Sum of the reciprocals of all Mersenne prime exponents
 

The sum of the reciprocals of all known Mersenne prime exponents converges to [B]1.4481818[/B]...

(* 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};
sum = 0; n = 1; While[n <= 51, sum = sum + 1/Mexponent[[n]]; Print[n, ", ", Mexponent[[n]], ", ", SetPrecision[N[sum], 10]]; n++];

#, Mexponent, Sum of the reciprocals
1, 2, 0.5000000000
2, 3, 0.8333333333
3, 5, 1.033333333
4, 7, 1.176190476
5, 13, 1.253113553
6, 17, 1.311937083
7, 19, 1.364568661
8, 31, 1.396826726
9, 61, 1.413220169
10, 89, 1.424456124
11, 107, 1.433801918
12, 127, 1.441675934
13, 521, 1.443595320
14, 607, 1.445242766
15, 1279, 1.446024627
16, 2203, 1.446478553
17, 2281, 1.446916958
18, 3217, 1.447227806
19, 4253, 1.447462934
20, 4423, 1.447689025
21, 9689, 1.447792235
22, 9941, 1.447892829
23, 11213, 1.447982011
24, 19937, 1.448032169
25, 21701, 1.448078250
26, 23209, 1.448121336
27, 44497, 1.448143810
28, 86243, 1.448155405
29, 110503, 1.448164454
30, 132049, 1.448172027
31, 216091, 1.448176655
32, 756839, 1.448177976
33, 859433, 1.448179140
34, 1257787, 1.448179935
35, 1398269, 1.448180650
36, 2976221, 1.448180986
37, 3021377, 1.448181317
38, 6972593, 1.448181460
39, 13466917, 1.448181535
40, 20996011, 1.448181582
41, 24036583, 1.448181624
42, 25964951, 1.448181662
43, 30402457, 1.448181695
44, 32582657, 1.448181726
45, 37156667, 1.448181753
46, 42643801, 1.448181776
47, 43112609, 1.448181800
48, 57885161, 1.448181817
49, 74207281, 1.448181830
50, 77232917, 1.448181843
51, 82589933, 1.448181855

[QUOTE=Dobri;585672]The sum of the reciprocals of all known Mersenne prime exponents converges to [B]1.4481818[/B]...[/QUOTE]

The sum of reciprocals of [TEX]83\pi\over{296x^2}[/TEX] converges to a similar number.

Coincidence? I don't think so /s

It seems "obvious" that the Mersenne prime exponents (primes p such that 2[sup]p[/sup] - 1 is prime) are sufficiently rare that the sum of their reciprocals converges. However, I am not aware of any proof of this.

Interesting.

Per [URL]https://primes.utm.edu/notes/faq/NextMersenne.html[/URL] re Lenstra and Pomerance's conjecture, the geometric mean of the ratio of successive Mersenne exponents is expected to be R ~1.4757614.
Let r=1/R = 0.677616314.

I think without knowing at least certain bounds on the distribution that determines the series' terms, one can not prove whether a series converges or diverges.
We know r < 1 for the real Mersenne primes exponent series (known and unknown), because we will sort them into ascending order.
We know r > 0.
Under these conditions a geometric series sum will converge.
Summing the series for that, from the first term, as an infinite series,
beginning with 1/2, 1/2 (1+r+r^2+r^3+...) = 1/2 /(1-r) = 1.550946966...,
~7.1% larger than the sum of reciprocals of the known Mersenne primes' exponents to date, ~1.448181855...
Some might point to that difference as indication of "missed" primes. I think that's wrong.
It could be that a=1/2 is not the correct multiplier value for the series.
(Choosing to match terms at the first Mersenne prime exponent is arbitrary. Matching at the second or later instead is also arbitrary and yields a mildly better comparison in a few cases I tried.)
The geometric series sum proves nothing. It only shows that certain assumptions regarding the conjecture yield results somewhat consistent with the known data.
The reciprocals of known Mersenne primes' exponents do not constitute a geometric series.

