
Re: abundant numbers, Lagarias criterion for the Riemann Hypothesis
Posted:
Jul 18, 2013 2:13 AM


On 06/21/2013 02:38 PM, David Bernier wrote: > On 06/19/2013 08:56 AM, 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: >>>> >>>> whenener 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 ... >>> >>> 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 >> >> I've used PARI/gp to find whole numbers with as large >> a "quotient of abundance" Q as possible, and it has >> taken a while... >> >> a14:= >> >> primorial(3358)*primorial(53)*13082761331670030*510510*210*210*30*1296*128. >> >> >> >> a14 has 13559 digits. The number a14 has a large sigma_1 value >> relative to itself: >> >> sigma(a14)/(harmonic(a14)+log(harmonic(a14))*exp(harmonic(a14))) >> >> >> ~= 0.99953340717845609264672369120283054134 . >> >> // The expression in 'a14' is related to >> // the ratio in the Lagarias RH criterion. >> >> Cf: >> >> "Lagarias discovered an elementary >> problem that is equivalent to the [...]" >> >> at: >> >> < http://en.wikipedia.org/wiki/Jeffrey_Lagarias > . > > Update after more experimentation: > > > a30 = primorial(8555)*primorial(66)*primorial(16) [continued] > *primorial(8)*primorial(5)*primorial(4) [continued] > *primorial(3)^2*primorial(2)^4*2^8; > > Qr(a30) ~= 0.9997306665 . > > Qr(W) := sigma(W)/(harmonic(W)+log(harmonic(W))*exp(harmonic(W))) [...]
Copied output from PARI/gp with comments.
Best: 1 Score: 9999.750775 007375 490174 519248 3295500557 [306738, 418, 50, 18, 9, 6, 5, 4, 3, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1]
 The number 'n' is a product of 25 primorials represented in the vector:
n := p#(306738)*p#(418) ... * (p#(1))^9.
Q(n):= sigma(n)/[ H_n + log(H_n)*exp(H_n) ] ,
sigma() is the sum of divisors function, H_n is the n'th harmonic number: H_n = sum_{k = 1...n} 1/k .
Jeffrey Lagarias showed: RH <==> Q(n) < 1 for all integers n > 2. Gronwall first worked out the asymptotic extreme behaviour of sigma() to "first order" .
The score of 'n' is 10000*Q(n).
Try to replace p#(m_j) by p#(m_j +1), j = 1 ... 25 in succession. The winning 'j' is the one for which the Qvalue is the largest.
m_1 = 306738, m_2 = 418, m_3 = 50, ... m_17 = m_18 = ... = m_24 = m_25 = 1 .
Very occasionally, the best 'j' means a _lower_ Qvalue:
9999.750775004441 < 9999.750775007375 : 
1 9999.750775 004441 457325 932065 3334422049 [ Best is from using j=1] Best: 1 Score: 9999.7507750044414573259320653334422049
1 9999.750774 999987 661808 339470 2904857568 [ Best is from using j=1] Best: 1 Score: 9999.7507749999876618083394702904857568
1 9999.750775 001859 641379 985230 1801404019 [ Best is from using j=1] Best: 1 Score: 9999.7507750018596413799852301801404019
1 9999.750775 004453 408349 110566 0971511393 [ Best is from using j=1] Best: 1 Score: 9999.7507750044534083491105660971511393
1 9999.750775 003285 856812 854132 7209233142 [ Best is from using j=1] Best: 1 Score: 9999.7507750032858568128541327209233142 [306743, 418, 50, 18, 9, 6, 5, 4, 3, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1]
 The new m_1 = 306743 = 306738 + 5.
Score 5 ago: Score: 9999.750775 007375 490174 519248 329550 0557
current score: Score: 9999.750775 003285 856812 854132 720923 3142
Current score < (Score 5 ago). 
It's a temporary dip, no doubt ....
david bernier
 On Hypnos, http://messagenetcommresearch.com/myths/bios/hypnos.html

