernel of n ≥ 2, that is γ(n) = p|n p (with γ(1) = 1). Then, given an arbitrary positive
constant c, an arbitrary real number α > 0 and a multiplicative function κ(n) satisfying
c
|κ(n)| ≤
for all n ≥ 2, let R ∈ R be defined by
γ(n)α
X
Y
R(n) = Rκ,c,α (n) := τ (n)
κ(d) = τ (n)
(1 + κ(pa )).
(3)
dkn
2
pa kn
Here dkn means that the sum runs over the unitary divisors of n, that is over all divisors d
of n for which (d, n/d) = 1.
a/(a + 1)
It is easily seen that if we let κ(pa ) = −
, then the corresponding function R(n)
p
is precisely P (n)/n.
As for the class of functions U, it is made of the functions
X
h(d),
(4)
U (n) = Uh,c,α (n) := 2ω(n)
d|n
c
for each integer d ≥ 2, where
γ(d)α
1
and h(pa ) = 0 for a ≥ 2,
α > 0 is a given number. It is easily seen that by taking h(p) = − 2p
we obtain the particular case U (n) = Pe(n)/n.
Throughout this paper, c1 , c2 , . . . denote absolute positive constants.
where h is a multiplicative function satisfying |h(d)| ≤
2
Main results
Theorem 1. Let R be as in (3). For any arbitrary ε > 0, as x → ∞,
X
T (x) :=
R(n) = A0 x log x + B0 x + O xβ ,
n≤x
with
θ + ε,
if α ≥ 1 − θ;
1 − α + ε, if α < 1 − θ;
where θ is the number mentioned in Lemma 5 below and where
X λ(d)
X λ(d)
and
B0 =
(2γ − 1 − log d),
A0 =
d
d
d≥1
d≥1
β=
(5)
the function λ being defined below in (20) and (21).
Theorem 2. Let U be as in (4). As x → ∞,
X
x
,
S(x) :=
U (n) = t1 x log x + t2 x + O
log
x
n≤x
where
t1
∞
∞
X
h(δ)2ω(δ) B1 (δ)ηδ X 1
=
,
δ
n
n=1
δ=1
γ(n)|δ
t2
∞
X
h(δ)2ω(δ)
=
B2 (δ)ηδ − 2B1 (δ)µδ − B1 (δ)ηδ log(δb) ,
δb
δ=1
b∈Gδ
where B1 (δ) and B2 (δ) are defined below in Lemma 7, while ηδ and µδ are defined respectively
in (35) and (36).
3
Theorem 3. Let R be as in (3). As x → ∞,
X
M (x) :=
R(p − 1) = K1 x + O
p≤x
where K1 =
x
log log x
,
(6)
∞
1 X κ(d)τ (d)cd
, where cd is itself defined in Lemma 15.
2 d=1
d
Theorem 4. Let U be as in (4). As x → ∞,
N (x) :=
X
p≤x
U (p − 1) = K2 x + O
x
log log x
,
(7)
where K2 is a positive constant which may depend on the function κ.
3
Preliminary results
Lemma 5. As x → ∞,
X
D(x) :=
τ (n) = x log x + (2γ − 1)x + O(xθ )
(x → ∞),
n≤x
(8)
for some positive constant θ < 1/3.
Proof. A proof can be found in the book of Ivi´c [4], where one can also find a history of the
improvements concerning the size of θ.
Lemma 6. Given 1 ≤ ℓ < k with gcd(ℓ, k) = 1,
X
τ (n) = A1 (k)x log x + A2 (k)x + O k 7/3 x1/3 log x ,
n≡ℓ
n≤x
(mod k)
where
A1 (k) =
ϕ(k)
,
k2
A2 (k) = (2γ − 1)
ϕ(k) 2 X µ(d) log d
−
.
k2
k
d
d|k
Proof. This result is due to Tolev [9].
Lemma 7. Given a positive integer k,
X
τ (n) = B1 (k)x log x + B2 (k)x + O k 10/3 x1/3 log x ,
n≤x
(n,k)=1
where B1 (k) =
ϕ(k)
k
2
and
B2 (k) = ϕ(k)A2 (k).
4
Proof. Observing that
X
τ (n) =
k
X
X
τ (n),
ℓ=1
n≤x
gcd(ℓ,k)=1 n≡ℓ (mod k)
n≤x
gcd(n,k)=1
and using the trivial fact that ϕ(k) ≤ k, the result follows immediately from Lemma 6.
Lemma 8. Given an arbitrary positive real number α < 1,
X
1
1
≪ α.
α
nγ(n)
x
n>x
Proof. Writing each number n as n = rm, where r and m are the square-full and square-free
parts of n respectively with (r, m) = 1, we may write
X
1
nγ(n)α
n>x
=
X
µ2 (m)
=
rγ(r)α m1+α
X
1 ϕ(r)
rγ(r)α r
rm>x
gcd(r,m)=1
r square-full
≪
=
r≥1
r square-full
Z
r≥1
r square-full
∞
x/r
X
dt
t1+α
1
rγ(r)α
1
= α
x
1
p−1
1 Y
≪ α.
1 + 2 1−α
α
x p
p (p
− 1)
x
X
m>x/r
gcd(m,r)=1
X
r≥1
r square-full
µ2 (m)
m1+α
ϕ(r)
r2
r
γ(r)
α
Lemma 9. Given any fixed number z > 0,
X
z ω(n) ≪ x logz−1 x.
(9)
n≤x
Remark 10.
This result is a weak form of the well known Selberg-Sathe formula
X
z ω(n) = C(t)x logt−1 x + O(x logt−2 x),
n≤x
an estimate which holds uniformly for x ≥ 2 and |t| ≤ 1, where C(t) is a constant depending
only on t (see Selberg [7]). Here we give a simple direct proof of (9).
Proof. For each positive integer k, let
πk (x) := #{n ≤ x : ω(n) = k}.