Assuming that whatever Mersenne primes are discovered in the GIMPS search up to 10[SUP]9[/SUP] are consistent with the conjecture, they are likely to add less than 10[SUP]-7[/SUP] to the sum of reciprocals of known Mersenne primes' exponents.

The empirical data looks persuasive that the sum converges.

An attempt to apply the Wynn's Epsilon Method [url]https://mathworld.wolfram.com/WynnsEpsilonMethod.html[/url] for convergence acceleration gives 1.448181825... which is of no use as it is less than 1.448181855...

(* 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};
SumArray = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
sum = 0; n = 1; While[n <= 51, sum = sum + 1/Mexponent[[n]]; SumArray[[n]] = sum; n++]; Print[SetPrecision[N[sum], 10]];
limit = ResourceFunction["SequenceLimit"][SumArray, Method -> "WynnEpsilon"]; Print[SetPrecision[N[limit], 10]];

The sum of the reciprocals of the number of primes [I]π[/I](Mexponent) less than or equal to the known Mersenne prime exponents converges to [B]2.81383[/B]...

An attempt to apply the Wynn's Epsilon Method [url]https://mathworld.wolfram.com/WynnsEpsilonMethod.html[/url] for convergence acceleration gives [B]2.813839[/B]133... which is greater than [B]2.813838[/B]890...

(*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};
sum = 0; n = 1; While[n <= 51, sn = PrimePi[Mexponent[[n]]]; sum = sum + 1/sn; Print[n, ", ", Mexponent[[n]], ", ", sn, ", ", SetPrecision[N[sum], 10]]; n++];

SumArray = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
sum = 0; n = 1; While[n <= 51, sum = sum + 1/PrimePi[Mexponent[[n]]]; SumArray[[n]] = sum; n++]; Print[SetPrecision[N[sum], 10]];
limit = ResourceFunction["SequenceLimit"][SumArray, Method -> "WynnEpsilon"]; Print[SetPrecision[N[limit], 10]];

#, Mexponent, [I]π[/I](Mexponent), Sum of the reciprocals of [I]π[/I](Mexponent)
1, 2, 1, 1.000000000
2, 3, 2, 1.500000000
3, 5, 3, 1.833333333
4, 7, 4, 2.083333333
5, 13, 6, 2.250000000
6, 17, 7, 2.392857143
7, 19, 8, 2.517857143
8, 31, 11, 2.608766234
9, 61, 18, 2.664321789
10, 89, 24, 2.705988456
11, 107, 28, 2.741702742
12, 127, 31, 2.773960806
13, 521, 98, 2.784164888
14, 607, 111, 2.793173897
15, 1279, 207, 2.798004815
16, 2203, 328, 2.801053595
17, 2281, 339, 2.804003448
18, 3217, 455, 2.806201250
19, 4253, 583, 2.807916516
20, 4423, 602, 2.809577645
21, 9689, 1196, 2.810413766
22, 9941, 1226, 2.811229426
23, 11213, 1357, 2.811966346
24, 19937, 2254, 2.812410002
25, 21701, 2435, 2.812820679
26, 23209, 2591, 2.813206631
27, 44497, 4624, 2.813422894
28, 86243, 8384, 2.813542169
29, 110503, 10489, 2.813637507
30, 132049, 12331, 2.813718603
31, 216091, 19292, 2.813770438
32, 756839, 60745, 2.813786900
33, 859433, 68301, 2.813801541
34, 1257787, 97017, 2.813811849
35, 1398269, 106991, 2.813821195
36, 2976221, 215208, 2.813825842
37, 3021377, 218239, 2.813830424
38, 6972593, 474908, 2.813832530
39, 13466917, 877615, 2.813833669
40, 20996011, 1329726, 2.813834421
41, 24036583, 1509263, 2.813835084
42, 25964951, 1622441, 2.813835700
43, 30402457, 1881339, 2.813836232
44, 32582657, 2007537, 2.813836730
45, 37156667, 2270720, 2.813837170
46, 42643801, 2584328, 2.813837557
47, 43112609, 2610944, 2.813837940
48, 57885161, 3443958, 2.813838231
49, 74207281, 4350601, 2.813838460
50, 77232917, 4517402, 2.813838682
51, 82589933, 4811740, [B]2.813838[/B]890

[QUOTE=jnml;585975]The sum of reciprocals of [TEX]83\pi\over{296x^2}[/TEX] converges to a similar number.

