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arXiv:1010.3239v2 [math.GM] 23 Oct 2010
Riemann hypothesis from the Dedekind psi function
Michel Planat
Institut FEMTO-ST, CNRS, 32 Avenue de l’Observatoire,
F-25044 Besan¸con, France.
Abstract. Let Pbe the set of all primes and ψ(n) = nQn∈P ,p|n(1 + 1/p) be the
Dedekind psi function. We show that the Riemann hypothesis is satisfied if and only
if f(n) = ψ(n)/n −eγlog log n < 0 for all integers n > n0= 30 (D), where γ≈0.577 is
Euler’s constant. This inequality is equivalent to Robin’s inequality that is recovered
from (D) by replacing ψ(n) with the sum of divisor function σ(n)≥ψ(n) and the lower
bound by n0= 5040. For a square free number, both arithmetical functions σand ψ
are the same. We also prove that any exception to (D) may only occur at a positive
integer nsatisfying ψ(m)/m < ψ(n)/n, for any m < n, hence at a primorial number
Nnor at one its multiples smaller than Nn+1 (Sloane sequence A060735). According to
a Mertens theorem, all these candidate numbers are found to satisfy (D): this implies
that the Riemann hypothesis is true.
PACS numbers: 11A41, 11N37, 11M32
1. Introduction
Riemann zeta function ζ(s) = Pn>1n−s=Qp∈P
1
1−p−s(where the product is taken over
the set Pof all primes) converges for R(s)>1. It has a analytic continuation to the
complex plane with a simple pole at s= 1. The Riemann hypothesis (RH) states that
non-real zeros all lie on the critical line R(s) = 1
2. RH has equivalent formulations,
many of them having to do with the distribution of prime numbers [1, 2].
Let d(n) be the divisor function. There exists a remarkable parallel between the
error term ∆(x) = Pn≤xd(n)−x(log x+ 2γ−1) in the summatory function of d(n)
(Dirichlet’s divisor problem) and the corresponding mean-square estimates |ζ(1
2) + it|of
ζ(s) on the critical line, see [3] for a review. This may explain Ramanujan’s interest for
highly composite numbers [4]. A highly composite number is a positive integer nsuch
that for any integer m < n,d(m)< d(n), i.e. it has more divisors than any positive
integer smaller than itself.
This landmark work eventually led to Robin’s formulation of RH in terms of the
sum of divisor function σ(n) [5, 6, 7]. More precisely,
RH is true iff g(n) = σ(n)
n−eγlog log n < 0 for any n > 5040.(1)
The numbers that do not satisfy (1) are in the set A={2,3,4,5,6,8,9,10,12,16,18,20,24,
30,36,48,60,72,84,120,180,240,360,720,840,2520,5040}.
2
If RH is false, the smallest value of n > 5040 that violates the inequality should be
a superabundant number [8, 9, 10], i.e. a positive number nsatisfying
σ(m)
m<σ(n)
nfor any m < n. (2)
No counterexample has been found so far.
In the present paper, Robin’s criterion is refined by replacing the sum of divisor
function by the Dedekind psi function ψ(n) = Qp∈P,p|n(1 + 1
p)‡. Since ψ(n)≤σ(n),
with equality when nis free of square, we establish the refined criterion
RH is true iff f(n) = ψ(n)
n−eγlog log n < 0 for any n > 30.(3)
The numbers that do not satisfy (3) are in the set B={2,3,4,5,6,8,10,12,18,30}.
A number that would possibly violate (3) should satisfy
ψ(m)
m<ψ(n)
nfor any m < n, (4)
and be a primorial number Nn=Qn
i=1 pi(the product of the first nprimes) or one its
multiples smaller than Nn+1 (Sloane sequence A060735).
According to a Mertens theorem [2], one has limn→∞ ψ(Nn)
Nn=eγ
ζ(2) log(pn) and we
show that none of the numbers larger than 30 in the sequence A060735 can violate (3).
As a result, RH may only be true.
In the next section, we provide the proof of (3) and compare it with the Robin’s
criterion (1). Furthermore, we investigate the structure of numbers satisfying (4) and
justify why they fail to provide counterexamples to RH.
Originally, Dedekind introduced ψ(n) as the index of the congruence subgroup Γ0(n)
in the modular group (see [11], p. 79). In our recent work, the Dedekind psi function
plays a role for understanding the commutation relations of quantum observables within
the discrete Heisenberg/Pauli group [12]. In particular, it counts the cardinality of the
projective line P1(Zn) of the lattice Zn×Zn. The relevance of the Dedekind psi function
ψ(n) in the context of RH is novel. For other works aiming at a refinement of Robin’s
inequality, we mention [13], [7] and [14].
