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from the Dedekind psi function Michel Planat

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Michel Planat. Riemann hypothesis from the Dedekind psi function. 2010. ￿hal-00526454v1￿

HAL Id: hal-00526454 https://hal.archives-ouvertes.fr/hal-00526454v1 Preprint submitted on 14 Oct 2010 (v1), last revised 22 Oct 2010 (v2)

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Michel Planat Institut FEMTO-ST, CNRS, 32 Avenue de l’Observatoire, F-25044 Besan¸con, France.

Abstract. Let P be the set of all primes and ψ(n) = n Qn∈P,p|n(1+1/p) be the Dedekind psi function. We prove that, under the assumption that the n-th prime pn is always larger than the first θ(pn), the Riemann hypothesis is γ satisfied if and only if f(n) = ψ(n)/n − e log 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 function σ(n) ≥ ψ(n) and the lower bound by n0 = 5040. For a square free number, both arithmetical functions σ and ψ are the same. We also prove that any exception to (D) may only occur at a positive n satisfying ψ(m)/m < ψ(n)/n, for any m

PACS numbers: 11A41, 11N37, 11M32

1. Introduction

−s 1 ζ(s)= n = − (where the product is taken over Pn>1 Qp∈P 1−p s the set P of 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 1 non-real zeros all lie on the critical line R(s) = 2 . RH has equivalent formulations, many of them having to do with the distribution of prime numbers [1, 2]. Let d(n) be the . There exists a remarkable parallel between the

error term ∆(x) = Pn≤x d(n) − x(log x +2γ − 1) in the summatory function of d(n) 1 (Dirichlet’s divisor problem) and the corresponding mean-square estimates |ζ( 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 is a positive integer n such 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, σ(n) RH is true iff g(n)= − eγ log log n< 0 for any n> 5040. (1) n 2

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}. 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 n satisfying σ(m) σ(n) < for any m < n. (2) m n No counterexample has been found so far. In the present paper, Robin’s criterion is refined by replacing the sum of divisor 1 function by the Dedekind psi function ψ(n) = Qp∈P,p|n(1 + p ) ‡. Since ψ(n) ≤ σ(n), with equality when n is free of square, we establish the refined criterion ψ(n) RH is true iff f(n)= − eγ log log n< 0 for any n> 30. (3) n 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) ψ(n) < for any m < n, (4) m n n and be a primorial number Nn = Qi=1 pi (the product of the first n primes) or one its multiples smaller than Nn+1 (Sloane sequence A060735). γ According to a Mertens theorem [2], one has lim ψ(Nn) = e log(p ) and we n→∞ Nn ζ(2) n 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 S of all positive numbers satisfying (4). For 2

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 Nn and their multiples up to the next primorial Nn+1.

‡ The Dedekind function ψ(n) should not be confused with the second Chebyshev function ψT (n) = Ppk≤n log p. 3

It is straightforward to check that about half of the numbers in S are not superabundant (compare to Sloane sequence A004394). 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). ˜ ψ(n) Proof: One may use f(n)= n log log n instead of f(n). 2 One observes that lNn = p1p2 ··· l ··· pn for some pj = l. The corresponding Dedekind psi function is evaluated as

2 ψ(lNn)=(pj + pj) Y ψ(pi)= lψ(Nn). i6=j Then, with ψ(lNn) ψ(Nn) f˜(lNn)= = , lNn log log(lNn) Nn log log(lNn) one concludes that

f˜(lNn) log log Nn = < 1 n the required range 1

Proposition 2: Given l ≥ 1, for any m such that lNn

Proof: This proposition is proved by using inequality (4) at n =(l + 1)Nn

ψ(m) ψ[(l + 1)Nn] < for any m< (l + 1)Nn m (l + 1)Nn

and the relation ψ[(l + 1)Nn]=(l + 1)ψ(Nn). As a result

ψ(Nn) log log Nn f˜(m) < = f˜(Nn) < f˜(Nn) since m > Nn. Nn log log m log log m

Proposition 3: For any n > 1 one has (i) pn+1 > θ(pn) and as result (ii) n f(Nn+1) < f(Nn), where θ(pi)= Pi=1 log pi is the first Chebyshev function at pi. This statement is reinforced by numerical calculations: an excerpt is in Table 1.

