mersenneforum.org Sum of the reciprocals of all Mersenne prime exponents
 Register FAQ Search Today's Posts Mark Forums Read

2021-08-20, 22:54   #23
a1call

"Rashid Naimi"
Oct 2015
Remote to Here/There

3·757 Posts

Quote:
 Originally Posted by charybdis 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

2021-08-20, 23:13   #24
charybdis

Apr 2020

5·163 Posts

Quote:
 Originally Posted by a1call 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

 2021-08-20, 23:24 #25 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 1000110111112 Posts 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
2021-08-21, 00:12   #26
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

3·52·89 Posts

Quote:
 Originally Posted by a1call 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)

 2021-08-21, 01:52 #27 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 1000110111112 Posts 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
 2021-08-21, 11:28 #28 Dobri   "Καλός" May 2018 73 Posts 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.
 2021-08-21, 13:13 #29 Dobri   "Καλός" May 2018 1010101112 Posts 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
 2021-08-21, 14:19 #30 Dr Sardonicus     Feb 2017 Nowhere 23×739 Posts 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
2021-08-21, 15:50   #31
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

150238 Posts

Quote:
 Originally Posted by Dr Sardonicus 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

Last fiddled with by kriesel on 2021-08-21 at 16:08

2021-08-22, 20:22   #32
Dobri

"Καλός"
May 2018

73 Posts

Quote:
 Originally Posted by Dobri 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];

 2021-08-22, 21:12 #33 Dobri   "Καλός" May 2018 15716 Posts 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"];];

 Similar Threads Thread Thread Starter Forum Replies Last Post drkirkby Miscellaneous Math 43 2022-07-17 03:34 biwema Math 10 2021-11-28 04:05 ProximaCentauri Miscellaneous Math 22 2014-12-05 13:07 science_man_88 Miscellaneous Math 3 2010-10-13 14:32 mart_r Math 10 2009-04-05 07:29

All times are UTC. The time now is 23:06.

Tue Aug 16 23:06:46 UTC 2022 up 40 days, 17:54, 1 user, load averages: 1.93, 1.38, 1.23