Exponential Carmichael function

0.17 MB·

by  Andrew V. Lelechenko

EXPONENTIAL CARMICHAEL FUNCTION
ANDREWV.LELECHENKO
Abstract. Consider exponential Carmichael function λ(e) such that λ(e) is
multiplicativeandλ(e)(pa)=λ(a),whereλisusualCarmichaelfunction. We
4 discuss the value of Pλ(e)(n), where n runs over certain subsets of [1,x],
1
andprovidebounds ontheerror term, usinganalyticmethods andespecially
0 T m
estimatesofR (cid:12)ζ(σ+it)(cid:12) dt.
2 1 (cid:12) (cid:12)

y
a
M
1. Introduction

9 Consider an operator E over arithmetic functions such that for every f the
2
function Ef is multiplicative and

] (Ef)(pa)=f(a), p is prime.
T
N Forvariousfunctionsf (suchasthedivisorfunction,thesum-of-divisorfunction,
M¨obius function, the totient function and so on) the behaviour of Ef was studied
.
h by many authors, starting from Subbarao [13]. The bibliography can be found
t
a in [11].
m The notation for Ef, established by previous authors, is f(e).
[
Carmichael function λ is an arithmetic function such that

2 φ(pa), p>2 or a=1,2,
v λ(pa)=
6 (φ(pa)/2, p=2 and a>2,
6 and if n=pa1 pam is a canonical representation, then
1 1 ··· m
3 λ(n)=lcm λ(pa1),...,λ(pam) .
1 m
.
1 This function was introduced at(cid:0)the beginning of t(cid:1)he XX century in [2], but
0
intense studies started only in 1990-th, e. g. [3]. Carmichael function finds appli-
4
cations in cryptography, e. g. [4].
1
v: Consider also the family of multiplicative functions
i
X 0, a<r,
δ (pa)= r is integer.
r r (1, a>r,
a
Functionδ is acharacteristicfunctionofthe setofsquare-fullnumbers,δ — of
2 3
cube-full numbers and so on. Of course, δ 1.
1
≡
Denote λ(e) for the product of δ and λ(e):
r r
λ(e)(n)=δ (n)λ(e)(n).
r r
The aim of our paper is to study asymptotic properties of λ(e) λ(e), λ(e), λ(e)
≡ 1 2 3
and λ(e).
4
Note that all proofs below remains valid for φ(e)(n) = δ (n)φ(e)(n) instead
r r
of λ(e)(n) for r=1,2,3,4.
r
2010 Mathematics Subject Classification. 11A2511M06,11N37,11N56.
Keywordsandphrases. Exponentialdivisors,Carmichaelfunction,momentsofRiemannzeta-
function.
1
2 ANDREWV.LELECHENKO
2. Notations
Letter p with or without indexes denotes a prime number.
We write f ⋆g for Dirichlet convolution
(f ⋆g)(n)= f(d)g(n/d).
Xd|n
Denote
τ(a ,...,a ;n):= 1.
1 k
da11·X··dakk=n
In asymptotic relations we use , , Landau symbols O and o, Vinogradov
∼ ≍
symbols and in their usual meanings. All asymptotic relations are given as
≪ ≫
an argument (usually x) tends to the infinity.
Everywhere ε > 0 is an arbitrarily small number (not always the same even in
one equation).
As usual ζ(s) is Riemann zeta-function. Real and imaginary components of the
complex s are denoted as σ := s and t:= s, so s=σ+it.
ℜ ℑ
For a fixed σ [1/2,1] define
∈
T
m(σ):=sup m ζ(σ+it) mdt T1+ε .
≪
(cid:26) (cid:12)Z1 (cid:27)
(cid:12) (cid:12) (cid:12)
and (cid:12) (cid:12) (cid:12)
(cid:12) log ζ(σ+it)
µ(σ):=limsup .
logt
t→∞ (cid:12) (cid:12)
Below H = (32/205 + ε,269/410+ ε(cid:12)) stands (cid:12)for Huxley’s exponent pair
2005
from [6].
3. Preliminary lemmas
Lemma 1. Let F: Z C be a multiplicative function such that F(pa) = f(a),
→
where f(n) nβ for some β >0. Then
≪
logF(n)llogn logf(n)
limsup =sup .
logn n
n→∞ n>1
Proof. See [14]. (cid:3)
Lemma 2. Let f(t)>0. If
T
f(t)dt g(T),
≪
Z1
where g(T)=TαlogβT, α>1, then
T f(t) logβ+1T if α=1,
I(T):= dt
t ≪ Tα−1logβT if α>1.