3 5 7 pn 10 10 10 10 θ(pn) 0.779 0.986 0.99905 0.999958 pn ˜ f(Nn+1) 0.987 0.9999980 0.99999999921 0.99999999999975 f˜(Nn)

f˜(Nn+1) Table 1. An excerpt of values of θ(pn)/pn and versus the number of primes f˜(Nn) in the primorial Nn. 4

Justification n : At a primorial n = Nn, ψ(Nn) = Qi=1(1 + pi) so that ψ(Nn+1) = (1 + pn+1)ψ(pn). One would like to show that, for any pn > 2

1 f˜(N ) 1 log θ(p ) 1+ n+1 =(1+ ) n = pn+1 < 1 log pn+1 f˜(N ) pn+1 log θ(pn+1) 1 + log(1 + )/ log θ(p ) n θ(pn) n

This is equivalent to

log θ(p ) log p n < log(1 + n+1 ), pn+1 θ(pn) and this follows from the weaker inequality

pn+1 log pn+1 > θ(pn) log θ(pn),

that is satisfied if n> 2, pn+1 >pn > θ(pn). Thus (i) ⇒ (ii). Proving (i) is not an easy task. From the theorem [1, 2], one knows that

θ(pn)= π(pn) log(pn) ∼ pn, where π(n) is the prime counting function. In addition, it has been proved that [16] (see also table 2 in [17])

θ(n) < n for 0 < n ≤ 1011, but no result is known to us concerning general values of n. 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 stipulates that the function f(Nn) versus

Nn is a strictly decreasing function. Let us first show that RH ⇒ (3). This is easy because if RH is true, then Robin’s inequality (1) is true, for any

n > 5040, including at primorials n = Nn [5]. Since Nn is free of square, one has

ψ(Nn)= σ(Nn) so that the refined inequality (3) is satisfied at n = Nn and, according to propositions 1 to 3, it is also satisfied at any n. The reverse implication (3) ⇒ RH is similar to that for Robin’s inequality (theorem 2 in [5]). By contradiction, if RH is false, there exists an infinity of numbers n such that γ ψ(n) g(n) > e and the bound on n is weakened as ψ(n) σ(n) 0.6482 for n ≥ 3, ≤ ≤ eγ log log n + . n n log log n We have proven that the modified Robin’s criterion is equivalent to RH. To prove that RH is true, it is sufficient to prove that no exception to this criterion can be found. 5

At large value of primorials Nn, we use Mertens theorem about the density of primes n 1 −γ ln pn (1 − ) ∼ e , or the equivalent relation § Qk=1 pk n γ 1 ψ(Nn) 1 1 e ≡ Y(1 + ) ∼ , (5) ln pn Nn ln pn pk ζ(2) k=1 and the lower bound given p. 206 of [5] 0.1235 for pn ≥ 20000, log log Nn > log pn − . (6) log pn Using (5) and (6), and with eγ (ζ(2)−1 − 1) ≈ −0.6983, one obtains a lower bound 0.1235 for pn ≥ 20000, f(n) < −0.6983 log pn + ∼ −6.90. (7) log pn

For values of 210 ≤ Nn ≤ 200000, computer calculations can be performed. The

calculation of f(Nn) = g(Nn) < 0 is fast using the multiplicative property of σ(n), i.e. n using σ(Nn) = Qi=1(1 + pi). One find a decreasing function g(pn) that is negative if pn > 2 as expected (and in agreement with the exceptions in the set A).

pn 2 20 200 2000 20000 200000 f(Nn)= g(Nn) 0.96 −2.57 −4.87 -6.79 -8.60 -10.35

Table 2. The approximate values of the function f(Nn)= g(Nn) versus the number of 2748049 primes in the primorials Nn. The highest primorial in the table is N200000 ≈ 10 .

According to our calculations above and the bound (7), there is no exception to the refined Robin’s criterion. To summarize, if it can be proved that the n-th prime pn is always larger that the first Chebyshev function θ(pn), then the refined Robin’s criterion is equivalent to RH, and RH must be true.

Acknowledgements

The author thanks P. Sol´efor pointing out his paper [7], following the presentation of [12] at QuPa meeting in Paris on 09/23/2010. He also thanks Fabio Anselmi for his feedback on this subject.

Bibliography

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