5
From a classical result of Hardy and Ramanujan [3], we know that
πk (x) ≤
c1 x
(log log x + c2 )k−1
(k − 1)! log x
(x ≥ 3),
for all k ≥ 1, where c1 and c2 are some absolute constants. Using this estimate, it follows
that
X
X
x X (z(log log x + c2 ))k−1
log x k≥1
(k − 1)!
k≥1
x z log log x+zc2
x z log log x
= c1 z
e
e
= x logz−1 x,
≪
log x
log x
z ω(n) =
n≤x
πk (x)z k < c1 z
as required.
Lemma 11. (Brun-Titchmarsh Theorem) Let π(y; k, ℓ) := #{p ≤ y : p ≡ ℓ (mod k)}.
Given a fixed positive number β < 1, then, uniformly for k ∈ [1, xβ ], there exists a positive
constant ξ1 such that
x
.
π(x; k, ℓ) < ξ1
ϕ(k) log(x/k)
Proof. For a proof, see Titchmarsh [8].
Lemma 12. (Bombieri-Vinogradov Theorem). Given an arbitrary positive constant
A, and let B = 32 A + 17. Then
X
li(x)
≪ x .
max max π(y; k, ℓ) −
y≤x
gcd(k,ℓ)=1
ϕ(k)
√
logA x
x
k≤
logB x
Proof. For a proof, see Cheng-Dong [1].
Lemma 13. Given a fixed positive number β < 1, then, uniformly for d ∈ [1, x1−β ], there
exists a positive constant ξ3 such that
X
p≡1
p≤x
mod d
τ (p − 1) ≤ ξ3 x
6
τ (d)
.
ϕ(d)
Proof. Since τ (n) ≤ 2
that
X
d|n
√
d≤ n
1 and τ (mn) ≤ τ (m)τ (n) for all positive integers m, n, it follows
E: =
p≡1
X
p≤x
(mod d)
≤ τ (d)
p≡1
≤ 2τ (d)
X
√
τ(
p≤x
(mod d)
X
u≤
On the other hand, since du =
of Lemma 11, that
τ (p − 1)
√
p−1
)
d
π(x; du, 1).
(10)
x/d
√
√ √
β
β
1
1
d · u d ≤ d · x ≤ x 2 − 2 · x 2 = x1− 2 , we have, in light
π(x; du, −1) ≤ ξ1
x
x
≤ ξ2
,
ϕ(du) log(x/du)
ϕ(du) log x
(11)
with ξ2 = 2ξ1 /β. Using (11) in (10) and keeping in mind that ϕ(d)ϕ(u) ≤ ϕ(du) for positive
integers d, u, we obtain that
E ≤ 2ξ2
X
x
τ (d)
log x
√
u≤
≤ 2ξ2
x/d
x τ (d) X
log x ϕ(d) √
u≤
≤ ξ3 x
1
ϕ(du)
1
ϕ(u)
x/d
τ (d)
,
ϕ(d)
for some positive constant ξ3 , where we used the fact that
X
n≤y
the proof of the lemma.
Lemma 14. Given fixed numbers A > 0 and κ < α, then
X
d≥logA x
1
τ (d)
≪
.
dγ(d)α
logAκ x
7
1
≪ log y, thus completing
ϕ(n)
Proof. Clearly we have
X
A
d≥log x
τ (d)
dγ(d)α
X
≤
d≥logA x
τ (d)
dγ(d)α
d
logA x
κ
X
τ (d)
1
Aκ
1−κ
d γ(d)α
log x
d≥logA x
Y
2
2
1
1 + 1−κ α + 2(1−κ) α + . . .
p p
p
p
logAκ x p
=
<
≪
1
,
logAκ x
since the above infinite product converges in light of the fact that κ < α.
Lemma 15. Given a fixed positive integer D, then
X
1
= cD log x + O(1),
ϕ(n)
n≤x
gcd(n,D)=1
where
cD =
Y
p
1
1+
p(p − 1)
Y
1+
·
p|D
p
(p − 1)2
−1
.
Remark 16.
Note that the result of Lemma 15 is known in a more precise form (see the
book of Montgomery and Vaughan [5, pp. 42–43]).
Proof. We first compute the generating series of n/ϕ(n). We have
∞
X
Y
n/ϕ(n)
(1 − 1/p)−1 (1 − 1/p)−1
1+
=
+
+ ...
ns
ps
p2s
n=1
gcd(n,D)=1
p6|D
Q
(1−1/p)−1
(1−1/p)−1
+
+
.
.
.
1
+
s
2s
p
p
p
= Q
−1
(1−1/p)
(1−1/p)−1
1
+
+
+
.
.
.
p|D
ps
p2s
Q
(1−1/p)−1
(1−1/p)−1
1
+
+
+
.
.
.
Y
p
ps
p2s
1
= ζ(s)
1− s Q
(1−1/p)−1
(1−1/p)−1
p
1+
+
+ ...
p
p|D
= ζ(s)
Y
p
1
1+ s
p (p − 1)
Y
p|D
ps
p
1+
(p − 1)(ps − 1)
which by Wintner’s Theorem yields
X
n
= cD x + O(x1/2 log x),
ϕ(n)
n≤x
gcd(n,D)=1
8
p2s
−1
,
(12)
where
cD =
Y
p
1
1+
p(p − 1)
Y
1+
·
p|D
p
(p − 1)2
−1
.
Then, using partial summation, we get that
Z x
X
1
dt
log x
cD + O(1) = cD log x + O(1),
+
= cD + O
1/2
ϕ(n)
x
t
1
n≤x
gcd(n,D)=1
as required.
