20190520, 02:59  #23 
"Sam Laur"
Dec 2018
Turku, Finland
475_{8} Posts 
About the manual assignment progress reports. Different programs have wildly different progress logs. Example from Mlucas:
[May 20 03:38:41] M50346019 Iter# = 24280000 [48.23% complete] clocks = 00:26:42.587 [160.2587 msec/iter] Res64: 54912F2731A45A41. AvgMaxErr = 0.042192089. MaxErr = 0.062500000. Residue shift count = 12144693. But minimum common requirements should be current iteration count and Res64/shift so that interim residues could be stored, if needed. I propose that instead of having the manual results submission form decipher many different log formats, there should be a simple common format, and let the client submission script do the conversion to that. 
20190521, 02:20  #24  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
1011111111011_{2} Posts 
Quote:
I requested only CUDALucas progress reporting support, initially. But what client submission script? Some gpu applications lack any client submission script, so all work reservation and submission is manual. Manual progress updates will be done only as needed in that use case. CUDALucas, for example, and CUDAPm1, have no client submission scripts. (Also many earlier versions of gpuowl, but there are so many log styles in gpuowl versus version, and many different gpuowl versions are fastest depending on the exponent / fft length, so I think supporting all the older useful versions of gpuowl progress reporting by the manual submission page is too much to ask.) 

20190521, 04:08  #25  
"Sam Laur"
Dec 2018
Turku, Finland
317 Posts 
Quote:
Quote:
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20190521, 17:17  #26  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
7×877 Posts 
Quote:
Quote:
A possible benefit of CUDALucas or cllucas progress reporting, that is not very relevant to the other gpu applications, is early detection of pathological runs that are producing all 0x0 or 0x02 or 0xfffffff800000000 interim res64, with potential savings of days or weeks of wasted run time. Some users don't recognize those as an issue, but if they could be persuaded to do interim progress reports, the server might educate them! Mainly though, manual progress reporting is about preventing expiration of computations that are progressing, and reducing the chafing of other users who see no indicated progress at server reports and imagine wrongly that means no progress is occurring. Last fiddled with by kriesel on 20190521 at 17:19 

20190713, 14:47  #27 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
7·877 Posts 
Background
The purpose of this post is to describe a shared foundation. Posting claims counter to what's here may indicate someone is an unaware novice, or is a troll. (Once I'm done weeding out my errors, with some help from others, that is.) This necessarily covers some points and leaves out much other material; whole books have been written about individual aspects, such as factorization. It's probably a good idea to read https://www.mersenne.org/various/math.php and then maybe return and resume here.
"A Friendly Introduction to Number Theory", Joseph H. Silverman, https://www.math.brown.edu/~jhs/frint.html The Prime Pages https://primes.utm.edu/ Knuth's Algorithms, Donald Knuth Prime Numbers and Computer Methods for Factorization, Hans Riesel Number Theory Web http://www.numbertheory.org/ntw/ "Prime Numbers: A Computational Perspective", Crandall and Pomerance "Humble Pi", Matt Parker "The C Programming Language", Kernighan and Ritchie https://en.wikipedia.org/wiki/The_C_...mming_Language (Thanks to Batalov, Dylan14, and LaurV for contributing to the accuracy and readability of this post.) Last fiddled with by kriesel on 20190715 at 02:47 
20190714, 15:11  #28  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
7·877 Posts 
Quote:
I'm open to constructive suggestions for the title, organization, and content. Re https://www.mersenneforum.org/showpo...32&postcount=2, I saw M6 != M(6) and previously missed the missing parentheses, since added. (In the usual notation, M6=2^{17}1; for other readers, https://www.mersenne.org/primes/) 

20190714, 15:54  #29 
"Dylan"
Mar 2017
2^{2}·3·7^{2} Posts 
A few suggestions:
For 5. it should be that a single nontrivial factor of a natural number proves the number composite (meaning a number other than 1 or the number itself). For 20. link to the full list of programs (which is https://www.mersenneforum.org/showpo...91&postcount=2). And finally, perhaps after 28., we should note that some tests need to be done 3 times (or more) due to mismatches, and that there is a dedicated effort to handle these (https://www.mersenneforum.org/showthread.php?t=24148). 
20190714, 18:26  #30  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
7×877 Posts 
Quote:
Thanks also to Dylan14. 

20190714, 20:56  #31  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{5}×3×101 Posts 
Parentheses are important but that is only the first problem that comes up with that explanation.
Quote:
"Abracadabra, 63 = 3 * 3 * 7 = {and because of that} = (2^{2}1)^{2} (2^{3}1)" No, the "3"s here are not (2^{2}1)^{2}. they are (2+1)(2^{2}2^{1}+1) Straightforward explanation: 2^{6}  1 = (2^{3})^{2}  1 = (2^{3}+1)(2^{3}1), because this is a x^{2}  1 and you learned this at school. Or 2^{6}  1 = (2^{2})^{3}  1 = (2^{2}1)(2^{4}+2^{2}+1), because this is a x^{3}  1 and you learned this at school or you learned this now. Take a piece of paper, open the parentheses, and observe how everything cancels. That is sufficient. If you want to get fancy, you can continue and say, "but there is more, (2^{3}+1) = (2+1)(2^{2}2^{1}+1) and that equals 3 *3 " Quote:
Instead: 2^{15}  1 = (2^{5})^{3}  1 = (2^{5}1)(2^{10}+2^{5}+1) Or 2^{15}  1 = (2^{3})^{5}  1 = (2^{3}1)(2^{12}+2^{9}+2^{6}+2^{3}+1) What's more for every 2^{(r*s)}  1 (with s>1), have a look at how the above factorization repeats every time: https://primes.utm.edu/notes/proofs/Theorem2.html And finally, there is a much better way. Just say  "you will benefit from reading this ( https://www.mersenne.org/various/math.php ). Ask questions if you will have them." P95's webpage is coherent, brief and convincing, and has links to more detailed explanations. Yours leaves the reader more confused that they were when they started reading it. 

20190715, 12:57  #32 
Feb 2017
Nowhere
14FE_{16} Posts 
Composite exponents
As is well known,
(x^{n}  1) /(x  1) = x^{n1} + x^{n2} + ... + 1. Now let a > 1 and b > 1 be integers. We apply the above formula to x^{ab}  1. If we substitute x^{a} for x and b for n, we obtain (x^{ab}  1)/(x^{a}  1) = x^{ab  a} + x^{ab2a} + ... + 1. Similarly, substituting x^b for x and a for n, (x^{ab}  1)/(x^{b}  1) = x^{ab  b} + x^{ab2b} + ... + 1. The ne plus ultra of this is the "cyclotomic factorization" corresponding to the factorization into cyclotomic polynomials Examples of its application may be found in this Forum, e.g. here For more on the application of cyclotomic polynomials to integer factorization I suggest this paper. 
20190715, 17:55  #33  
Apr 2019
5·41 Posts 
From new participant reference:
Quote:
Code:
2[sup]p[/sup]1 

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