[QUOTE=James Heinrich;600190]B1 used: 1,000,000 but you could've found it in stage1 with B1 = 1,058,921 :rogue:[/QUOTE]
Killjoy. :davieddy: 
[QUOTE=James Heinrich;600190][QUOTE=Xyzzy;600188][M]M10674101[/M] has a 128.559bit (39digit) factor (P1,B1=1000000,B2=683234370)[/QUOTE]B1 used: 1,000,000 but you could've found it in stage1 with B1 = 1,058,921 :rogue:[/QUOTE]Hmm, why is that? Oh, wait...[code]? q=501429461099632378102024867763831999287;print(factor(q1))
[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... 
[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.

[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?

[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. 
[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: 
[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. 
M14 653 103 has 76.7 bit factor :122734318525096261632697
[url]https://www.mersenne.ca/exponent/14653103[/url] This was missed by an earlier P1 
[M]M10694611[/M] has a 131.385bit (40digit) factor: [URL="https://www.mersenne.ca/M10694611"]3554057355364082639398093556551855458167[/URL] (P1,B1=1000000,B2=683234370)
:mike: 
[M]M13799479[/M] has a 197.019bit (60digit) [b]composite[/b] (P28+P33) factor: [url=https://www.mersenne.ca/M13799479]203567302892764010174130301907774551633686861729505640138639[/url] (P1,B1=531000,B2=492104340)

M[M]65330107[/M] has a 74.99bit 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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