2. A proof of the refined Robin’s inequality
Let us compute the sequence Sof all positive numbers satisfying (4). For 2 < n < 105,
one obtains S={N1= 2,4, N2= 6,12,18,24, N3= 30,60,90,120,150,180, N4=
210,420,630,840,1050,1260,1470,1680,1890,2100, N5= 2310,4620,6930,9240}, which
are the first terms of Sloane sequence A060735, consisting of the primorials Nnand their
multiples up to the next primorial Nn+1.
It is straightforward to check that about half of the numbers in Sare not
superabundant (compare to Sloane sequence A004394).
‡The Dedekind function ψ(n) should not be confused with the second Chebyshev function ψT(n) =
Ppk≤nlog p.
3
Based on calculations performed on the numbers in the finite sequence S, we are
led to three properties that the infinite sequence A060735 should satisfy
Proposition 1: For any l > 1 such that Nn< lNn< Nn+1 one has f(lNn)<
f(Nn).
Proof: One may use ˜
f(n) = ψ(n)
nlog log ninstead of f(n) to simplify the proof.
When lis prime, one observes that lNn=p1p2···l2···pnfor some pj=l. The
corresponding Dedekind psi function is evaluated as
ψ(lNn) = (p2
j+pj)Y
i6=j
ψ(pi) = lψ(Nn).
Then, with
˜
f(lNn) = ψ(lNn)
lNnlog log(lNn)=ψ(Nn)
Nnlog log(lNn),
one concludes that
˜
f(lNn)
˜
f(Nn)=log log Nn
log log(lNn)<1 in the required range 1 < l < pn+1 .
When lis not prime, a similar calculation is performed by decomposing linto a
product of primes and by using the multiplicative property of ψ(n).
Proposition 2: Given l≥1, for any msuch that lNn< m < (l+ 1)Nn< Nn+1
one has f(m)< f(Nn).
Proof: This proposition is proved by using inequality (4) at n= (l+ 1)Nn
ψ(m)
m<ψ[(l+ 1)Nn]
(l+ 1)Nn
for any m < (l+ 1)Nn
and the relation ψ[(l+ 1)Nn] = (l+ 1)ψ(Nn). As a result
˜
f(m)<ψ(Nn)
Nnlog log m=log log Nn
log log m˜
f(Nn)<˜
f(Nn) since m > Nn.
Proposition 3: There exist infinitely many prime numbers pnsuch that f(Nn+1)>
f(Nn).
Remarks on the proposition 3: Based on experimental evidence in table 1, one
would expect that f(Nn+1)< f (Nn) and that the proposition 3 is false.
Similarly, one would expect that θ(pn)< pnfor any n. For instance it is known [15]
that
θ(n)< n for 0 < n ≤1011.
Many oscillating functions were studied in the context of the prime number theorem.
It was believed in the past that, for any real number x, the function ∆1(x) = π(x)−li(x)
(where li(x) is the logarithmic integral) is always negative. However, J E. Littlewood
has shown that ∆1(x) changes sign infinitely often at some large values x > x0[16]. The
4
n10 103105107
θ(pn)
pn0.779 0.986 0.99905 0.999958
˜
f(Nn+1)
˜
f(Nn)0.987 0.9999980 0.99999999921 0.99999999999975
knlog kn
pn+1 log pn+1 0.938 1.00378 1.000447 1.0000423
Table 1. An excerpt of values of θ(pn)/pnand ˜
f(Nn+1)
˜
f(Nn)versus the number of primes
in the primorial Nn.
smallest value x0such that for the first time π(x0)>li(x0) holds is called the Skewes
number. The lowest present day value of the Skewes number is around 10316.
In what concerns the function ∆4(x) = θ(x)−x, according to theorem 1 in [17]
(see also [18], Lemme 10.1), there exists a positive constant c4such that for sufficiently
large T, the number of sign changes of ∆4(x) in the interval [2, T ] is
V4(T)≥c4log T.
Justification of proposition 3
According to theorem 34 in [19] (also used in [18])
θ(x)−x= Ω±(x1/2log3x) when x→ ∞,
where log3x= log log log x. The omega notation means that there exist infinitely
many real numbers xsatisfying
θ(x)≥x+1
2√xlog3x=kx,(5)
and θ(x)≤x−1
2√xlog3x
If x=pn, for some nthen
θ(pn)≥pn+1
2√pnlog3pn=kn.(6)
Otherwise, let us denote pnthe first prime number preceeding x, one has
θ(pn) = θ(x)≥kx≥kn,
that is similar to (6).
At a primorial n=Nn,ψ(Nn) = Qn
i=1(1 + pi) so that ψ(Nn+1 ) = (1 + pn+1)ψ(pn).