Lemma 17. For each integer δ ≥ 2, let Gδ be the semigroup generated by the prime factors
of δ, i.e. for δ = q1α1 . . . qrαr , let Gδ = {q1β1 . . . qrβr : βi ≥ 0}. Then
∞
X
X 1
1
=
≪ log log δ.
n
n
n=1
n∈G
δ
γ(n)|δ
Proof. Using the well known result of Landau
lim sup
n→∞
n
= eγ ,
ϕ(n) log log n
we certainly have that there exists some constant C > 0 such that
n
< C log log n
ϕ(n)
(n ≥ 3),
from which is follows easily that
−1
Y
∞
X
Y
1
1
1
1
1−
1 + + 2 + ... =
=
n
p p
p
n=1
p|δ
γ(n)|δ
=
p|δ
δ
< C log log δ,
ϕ(δ)
which proves our result.
Lemma 18. Let Gδ be as in Lemma 17. Then
∞
X
|h(δ)|2ω(δ) X 1
< +∞.
δ
b
b∈G
δ=1
δ
9
Proof. In light of Lemma 17, the fact that |h(δ)| ≤
c
and since given any ε > 0, log δ < δ ε
γ(δ)α
provided δ ≥ δ0 (ε), we have
∞
∞
X
X
|h(δ)|2ω(δ) X 1
|h(δ)|2ω(δ) log log δ
≪
δ
b
δ
δ=1
b∈G
δ=1
δ
≪
∞
X
2ω(δ) log log δ
δγ(δ)α
δ=1
∞
X
2ω(δ)
δ 1−ε γ(δ)α
δ=δ0
Y
2
2
<
1 + 1−ε α + 2−2ε α + . . .
p p
p
p
p
Y
2
< +∞,
=
1 + α 1−ε
p (p − 1)
p
≪
provided ε < α.
Lemma 19. Let Gδ be as in Lemma 17 and let ρ(x) be a real function which tends to +∞
as x → ∞. Given any fixed positive constant κ < α, we have
Z(x) :=
X
ρ(x)ρ(x)
δb>ρ(x)
δb>ρ(x)
b∈Gδ
δρ(x)/δ
b∈G
δx
Substituting (24) and (25) in (23), one obtains
T (x) = A0 x log x + B0 x + O x
where A0 =
Since
∞
X
λ(d)
d=1
d
it follows that
d
d=1
X |λ(d)|
d≤x
∞
X
λ(d)
and B0 =
dθ
x
θ
1−α+ε
+ O xθ
X |λ(d)|
d≤x
dθ
!
,
(26)
(2γ − 1 − log d).
X cτ (d)c4
xε ,
if θ + α ≥ 1;
≤
≤
1−(θ+α)
θ
α
x
, if θ + α < 1;
d γ(d)
d≤x
X |λ(d)|
d≤x
≤
dθ
xθ+ε , if θ + α ≥ 1;
x1−α , if θ + α < 1.
Using this last estimate in (26) completes the proof of Theorem 1.
5
Proof of Theorem 2
Since
∞
X
2ω(n)
n=1
we have that
ns
=
ζ 2 (s)
ζ(2s)
2ω(n) =
X
(ℜ(s) > 1),
µ(d)τ (e).
(27)
d2 e=n
Let Gδ be as in the statement of Lemma 17. If δ|n, then let m be the largest divisor of n/δ
which is co-prime with δ, and let b = b(δ) ∈ Gδ be defined implicitly by n = δ · b · m. Note
that with this setup, the numbers m and b are uniquely determined, and we therefore have
2ω(n) = 2ω(δ) · 2ω(m) . Hence,
X X
X
S(x) =
h(δ)2ω(δ)
2ω(m)
b≤x/δ
m≤ x
δb
b∈Gδ gcd(m,δ)=1
δ≤x
X
=
ω(δ)
h(δ)2
δ≤x
X
Eδ
b≤x/δ
b∈Gδ
x
say. We therefore need to estimate
Ek (X) :=
X
n≤X
gcd(n,k)=1
13
2ω(n) .
δb
,
(28)
Observe that it follows from (27) that
X
Ek (X) =
µ(d)
√
d≤ X
gcd(d,k)=1
X
X
τ (e) =
where
X
Vk (y) :=
µ(d)Vk
√
d≤ X
gcd(d,k)=1
d2 e≤X
gcd(e,k)=1
X
d2
,
(29)
τ (n),
n≤y
gcd(n,k)=1
while we trivially have
Ek (X) ≪ X log X.
(30)
We shall now make use of a function ρ(x) satisfying
np
o
√
exp
log x ≤ ρ(x) ≤ x
(31)
and which we will later determine more precisely.
We first write (29) as
X
Ek (X) =
µ(d)Vk
d≤ρ(X)
gcd(d,k)=1
X
d2
+
X
µ(d)Vk
√
ρ(X)x1/10
K,L∈Gd
1
KL
KL
x1/10
1/2
(log x)ε
1
2 < 1/20 .
x
1 − √1p
(71)
Estimates (68), (70) and (71) therefore establish that we can drop from the sum in (67)
those K,L ∈ Gd forwhich KL > logD x for a large D, the error thus created being no larger
li(x)
. In light of these observations and using the fact that ϕ(δ 2 KLde
u) =
than O
(log x)D/3
KLϕ(δ 2 de
u) and that
X
K,L∈Gd
KL(log x) 0
≪ (log x)
X
1
1
2
ϕ(δ1 d) ϕ(δ22 )
X
u
f
1 ∈Gδ2
∆
u
f
1 >(log x) 0
1
ue1
X 1
log x
1
√
ϕ(δ12 d)ϕ(δ22 ) (log x)∆0 /2
ue1
u
f1 ∈Gδ2
−1
Y
1
1
1−∆0 /2
1− √
≪ (log x)
δ12 ϕ(d)ϕ(δ22 )
p
≪
p|δ2
≪ (log x)1−∆0 /2
ω(δ2 )
1 2
.