Z1 (cid:26)
Proof. Let us divide the interval of integration into parts:
log2T T/2k f(t) log2T 1 T/2k log2T g(T/2k)
I(T)6 dt< f(t)dt .
k=0 ZT/2k+1 t k=0 T/2k+1 Z1 ≪ k=0 T/2k+1
X X X
Now the lemma’s statement follows from elementary estimates. (cid:3)
Lemma 3. For σ > 1/2 and for any exponent pair (k,l) such that l k > σ we
−
have
k+l σ
µ(σ)6 − +ε.
2
Proof. See [7, (7.57)]. (cid:3)
EXPONENTIAL CARMICHAEL FUNCTION 3
A well-known application of Lemma 3 is
(1) µ(1/2)632/205,
following from the choice (k,l)=H . Another (maybe new) application is
2005
(2) µ(3/5)61409/12170,
following from
269 1755
(k,l)= , =ABAH ,
2005
2434 2434
(cid:18) (cid:19)
where A and B stands for usual A- and B-processes [8, Ch. 2].
Lemma 4. Let η >0 be arbitrarily small. Then for growing t >3
| |
t1/2−(1−2µ(1/2))σ, σ [0,1/2],
| | ∈
t2µ(1/2)(1−σ), σ [1/2,1 η],
(3) ζ(s) | | ∈ −
≪|t|2µ(1/2)(1−σ)log2/3|t|, σ ∈[1−η,1],
log2/3 t, σ >1.
| |
More exact estimates for σ [1/2,1 η] are also available, e. g.
∈ −
10 µ(3/5) µ(1/2) σ+ 6µ(1/2) 5µ(3/5) , σ [1/2,3/5],
(4) µ(σ) − − ∈
≪(5µ(cid:0)(3/5)(1−σ)/2, (cid:1) (cid:0) (cid:1) σ ∈[3/5,1−η],
Proof. EstimatesfollowfromPhragm´en—Lindel¨ofprinciple,exactandapproximate
functional equations for ζ(s) and convexity properties. See [15, Ch. 5] and [7, Ch.
7.5] for details. (cid:3)
Lemma 5. For any integer r
maxλ(e)(n) xε.
n6x r ≪
Proof. Surely λ(e)(n)6λ(e)(n). By Lemma 1 we have
r
logλ(e)(n)loglogn logλ(m) log4
limsup =sup = =:c,
logn m 5
n→∞ m
because λ(m)6m 1. It implies
−
maxλ(e)(n) xc/loglogn xε.
n6x ≪ ≪
(cid:3)
Lemma 6. Let L (s) be the Dirichlet series for λ(e):
r r
∞
L (s):= λ(e)(n)n−s.
r r
n=1
X
Then for r =1,2,3,4 we have L (s)=Z (s)G (s), where
r r r
(5) Z (s)=ζ(s)ζ(3s)ζ2(5s),
1
(6) Z (s)=ζ(2s)ζ2(3s)ζ(4s)ζ2(5s),
2
(7) Z (s)=ζ2(3s)ζ2(4s)ζ4(5s),
3
(8) Z (s)=ζ2(4s)ζ4(5s)ζ2(6s)ζ6(7s),
4
Dirichlet series G (s), G (s), G (s) converge absolutely for σ > 1/6 and G (s)
1 2 3 4
converges absolutely for σ >1/8.
4 ANDREWV.LELECHENKO
Proof. Follows from the identities
1+ λ(e)(pa)xa =1+x+x2+2x3+2x4+4x5+2x6+6x7+O(x8)
a>1
X
1+O(x8)
= ,
(1 x)(1 x3)(1 x5)2
− − −
1+O(x6)
1+ λ(e)(pa)xa = ,
(1 x2)(1 x3)2(1 x4)(1 x5)2
a>2 − − − −
X
1+O(x6)
1+ λ(e)(pa)xa = ,
(1 x3)2(1 x4)2(1 x5)4
a>3 − − −
X
1+O(x8)
1+ λ(e)(pa)xa = .
(1 x4)2(1 x5)4(1 x6)2(1 x7)6
a>4 − − − −
X
(cid:3)
Lemma 7. Let ∆(x) be the error term in the well-known asymptotic formula for
τ(a ,a ,a ,a ;n), let A = a +a +a +a and let (k,l) be any exponent
n6x 1 2 3 4 4 1 2 3 4
pair. Suppose that the following conditions are satisfied:
P
(1) (k+l+2)a <(k+l)a +A .
4 1 4
(2) 2(k+l+1)a 6(2k+1)(a +a ).