One would like to show that there are infinitely many prime numbers pn>2 such that
˜
f(Nn+1)
˜
f(Nn)= (1 + 1
pn+1
)log θ(pn)
log θ(pn+1)=1 + 1
pn+1
1 + log(1 + log pn+1
θ(pn))/log θ(pn)>1.
By contradiction, let us assume that the reverse inequality holds for those prime numbers
x=pnsatisfying (6)
log θ(pn)
pn+1
<log(1 + log pn+1
θ(pn))
5
with θ(pn)≥pn+1
2√pnlog3pn.
(7)
Taking the development of the logarithm in the first equation of (7) one obtains
log kpn
pn+1 <log pn+1
kpn, that is
kpnlog kpn< pn+1 log pn+1 for kpn=pn+1
2√pnlog3pn.(8)
The inequality (8) contradicts the calculations performed in table 1 for 10 < n <
107. We conclude that our proposition 3 is correct in a finite range of pn’s. In addition,
since there are infinitely many prime numbers satisfying (6), proposition 3 is also satisfied
for a infinite range of values.
The Riemann hypothesis
Propositions 1 and 2 show that the worst case for the inequality (3), if not satisfied,
should occur at a primorial Nn. Proposition 3 deals about the distribution of values of
f(Nn) at large n. Propositions 1 and 2 are needed in the proof of RH, based on the
refined Robin inequality.
Let us first show that RH ⇒(3).
This is easy because if RH is true, then Robin’s inequality (1) is true, for any
m > 5040, including at primorials m=Nn,m=Nn+1 and so on [5], despite the
result established in proposition 3 that there are infinitely many values of nsuch that
f(Nn+1)> f (Nn)§. Since Nnis free of square, one has ψ(Nn) = σ(Nn) so that
the refined inequality (3) is satisfied at any m=Nn. This means that if RH is true
then, according to proposition 1, (3) is satisfied at lNn, where Nn< lNn< Nn+1 and,
according to proposition 2, it is also satisfied at any mbetween consecutive values lNn
and (l+ 1)Nnof the sequence A060735.
The reverse implication (3) ⇒RH is similar to that for the Robin’s inequality
(theorem 2 in [5]). We observe that there exists an infinity of numbers nsuch that
˜g(n) = σ(n)
nlog log n> eγand the bound on ψ(n)
nis such that
for n≥3,ψ(n)
n≤σ(n)
n≤eγlog log n+0.6482
log log n.
To prove that RH is true, it is sufficient to prove that no exception to the refined
Robin’s criterion may be found.
At large value of primorials Nn, we use Mertens theorem about the density of primes
log pnQn
k=1(1 −1
pk)∼e−γ, or the equivalent relation k
1
log pn
ψ(Nn)
Nn≡1
log pn
n
Y
k=1
(1 + 1
pk
)∼eγ
ζ(2),(9)
§In the first version of this paper, it was expected that for any pn, one would have θ(pn)< pnand as
result f(Nn+1)< f (Nn). But this property is unnecessary for showing that the refined Robin inequality
is equivalent to RH.
kA better approximation could be obtained from proposition 9 in [20], i.e. from Qp≤x(1 + 1
p) =
eγ
ζ(2) log x+O(1/log x).
6
and the lower bound given p. 206 of [5]
for pn≥20000,log log Nn>log pn−0.123
log pn
.(10)
Using (9) and (10), and with eγ(ζ(2)−1−1) ≈ −0.698, one obtains a lower bound
for pn≥20000, f (Nn)<−0.698 log pn+0.220
log pn∼ −6.89.(11)
For values of 2 ≤n≤100000, computer calculations can be performed. The calculation
of f(Nn) = g(Nn)<0 is fast using the multiplicative property of σ(n), i.e. using
σ(Nn) = Qn
i=1(1 + pi). One find a decreasing function g(Nn) that is negative if n > 3,
i.e. Nn>30, as expected (and in agreement with the exceptions in the sets Aand B).
n3 10 102103104105
f(Nn) = g(Nn) 0.22 −1.67 −4.24 −6.23 -8.06 -9.83
Table 2. The approximate value of the function f(Nn) = g(Nn) versus the number
of primes in the primorial Nn. The smallest primorial in the table is Nn= 30 and the
highest one is N100000 ≈1.9×10563920.
According to the calculations illustated in the table II and the bound established
in (11), there are no exceptions to the refined Robin’s criterion. Since Robin’s criterion
has been shown to be equivalent to RH hypothesis, RH may only be true.
Acknowledgements
The author thanks P. Sol´e for pointing out his paper [7], following the presentation of [12]
at QuPa meeting in Paris on 09/23/2010. He acknowledges J. L. Nicolas for helping him
to establish the proposition 3. He also thanks Fabio Anselmi for his sustained feedback
on this subject.
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