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#45 | |
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"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
32×11×79 Posts |
Quote:
https://primes.utm.edu/notes/faq/NextMersenne.html https://en.wikipedia.org/wiki/Benford%27s_law For separate digit lengths of exponents in base 10: 1 digit: 2 3 5 7; 7 is as good as it gets since 8 and 9 are composite; average 4.25 2 digit: 13 17 19 31 61 89; 89 is maximal possible sum 17; average per digit sum in 4 8 10 4 7 17; sum 50 / 6 = 8.33 / 2 digits = 4.167/digit 3 digit: 107 127 521 607; 8 10 8 13; 39 / 4 = 9.75; 3.25/digit 4 digit: 1279 2203 2281 3217 4253 4423 9689 9941; 19 7 13 13 14 13 32 23; 134 / 8 = 16.75; 4.1875/digit 5 digit: 11213 19937 21701 23209 44497 86243; 8 29 11 16 28 23; 95 / 6 = 15.833; 3.167/digit 6 digit: 110503 132049 216091 756839 859433; 10 19 19 38 32; 118 / 5 = 23.6; 3.933/digit 7 digit: 1257787 1398269 2976221 3021377 6972593; 37 38 29 23 41; 168 / 5 = 33.6; 4.8/digit 8-digit: exponent digitsum 13466917 37 20996011 28 24036583 31 25964951 41 30402457 25 32582657 38 37156667 41 42643801 28 43112609 26 57885161 41 74207281 31 77232917 38 82589933 47 sum 452 / 13 exponent = 34.769 / 8 digits = 4.346/digit (apologies for any lingering math errors) So for 8-decimal-digit, the histogram of # of Mp vs. digitsum value is 25 1 26 1 28 2 31 2 37 1 38 2 41 3 47 1 Graphing that manually in black with only digitization noise +-0.5 counts atop Dobri's base 10 digit sum distributions, using the existing rulings for scale for convenience yields the attachment. The statistical sample size is terribly small. Last fiddled with by kriesel on 2021-07-17 at 17:49 |
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#46 |
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6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
3×7×17×31 Posts |
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#47 | |
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"ม้าไฟ"
May 2018
2·3·89 Posts |
Closing the thread at
https://mersenneforum.org/showthread.php?t=26997 was premature because the OP was not given a chance to respond, especially at a stage when a post was submitted in their favor. The post at https://mersenneforum.org/showpost.p...3&postcount=44 Quote:
Therefore, it appears one could select exponents for manual testing for a given digit sum as often as indicated by the corresponding expectancy in accordance with the LPW heuristic. Therefore, the OP would like to respectfully request the previous thread to be reopened and/or the discussion to be continued in this thread instead. |
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#48 |
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Apr 2020
3·353 Posts |
You have misunderstood the point of my post. It was not intended to support your views. All I did was calculate the probabilities given by the LPW heuristic for each prime up to 103M and add up the probabilities for each digit sum. My graph therefore shows what we expect if digit sum has no effect on the likelihood of being prime - and the actual distribution of primes is consistent with this.
5 is the digit sum with the highest expected number of primes because there are several very small primes with digit sum 5, namely 5, 23 and 41, and the heuristic gives high probabilities for these. For p=5 it gives the nonsensical probability of 1.748. This does not make higher exponents with digit sum 5 more likely to be prime! Please can a mod close this thread too? |
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#49 |
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Sep 2002
Database er0rr
5×937 Posts |
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#50 | |
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Romulan Interpreter
"name field"
Jun 2011
Thailand
41×251 Posts |
Quote:
![]() Thread closed. Edit: Whhops, sorry, it was closed already.
Last fiddled with by LaurV on 2021-07-18 at 07:40 |
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#51 |
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"ม้าไฟ"
May 2018
2·3·89 Posts |
The minimal bases bmin for which the digit sums of the prime exponents of the known Mersenne primes are distinct (different from each other) is given in the following list:
bmin = {2, 2, 3, 7, 7, 10, 10, 20, 20, 21, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 82, 82, 82, 82, 108, 108, 108, 108, 108, 108, 108, 108, 108, 108, 108, 108, 108, 108, 110, 110, 110, 110, 110, 110, 142, 142} for n = 1, 2, …, 51. Therefore, bmin = 142 for n = 51, and the distinct base-142 digit sums for the 51 prime exponents of known Mersenne primes are {2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 98, 43, 10, 88, 25, 115, 164, 52, 101, 71, 215, 197, 128, 85, 82, 92, 100, 214, 220, 233, 179, 208, 254, 134, 170, 284, 289, 124, 172, 224, 319, 236, 347, 325, 167, 290, 250, 308, 311}. If one assumes for manual selection that bmin remains equal to 142 also for n = 52, then prime exponents for finding M52 (if any) could be selected for further testing if bmin = 142. Code:
(* Wolfram code *)
MpData = {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};
nMpmax = Length[MpData]; bm = ConstantArray[0, nMpmax]; basemax = 1000; nMp = 0; While[nMp < nMpmax, nMp++; ds = ConstantArray[0, nMp]; check = 0; base = 1; While[(base < basemax) && (check == 0), base++; sum = 0; ic = 0; While[ic < nMp, ic++; ds[[ic]] = Total[IntegerDigits[MpData[[ic]], base]];]; dssort = Sort[ds]; check = 1; ic = 1; While[ic < nMp, ic++; If[(dssort[[ic]] == dssort[[ic - 1]]), check = 0;];]; If[check == 1, bm[[nMp]] = base;];];];
Print[bm]; Print["b_min = ", base]; Print[ds];
b = {224, 230, 278, 300, 314, 330, 334, 336, 342, 352, 374, 398, 402, 404, 412, 414, 418, 424, 432, 444, 447, 450, 458, 459, 467, 468, 473, 474, 488, 490, 498, 504, 507, 512, 518, 522, 541, 543, 545, 546, 548, 552, 555, 566, 572, 576, 584, 585, 588, 594, 608, 614, 620, 627, 630, 642, 654, 656, 658, 660, 674, 682, 683, 688, 690, 696, 699, 704, 706, 713, 718, 722, 723, 726, 738, 740, 742, 746, 750, 756, 762, 768, 770, 777, 778, 793, 797, 798, 800, 811, 812, 814, 816, 820, 826, 827, 830, 833, 835, 836, 847, 852, 854, 857, 863, 866, 868, 870, 878, 880, 882, 884, 887, 890, 894, 896, 897, 900, 905, 906, 910, 914, 922, 923, 924, 926, 929, 930, 933, 935, 936, 940, 941, 942, 944, 952, 954, 958, 960, 962, 974, 975, 978, 979, 980, 982, 984, 988, 993, 994, 996, 998, 1000,...}. |
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