mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Miscellaneous Math

Reply
 
Thread Tools
Old 2021-08-20, 22:54   #23
a1call
 
a1call's Avatar
 
"Rashid Naimi"
Oct 2015
Remote to Here/There

32×239 Posts
Default

Quote:
Originally Posted by charybdis View Post
Well yes, obviously it does. But if you read my post properly you'll see that I was talking about the sum of the reciprocals of that sequence.
Yes, you are correct. I erred as usual in entering the formula:
https://www.wolframalpha.com/input/?...%5En+converges
a1call is offline   Reply With Quote
Old 2021-08-20, 23:13   #24
charybdis
 
charybdis's Avatar
 
Apr 2020

1F416 Posts
Default

Quote:
Originally Posted by a1call View Post
Yes, you are correct. I erred as usual in entering the formula:
https://www.wolframalpha.com/input/?...%5En+converges
Shouldn't be any need for Wolfram Alpha here. Any geometric series with constant ratio r between -1 and 1 (not inclusive) converges, and if the first term is a then the sum is a/(1-r); see here for an explanation. I think this should be standard high-school knowledge?

Last fiddled with by charybdis on 2021-08-20 at 23:18
charybdis is offline   Reply With Quote
Old 2021-08-20, 23:24   #25
a1call
 
a1call's Avatar
 
"Rashid Naimi"
Oct 2015
Remote to Here/There

32·239 Posts
Default

Well, from that we could jump to

sum_(n=1)^∞ (a/(a+1))^n = a

https://www.wolframalpha.com/input/?...%2B1%29%29%5En

and

1/x=a/(a+1) solve for x => x = 1/a + 1 and a!=0 and a + 1!=0


https://www.wolframalpha.com/input/?...29+solve+for+x


So for the infinite sum:

sum_(n=1)^∞ (1/a)^n
as long as a is greater than 1, the sum will converge.

So putting that in perceptive of the OP, the sum would likely convege. I think the smoke from my ears just set out the smole detectors.

Thank you charybdis for your replies and humoring me. I do appreciate it very much.

Last fiddled with by a1call on 2021-08-20 at 23:24
a1call is offline   Reply With Quote
Old 2021-08-21, 00:12   #26
kriesel
 
kriesel's Avatar
 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

32·643 Posts
Default

Quote:
Originally Posted by a1call View Post
I have no way of proving any of the following, so I state them as virtual (as opposed to literal) wagers:

* There are infinite Mersenne primes
* The sum of the reciprocals of Mersenne-Prime-Exponents does not converge
Let Ri=pi/pi-1 (where pi is the exponent of the i'th known Mersenne prime). R is always > 1. Because we magnitude-sort the list of positive integer exponents into ascending order, and permit no duplication.
Let ri=1/Ri. r is then always less than 1, and > 0.

Sum of a(1/R)n =a rn converges for R > 1, -1 < r < 1, for a finite or infinite geometric series, n=1, 2, 3, ....
(Any decent math textbook above a certain level will contain a proof. See https://en.wikipedia.org/wiki/Geometric_series)
So the sum of the reciprocals of the exponents of Mersenne primes does converge, must converge.


It's an open question whether there are an infinite number of Mersenne primes.
But if they were infinitely many, and were mostly to fall near the spacing indicated as average for the Lenstra & Pomerance conjecture, R~1.47576139704882, that leaves considerable room for modest fluctuation from term to term relative to that convergence requirement, such as occasionally to R~1.0001 which would still converge even if all were so closely spaced, and even more so for R>>1.0001 or R>1.476.
The empirical evidence provided by the 51 known data points spread over ~108 is consistent with the Lenstra & Pomerance conjecture being correct. The sum of reciprocals over the 51 actual is actually ~7% lower than we get by summing what the conjecture predicts, beginning with a=1/2. The sum over the next ~6 we might expect to find below p~109 cannot be greater than 10-7, and so must fall far short of making up the difference. See post 4 this thread. So far the 50 observed ratios of successive exponents of Mersenne primes range ~1.011 to 4.1024. (43112609/42643801 to 521/127)
kriesel is online now   Reply With Quote
Old 2021-08-21, 01:52   #27
a1call
 
a1call's Avatar
 
"Rashid Naimi"
Oct 2015
Remote to Here/There

1000011001112 Posts
Default

Ok, to elaborate further:

* The infinite series sum:
sum_(n=1)^∞ (1/a)^n
converges for all real numbers a such that a>1

https://www.wolframalpha.com/input/?...%5En+converges

* There are infinite prime numbers

* The sum of reciprocals of all prime numbers diverges

* We can formulate a in the 1st statement infinitely smaller than any given real number greater than 1
** In this way we can make sure that the 1st few/many/given/finite-number=m addends of the series in 1st statement are larger than any given first-finite-number=m of the reciprocals-of-all-prime-numbers
*** This entails that it is guaranteed that at some point beyond m the addends of the reciprocals-of-all-prime-numbers will overtake/get-larger-than the addends of the formulated infinite series

* All this to say there absolutely can be no conclusion that can be based on comparing the OP-sum's convergence with a converging sum of a geometric series


Corrections/enlightenments/comments are as always appreciated.

