fond of a factor? Urn yourself to become remains

[Dubslow]

Found a P-1 factor in the standard LL range with k not having a factor of 2.
http://www.mersenne-aries.sili.net/e...tails=52996729

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

[Batalov]

Originally Posted by Dubslow

with k not having a factor of 2.

Wow! What are the chances of that? ... 50% ?

[Dubslow]

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

[axn]

Originally Posted by Batalov

Wow! What are the chances of that? ... 50% ?

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

[Batalov]

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

[LaurV]

Originally Posted by Dubslow

Found a P-1 factor in the standard LL range with k not having a factor of 2.
http://www.mersenne-aries.sili.net/e...tails=52996729

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

Originally Posted by Batalov

Wow! What are the chances of that? ... 50% ?

Originally Posted by axn

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

Originally Posted by Batalov

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

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 8zp+1. 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 8zp+sp+1, 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 f=2*[4*z]*p+1.
(2) factors of the form d=2*[4*z+t]*p+1, 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).

[axn]

This ...

Originally Posted by LaurV

The chance of k being odd is higher then 50%.

Does not follow from this ...

Originally Posted by LaurV

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).

[TObject]

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

[Stef42]

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 = 2834101254247 (21 digits)

[Batalov]

Just noticed this in the "Recent cleared"

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

(160.06 bits... Composite of course: = p24*p25)

[TObject]

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.