ϕ(d) δ12 δ22
On√the other hand, the contribution of those ue1 ≤ (log x)∆0 in Yd (x), as ue2 runs up to
y0 := x/((log x)C ue1 ), and also using Lemma 15 so that
X
1
1
= cdδ log x + O(log log x),
ϕ(ue2 )
2
gcd(u
f ,dδ)=1
2
u
f
2 ≤y0
we get
Yd (x) =
X 1
1
1
1
cdδ (log x + log log x)
2
2
2
ϕ(δ1 d) ϕ(δ2 )
ue1
u
f1 ∈Gδ2
ω(δ2 )
2
1−C0 /2
.
(log x)
+O
δ 2 ϕ(d)
(74)
Observing that ϕ(δ12 d) = δ12 ϕ(d) and that
−1
Y
X 1
1
δ2
=
,
1−
=
u
e
p
ϕ(δ2 )
u
e∈Gδ2
(74) becomes
Yd (x) =
p|Gδ2
1 cdδ 1
δ2
(log x + log log x)
2 2
2 ϕ(d) δ1 δ2 ϕ(δ2 )
ω(δ2 )
2
1−C0 /2
+O
.
(log x)
δ 2 ϕ(d)
24
(75)
Using (75) in (72), we obtain
cd δ
B(x) = (li(x) − li(x/2)) 2
(log x + log log x)
δ1 δ2 ϕ(δ2 )
ω(δ2 )
2
1−C0 /2
+O
.
(log x)
δ 2 ϕ(d)
Finally, summing over d ≤ (log x)A and δ ≤ (log x)B , the theorem follows.
7
Final remarks
It is interesting to inquire about the solutions to the equation
or equivalently
Pe(n) = Pe(n + 1),
n Y
n+1 Y
(2p − 1) =
(2q − 1).
γ(n)
γ(n + 1)
p|n
(76)
(77)
q|n+1
The first 13 solutions are 45, 225, 1125, 2025, 3645, 140 625, 164 025, 257 174, 703 125,
820 125, 1 265 625, 2 657 205 and 3 515 625.
a b
One can easily check that if for some positive integers a and b, the number p = 3 ·52 +1 is
a prime number, then n = 3a · 5b is a solution of (76), with Pe(n) = 3a+1 · 5b . If one could
prove that there exist infinitely many primes of this form, it would follow that equation (76)
has infinitely many solutions. Interestingly, a computer search establishes that (76) has 37
solutions n < 1010 and that 21 of them are of the form n = 3a · 5b .
Moreover, all 37 solutions n0 are such that 33 · 5|Pe(n0 ). To see that 3|n0 , proceed as
follows. Suppose that 3 6 | Pe(n0 ), then, in light of (77), 3 6 | (2p − 1) for each prime divisor p
of n and 3 6 | (2q − 1) for each prime divisor q of n + 1. This would imply that p ≡ 1 (mod 3)
for all p|n (implying that n ≡ 1 (mod 3)) and q ≡ 1 (mod 3) for all q|n + 1 (implying that
n + 1 ≡ 1 (mod 3)), a contradiction.
The fact that 3|Pe(n0 ) clearly implies that 5|Pe(n0 ).
8
Acknowledgement
The authors would like to thank the referee for some very helpful suggestions which improved
the quality of this paper.
References
[1] P. Cheng-Dong, A new mean value theorem and its applications, in Recent Progress in
Analytic Number Theory, Vol. 1, (Durham, 1979), Academic Press, 1981, pp. 275–287.
25
[2] J. Chidambaraswamy and R. Sitaramachandrarao, Asymptotic results for a class of
arithmetical functions, Monatsh. Math. 99 (1985), 19–27.
[3] G. H. Hardy and S. Ramanujan, The normal number of prime factors of a number n,
Quart. J. Math. 48 (1920), 76–92.
[4] A. Ivi´c, The Riemann Zeta Function, John Wiley & Sons, New York, 1985.
[5] H. L. Montgomery and R. C. Vaughan, Multiplicative Number Theory I. Classical Theory, Cambridge University Press, 2007.
[6] S. S. Pillai, On an arithmetic function, J. Annamalai Univ. 2 (1933), 243–248.
[7] A. Selberg, Note on a paper by L. G. Sathe, J. Indian Math. Soc. 18 (1954), 83–87.
[8] E. C. Titchmarsh, A divisor problem, Rend. Palermo 54 (1930), 414–429 and 57 (1933),
478–479.
[9] D. I. Tolev, On the divisor problem in arithmetic progressions, C. R. Acad. Bulgare Sci.
41 (1988), 33–36.
[10] L. Toth, A gcd-sum function over regular integers modulo n, J. Integer Sequences 12
(2009), Article 09.2.5.
2000 Mathematics Subject Classification: Primary 11A25; Secondary 11N37.
Keywords: gcd-sum function, Dirichlet divisor problem, shifted primes.
Received August 31 2009; revised version received December 29 2009. Published in Journal
of Integer Sequences, December 31 2009.
Return to Journal of Integer Sequences home page.
26
2
3
47
6
Journal of Integer Sequences, Vol. 13 (2010),
Article 10.1.2
23 11
Some Remarks on a Paper of L. Toth
Jean-Marie De Koninck
D´ep. de math´ematiques
Universit´e Laval
Qu´ebec, PQ G1V 0A6
Canada
jmdk@mat.ulaval.ca
Imre K´atai
Computer Algebra Department
E¨otv¨os Lor´and University
P´azm´any P´eter S´et´any I/C
1117 Budapest
Hungary
katai@compalg.inf.elte.hu
Abstract
Pn
Consider the functions P (n) :=
k=1 gcd(k, n) (studied by Pillai in 1933) and
Q
e
P (n) := n p|n (2 − 1/p) (studied by Toth in 2009). From their results, one can
P
P
obtain asymptotic expansions for n≤x P (n)/n and n≤x Pe(n)/n. We consider two
wide classes of functions R and U of arithmetical functions which include P (n)/n and
Pe(n)/n
given R ∈ R and U ∈PU, we obtain asymptotic expansions
P For any P
P respectively.
for n≤x R(n), n≤x U (n), p≤x R(p − 1) and p≤x U (p − 1).
