Found a factor? Post it here. Or forever fold your crease. - Page 86

Jwb52z · 2015-04-29, 01:45

UID: Jwb52z/Clay, M76106587 has a factor: 2253706211548544927116376383 (P-1, B1=690000, B2=12937500)

90.864 bits

Jwb52z · 2015-05-10, 14:01

P-1 found a factor in stage #2, B1=715000, B2=13227500.
UID: Jwb52z/Clay, M78208159 has a factor: 5980644118600514360767129 (P-1, B1=715000, B2=13227500)

82.307 bits

Jwb52z · 2015-05-15, 01:54

P-1 found a factor in stage #2, B1=690000, B2=12937500.
UID: Jwb52z/Clay, M76097309 has a factor: 160406892606660651307306068007 (P-1, B1=690000, B2=12937500)

97.018 bits.

tha · 2015-05-18, 20:54

I found a factor for the exponent 12973951 which has 2^11 as part of the value for 'k'. Any exponent with a 'k' that includes a higher power than 11?

TheJudger · 2015-05-18, 22:30

Hi tha,

you've asked for it, perhaps my highscore:

http://www.mersenne.ca/exponent/62419603

and some slightly lower powers of 2:

and some even lower powers of 3:

and for 5:

http://www.mersenne.ca/exponent/68938921

Oliver

LaurV · 2015-05-19, 06:07

If p is 1 (mod 4), then k can be 0 or 3 (mod 4), and if p is 3 (mod 4), then k can be 0 or 1 (mod 4). There is no k which is 2 (mod 4) because in this case 2kp+1 would be either 3 or 5 (mod 8) which is not possible for a factor.

Therefore, considering that 50% of the factors have a k which is 0 (mod 4), (i.e. 50% of the k's are multiple of 2^2 already) then about one in 1000 will have a k which is 0 (mod 2^11). As we have few millions of them...

TObject · 2015-05-26, 22:02

M78098261 has a factor q=2715126139938490881846927116066757839; log2(q)=121.03...
k = 11*13*3911*7321*81547*1802039*28890311 = 17382756704009650623635058379; log2(k)=93.812...

M78098261 has a factor q=56873105138451031651367; log2(q)=75.59...
k = 19 * 23^2 * 29 * 163 * 7663739 = 364112493736903; log2(k)=48.371...

TheJudger · 2015-05-27, 21:25

indeed not a every day factor, congratulations!

Oliver

P.S. I know there is nothing specials about these factors, small or big, B1-smooth or not, a factor is a factor. But bigger factors feel better than smaller ones (bigger is better!).

Gordon · 2015-05-28, 21:27

Quote:

Originally Posted by TObject

M78098261 has a factor q=2715126139938490881846927116066757839; log2(q)=121.03...
k = 11*13*3911*7321*81547*1802039*28890311 = 17382756704009650623635058379; log2(k)=93.812...

M78098261 has a factor q=56873105138451031651367; log2(q)=75.59...
k = 19 * 23^2 * 29 * 163 * 7663739 = 364112493736903; log2(k)=48.371...

Which leads to the interesting thought, if we had run TF to 76 bits and found the factor, how long would it have been before we found the 121 bit one...

TObject · 2015-05-28, 22:53

True.

Also, with slightly higher B1, the 76-bit factor could have shown up in the Stage 1 and the Stage 2 would not run at all.

Jwb52z · 2015-06-01, 02:04

P-1 found a factor in stage #2, B1=700000, B2=12950000.
UID: Jwb52z/Clay, M77149753 has a factor: 100126154465201540478799 (P-1, B1=700000, B2=12950000)

76.406 bits.

2015-05-18, 22:30	#940
TheJudger "Oliver" Mar 2005 Germany 10001011011₂ Posts	Hi tha, you've asked for it, perhaps my highscore: http://www.mersenne.ca/exponent/62419603 and some slightly lower powers of 2: http://www.mersenne.ca/exponent/70213463 http://www.mersenne.ca/exponent/60967087 http://www.mersenne.ca/exponent/65490281 http://www.mersenne.ca/exponent/72541463 and some even lower powers of 3: http://www.mersenne.ca/exponent/66150701 http://www.mersenne.ca/exponent/59890121 http://www.mersenne.ca/exponent/71694323 and for 5: http://www.mersenne.ca/exponent/68938921 Oliver

2015-05-19, 06:07	#941
LaurV Romulan Interpreter "name field" Jun 2011 Thailand 41·251 Posts	If p is 1 (mod 4), then k can be 0 or 3 (mod 4), and if p is 3 (mod 4), then k can be 0 or 1 (mod 4). There is no k which is 2 (mod 4) because in this case 2kp+1 would be either 3 or 5 (mod 8) which is not possible for a factor. Therefore, considering that 50% of the factors have a k which is 0 (mod 4), (i.e. 50% of the k's are multiple of 2^2 already) then about one in 1000 will have a k which is 0 (mod 2^11). As we have few millions of them... Last fiddled with by LaurV on 2015-05-19 at 06:43

2015-05-26, 22:02	#942
TObject Feb 2012 110010101₂ Posts	Nice couple of factors for an exponent M78098261 has a factor q=2715126139938490881846927116066757839; log2(q)=121.03... k = 11133911732181547180203928890311 = 17382756704009650623635058379; log2(k)=93.812... M78098261 has a factor q=56873105138451031651367; log2(q)=75.59... k = 19 * 23^2 * 29 * 163 * 7663739 = 364112493736903; log2(k)=48.371...

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2015-04-29, 01:45	#936
Jwb52z Sep 2002 35F₁₆ Posts	UID: Jwb52z/Clay, M76106587 has a factor: 2253706211548544927116376383 (P-1, B1=690000, B2=12937500) 90.864 bits

2015-05-10, 14:01	#937
Jwb52z Sep 2002 863 Posts	P-1 found a factor in stage #2, B1=715000, B2=13227500. UID: Jwb52z/Clay, M78208159 has a factor: 5980644118600514360767129 (P-1, B1=715000, B2=13227500) 82.307 bits

2015-05-15, 01:54	#938
Jwb52z Sep 2002 863 Posts	P-1 found a factor in stage #2, B1=690000, B2=12937500. UID: Jwb52z/Clay, M76097309 has a factor: 160406892606660651307306068007 (P-1, B1=690000, B2=12937500) 97.018 bits.

2015-05-18, 20:54	#939
tha Dec 2002 881 Posts	I found a factor for the exponent 12973951 which has 2^11 as part of the value for 'k'. Any exponent with a 'k' that includes a higher power than 11?

2015-05-27, 21:25	#943
TheJudger "Oliver" Mar 2005 Germany 5·223 Posts	indeed not a every day factor, congratulations! Oliver P.S. I know there is nothing specials about these factors, small or big, B1-smooth or not, a factor is a factor. But bigger factors feel better than smaller ones (bigger is better!).

2015-05-28, 22:53	#945
TObject Feb 2012 110010101₂ Posts	True. Also, with slightly higher B1, the 76-bit factor could have shown up in the Stage 1 and the Stage 2 would not run at all.

2015-06-01, 02:04	#946
Jwb52z Sep 2002 863 Posts	P-1 found a factor in stage #2, B1=700000, B2=12950000. UID: Jwb52z/Clay, M77149753 has a factor: 100126154465201540478799 (P-1, B1=700000, B2=12950000) 76.406 bits.