Found a P-1 factor in the standard LL range with k not having a factor of 2.
[url]http://www.mersenne-aries.sili.net/exponent.php?exponentdetails=52996729[/url]

3 × 5 × 53 × 4397 × 18911 × 61231
78.* bits.

[QUOTE=Dubslow;295980]with k not having a factor of 2.[/QUOTE]
Wow! What are the chances of that? [SPOILER]... 50% ?[/SPOILER]

It's the first one I can recall, out of 25.

[QUOTE=Batalov;296001]Wow! What are the chances of that? [SPOILER]... 50% ?[/SPOILER][/QUOTE]

Across all factors, it should be 50%. But across P-1 factors, it should be much less. Why? :smile:

In Stage 1 factors, should be 100%! [SPOILER]Proof: I have only one of these in results.txt and it has an odd k: M52361579 has a factor: 3833960913376723923372391[/SPOILER] :missingteeth:

[QUOTE=Dubslow;295980]Found a P-1 factor in the standard LL range with k not having a factor of 2.
[URL]http://www.mersenne-aries.sili.net/exponent.php?exponentdetails=52996729[/URL]

3 × 5 × 53 × 4397 × 18911 × 61231
78.* bits.[/QUOTE]

[QUOTE=Batalov;296001]Wow! What are the chances of that? [SPOILER]... 50% ?[/SPOILER][/QUOTE]

[QUOTE=axn;296028]Across all factors, it should be 50%. But across P-1 factors, it should be much less. Why? :smile:[/QUOTE]

[QUOTE=Batalov;296044]In Stage 1 factors, should be 100%! [SPOILER]Proof: I have only one of these in results.txt and it has an odd k: M52361579 has a factor: 3833960913376723923372391[/SPOILER] :missingteeth:[/QUOTE]

The chance of k being odd is higher then 50%. Factors fall in two categories:
(1) of the form f=2kp+1 and f=8x+1
(2) of the form d=2kp+1 and d=8x-1

By solving each pair of conditions, factors of the form (1) are always 2kp+1=8x+1, so k is a multiple of 4 and we have in fact only factors of the form [B]8zp+1[/B]. These are the only factors with "even k" in the "classical" sense, not only even, but "quadruple k" too. There is no factor where k is equal to 2 (mod 4).

Same as above, factors of the form (2) will always be (by re-notation of k) of the form [B]8zp+sp+1[/B], where s=8-(p (mod 4)). So s=6 if p=4q+1, but s=2 if p=4q+3 for some q. We can factor a two out of it and we get the "k in the classical sense" is always odd, and the form 4z+t, where t=-p (mod 4).

So we have:
(1) factors of the form [B]f=2*[4*z]*p+1[/B].
(2) factors of the form [B]d=2*[4*z+t]*p+1[/B], t=-p (mod 4).
where the brackets were used to show the decomposition of k.

So, the "d-factors" always exists, for any composite Mp=2^p-1, for an odd prime p, because in this case Mp is 7 (mod 8), and it can't have only factors of the form f, because f is 1 (mod 8) and their set is close to multiplication (their product is always 1 (mod 8)). The conclusion is that a composite Mp may have any number of f-factors, but it MUST have an ODD number of d-factors, as the product of an even number of d-factors is also 1 (mod 8).

This shows that all composite Mp will have a factor with odd k, but some composites may exists which have no f-factors (they can be a product of 3, 5, 7, etc d-factors). There are more d-factors (odd k) then f-factors (quadruple k). There is no factor where k is 2 (mod 4).

This ...
[QUOTE=LaurV;296082]The chance of k being odd is higher then 50%. [/quote]
Does not follow from this ...
[QUOTE=LaurV;296082]So, the "d-factors" always exists, for any composite Mp=2^p-1, for an odd prime p, because in this case Mp is 7 (mod 8), and it can't have only factors of the form f, because f is 1 (mod 8) and their set is close to multiplication (their product is always 1 (mod 8)). The conclusion is that a composite Mp may have any number of f-factors, but it MUST have an ODD number of d-factors, as the product of an even number of d-factors is also 1 (mod 8).

This shows that all composite Mp will have a factor with odd k, but some composites may exists which have no f-factors (they can be a product of 3, 5, 7, etc d-factors). There are more d-factors (odd k) then f-factors (quadruple k). There is no factor where k is 2 (mod 4).[/QUOTE]

M343111009 has a factor: 101147794026026459897
k = 2^2 * 103 * 357762281 = 147398059772

M36118457 has a factor: 204726728570332673759 [TF:67:68:mfaktc 0.18 barrett79_mul32]
found 1 factor for M36118457 from 2^67 to 2^68 [mfaktc 0.18 barrett79_mul32]

k = [COLOR="Red"]2834101254247[/COLOR] (21 digits)

Just noticed this in the "Recent cleared"
[CODE]Member Name Computer Name Exponent Type UTC Time Received Days GHz-days Result
-------------------- ---------------- --------- -------- ------------------- ----- -------- -------------------------------------------------
PPed72 Unimib 56261729 F-PM1 Apr 17 2012 9:20PM 10.9 4.0290 1531076005907436082137874576376865534182896705073
[/CODE]
(160.06 bits... Composite of course: = p24*p25)

M55255747 has a factor: 2214689268597166059044783
k = 59 * 101 * 257 * 12323 * 1061897 = 20040352260527453

A tiny Intel Atom CPU powered computer found it on its first PM1 assignemnt.