Coincidence? I don't think so /s[/QUOTE]
The sum of the reciprocals of [TEX]{296n^2}\over{83\pi}[/TEX] tends to converge to [B]1.4[/B]...

An attempt to apply the Wynn's Epsilon Method [url]https://mathworld.wolfram.com/WynnsEpsilonMethod.html[/url] for convergence acceleration gives just one more true digit [B]1.44[/B]... for this very slowly converging infinite series.

(* Wolfram code *)
sum = 0; n = 1; While[n <= 51, sn = (296*n^2)/(83*Pi); sum = sum + 1/sn; Print[n, ", ", SetPrecision[sn, 10], ", ", SetPrecision[1/sn, 10], ", ", SetPrecision[N[sum], 10]]; n++];

SumArray = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
sum = 0; n = 1; While[n <= 51, sum = sum + 1/((296*n^2)/(83*Pi)); SumArray[[n]] = sum; n++]; Print[SetPrecision[N[sum], 10]];
limit = ResourceFunction["SequenceLimit"][SumArray, Method -> "WynnEpsilon"]; Print[SetPrecision[N[limit], 10]];

[I]n[/I], [TEX]{296n^2}\over{83\pi}[/TEX], Reciprocals [TEX]83\pi\over{296n^2}[/TEX], Sum of the reciprocals [TEX]83\pi\over{296n^2}[/TEX]
1, 1.135177425, 0.880919562, 0.8809195616
2, 4.540709702, 0.2202298904, 1.101149452
3, 10.21659683, 0.0978799513, 1.199029403
4, 18.16283881, 0.0550574726, 1.254086876
5, 28.37943564, 0.03523678247, 1.289323658
6, 40.86638732, 0.02446998782, 1.313793646
7, 55.6236938, 0.01797795024, 1.331771596
8, 72.6513552, 0.01376436815, 1.345535965
9, 91.9493715, 0.01087555014, 1.356411515
10, 113.5177425, 0.00880919562, 1.365220710
11, 137.3564685, 0.00728032696, 1.372501037
12, 163.4655493, 0.00611749696, 1.378618534
13, 191.8449849, 0.00521254178, 1.383831076
14, 222.4947754, 0.004494487559, 1.388325564
15, 255.4149207, 0.003915198052, 1.392240762
16, 290.6054209, 0.003441092038, 1.395681854
17, 328.0662759, 0.003048164573, 1.398730018
18, 367.7974858, 0.002718887536, 1.401448906
19, 409.7990506, 0.002440220392, 1.403889126
20, 454.0709702, 0.002202298904, 1.406091425
21, 500.613245, 0.001997550026, 1.408088975
22, 549.425874, 0.001820081739, 1.409909057
23, 600.508858, 0.001665254370, 1.411574311
24, 653.862197, 0.001529374239, 1.413103686
25, 709.485891, 0.001409471299, 1.414513157
26, 767.379940, 0.001303135446, 1.415816292
27, 827.544343, 0.001208394460, 1.417024687
28, 889.979102, 0.001123621890, 1.418148309
29, 954.684215, 0.001047466780, 1.419195775
30, 1021.659683, 0.000978799513, 1.420174575
31, 1090.905506, 0.000916669679, 1.421091245
32, 1162.421684, 0.000860273009, 1.421951518
33, 1236.208216, 0.000808925217, 1.422760443
34, 1312.265104, 0.000762041143, 1.423522484
35, 1390.592346, 0.000719118010, 1.424241602
36, 1471.189943, 0.000679721884, 1.424921324
37, 1554.057895, 0.000643476670, 1.425564801
38, 1639.196202, 0.000610055098, 1.426174856
39, 1726.604864, 0.000579171309, 1.426754027
40, 1816.283881, 0.000550574726, 1.427304602
41, 1908.233252, 0.000524044950, 1.427828647
42, 2002.452978, 0.0004993875066, 1.428328034
43, 2098.943060, 0.0004764302659, 1.428804464
44, 2197.703496, 0.0004550204347, 1.429259485
45, 2298.734286, 0.0004350220058, 1.429694507
46, 2402.035432, 0.0004163135925, 1.430110820
47, 2507.606933, 0.0003987865829, 1.430509607
48, 2615.448788, 0.0003823435597, 1.430891951
49, 2725.560998, 0.0003668969436, 1.431258847
50, 2837.943564, 0.0003523678247, 1.431611215
51, 2952.596484, 0.0003386849526, [B]1.4[/B]31949900

