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Topic: abundant numbers, Lagarias criterion for the Riemann Hypothesis
Replies: 13   Last Post: Jul 18, 2013 2:13 AM

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David Bernier

Posts: 3,276
Registered: 12/13/04
Re: abundant numbers, Lagarias criterion for the Riemann Hypothesis
Posted: Jun 14, 2013 3:01 PM
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On 06/14/2013 01:15 PM, David Bernier wrote:
> On 06/14/2013 12:08 PM, James Waldby wrote:
>> On Thu, 13 Jun 2013 16:45:10 -0400, David Bernier wrote:
>>> On 06/13/2013 12:52 PM, David Bernier wrote:
>>>> On 06/13/2013 10:38 AM, David Bernier wrote:
>>>>> I've been looking for abundant numbers, a number `n' whose
>>>>> sum of divisors sigma(n):= sum_{d dividing n} d
>>>>> is large compared to `n'.
>>>>>
>>>>> One limiting bound, assuming the Riemann Hypothesis,
>>>>> is given by a result of Lagarias:
>>>>> whenever n>1, sigma(n) < H_n + log(H_n)*exp(H_n) ,
>>>>> where H_n := sum_{k=1 ... n} 1/k .
>>>>> Cf.:
>>>>> <http://en.wikipedia.org/wiki/Harmonic_number#Applications> .
>>>>>
>>>>> The measure of "abundance" I use, for an integer n>1, is
>>>>> therefore:
>>>>> Q = sigma(n)/[ H_n + log(H_n)*exp(H_n) ].
>>>>>
>>>>> For n which are multiples of 30, so far I have the
>>>>> following `n' for which the quotient of "abundance"
>>>>> Q [a function of n] surpasses 0.958 :
>>>>>
>>>>> n Q
>>>>> -----------------------
>>>>> 60 0.982590
>>>>> 120 0.983438
>>>>> 180 0.958915
>>>>> 360 0.971107
>>>>> 840 0.964682
>>>>> 2520 0.978313
>>>>> 5040 0.975180
>>>>> 10080 0.959301
>>>>> 55440 0.962468
>>>>> 367567200 0.958875
>>>>>
>>>>> What is known about lower bounds for
>>>>> limsup_{n-> oo} sigma(n)/[ H_n + log(H_n)*exp(H_n) ] ?

>>>>
>>>> I know there's Guy Robin earlier and, I believe, Ramanujan
>>>> who worked on "very abundant" numbers ...

>>>
>>> limsup_{n-> oo} sigma(n)/( n log(log(n)) ) = exp(gamma), (***)
>>> gamma being the Euler-Mascheroni constant.
>>>
>>> This result above, (***), is known as
>>> Grönwall's Theorem, dated in the literature to 1913.

>> ...
>>>> n = 2021649740510400 with Q = 0.97074586,
>>>> almost as "abundantly abundant" as n=360, with Q = 0.971107
>>>> sigma(2,021,649,740,510,400) = 12,508,191,424,512,000

>>
>> Q values (with H_n approximated by gamma + log(n+0.5))) for
>> "colossally abundant numbers" <http://oeis.org/A004490>
>> where Q exceeds 0.958 include the following.
>> k Q(a_k) a_k sigma(a_k) Exponents
>> of prime factors of a_k
>> 13. 0.958875 367567200 1889879040 [5, 3, 2, 1,
>> 1, 1, 1, 0, 0, 0, 0, 0, 0]
>> 14. 0.965887 6983776800 37797580800 [5, 3, 2, 1,
>> 1, 1, 1, 1, 0, 0, 0, 0, 0]
>> 15. 0.968911 160626866400 907141939200 [5, 3, 2, 1,
>> 1, 1, 1, 1, 1, 0, 0, 0, 0]
>> 16. 0.968922 321253732800 1828682956800 [6, 3, 2, 1,
>> 1, 1, 1, 1, 1, 0, 0, 0, 0]
>> 17. 0.967932 9316358251200 54860488704000 [6, 3, 2, 1,
>> 1, 1, 1, 1, 1, 1, 0, 0, 0]
>> 18. 0.968838 288807105787200 1755535638528000 [6, 3, 2, 1,
>> 1, 1, 1, 1, 1, 1, 1, 0, 0]
>> 19. 0.970746 2021649740510400 12508191424512000 [6, 3, 2, 2,
>> 1, 1, 1, 1, 1, 1, 1, 0, 0]
>> 20. 0.970641 6064949221531200 37837279059148800 [6, 4, 2, 2,
>> 1, 1, 1, 1, 1, 1, 1, 0, 0]
>> 21. 0.971747 224403121196654400 1437816604247654400 [6, 4, 2, 2,
>> 1, 1, 1, 1, 1, 1, 1, 1, 0]
>>
>> Similar results arise for "superior highly composite numbers"
>> <http://en.wikipedia.org/wiki/Superior_highly_composite_number>
>> and <http://oeis.org/A002201>.
>>
>> I haven't computed Q values for other numbers than the SHCN's
>> and CAN's shown in OEIS. However, it would be quite easy to
>> compute Q values for the first 124260 HCN's, because a 2MB
>> compressed file of them is available which contains ln h_k and
>> ln(sigma(h_k)) at the front of line k, followed by a list of
>> exponents of h_k's prime factorization. The link to the file,
>> <http://wwwhomes.uni-bielefeld.de/achim/HCNs.gz>, is shown at
>> the end of <http://wwwhomes.uni-bielefeld.de/achim/highly.html>.
>>

