20210815, 06:17  #1 
"刀比日"
May 2018
317 Posts 
Sum of the reciprocals of all Mersenne prime exponents
The sum of the reciprocals of all known Mersenne prime exponents converges to 1.4481818...
(* 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 
20210818, 12:40  #2 
Feb 2012
Prague, Czech Republ
5·37 Posts 

20210818, 13:35  #3 
Feb 2017
Nowhere
3·5^{2}·7·11 Posts 
It seems "obvious" that the Mersenne prime exponents (primes p such that 2^{p}  1 is prime) are sufficiently rare that the sum of their reciprocals converges. However, I am not aware of any proof of this.

20210818, 19:03  #4 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2^{2}×1,619 Posts 
Interesting.
Per https://primes.utm.edu/notes/faq/NextMersenne.html 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 /(1r) = 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^{9} are consistent with the conjecture, they are likely to add less than 10^{7} to the sum of reciprocals of known Mersenne primes' exponents. The empirical data looks persuasive that the sum converges. Last fiddled with by kriesel on 20210818 at 19:04 
20210818, 21:42  #5 
"刀比日"
May 2018
317_{10} Posts 
An attempt to apply the Wynn's Epsilon Method https://mathworld.wolfram.com/WynnsEpsilonMethod.html 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]]; 
20210818, 22:11  #6 
"刀比日"
May 2018
317 Posts 
The sum of the reciprocals of the number of primes π(Mexponent) less than or equal to the known Mersenne prime exponents converges to 2.81383...
An attempt to apply the Wynn's Epsilon Method https://mathworld.wolfram.com/WynnsEpsilonMethod.html for convergence acceleration gives 2.813839133... which is greater than 2.813838890... (*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, π(Mexponent), Sum of the reciprocals of π(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, 2.813838890 
20210819, 07:44  #7  
"刀比日"
May 2018
317 Posts 
Quote:
An attempt to apply the Wynn's Epsilon Method https://mathworld.wolfram.com/WynnsEpsilonMethod.html for convergence acceleration gives just one more true digit 1.44... 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]]; n, , Reciprocals , Sum of the reciprocals 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, 1.431949900 

20210819, 11:54  #8  
May 2007
Kansas; USA
7·1,531 Posts 
Quote:
Referring to the convergence sum of 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. One more thing: sum(P) seems to converge around ~1.449054. It is at ~1.44905371619 at the 1,000,000th term. Last fiddled with by gd_barnes on 20210819 at 12:14 

20210819, 12:48  #9  
"刀比日"
May 2018
317 Posts 
Quote:
(83π/296)(π^{2}/6) = 1.4490545971086064609034539136089052937976908507705314969673404631846457084175977483772726270867339831202140331140224885456879134375811896604085561457304657798900869489793277021920199145038060189372423... Last fiddled with by Dobri on 20210819 at 13:28 

20210819, 13:00  #10 
Apr 2020
2×3^{3}×13 Posts 
I don't think it's even known that there are infinitely many composite 2^p1  which makes proving the infinitude of Mersenne primes seem pretty unrealistic.

20210819, 13:03  #11  
Feb 2017
Nowhere
3·5^{2}·7·11 Posts 
Quote:
Code:
? 83*Pi^3/1776 %1 = 1.4490545971086064609034539136089052938 

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