1 2 3
(3.1) la 6ka and (k+l+1)a >k(a +a )
1 2 1 2 3
or
(3.2) la >ka and(l k)(2k+1)a 6(2l 2l 1)(k+l+1)a + 2k(k l+1)+1 a .
1 2 3 1 2
− − − −
Proof. This is [9, Th. 3] with p=4. (cid:0) (cid:1) (cid:3)
Lemma 8.
4/(3 4σ), 1/2 6σ 6 5/8,
−
10/(5 6σ), 5/8 6σ 635/54,

−
m(σ)>12214110928///(((68455−937−6−σ9),4488σ90),σ), 43315///46504666σσσ 6664315///466,,0,
4324/(1031 1044σ), 5/6 6σ 6 7/8,
−
Proof. See [7, Th. 8(.244]σ.98−/(93)1/−(4σ32−σ)1,)(1−σ), 07.9/185961.σ..660σ.9615191−.ε..., (cid:3)
4. Main results
Theorem 1.
λ(e)(n)=c x+c x1/3+(c′ logx+c )x1/5+O(x1153/6073+ε),
11 13 15 15
n6x
X
where c , c , c and c′ are computable constants.
11 13 15 15
Proof. Lemma 6 and equation (5) implies that λ(e) = τ(1,3,5,5; ) ⋆ g , where
1
·
g (n) x1/6+ε. Due to [8]
n6x 1 ≪
P τ(1,3,5,5;n)=xζ(3)ζ2(5)resζ(s)+3x1/3ζ(1/3)ζ2(5/3) res ζ(3s)+
s=1 s=1/3
n6x
X
+5x1/5ζ(1/5)ζ(3/5) res ζ2(5s)+R(x).
s=1/5
EXPONENTIAL CARMICHAEL FUNCTION 5
To estimate R(x) we use Lemma 7 with a = 1, a = 3, a = a = 5. Exponent
1 2 3 4
pair (k,l)=H satisfies conditions 1, 2 and 3.2 and thus
2005
R(x) x(k+l+2)/(k+l+14) =x1153/6073+ε, 1/6<1153/6073<1/5.
≪
Now the convolution argument completes the proof. (cid:3)
Exponential totient function φ(e) has similar to λ(e) Dirichlet series:
∞
φ(e)(n)=ζ(s)ζ(3s)ζ2(5s)H(s),
n=1
X
where H(s) converges absolutely for σ > 1/6. Theorem 1 can be extended to this
case without any changes, so
φ(e)(n)=c x+c x1/3+(c′ logx+c )x1/5+O(x1153/6073+ε).
11 13 15 15
n6x
X
This improves the result of P´etermann [12], who obtained φ(e)(n) = c x+
n6x 11
+c x1/3+O(x1/5logx).
13 P
Update from 23.05.2014: RecentlyCaoandZhai[1,(1.13)]obtainedR(x)
≪
x18/95+ε,whichisbetter thanbothP´etermann’sx1/5logxandourx1153/6073+ε.
≪
Theorem 2.
λ(e)(n)=c x1/2+(c′ logx+c )x1/3+c x1/4+O(x1153/5586+ε),
2 22 23 23 24
n6x
X
where c , c , c′ and c are computable constants.
22 23 23 24
Proof. Similar to Theorem 1 with following changes: now by (6)
λ(e) =τ(2,3,3,4; )⋆g ,
2 · 2
where g (n) x1/6+ε. But
n6x 2 ≪
Pτ(2,3,3,4;n)=2x1/2ζ2(3/2)ζ(2) res ζ(2s)+
s=1/2
n6x
X
+3x1/3ζ(2/3)ζ(4/3) res ζ2(3s)+4x1/4ζ(1/2)ζ2(3/4) res ζ(4s)+R(s).
s=1/3 s=1/4
Again by Lemma 7 with a =2, a =a =3, a =4, (k,l)=H we get
1 2 3 4 2005
R(x) x(k+l+2)/(k+l+12) =x1153/5586+ε, 1/5<1153/5586<1/4.
≪
(cid:3)
Theorem 3.
(9) λ(e)(n)=(c′ logx+c )x1/3+(c′ logx+c )x1/4+
3 33 33 34 34
n6x
X
+P (logx)x1/5+O(x1/6+ε),
35
where c , c′ , c and c′ are computable constants, P is a polynomial of degree
33 33 34 34 35
3 with computable coefficients.
Proof. Lemma 6 and equation (7) implies that λ(e) = z ⋆g , where z is defined
3 3 3 3
implicitly by
∞
z (n)n−s =Z (s)=ζ2(3s)ζ2(4s)ζ4(5s),
3 3
n=1
X
and g is a multiplicative function such that g (n) x1/6+ε.