1
Introduction
In 1933, Pillai [6] introduced the function
P (n) :=
n
X
k=1
1
gcd(k, n)
and showed that
P (n) =
X
dϕ(n/d)
d|n
and
X
P (d) = nτ (n) =
d|n
X
σ(d)ϕ(n/d),
d|n
where ϕ stands for the Euler function and where τ (n) and σ(n) stand for the number of
divisors of n and the sum of the divisors of n respectively.
It is easily shown that
Y
a/(a + 1)
P (n) = nτ (n)
1−
.
p
a
p kn
In 1985, Chidambaraswamy and Sitaramachandrarao [2] showed that, given an arbitrary
ε > 0,
X
(1)
P (n) = e1 x2 log x + e2 x2 + O x1+θ+ε ,
n≤x
1
1 ζ ′ (2)
1
where e1 =
2γ − −
, and where θ is the constant appearing
and e2 =
2ζ(2)
2ζ(2)
2
ζ(2)
below in Lemma 5, ζ stands for the Riemann Zeta Function and γ stands for Euler’s constant.
Using partial summation, one easily deduces from (1) that
X P (n)
n≤x
n
= e1 x log x + (2e2 − e1 )x + O xθ+ε .
In [10], Toth introduced the function
Y
Y
1
1
ω(n)
e
1−
2−
P (n) = n
=n·2
,
p
2p
p|n
(2)
p|n
where ω(n) stands for the number of distinct prime factors of n. Toth obtained an estiP
mate for n≤x Pe(n), analogous to (1) and from which one can easily derive an asymptotic
P
expansion for n≤x Pe(n)/n.
In this paper, we consider two wide classes of arithmetical functions R and U, the first
of which includes the function P (n)/n, and theP
second of which includes Pe(n)/n. Given
R ∈ R, we obtain an asymptotic
expansion P
for n≤x R(n); similarly for U ∈ U. We then
P
examine the behavior of p≤x R(p − 1) and p≤x U (p − 1).
More precisely, the class R is made
Q of the following functions R. First, let γ(n) stand for
the kernel of n ≥ 2, that is γ(n) = p|n p (with γ(1) = 1). Then, given an arbitrary positive
constant c, an arbitrary real number α > 0 and a multiplicative function κ(n) satisfying
c
|κ(n)| ≤
for all n ≥ 2, let R ∈ R be defined by
γ(n)α
X
Y
R(n) = Rκ,c,α (n) := τ (n)
κ(d) = τ (n)
(1 + κ(pa )).
(3)
dkn
2
pa kn
Here dkn means that the sum runs over the unitary divisors of n, that is over all divisors d
of n for which (d, n/d) = 1.
a/(a + 1)
It is easily seen that if we let κ(pa ) = −
, then the corresponding function R(n)
p
is precisely P (n)/n.
As for the class of functions U, it is made of the functions
X
h(d),
(4)
U (n) = Uh,c,α (n) := 2ω(n)
d|n
c
for each integer d ≥ 2, where
γ(d)α
1
and h(pa ) = 0 for a ≥ 2,
α > 0 is a given number. It is easily seen that by taking h(p) = − 2p
we obtain the particular case U (n) = Pe(n)/n.
Throughout this paper, c1 , c2 , . . . denote absolute positive constants.
where h is a multiplicative function satisfying |h(d)| ≤
2
Main results
Theorem 1. Let R be as in (3). For any arbitrary ε > 0, as x → ∞,
X
T (x) :=
R(n) = A0 x log x + B0 x + O xβ ,
n≤x
with
θ + ε,
if α ≥ 1 − θ;
1 − α + ε, if α < 1 − θ;
where θ is the number mentioned in Lemma 5 below and where
X λ(d)
X λ(d)
and
B0 =
(2γ − 1 − log d),
A0 =
d
d
d≥1
d≥1
β=
(5)
the function λ being defined below in (20) and (21).
Theorem 2. Let U be as in (4). As x → ∞,
X
x
,
S(x) :=
U (n) = t1 x log x + t2 x + O
log
x
n≤x
where
t1
∞
∞
X
h(δ)2ω(δ) B1 (δ)ηδ X 1
=
,
δ
n
n=1
δ=1
γ(n)|δ
t2
∞
X
h(δ)2ω(δ)
=
B2 (δ)ηδ − 2B1 (δ)µδ − B1 (δ)ηδ log(δb) ,
δb
δ=1
b∈Gδ
where B1 (δ) and B2 (δ) are defined below in Lemma 7, while ηδ and µδ are defined respectively
in (35) and (36).
3
Theorem 3. Let R be as in (3). As x → ∞,
X
M (x) :=
R(p − 1) = K1 x + O
p≤x
where K1 =
x
log log x
,
(6)
∞
1 X κ(d)τ (d)cd
, where cd is itself defined in Lemma 15.
2 d=1
d
Theorem 4. Let U be as in (4). As x → ∞,
N (x) :=
X
p≤x
U (p − 1) = K2 x + O
x
log log x
,
(7)
where K2 is a positive constant which may depend on the function κ.
3
Preliminary results
Lemma 5. As x → ∞,
X
D(x) :=
τ (n) = x log x + (2γ − 1)x + O(xθ )
(x → ∞),
n≤x
(8)
for some positive constant θ < 1/3.
Proof. A proof can be found in the book of Ivi´c [4], where one can also find a history of the
improvements concerning the size of θ.
