notes on the riemann hypothesis and 1ex abundant...

Notes on the Riemann hypothesisand

abundant numbers

Keith Briggs

[email protected]

more.btexact.com/people/briggsk2/

2005 April 21 14:27

RH abundant.tex (‘notes’ option) typeset in pdfLATEX on a linux system

RH and abundant numbers 1 of 22

http://more.btexact.com/people/briggsk2/

Introduction

F In [1], Lagarias showed the equivalence of the Riemann hy-pothesis (RH) to a condition on harmonic sums, namely

RH⇔ σ(n) 6 Hn+exp(Hn) log(Hn) ∀n

F Robin [3] had already shown that

RH⇔ σ(n)/n < eγ log log (n) for n > 5040

which is probably more convenient for numerical tests

F so to disprove RH, we need to find n with large σ(n)/n

Keith Briggs RH and abundant numbers 2 of 22

Notation

F n is always a positive integer; p is always a prime

F pi = ith prime (p1=2, p2=3, . . . )

F harmonic sum for n > 1 is Hn =∑n

i=1 i−1

F sum of divisors is σ(n) =∏

ip

ai+1i −1pi−1 for n =

∏i pai

i , (ai > 0)

F σ(n)/n =∏

ipi−p

−aii

pi−1

F I define ρ(n) ≡ σ(n)/n

F eγ = 1.78107241799019798523650410310717954 . . . , where γ isEuler's constant

F we will deal with n too large to represent in the computer(≈ 10(1010)), but luckily we have its prime factorization whichsuffices


Superabundant numbers [2]

F n is superabundant (SA) if σ(k)/k < σ(n)/n for all k < n

F if n = 2a2 3a3 . . . mam is SA, where m is the maximal primefactor, then a2 > a3 > · · · > am

F if 1 < j < i 6 m, then |ai−baj logi jc| 6 1

F am = 1 unless n = 4 or 36

F iai < 2a2+1, i > 2

F m ∼ log(n)

F SA numbers, with CA numbers in red:2, 22, 2.3, 22.3, 23.3, 22.32, 24.3, 22.3.5 . . .


The work of Robin [3]

F Theorem: RH⇔ ρ(n) < eγ log log (n) for n > 5040

F Theorem: independently of RH, except for n = 1, 2, 12:

ρ(n) < eγ log log (n)+[7/3−eγ log log (12)] log log (12)

log log (n)

The numerator in the last term is about 0.6482

F But note lim supn→∞σ(n)

n log log (n) = eγ


The structure of the set of CA numbers

F Defn: n is colossally abundant (CA) if there exists ε > 0 suchthat

σ(n)/n1+ε > σ(k)/k1+ε ∀k > 1

F Robin [3] has given the following precise description, dueoriginally to Erd ′′os & Nicolas [6]

F we first form the set E of critical ε values:

Ep ≡⋃

α=1,2,3,...

{ logp

(1+

1∑αi=1 pi

)}

E ≡⋃p

Ep

F we label the elements of E in decreasing order:ε1 = log2(3/2) > ε2 = log3(4/3) > ε3 = log2(7/6) > . . .


The structure of the set of CA numbers cotd.

(a) if ε /∈ E, σ(n)/n1+ε has a unique maximum attained at thenumber nε with prime exponents given by

ap(ε) =⌊logp

(p1+ε−1pε−1

)⌋−1

(b) if ε satisfies εi+1 < ε < εi for i = 1, 2, 3, . . . , then nε isconstant and we call it Ni. We have N1 = 2, N2 = 6, . . .

(c) if the sets Ep are pairwise disjoint (which is likely, but notcertainly known), then the set of CA numbers is equal to theset of Ni, i = 1, 2, 3, . . . . If this is the case, σ(n)/n1+ε attainsits maximum at the two points Ni and Ni+1

(d) if the sets Ep are not pairwise disjoint, then for eachεi ∈ Eq∩Ep, σ(n)/n1+εi attains its maximum at the four pointsNi, qNi, rNi and Ni+1 = qrNi


Alaoglu & Erd ′′os [2]

F these authors have a slightly stronger definition:

F Defn: n is colossally abundant if

σ(n)/n1+ε > σ(k)/k1+ε for 1 6 k < n

σ(n)/n1+ε > σ(k)/k1+ε for k > n

F the effect of this is to make a unique choice from the 2 or 4possibilities in cases (c) and (d) above. But I will perform mycomputations with the Robin definition

F we will call these numbers strongly colossally abundant (SCA)if it is necessary to distinguish them from ordinary CA numbers


Position of critical ε values

critical ε values

0 2 4 6 8 100

2

4

6

8

10

12

−log(ε)

p

The vertical lines mark thecritical ε values arising from thesmall primes on the y axis


Density of critical ε values

strongly colossally abundant numbers

1 2 3 4 5 6 70

1

2

3

4

5

6

−log(ε)

log

log

(n)

The vertical lines markthe critical ε values


Dependence of maximal prime on n

0 2 4 6 8 10 12 14100

101

102

103

104

105

106

loglog(n)

p max

This shows the maximalprime needed as a functionof log log (n)


Method

F it is known that if RH is false, there will be a violation ofRobin's inequality which is a CA number . . .

