Sum of the reciprocals of all Mersenne prime exponents

[a1call]

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

[charybdis]

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?

[a1call]

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.

[kriesel]

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 R_i=p_i/p_i-1 (where p_i 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 r_i=1/R_i. 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)

[a1call]

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.

[Dobri]

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:
- M_Σ1/p for the sum of the reciprocals of Mersenne prime exponents;
- M_Σ1/π(p) for the sum of the reciprocals of π(p) of Mersenne prime exponents;
- M_Σ±1/p for the alternating sum of the reciprocals of Mersenne prime exponents;
- M_Σ±1/π(p) for the alternating sum of the reciprocals of π(p) of Mersenne prime exponents;
- M_{Σ1/(p[n+1]-p[n])} for the sum of the reciprocals of the difference between consecutive Mersenne prime exponents;
- M_{Σ1/(π(p[n+1])-π(p[n]))} for the sum of the reciprocals of the difference between consecutive π(p) of Mersenne prime exponents;
- M_{Σ±1/(p[n+1]-p[n])} for the alternating sum of the reciprocals of the difference between consecutive Mersenne prime exponents;
- M_{Σ±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]

On the basis of the 51 known Mersenne prime exponents:
M_Σ1/p = 1.448181855...
M_Σ1/π(p) = 2.813838890...
M_Σ±1/p = 0.2697096036...
M_Σ±1/π(p) = 0.6628596647...
M_{Σ1/(p[n+1]-p[n])} = 3.218584673...
M_{Σ1/(π(p[n+1])-π(p[n]))} = 7.055035487...
M_{Σ±1/(p[n+1]-p[n])} = 0.6684789674...
M_{Σ±1/(π(p[n+1])-π(p[n]))} = 0.9880482016...

[Dr Sardonicus]

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 S_p is the sum of reciprocals up to the Mersenne prime exponent p, and S is the sum over all Mersenne prime exponents,

S ~ S_p + 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

[kriesel]

Originally Posted by Dr Sardonicus

S ~ S_p + 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)

[Dobri]

Originally Posted by Dobri

Assuming that Sum_EPSILON is close to the unknown Sum_TOTAL, 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) + ...) / (Sum_EPSILON - Sum₅₁)

for some π(M₅₂)/π(M₅₃) < 1, π(M₅₂)/π(M₅₄) < 1,...

Let k_n = π(M_n)/π(M_n+1) + π(M_n)/π(M_n+2) + ...
Let's also assume that the number of Mersenne primes is infinite.

Corollary: M_Σ1/π(p) is obtained for the k-distribution having min(k_max - k_min).

The corollary implies that the k-distribution has to be the most 'flat'-like one as compared to other distributions resulting in other Sum_TOTAL values.
Then M_Σ1/π(p) = 2.813839146... > 2.813838890... for min(k_max - k_min) = 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 M_Σ1/p, etc.
M_Σ1/π(p) was chosen as an illustration because the Wynn's Epsilon Method has a better convergence for it as compared to M_Σ1/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]

Assuming that M_Σ1/π(p) = 2.813839146..., one could roughly estimate that 154,900,853 <= M₅₂ <= 540,044,243 for k_min = 1.236136005... and k_max = 6.283271135...
Even if k_min = 1, then 137,615,273 <= M₅₂.

#, M_exponent, π(M_exponent), k
52, 154900853, 8704305, 1.236136005 <- k_min
52, 540044243, 28350608, 6.283271135 <- k_max

(* 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"];];