Lemma 6. Given 1 ≤ ℓ < k with gcd(ℓ, k) = 1,
X
τ (n) = A1 (k)x log x + A2 (k)x + O k 7/3 x1/3 log x ,
n≡ℓ
n≤x
(mod k)
where
A1 (k) =
ϕ(k)
,
k2
A2 (k) = (2γ − 1)
ϕ(k) 2 X µ(d) log d
−
.
k2
k
d
d|k
Proof. This result is due to Tolev [9].
Lemma 7. Given a positive integer k,
X
τ (n) = B1 (k)x log x + B2 (k)x + O k 10/3 x1/3 log x ,
n≤x
(n,k)=1
where B1 (k) =
ϕ(k)
k
2
and
B2 (k) = ϕ(k)A2 (k).
4
Proof. Observing that
X
τ (n) =
k
X
X
τ (n),
ℓ=1
n≤x
gcd(ℓ,k)=1 n≡ℓ (mod k)
n≤x
gcd(n,k)=1
and using the trivial fact that ϕ(k) ≤ k, the result follows immediately from Lemma 6.
Lemma 8. Given an arbitrary positive real number α < 1,
X
1
1
≪ α.
α
nγ(n)
x
n>x
Proof. Writing each number n as n = rm, where r and m are the square-full and square-free
parts of n respectively with (r, m) = 1, we may write
X
1
nγ(n)α
n>x
=
X
µ2 (m)
=
rγ(r)α m1+α
X
1 ϕ(r)
rγ(r)α r
rm>x
gcd(r,m)=1
r square-full
≪
=
r≥1
r square-full
Z
r≥1
r square-full
∞
x/r
X
dt
t1+α
1
rγ(r)α
1
= α
x
1
p−1
1 Y
≪ α.
1 + 2 1−α
α
x p
p (p
− 1)
x
X
m>x/r
gcd(m,r)=1
X
r≥1
r square-full
µ2 (m)
m1+α
ϕ(r)
r2
r
γ(r)
α
Lemma 9. Given any fixed number z > 0,
X
z ω(n) ≪ x logz−1 x.
(9)
n≤x
Remark 10.
This result is a weak form of the well known Selberg-Sathe formula
X
z ω(n) = C(t)x logt−1 x + O(x logt−2 x),
n≤x
an estimate which holds uniformly for x ≥ 2 and |t| ≤ 1, where C(t) is a constant depending
only on t (see Selberg [7]). Here we give a simple direct proof of (9).
Proof. For each positive integer k, let
πk (x) := #{n ≤ x : ω(n) = k}.
5
From a classical result of Hardy and Ramanujan [3], we know that
πk (x) ≤
c1 x
(log log x + c2 )k−1
(k − 1)! log x
(x ≥ 3),
for all k ≥ 1, where c1 and c2 are some absolute constants. Using this estimate, it follows
that
X
X
x X (z(log log x + c2 ))k−1
log x k≥1
(k − 1)!
k≥1
x z log log x+zc2
x z log log x
= c1 z
e
e
= x logz−1 x,
≪
log x
log x
z ω(n) =
n≤x
πk (x)z k < c1 z
as required.
Lemma 11. (Brun-Titchmarsh Theorem) Let π(y; k, ℓ) := #{p ≤ y : p ≡ ℓ (mod k)}.
Given a fixed positive number β < 1, then, uniformly for k ∈ [1, xβ ], there exists a positive
constant ξ1 such that
x
.
π(x; k, ℓ) < ξ1
ϕ(k) log(x/k)
Proof. For a proof, see Titchmarsh [8].
Lemma 12. (Bombieri-Vinogradov Theorem). Given an arbitrary positive constant
A, and let B = 32 A + 17. Then
X
li(x)
≪ x .
max max π(y; k, ℓ) −
y≤x
gcd(k,ℓ)=1
ϕ(k)
√
logA x
x
k≤
logB x
Proof. For a proof, see Cheng-Dong [1].
Lemma 13. Given a fixed positive number β < 1, then, uniformly for d ∈ [1, x1−β ], there
exists a positive constant ξ3 such that
X
p≡1
p≤x
mod d
τ (p − 1) ≤ ξ3 x
6
τ (d)
.
ϕ(d)
Proof. Since τ (n) ≤ 2
that
X
d|n
√
d≤ n
1 and τ (mn) ≤ τ (m)τ (n) for all positive integers m, n, it follows
E: =
p≡1
X
p≤x
(mod d)
≤ τ (d)
p≡1
≤ 2τ (d)
X
√
τ(
p≤x
(mod d)
X
u≤
On the other hand, since du =
of Lemma 11, that
τ (p − 1)
√
p−1
)
d
π(x; du, 1).
(10)
x/d
√
√ √
β
β
1
1
d · u d ≤ d · x ≤ x 2 − 2 · x 2 = x1− 2 , we have, in light
π(x; du, −1) ≤ ξ1
x
x
≤ ξ2
,
ϕ(du) log(x/du)
ϕ(du) log x
(11)
with ξ2 = 2ξ1 /β. Using (11) in (10) and keeping in mind that ϕ(d)ϕ(u) ≤ ϕ(du) for positive
integers d, u, we obtain that
E ≤ 2ξ2
X
x
τ (d)
log x
√
u≤
≤ 2ξ2
x/d
x τ (d) X
log x ϕ(d) √
u≤
≤ ξ3 x
1
ϕ(du)
1
ϕ(u)
x/d
τ (d)
,
ϕ(d)
for some positive constant ξ3 , where we used the fact that
X
n≤y
the proof of the lemma.
Lemma 14. Given fixed numbers A > 0 and κ < α, then
X
d≥logA x
1
τ (d)
≪
.
dγ(d)α
logAκ x
7
1
≪ log y, thus completing
ϕ(n)
Proof. Clearly we have
X
A
d≥log x
τ (d)
dγ(d)α
X
≤
d≥logA x
τ (d)
dγ(d)α
d
logA x
κ
X
τ (d)
1
Aκ
1−κ
d γ(d)α
log x
d≥logA x
Y
2
2
1
1 + 1−κ α + 2(1−κ) α + . . .
p p
p
p
logAκ x p
=
<
≪
1
,
logAκ x
since the above infinite product converges in light of the fact that κ < α.