>
>
> Ok. so I would go to:
> http://wwwhomes.uni-bielefeld.de/achim/highly.html
> and from there, download and "gunzip" the file HCNs.gz ...
>
> Thanks for the feedback.
>
> I'm looking at finding smallish numbers `n' with unusually high
> Q ratio ...
>
> I get a Q ratio of 0.9925 for some 148-digit number.
> Then, 1 - Q is about 0.0075, and I wonder how this relates,
> say, to 1/log(n) , so I want a "Gold standard" to
> single-out exceptionally abundant numbers ...
>
>
> ? A = 2^10*3^6*5^4*7^3*11^2*13^2*17^2*19^2*23^2;
> ? B = 29*31*37*41*43*47*53*59*61*67*71*73*79*83;
> ? C = 89*97*101*103*107*109*113*127*131*137*139;
> ? D = 149*151*157*163*167*173*179*181*191*193;
> ? E = 197*199*211*223*227*229*233*239*241*251;
> ? F = 257*263*269*271*277*281*283*293*307*311;
> ? G = 313*317*331;
> ? n = A*B*C*D*E*F*G; // n is the 148-digit number ...
>
>
> ? harmonic(Z) = Euler+psi(Z+1); // Function definition ...
> // Euler = 0.577... and psi is the
> digamma function.
>
>
> ? hh = harmonic(n); // sum_{k = 1... n} 1/k using fn. def. above
>
> ? dd = hh + log(hh)*exp(hh); // Expression based on n'th
> // harmonic number
>
>
> ? Q = sigma(n)/dd; // the ratio, Q, for 148-digit `n' gets defined
>
>
> ? Q
> %13 = 0.99251022615763635838615903736818502634
>
>
> ? 1 + floor( log(n)/log(10) ) // `n' has 148 digits
> %14 = 148


I got the HCNs text file.

One of the highly composite numbers I'll call `m' ,
whose logarithm differs very little from the `n'
defined above.

We have:

sigma(n)/[ exp(gamma)*n*log(log(n)) ] ~= 0.992798

and

sigma(m)/[ exp(gamma)*m*log(log(m)) ] ~= 0.991642 ,

so the 'n' from the epirical work on looking for large
Q is somewhat better than the highly-composite 'm'.

Actually, m and n have the same number of divisors,
(for whatever reason), although 'n' has 67 distinct
prime factors, and 'm' has 66 distinct prime factors.

n/m = 331/319, and 319 = 11*29.

So, n = 331*m/(11*29) , and 331 is prime, just like 11 and 29.

The mystery is why, going from 'm' to 'n',
switch a factor of 11 and a factor
of 29, to be replaced by one prime factor of 331,
to get a higher 'Q' ratio?

The first five primes after 300 are
307, 311, 313, 317 and 331 .

They are all factors of 'n'.

So, 307, 311, 313 and 317 are factors of the highly composite
number 'm', but not 331.

Instead, 'm' can be obtained as: (n/331)*11*29 .

David Bernier


Work:

? (sigma(n, 1)/(n*log(log(n))))/exp(Euler)
%53 = 0.99279892618630412257048797008815184714

? (sigma(m, 1)/(m*log(log(m))))/exp(Euler)
%54 = 0.99164208754504011740275423920309459284


? n/m
%55 = 331/319



--
On Hypnos,
http://messagenetcommresearch.com/myths/bios/hypnos.html



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