Thank you very much for your time.

Last fiddled with by a1call on 2021-08-21 at 01:55
a1call is offline   Reply With Quote
Old 2021-08-21, 11:28   #28
Dobri
 
May 2018

2×97 Posts
Default Mersenne Constant

Let's introduce the term Mersenne constant for constants obtained from converging sums of reciprocals involving the Mersenne prime exponents.

The following notations could prove useful when dealing with such Mersenne constants:
- MΣ1/p for the sum of the reciprocals of Mersenne prime exponents;
- MΣ1/π(p) for the sum of the reciprocals of π(p) of Mersenne prime exponents;
- MΣ±1/p for the alternating sum of the reciprocals of Mersenne prime exponents;
- MΣ±1/π(p) for the alternating sum of the reciprocals of π(p) of Mersenne prime exponents;
- MΣ1/(p[n+1]-p[n]) for the sum of the reciprocals of the difference between consecutive Mersenne prime exponents;
- MΣ1/(π(p[n+1])-π(p[n])) for the sum of the reciprocals of the difference between consecutive π(p) of Mersenne prime exponents;
- MΣ±1/(p[n+1]-p[n]) for the alternating sum of the reciprocals of the difference between consecutive Mersenne prime exponents;
- MΣ±1/(π(p[n+1])-π(p[n])) for the alternating sum of the reciprocals of the difference between consecutive π(p) of Mersenne prime exponents;
etc.
Dobri is offline   Reply With Quote
Old 2021-08-21, 13:13   #29
Dobri
 
May 2018

C216 Posts
Default

On the basis of the 51 known Mersenne prime exponents:
MΣ1/p = 1.448181855...
MΣ1/π(p) = 2.813838890...
MΣ±1/p = 0.2697096036...
MΣ±1/π(p) = 0.6628596647...
MΣ1/(p[n+1]-p[n]) = 3.218584673...
MΣ1/(π(p[n+1])-π(p[n])) = 7.055035487...
MΣ±1/(p[n+1]-p[n]) = 0.6684789674...
MΣ±1/(π(p[n+1])-π(p[n])) = 0.9880482016...

Last fiddled with by Dobri on 2021-08-21 at 13:15
Dobri is offline   Reply With Quote
Old 2021-08-21, 14:19   #30
Dr Sardonicus
 
Dr Sardonicus's Avatar
 
Feb 2017
Nowhere

22·29·43 Posts
Default

So far, the most reasonable approach IMO is to apply the best guess about Mersenne prime exponents (that they increase approximately geometrically with ratio R ~ 1.47576139704882). This gives the simple estimate for the "tail" of the series of reciprocals after a given exponent p, of 1/(p*(R-1)), or about 2.1/p, so if Sp is the sum of reciprocals up to the Mersenne prime exponent p, and S is the sum over all Mersenne prime exponents,

S ~ Sp + 2.10189393/p.

This thread has led me to the discovery of a feature of the calculator app on my computer: Instead of typing input and transcribing output one digit at a time, I can copy-paste plain text directly into the calculator, and copy the number in the calculator's display, and then paste it into a text file. Given my proficiency at making typos, this may be the most significant result of this thread so far
Dr Sardonicus is offline   Reply With Quote
Old 2021-08-21, 15:50   #31
kriesel
 
kriesel's Avatar
 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

32×643 Posts
Default

Quote:
Originally Posted by Dr Sardonicus View Post
S ~ Sp + 2.10189393/p.
... Given my proficiency at making typos, this may be the most significant result of this thread so far



So, S ~ 1.4481818554213 + 2.10189393 / 82589933 ~ 1.448181880871059...

(On Windows 7 or 10 calculator, one can copy/paste multiple numbers and operators and spaces in a single go, including from web pages or pdf documents:
try 1.4481818554213 + 2.10189393 / 82589933= )


(Yay, both Mersenne prime & exponent & sequence #, post number! That's only happened, um, at # 3, 7 and here in this thread)
Attached Thumbnails
Click image for larger version

Name:	win7 calc copy paste.png
Views:	29
Size:	19.4 KB
ID:	25511   Click image for larger version

Name:	win10 calc copy paste.png
Views:	23
Size:	317.9 KB
ID:	25512  

Last fiddled with by kriesel on 2021-08-21 at 16:08
kriesel is online now   Reply With Quote
Old 2021-08-22, 20:22   #32
Dobri
 
May 2018

2·97 Posts
Default

Quote:
Originally Posted by Dobri View Post
Assuming that SumEPSILON is close to the unknown SumTOTAL, one could entertain the idea of estimating a wide range for M52 as follows:

π(M52) (1 + π(M52)/π(M53) + π(M52)/π(M54) + ...) / (SumEPSILON - Sum51)

for some π(M52)/π(M53) < 1, π(M52)/π(M54) < 1,...
Let kn = π(Mn)/π(Mn+1) + π(Mn)/π(Mn+2) + ...
Let's also assume that the number of Mersenne primes is infinite.