Lemma 15. Given a fixed positive integer D, then
X
1
= cD log x + O(1),
ϕ(n)
n≤x
gcd(n,D)=1
where
cD =
Y
p
1
1+
p(p − 1)
Y
1+
·
p|D
p
(p − 1)2
−1
.
Remark 16.
Note that the result of Lemma 15 is known in a more precise form (see the
book of Montgomery and Vaughan [5, pp. 42–43]).
Proof. We first compute the generating series of n/ϕ(n). We have
∞
X
Y
n/ϕ(n)
(1 − 1/p)−1 (1 − 1/p)−1
1+
=
+
+ ...
ns
ps
p2s
n=1
gcd(n,D)=1
p6|D
Q
(1−1/p)−1
(1−1/p)−1
+
+
.
.
.
1
+
s
2s
p
p
p
= Q
−1
(1−1/p)
(1−1/p)−1
1
+
+
+
.
.
.
p|D
ps
p2s
Q
(1−1/p)−1
(1−1/p)−1
1
+
+
+
.
.
.
Y
p
ps
p2s
1
= ζ(s)
1− s Q
(1−1/p)−1
(1−1/p)−1
p
1+
+
+ ...
p
p|D
= ζ(s)
Y
p
1
1+ s
p (p − 1)
Y
p|D
ps
p
1+
(p − 1)(ps − 1)
which by Wintner’s Theorem yields
X
n
= cD x + O(x1/2 log x),
ϕ(n)
n≤x
gcd(n,D)=1
8
p2s
−1
,
(12)
where
cD =
Y
p
1
1+
p(p − 1)
Y
1+
·
p|D
p
(p − 1)2
−1
.
Then, using partial summation, we get that
Z x
X
1
dt
log x
cD + O(1) = cD log x + O(1),
+
= cD + O
1/2
ϕ(n)
x
t
1
n≤x
gcd(n,D)=1
as required.
Lemma 17. For each integer δ ≥ 2, let Gδ be the semigroup generated by the prime factors
of δ, i.e. for δ = q1α1 . . . qrαr , let Gδ = {q1β1 . . . qrβr : βi ≥ 0}. Then
∞
X
X 1
1
=
≪ log log δ.
n
n
n=1
n∈G
δ
γ(n)|δ
Proof. Using the well known result of Landau
lim sup
n→∞
n
= eγ ,
ϕ(n) log log n
we certainly have that there exists some constant C > 0 such that
n
< C log log n
ϕ(n)
(n ≥ 3),
from which is follows easily that
−1
Y
∞
X
Y
1
1
1
1
1−
1 + + 2 + ... =
=
n
p p
p
n=1
p|δ
γ(n)|δ
=
p|δ
δ
< C log log δ,
ϕ(δ)
which proves our result.
Lemma 18. Let Gδ be as in Lemma 17. Then
∞
X
|h(δ)|2ω(δ) X 1
< +∞.
δ
b
b∈G
δ=1
δ
9
Proof. In light of Lemma 17, the fact that |h(δ)| ≤
c
and since given any ε > 0, log δ < δ ε
γ(δ)α
provided δ ≥ δ0 (ε), we have
∞
∞
X
X
|h(δ)|2ω(δ) X 1
|h(δ)|2ω(δ) log log δ
≪
δ
b
δ
δ=1
b∈G
δ=1
δ
≪
∞
X
2ω(δ) log log δ
δγ(δ)α
δ=1
∞
X
2ω(δ)
δ 1−ε γ(δ)α
δ=δ0
Y
2
2
<
1 + 1−ε α + 2−2ε α + . . .
p p
p
p
p
Y
2
< +∞,
=
1 + α 1−ε
p (p − 1)
p
≪
provided ε < α.
Lemma 19. Let Gδ be as in Lemma 17 and let ρ(x) be a real function which tends to +∞
as x → ∞. Given any fixed positive constant κ < α, we have
Z(x) :=
X
ρ(x)ρ(x)
δb>ρ(x)
δb>ρ(x)
b∈Gδ
δρ(x)/δ
b∈G
δx
Substituting (24) and (25) in (23), one obtains
T (x) = A0 x log x + B0 x + O x
where A0 =
Since
∞
X
λ(d)
d=1
d
it follows that
d
d=1
X |λ(d)|
d≤x
∞
X
λ(d)
and B0 =
dθ
x
θ
1−α+ε
+ O xθ
X |λ(d)|
d≤x
dθ
!
,
(26)
(2γ − 1 − log d).
X cτ (d)c4
xε ,
if θ + α ≥ 1;
≤
≤
1−(θ+α)
θ
α
x
, if θ + α < 1;
d γ(d)
d≤x
X |λ(d)|
d≤x
≤
dθ
xθ+ε , if θ + α ≥ 1;
x1−α , if θ + α < 1.
Using this last estimate in (26) completes the proof of Theorem 1.
5
Proof of Theorem 2
Since
∞
X
2ω(n)
n=1
we have that
ns
=
ζ 2 (s)
ζ(2s)
2ω(n) =
X
(ℜ(s) > 1),
µ(d)τ (e).
(27)
d2 e=n
Let Gδ be as in the statement of Lemma 17. If δ|n, then let m be the largest divisor of n/δ
which is co-prime with δ, and let b = b(δ) ∈ Gδ be defined implicitly by n = δ · b · m. Note
that with this setup, the numbers m and b are uniquely determined, and we therefore have
2ω(n) = 2ω(δ) · 2ω(m) . Hence,
X X
X
S(x) =
h(δ)2ω(δ)
2ω(m)
b≤x/δ
m≤ x
δb
b∈Gδ gcd(m,δ)=1
δ≤x
X
=
ω(δ)
h(δ)2
δ≤x
X
Eδ
b≤x/δ
b∈Gδ
x
say. We therefore need to estimate
Ek (X) :=
X
n≤X
gcd(n,k)=1
13
2ω(n) .
