6- table

[Batalov]

6, 341- C224 = p94 . p130
(exp divisible by 11, and therefore a quintic with difficulty 241)

A tongue-in-cheek recipe for 'success':
"If you only want long enough, any number will become the 1st hole."

Batalov+Dodson snfs

[Raman]

6,355- is an ECM miss rather?

6,355- c206 = p58 * p148

You know that every prime number of form 1 (mod 4) can be uniquely represented up as sum of two squares, right?
p58 = a²+b²
where

a = 26954637581188276770322320890
b = 53660966062879867364046240361

Expecting up the factors of 2,935- on February 20, 2010 itself.
That is right now being the expected completion time of that number only.

Due to post regarding this number only on 5 February 2010, I got extremely late
for my cousin sister's marriage betrothal (actually reached there when all the
function was over), and already then they have taken up the photos and videos
of all my other beloved relatives except me and my parents (family).

Very frustrating it is.

Marriage is upon the summer solstice day only. But, actually in fact that I slept off for 3 hours
before writing up that post, though, due to lack of patience in writing it up. Should
censor up irregular time sleep from now onwards. I wish that I would have gone there
earlier, before itself, instead of rather lying down, getting to sleep, and then that post could
have been done later on.
The photographs of me and then my cousins have been attached up hereby itself, only.

[Raman]

Mystery number - find out that candidate
by using this following hint:
How many days are there within any given year?

mystery number
number of days within any given year = 365
365 = 5 * 73
= (2^2 + 1^2) * (8^2 + 3^2)
= 19^2 + 2^2
= 14^2 + 13^2

What about that for 689, 1457, 1001, 1009...

[ET_]

Originally Posted by Raman

Mystery number - find out that candidate
by using this following hint:
How many days are there within any given year?

mystery number
number of days within any given year = 365
365 = 5 * 73
= (2^2 + 1^2) * (8^2 + 3^2)
= 19^2 + 2^2
= 14^2 + 13^2

What about that for 689, 1457, 1001, 1009...

To be honest, there are 365.25 days in a year...

Luigi

[Raman]

Originally Posted by ET_

To be honest, there are 365.25 days in a year...

Luigi

I meant that calendar year, not the Earth's rotation

Earth rotates in 23 hours 56 minutes 4.09 seconds
Revolves around the sun within precisely
365 days 5 hours 48 minutes 45.51 seconds
= 365.2422 days

Earth's axial tilt = 23.44 degrees

How many days does Gregorian calendar have so accurately within 10000 years
Absolutely, that is 3652425 days, no?

Thus, how many days do you think that year 10000 February should have so
in order to synchronize up with that Earth's rotation calendar?

Of course, it is true that earth's rotation is being slowed down regularly due to that
tidal friction from that moon, moon goes into a further orbit around earth, at the rate
of around 3 cm per year rather? This will continue until Earth is tidally locked with
moon as is moon with earth right now. At that point of time, rotation period of Earth
will be equal to revolution period of moon around earth at 47 days, right now from
27.3 days. This will take upto 50 billion years, but within another 5 billion years, that
Sun as a red giant star will rather swallow up both earth and moon then? If not, once
earth moon are both tidally locked up with each other, as is Pluto Charon system, then that moon
will start up moving closer to earth. Once it crosses up within that Roche limit, earth's gravity
can break up that moon into millions of fragments that will rather orbit planet in the form of
that rings.

Earth's rotation is right now being rather slowed down at the rate of about 1 second within every
500000 years.

[bdodson]

Originally Posted by Raman

...

For those of us that don't open zip-files

5910  6, 365- c185 1552875106954286892749964394710986213899516972480933644523600802712902591
. p113 Raman snfs

Also Batalov+Dodson's c171 gnfs, p83*p89. Supose the above p73
counts as a miss ..., ah, no; not a Mersenne number, guess not.
-Bruce

[Raman]

Originally Posted by bdodson

For those of us that don't open zip-files

5910  6, 365- c185 1552875106954286892749964394710986213899516972480933644523600802712902591
. p113 Raman snfs

Also Batalov+Dodson's c171 gnfs, p83*p89. Supose the above p73
counts as a miss ..., ah, no; not a Mersenne number, guess not.
-Bruce

No chance to be any ECM miss at all
For a p73 factor, imagine how many curves you need to run at B1=3*10⁹
Certainly that it is a lot easier to factor by using SNFS
my pet algorithm that has become right now, by now itself

p54 factors like that from 5,427+ can only be called as an ECM miss
I won't accept even p65 factors as an ECM miss at all.

By the way, is that Bos+Kleinjung ECM parallelization trick only applicable
to that list of Mersenne numbers, actually?
I want to know more about that case, much within that fact

[bdodson]

Originally Posted by Raman

...
By the way, is that Bos+Kleinjung ECM parallelization trick only applicable
to that list of Mersenne numbers, actually?...

Alex posted Thorsten's announcement:

Originally Posted by Thorsten

Stage 1: we implemented arithmetic functions for Playstation3s for
Mersenne numbers. Stage 1 for 24 curves in parallel and for B1=3*10^9
took less than 23 hours on one PS3, i.e., less than one hour per curve
per PS3.

which appears to me to say that B1=3*10^9 only works for Mersenne
numbers. I'd be happy to be wrong; perhaps there's a work-around,
or the timings for non-Mersenne numbers isn't off by a full magnitude
(like 1e9, instead of 3e9?). Until we hear otherwise, p73's are only for
Mersenne numbers, due to B1=3e9 being only for Mersenne numbers.

Meanwhile, looking at the "who's" list, are we going to have a new
Smaller-but-Needed above C180 in another page or two? -Bruce

PS - Here's another version from Nmbrthry

Originally Posted by Thorsten

Stage 1: We implemented arithmetic functions for PS3s for Mersenne
numbers. We used a recently developed 4-way SIMD carry-less
Karatsuba multiplier based on radix 212 signed digit
representation, thereby obtaining a speed-up of approximately a
factor of 2 over our previous unoptimized 4-way SIMD PS3 multiplier.
Stage 1 for 24 curves in parallel and for B1=3*10^9 took less than
23 hours on one PS3, i.e., less than one hour per curve per PS3.

Also, the parallelization matters. They did 30,000 step 1s (and then found
that first p73 after just 8800 of the step 2s), so without any parallelizing
(an extreme case), they'd only have had 30000/24 = c. 300*4 = 1200 curves.

Our only reason for not objecting that the entire method looks too ad-hoc
to consider seriously is that they found the second one. I wonder whether
we'll get another 70-digit+ any time soon.

[frmky]

NFS@Home has finished 6,385- by SNFS. Log is attached.

prp84 factor: 848309686035087642620840193724699651536186090795063231766401545260286503867205343001
prp103 factor: 8069925129421512315078607836272335020663433851059267139141811772952369600025478699376883002767164883851

[frmky]

And 6,377- is finished.

prp64 factor: 4163738647660343644736593693349157640855934319038312150518137743
prp77 factor: 10225045916601248647752444357593002700492832183893285320674260124721671052319
prp109 factor: 3180086558308793081180085210976696863166710882711790023757500947744933108800529269662639263040161216121494069

[frmky]

Add 6,349- to the finished list. Actually a while ago as you can see by the date. Finals week intervened, though!

Fri May 13 01:03:56 2011  prp68 factor: 42718986495841359540531087907270796421704343217605115730109486979483
Fri May 13 01:03:56 2011  prp141 factor: 271696650711235713248500256972633590883334525976176892387276364981832992375182426818192208801216461941493240842717505382975945252028996771663