3 n6x 3 ≪
P
6 ANDREWV.LELECHENKO
The main term at the right side of (9) equals to
M (x):= res + res + res ζ2(3s)ζ2(4s)ζ4(5s)xss−1 .
3
s=1/3 s=1/4 s=1/5
(cid:18) (cid:19)
(cid:0) (cid:1)
To obtain the desirable error term it is enough to prove that
z (n)=M (x)+O(x1/6+ε).
3 3
n6x
X
By Perron formula for c:=1/3+1/logx we have
1 c+iT
z (n)= Z (s)xss−1ds+O(x1+εT−1).
3 3
2πi
n6x Zc−iT
X
SubstitutingT =xandmovingthecontouroftheintegrationtill[1/6 ix,1/6+ix]
−
we get
f (n)=M (x)+O(I +I +I +xε),
3 3 0 − +
n6x
X
where
1/6+ix c±ix
I := Z (s)xss−1ds, I := Z (s)xss−1ds.
0 3 ± 3
Z1/6−ix Z1/6±ix
Firstly,
c
I x−1 Z (σ+ix)xσdσ.
+ 3
≪
Z1/6
Let α(σ) be a function such that Z (σ+ix) xα(σ)+ε. By (3) we have
3
≪
(16 68σ)µ(1/2)<4/5, σ [1/6,1/5),
− ∈
(8 28σ)µ(1/2)<3/4, σ [1/5,1/4),
α(σ)6 − ∈
 (4−12σ)µ(1/2)<2/3, σ ∈[1/4,1/3),
0, σ [1/3,c].
∈
This means that I+ xε. Plainly, the same estimate holds for I−.
≪
Secondly, it remains to prove that I x1/6+ε. Here
0
≪
x
I x1/6 Z (1/6+it)t−1dt
0 3
≪
Z1
and taking into account Lemma 2 it is enough to show xZ (1/6+it)dt x1+ε.
1 3 ≪
Applying Cauchy inequality twice we obtain
R
x x 1/2
Z (1/6+it)dt ζ4(1/2+it) dt
3
≪ ×
Z1 (cid:18)Z1 (cid:19)
x (cid:12)(cid:12) 1/4 (cid:12)(cid:12) x 1/4
ζ8(2/3+it) dt ζ16(5/6+it) dt
× ≪
(cid:18)Z1 (cid:19) (cid:18)Z1 (cid:19)
(cid:12)(cid:12) (cid:12)(cid:12) x(1(cid:12)(cid:12)+ε)·1/2x(1+ε)·(cid:12)(cid:12)1/4x(1+ε)·1/4 x1+ε
≪ ≪
since by Lemma 8 m(1/2)>4, m(2/3)>8 and m(5/6)>16. (cid:3)
Theorem 4.
λ(e)(n)=(c′ logx+c )x1/4+P (logx)x1/5+(c′ logx+c )x1/6+
4 44 44 45 46 46
n6x
X
+P (logx)x1/7+O(xC4+ε),
47
EXPONENTIAL CARMICHAEL FUNCTION 7
where c , c′ , c and c′ are computable constants, P and P are computable
44 44 46 46 45 47
polynomials, degP =3, degP =5,
45 47
7863059 √13780693090921
(10) C = − =0.134656..., 1/8<C <1/7.
4 4
85962240
Proof. We shall follow the outline of Theorem 3. Let us prove that for c :=1/4+
+1/logx we can estimate
c+ix
I := Z (s)xss−1ds xC4+ε
+ 4
≪
ZC4+ix
and
C4+ix
I := Z (s)xss−1ds xC4+ε.
0 4
≪
ZC4−ix
We start with I x−1 c Z (σ+ix)xσdσ. Now let α(σ) be a function such
+ ≪ C4 4
that Z (σ+ix) xα(σ)+ε. By (3) and (8) we have
4 ≪ R
(16 80σ)µ(1/2)<5/6, σ [1/7,1/6),
− ∈
(12 56σ)µ(1/2)<4/5, σ [1/6,1/5),
α(σ)6 − ∈
 (4−16σ)µ(1/2)<3/4, σ ∈[1/5,1/4),
0, σ [1/4,c].
∈
So c Z (σ+ix)xσ−1dσ xεandtheonlycasethatrequiresfurtherinvestigations
1/7 4 ≪
is σ [C ,1/7). Instead of (3) we apply (4) together with (1) and (2) to obtain
R∈ 4
1045018 2459357
α(σ)6 σ, σ [1/8,1/7],
249485 − 99794 ∈
which implies 1/7xα(σ)+σ−1dσ xC4+ε as soon as
C4 ≪
R C >1591066/12296785=0.129388...