δb
,
(28)
Observe that it follows from (27) that
X
Ek (X) =
µ(d)
√
d≤ X
gcd(d,k)=1
X
X
τ (e) =
where
X
Vk (y) :=
µ(d)Vk
√
d≤ X
gcd(d,k)=1
d2 e≤X
gcd(e,k)=1
X
d2
,
(29)
τ (n),
n≤y
gcd(n,k)=1
while we trivially have
Ek (X) ≪ X log X.
(30)
We shall now make use of a function ρ(x) satisfying
np
o
√
exp
log x ≤ ρ(x) ≤ x
(31)
and which we will later determine more precisely.
We first write (29) as
X
Ek (X) =
µ(d)Vk
d≤ρ(X)
gcd(d,k)=1
X
d2
+
X
µ(d)Vk
√
ρ(X)x1/10
K,L∈Gd
1
KL
KL
x1/10
1/2
(log x)ε
1
2 < 1/20 .
x
1 − √1p
(71)
Estimates (68), (70) and (71) therefore establish that we can drop from the sum in (67)
those K,L ∈ Gd forwhich KL > logD x for a large D, the error thus created being no larger
li(x)
. In light of these observations and using the fact that ϕ(δ 2 KLde
u) =
than O
(log x)D/3
KLϕ(δ 2 de
u) and that
X
K,L∈Gd
KL(log x) 0
≪ (log x)
X
1
1
2
ϕ(δ1 d) ϕ(δ22 )
X
u
f
1 ∈Gδ2
∆
u
f
1 >(log x) 0
1
ue1
X 1
log x
1
√
ϕ(δ12 d)ϕ(δ22 ) (log x)∆0 /2
ue1
u
f1 ∈Gδ2
−1
Y
1
1
1−∆0 /2
1− √
≪ (log x)
δ12 ϕ(d)ϕ(δ22 )
p
≪
p|δ2
≪ (log x)1−∆0 /2
ω(δ2 )
1 2
.
ϕ(d) δ12 δ22
On√the other hand, the contribution of those ue1 ≤ (log x)∆0 in Yd (x), as ue2 runs up to
y0 := x/((log x)C ue1 ), and also using Lemma 15 so that
X
1
1
= cdδ log x + O(log log x),
ϕ(ue2 )
2
gcd(u
f ,dδ)=1
2
u
f
2 ≤y0
we get
Yd (x) =
X 1
1
1
1
cdδ (log x + log log x)
2
2
2
ϕ(δ1 d) ϕ(δ2 )
ue1
u
f1 ∈Gδ2
ω(δ2 )
2
1−C0 /2
.
(log x)
+O
δ 2 ϕ(d)
(74)
Observing that ϕ(δ12 d) = δ12 ϕ(d) and that
−1
Y
X 1
1
δ2
=
,
1−
=
u
e
p
ϕ(δ2 )
u
e∈Gδ2
(74) becomes
Yd (x) =
p|Gδ2
1 cdδ 1
δ2
(log x + log log x)
2 2
2 ϕ(d) δ1 δ2 ϕ(δ2 )
ω(δ2 )
2
1−C0 /2
+O
.
(log x)
δ 2 ϕ(d)
24
(75)
Using (75) in (72), we obtain
cd δ
B(x) = (li(x) − li(x/2)) 2
(log x + log log x)
δ1 δ2 ϕ(δ2 )
ω(δ2 )
2
1−C0 /2
+O
.
(log x)
δ 2 ϕ(d)
Finally, summing over d ≤ (log x)A and δ ≤ (log x)B , the theorem follows.
7
Final remarks
It is interesting to inquire about the solutions to the equation
or equivalently
Pe(n) = Pe(n + 1),
n Y
n+1 Y
(2p − 1) =
(2q − 1).
γ(n)
γ(n + 1)
p|n
(76)
(77)
q|n+1
The first 13 solutions are 45, 225, 1125, 2025, 3645, 140 625, 164 025, 257 174, 703 125,
820 125, 1 265 625, 2 657 205 and 3 515 625.
a b
One can easily check that if for some positive integers a and b, the number p = 3 ·52 +1 is
a prime number, then n = 3a · 5b is a solution of (76), with Pe(n) = 3a+1 · 5b . If one could
prove that there exist infinitely many primes of this form, it would follow that equation (76)
has infinitely many solutions. Interestingly, a computer search establishes that (76) has 37
solutions n < 1010 and that 21 of them are of the form n = 3a · 5b .
Moreover, all 37 solutions n0 are such that 33 · 5|Pe(n0 ). To see that 3|n0 , proceed as
follows. Suppose that 3 6 | Pe(n0 ), then, in light of (77), 3 6 | (2p − 1) for each prime divisor p
of n and 3 6 | (2q − 1) for each prime divisor q of n + 1. This would imply that p ≡ 1 (mod 3)
for all p|n (implying that n ≡ 1 (mod 3)) and q ≡ 1 (mod 3) for all q|n + 1 (implying that
n + 1 ≡ 1 (mod 3)), a contradiction.
The fact that 3|Pe(n0 ) clearly implies that 5|Pe(n0 ).
8
Acknowledgement
The authors would like to thank the referee for some very helpful suggestions which improved
the quality of this paper.
References
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25
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2000 Mathematics Subject Classification: Primary 11A25; Secondary 11N37.
Keywords: gcd-sum function, Dirichlet divisor problem, shifted primes.
Received August 31 2009; revised version received December 29 2009. Published in Journal
of Integer Sequences, December 31 2009.
Return to Journal of Integer Sequences home page.
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