Corollary: MΣ1/π(p) is obtained for the k-distribution having min(kmax - kmin).

The corollary implies that the k-distribution has to be the most 'flat'-like one as compared to other distributions resulting in other SumTOTAL values.
Then MΣ1/π(p) = 2.813839146... > 2.813838890... for min(kmax - kmin) = 5.047135130... on the basis of the 51 known Mersenne prime exponents.

Note: Similar results using the same approach can be obtained also for the other Mersenne constants, like MΣ1/p, etc.
MΣ1/π(p) was chosen as an illustration because the Wynn's Epsilon Method has a better convergence for it as compared to MΣ1/p.

(* Wolfram code *)
Mexponent = {2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, 30402457, 32582657, 37156667, 42643801, 43112609, 57885161, 74207281, 77232917, 82589933, 0};
Mmax = 51; SumArray = ConstantArray[0, Mmax]; k = ConstantArray[0, Mmax];
sum = 0; m = 1; While[m <= Mmax, sum = sum + 1/PrimePi[Mexponent[[m]]]; SumArray[[m]] = sum; m++];
limitmin = SumArray[[Mmax]]; limit = limitmin; limitlbest = limitmin; kbest = 100; nbest = 0;
n = 0; While[n <= 10000, k[[1]] = limit; m = 2; While[m <= Mmax, k[[m]] = PrimePi[Mexponent[[m]]]*(limit - SumArray[[m - 1]]) - 1; m++];
kmin = Min[k[[1 ;; Mmax]]]; kmax = Max[k[[1 ;; Mmax]]];
If[kmax - kmin < kbest, limitbest = limit; kbest = kmax - kmin; nbest = n;];
limit = limitmin + 0.0000000001*n; n++];
Print[SetPrecision[N[limitbest], 10], ", ", SetPrecision[N[kbest], 10], ", ", nbest];
Dobri is offline   Reply With Quote
Old 2021-08-22, 21:12   #33
Dobri
 
May 2018

2·97 Posts
Default

Assuming that MΣ1/π(p) = 2.813839146..., one could roughly estimate that 154,900,853 <= M52 <= 540,044,243 for kmin = 1.236136005... and kmax = 6.283271135...
Even if kmin = 1, then 137,615,273 <= M52.

#, Mexponent, π(Mexponent), k
52, 154900853, 8704305, 1.236136005 <- kmin
52, 540044243, 28350608, 6.283271135 <- kmax

(* Wolfram code *)
limit = 2.81383914655656131031946642906;
Mexponent = {2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, 30402457, 32582657, 37156667, 42643801, 43112609, 57885161, 74207281, 77232917, 82589933, 0};
Mmax = 51; SumArray = ConstantArray[0, Mmax + 1]; k = ConstantArray[0, Mmax + 1];
sum = 0; m = 1; While[m <= Mmax, sum = sum + 1/PrimePi[Mexponent[[m]]]; SumArray[[m]] = sum; m++];
k[[1]] = limit; m = 2; While[m <= Mmax, k[[m]] = PrimePi[Mexponent[[m]]]*(limit - SumArray[[m - 1]]) - 1; m++];
kmin = Min[k[[1 ;; Mmax]]]; kmax = Max[k[[1 ;; Mmax]]];
m = Mmax; If[(limit - SumArray[[m]]) > 0, k[[m + 1]] = kmin;
Mexponent[[m + 1]] = Prime[Round[(1 + k[[m + 1]])/(limit - SumArray[[m]])]];
SumArray[[m + 1]] = SumArray[[m]] + 1/PrimePi[Mexponent[[m + 1]]];
Print[m + 1, ", ", Mexponent[[m + 1]], ", ", PrimePi[Mexponent[[m + 1]]], ", ", SetPrecision[N[k[[m + 1]]], 10], " <- kmin"];
k[[m + 1]] = kmax; Mexponent[[m + 1]] = Prime[Round[(1 + k[[m + 1]])/(limit - SumArray[[m]])]];
SumArray[[m + 1]] = SumArray[[m]] + 1/PrimePi[Mexponent[[m + 1]]];
Print[m + 1, ", ", Mexponent[[m + 1]], ", ", PrimePi[Mexponent[[m + 1]]], ", ", SetPrecision[N[k[[m + 1]]], 10], " <- kmax"];];
Dobri is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
More twin primes below Mersenne exponents than above Mersenne exponents. drkirkby Miscellaneous Math 39 2021-08-24 21:08
Another interesting pattern of Mersenne exponents: they are prime!!!!1111 ProximaCentauri Miscellaneous Math 22 2014-12-05 13:07
Mersenne(prime exponents) factorization science_man_88 Miscellaneous Math 3 2010-10-13 14:32
Sum of reciprocals of prime k-tuplets mart_r Math 10 2009-04-05 07:29
ECM on Mersenne numbers with prime exponents biwema Math 5 2004-04-21 04:44

All times are UTC. The time now is 00:54.


Fri Oct 22 00:54:05 UTC 2021 up 90 days, 19:23, 2 users, load averages: 1.61, 1.40, 1.42

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.