4
Our choice of C in (10) is certainly the case.
4
Let us move on I and prove that xZ (C +it)dt x1+ε. For q , q , q , q
0 1 4 4 ≪ 1 2 3 4
such that
R
(11) 1/q +1/q +1/q +1/q =1 and q ,q ,q ,q >1
1 2 3 4 1 2 3 4
by H¨older inequality we have
x x 1/q1 x 1/q2
Z (C +it)dt ζ2q1(4s+it) dt ζ4q2(5s+it) dt
4 4
≪ ×
Z1 (cid:18)Z1 (cid:19) (cid:18)Z1 (cid:19)
(cid:12)(cid:12) x (cid:12)(cid:12) 1/q3 (cid:12)(cid:12) x (cid:12)(cid:12) 1/q4
ζ2q3(6s+it) dt ζ6q4(7s+it) dt .
×
(cid:18)Z1 (cid:19) (cid:18)Z1 (cid:19)
(cid:12) (cid:12) (cid:12) (cid:12)
Choose (cid:12) (cid:12) (cid:12) (cid:12)
(12) q =m(4C )/2, q =m(5C )/4, q =m(6C )/2, q =m(7C )/6
1 4 2 4 3 4 4 4
One can make sure by substituting the value of C from (10) into Lemma 8 that
4
such choice of q satisfies (11). Thus we obtain
k
x
Z (C +it)dt x(1+ε)/q1x(1+ε)/q2x(1+ε)/q3x(1+ε)/q4 x1+ε,
4 4
≪ ≪
Z1
which finishes the proof. (cid:3)
8 ANDREWV.LELECHENKO
5. Decrease of C
4
In this section we obtain lower value of C by improving lower bounds of m(σ)
4
from Lemma 8.
Estimates below depend on values of
ak+bl+c
(13) inf ,
(k,l)dk+el+f
where (k,l)runs overthe setofexponentpairs andsatisfiescertainlinearinequali-
ties. Amethodtoestimate(13)withoutlinearconstrainswasgivenbyGraham[5].
In the recent paper [10] we have presented an effective algorithm to deal with (13)
under a nonempty set of linear constrains.
Let c be an arbitrary function such that c(σ) > µ(σ). Define θ by an implicit
equation
2c θ(σ) +1+θ(σ) 2 1+c θ(σ) σ =0.
−
Finally, define (cid:0) (cid:1) (cid:0) (cid:0) (cid:1)(cid:1)
1+c θ(σ)
f(σ)=2 .
c θ(σ)
(cid:0) (cid:1)
Due to Lemma 3 one can take c(σ)=infl−(cid:0)k>σ(k(cid:1)+l σ)/2, where (k,l) runs over
−
the set of exponent pairs. However even rougher choice of c leads to satisfiable
values of f such as in [7, (8.71)].
Lemma 9. Let σ >5/8. Compute
4 4σ 12 1 α
1
α = − , β = , m = − β ,
1 1 1 1
1+2σ −1+2σ µ(σ) −
4(1 σ)(k+l) 4(1+2k+2l)
α (k,l)= − , β (k,l)= ,
2 2
(2+4l)σ 1+2k 2l −(2+4l)σ 1+2k 2l
− − − −
1 α (k,l)
2
m (k,l)= − β (k,l), m = inf m (k,l),
2 2 2 2
µ(σ) − α2(k,l)61
where (k,l) runs over the set of exponent pairs. Then
m(σ)>min m ,m ,2f(σ) .
1 2
Note that for σ >2/3 the condition α ((cid:0)k,l)61 is alw(cid:1)ays satisfied.
2
Proof. Follows from [7, (8.97)] and from TαVβ TVβ+(α−1)/µ(σ) for α < 1
≪
and V Tµ(σ). (cid:3)
≪
Substituting pointwise estimates of m(σ) fromLemma 9 instead of segmentwise
from Lemma 8 into (12) we obtain following result.
Theorem 5. The statement of Theorem 4 remains valid for
C =0.133437785...
4
6. Conclusion
Wehaveobtainednontrivialerrortermsinasymptoticestimatesof λ(e)(n)
n6x r
forr =1,2,3,4. Casesofr =1andr =2dependonthemethodofexponentpairs.
P
Casesofr =3 andr =4 depend onlowerbounds of m(σ). Note that caseof r =4
may be improved under Riemann hypothesis up to C = 1/8, because Riemann
4
hypothesis implies µ(σ)=0 and m(σ)= for σ [1/2,1].
∞ ∈
EXPONENTIAL CARMICHAEL FUNCTION 9
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I. I. Mechnikov Odessa National University
E-mail address: [email protected]