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-   -   Found a factor, sunshine? Embalm and entomb it here! (RU "Выкрасить и выбросить")) (https://www.mersenneforum.org/showthread.php?t=13977)

 Prime95 2022-02-17 01:17

[QUOTE=James Heinrich;600190]B1 used: 1,000,000 but you could've found it in stage-1 with B1 = 1,058,921 :rogue:[/QUOTE]

Killjoy. :davieddy:

 Dr Sardonicus 2022-02-17 01:26

[QUOTE=James Heinrich;600190][QUOTE=Xyzzy;600188][M]M10674101[/M] has a 128.559-bit (39-digit) factor (P-1,B1=1000000,B2=683234370)[/QUOTE]B1 used: 1,000,000 but you could've found it in stage-1 with B1 = 1,058,921 :rogue:[/QUOTE]Hmm, why is that? Oh, wait...[code]? q=501429461099632378102024867763831999287;print(factor(q-1))
[2, 1; 3, 3; 13, 1; 109, 1; 139, 1; 397, 1; 1579, 1; 2593, 1; 8273, 1; 310169, 1; [color=red]1058921[/color], 1; 10674101, 1][/code]If only he'd known about that factor of q - 1 before he knew that q was a factor...

 James Heinrich 2022-02-17 01:40

[QUOTE=Dr Sardonicus;600195]If only he'd known about that factor of q - 1 before he knew that q was a factor...[/QUOTE]He should've consulted the secret table I keep that lists the k values of all undiscovered factors.

 Uncwilly 2022-02-17 01:44

[QUOTE=James Heinrich;600197]He should've consulted the secret table I keep that lists the k values of all undiscovered factors.[/QUOTE]Did Curtis sell that to you?

 Dr Sardonicus 2022-02-17 16:04

[QUOTE=James Heinrich;600197]He should've consulted the secret table I keep that lists the k values of all undiscovered factors.[/QUOTE]I would imagine that, for every positive integer k which is not congruent to 2 (mod 4), there are infinitely many primes p for which q = 2*k*p + 1 divides 2^p - 1.

Values of k congruent to 2 (mod 4) are ruled out because q would be congruent to 5 (mod 8), and therefore not a divisor of 2^p - 1.

 Viliam Furik 2022-02-17 18:06

[QUOTE=Dr Sardonicus;600219]I would imagine that, for every positive integer k which is not congruent to 2 (mod 4), there are infinitely many primes p for which q = 2*k*p + 1 divides 2^p - 1.

Values of k congruent to 2 (mod 4) are ruled out because q would be congruent to 5 (mod 8), and therefore not a divisor of 2^p - 1.[/QUOTE]

I guess it's not secret anymore :lol::razz:

 BudgieJane 2022-02-17 18:23

[QUOTE=Viliam Furik;600231]I guess it's not secret anymore :lol::razz:[/QUOTE]

Is this anything to do with Legendre's theorem: All primitive prime factors of a (homogeneous) Cunningham number have the form [I]kn[/I] + 1. If n is odd they have the form 2[I]kn[/I] + 1. Alternatively: all primitive prime factors are congruent to 1 modulo [I]n[/I] if [I]n[/I] is even and 2[I]n[/I] if [I]n[/I] is odd.

 firejuggler 2022-02-17 22:50

M14 653 103 has 76.7 bit factor :122734318525096261632697
[url]https://www.mersenne.ca/exponent/14653103[/url]

This was missed by an earlier P-1

 Xyzzy 2022-02-22 02:52

[M]M10694611[/M] has a 131.385-bit (40-digit) factor: [URL="https://www.mersenne.ca/M10694611"]3554057355364082639398093556551855458167[/URL] (P-1,B1=1000000,B2=683234370)

:mike:

 James Heinrich 2022-03-02 14:11

[M]M13799479[/M] has a 197.019-bit (60-digit) [b]composite[/b] (P28+P33) factor: [url=https://www.mersenne.ca/M13799479]203567302892764010174130301907774551633686861729505640138639[/url] (P-1,B1=531000,B2=492104340)

 techn1ciaN 2022-03-05 21:26

M[M]65330107[/M] has a 74.99-bit factor: 37530453286101721709273

For every instance of this:[QUOTE=chalsall;599558]Don't you just hate it when something like [URL="https://www.mersenne.ca/exponent/11786351"]this happens...[/URL][/QUOTE]there is also one of these :smile:

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