F for a range of small ε > 0, I compute n by the A&E formula

F I compute RHS-LHS of Robin's inequality (let's call this the de-viation δ(n) ≡ eγ log log (n)−ρ(n)), and look for any violations(i.e. δ < 0)

F we can also plot η(n) ≡ eγ−ρ(n)/ log log (n)

F the following plots show the behaviour observed so far (toabout ε = exp(−25))


Computational difficulties

F the exponents ap(ε) must be computed in interval arithmetic,to ensure they are correct and not corrupted by roundofferror. This means not just high precision arithmetic, butdynamically varying precision

F millions of primes are needed. Typically the n we deal withhave a huge tail of many primes to the power 1. It is fastestto precompute primes with a sieve, but then much storage isrequired.

F how do we vary ε to not miss any SCA numbers? (DONE)

F how to we compute explicit examples of WCA numbers?

F there are many other difficulties . . .


Computational strategy

We keep a list z of records, containing: a prime p, log p, its exponent a, and acritical εc, which is the value of ε at which this exponent will next change (as ε isdecreased). We exclude 1 exponents, which are counted by ones. We first initialize:

. fix 0 < εstart 6 1. Then, for each prime p, compute a =⌊logp

(p1+εstart−1pεstart−1

)⌋−1

and store it in the z list if a > 2. If a = 1, just increment the variableones. Stop when a = 0. During this p loop, also update log(n) and ρ(n)

Then each step of the main loop consists of determining which of possible eventsA, B, or C occurs:

. A: a new prime (with exponent 1) is added, so we increment `ones'. Thishappens when εext = logp (1+p) is maximal, where p is the new prime

. B: the first prime with exponent 1 has its exponent raised to 2. This happenswhen εinc = logp ((p+1+1/p)/(p+1)) is maximal, where p is the prime inquestion

. C: a prime with exponent > 2 has its exponent incremented. This happenswhen εmax = logp ((1−pa+1)/(p−pa+1)) is maximal, where p is the prime inquestion and a its exponent


Prime generation

For simple tests, just use a lookup list. With the BERNSTEIN option, crit eps14.cuses Bernstein’s quadratic sieve. Here we cannot just look up any prime; rather wehave a function to get the next prime and advance the internal state (also to peek atthe current prime without advancing). So the strategy is to use 2 prime generators- one (pg1) for the z list (which only grows slowly), and another (pg2) to see if thelist of 1s needs extending. pg2 goes a long way, but we rely on Bernstein’s code tokeep it efficient.


Questions

F how does n depend on ε?

F how does the largest prime needed depend on ε?

F how does δ depend on ε?

F interesting observation: when p is large and ε is small (thesituation we are interested in), in the exponent formula ofAlaoglu and Erd ′′os it is already sufficient to use the first termof a Taylor expansion in ε, namely as ε → 0+,

logp

(p1+ε−1pε−1

)= logp

(p−1log p

)−logp(ε)+O(ε)

already has error less than 1/2, so the floor is the correctinteger. How can we exploit this?

F in the following plots the computed data is in red and hy-pothesized fits or trends are in blue


Dependence of n on ε

strongly colossally abundant numbers

5 10 15 20 25 300

5

10

15

20

25

30

−log(ε)

log

log

(n( ε)

)

This shows that log log (n)at SCA numbers n ap-pears to be asymptoticallya linear function of − log ε.The line −1.779+0.947xis guesswork


Density of CA numbers

0 2 4 6 8 100.0009

0.0010

0.0011

0.0012

0.0013

0.0014

log(n)/1010

CA

num

ber

dens

ity


Dependence of δ on n

14 16 18 20 22 24 26 28−14

−13

−12

−11

−10

−9

−8

−7

loglog(n)

log

( δ)

This shows that log δ ap-pears to be asymptoticallya linear function of x =log log (n)


Dependence of δ on n

26.969 26.970 26.971 26.972 26.973 26.974 26.975−13.1495

−13.1490

−13.1485

−13.1480

−13.1475

−13.1470

−13.1465

−13.1460

loglog(n)

log

( δ)

This is the last 100000values of the previousplot


Deviation of log δ from a best-fit line

15 20 25−0.04

−0.02

0.00

0.02

0.04

0.06

0.08

loglog(n)

log

( δ)−

(0.3

23

33

6−

log

log

(n)/

2)

This shows the differencebetween log δ and a con-jectured best-fit line a−x/2, where x ≡ log log (n)


References

[1] J. C. Lagarias An elementary problem equivalent to the Rie-mann hypothesis, Amer. Math. Monthly, 109 (2002), 534-543www.math.lsa.umich.edu/∼lagarias/doc/elementaryrh.ps

[2] L. Alaoglu & P. Erd ′′os On highly composite and similar num-bers, Trans. Amer. Math. Soc. 56 (1944) 448-469

[3] G. Robin Grandes valeurs de la fonction somme des diviseurset hypothèse de Riemann, J. Math. pures appl. 63 (1984)187-213

[4] S. Ramanujan Highly composite numbers. Annotated and witha foreword by J.-L. Nicholas and G. Robin, Ramanujan J. 1(1997) 119-153

[5] J.-L. Nicolas Ordre maximal d'un élément du groupe despermutations et highly composite numbers, Bull. Math. Soc Fr.97 (1969), 129-191

[6] P. Erd ′′os and J.-L. Nicolas Répartition des nombres superabon-dants, Bull. Math. Soc Fr. 103 (1975), 65-90


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