[QUOTE=jnml;585975]The sum of reciprocals of [TEX]83\pi\over{296x^2}[/TEX] converges to a similar number.

Coincidence? I don't think so /s[/QUOTE]

Similar is very accurate but they are not quite the same.

Referring to the convergence sum of [TEX]83\pi\over{296x^2}[/TEX] as sum(P) and the convergence sum of Mersenne recpricols as sum(M) we have the following:

It appears that sum(P) eventually grows larger than sum(M).

After 1009 terms of sum(P) the sum is 1.44818196759933, already slightly larger than current sum(M). At that point the increase for the next term (term 1010) of sum(P) is .00000086356197, i.e. 8.6356*10^-7. That increase is dropping much more slowly than the sum(M) increase. The smallest possible increase for the next sum(M) term (if n=82589939 is prime) is .00000001210801, i.e. 1.2108*10^-8, and it will likely be < 10^-8 if the next Mersenne prime is n>100M.

Furthermore there are still 40 more terms of sum(P) that grow at > 8*10^-7. It finally dips below that increase at term 1050.

Therefore they are not quite equal but they are certainly similar. :smile:

One more thing: sum(P) seems to converge around ~1.449054. It is at ~1.44905371619 at the 1,000,000th term.

[QUOTE=jnml;585975]The sum of reciprocals of [TEX]83\pi\over{296x^2}[/TEX] converges to a similar number.

Coincidence? I don't think so /s[/QUOTE]
For completeness, let's write down the precise value (see [URL]https://en.wikipedia.org/wiki/Basel_problem[/URL]) of said sum with more true digits:
(83[I]π[/I]/296)([I]π[/I][SUP]2[/SUP]/6) = 1.4490545971086064609034539136089052937976908507705314969673404631846457084175977483772726270867339831202140331140224885456879134375811896604085561457304657798900869489793277021920199145038060189372423...

[QUOTE=Dr Sardonicus;585979]It seems "obvious" that the Mersenne prime exponents (primes p such that 2[sup]p[/sup] - 1 is prime) are sufficiently rare that the sum of their reciprocals converges. However, I am not aware of any proof of this.[/QUOTE]

I don't think it's even known that there are infinitely many composite 2^p-1 - which makes proving the infinitude of Mersenne primes seem pretty unrealistic.

[QUOTE=gd_barnes;586052]Similar is very accurate but they are not quite the same.

Referring to the convergence sum of [TEX]83\pi\over{296x^2}[/TEX] as sum(P) and the convergence sum of Mersenne recpricols as sum(M) we have the following:
<snip>
One more thing: sum(P) seems to converge around ~1.449054. It is at ~1.44905371619 at the 1,000,000th term.[/QUOTE]The series [tex]\sum_{n=1}^{\infty}\frac{1}{n^{2}}[/tex] converges to [tex]\frac{\pi^{2}}{6}[/tex] so the sum(P) has exact value [tex]\frac{83\pi^{3}}{1776}[/tex].

[code]? 83*Pi^3/1776
%1 = 1.4490545971086064609034539136089052938[/code]

As far as I can see, there is no known proof that sum(M) actually converges. It is well known that the sum of the reciprocals of [i]all[/i] primes diverges. Granted, it's not out to win any races. See, e.g. the Wolfram Mathworld page on the [url=https://mathworld.wolfram.com/MertensConstant.html]Mertens Constant[/url].