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 2012-01-09, 04:02 #1 antimath   Jan 2012 23 Posts MM127 does anyone know of a program to CALCULATE [ONLY!] MM127 = 2^(2^127 -1)-1)...i know factoring it is virtually impossible, but calculating it cant be...?? links and other stuff helps alot thx
2012-01-09, 04:25   #2
Uncwilly
6809 > 6502

"""""""""""""""""""
Aug 2003
101×103 Posts

3·29·109 Posts

Quote:
 Originally Posted by antimath does anyone know of a program to CALCULATE [ONLY!] MM127 = 2^(2^127 -1)-1)...i know factoring it is virtually impossible, but calculating it cant be...?? links and other stuff helps alot thx
That is 10^(10^(10^1.582413493301632)), go figure how much space it would take to store it, then get back to us.

 2012-01-09, 05:06 #3 LaurV Romulan Interpreter     Jun 2011 Thailand 33·347 Posts What do you want to calculate about it? This is just 2^127-1 bits, all of them being 1. So, first you look for about 1.7*10^38 bits of memory, that is not much, just 19807040628566084398385987584 gigabytes, then you fill all of them with 1. What is so difficult???
 2012-01-09, 06:40 #4 ixfd64 Bemusing Prompter     "Danny" Dec 2002 California 23·33·11 Posts I'm sure Curtis Cooper has it written down on a notepad somewhere...
2012-01-09, 09:03   #5
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

41·229 Posts

Quote:
 Originally Posted by antimath does anyone know of a program to CALCULATE [ONLY!] MM127 = 2^(2^127 -1)-1)...i know factoring it is virtually impossible, but calculating it cant be...?? links and other stuff helps alot thx
You can calculate any of its decimal digits if that will satisfy your curiosity: the last 1000, the first 1000, ... or any given digit, for example the 314159265th digit... or not.

2012-01-10, 00:16   #6
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

17×251 Posts

Here's Python code to calculate it, and print it:
Code:
MM127=2**(2**127-1)-1
print(MM127)
Note: no existing computer will be able to finish this program.

Quote:
 Originally Posted by LaurV What do you want to calculate about it? This is just 2^127-1 bits, all of them being 1. So, first you look for about 1.7*10^38 bits of memory, that is not much, just 19807040628566084398385987584 gigabytes, then you fill all of them with 1. What is so difficult???
Don't worry, it compresses very well. In fact, with the right algorithm, (i.e. "where the bytes of the file as a number are x, calculate MMx") you can store it in a single byte: 01111111.

Last fiddled with by Mini-Geek on 2012-01-10 at 00:27

2012-01-10, 01:49   #7
wblipp

"William"
May 2003
New Haven

3·787 Posts

Quote:
 Originally Posted by Batalov the last 1000
The last 1000 digits repeat cyclically, so you calculate MM127 modulo the cycle length.

Quote:
 Originally Posted by Batalov the first 1000
I haven't worked out the details, but I think you can track the first 1000+guard digits through successive doublings.

Quote:
 Originally Posted by Batalov for example the 314159265th digit
I don't know how to calculate this one, though.

2012-01-10, 04:56   #8
axn

Jun 2003

10011001100012 Posts

Quote:
 Originally Posted by wblipp I haven't worked out the details, but I think you can track the first 1000+guard digits through successive doublings.
High precision logs.

 2012-01-10, 21:34 #9 mart_r     Dec 2008 you know...around... 32·71 Posts IICC (If I calculated correctly), here's the first 1000+ digits (thanks to Pari): Code: 54543129001957378843931465896008210589349423685477667235404075275966293201609170424672529973992278171828573751806703589030869420288646093175468495080860272288857861747545862537756437098674222346365647119802401733149549254785201526438888502479253268057826437323564417964170617571980632548850088960037091796035202961108096679669489593720537183768359095819230433170516480756896976957583868764458099770918691408613562982373691771503874318596635831076763107059895504556733884228091178020321255026176317956600127015020809981773938409606830956146639230166636616192287793902321515403831972033102387976040468028125079678482416679257798154403144337229277566440573299804751002923089281341096461023036173103078419500018782121294711493545423257006683412552953435284142491984699944888780205024571856612842698394292710272603587159230032575847392878591995885731587973057009255468614011647081720403109593508461225776234622521599185438521239106860292082598601921729532323579501307651127212407305413121126823059540422578927614... The number itself has 51217599719369681875006054625051616349 digits, which itself has 38 digits.
 2012-01-10, 22:15 #10 wblipp     "William" May 2003 New Haven 236110 Posts Done with high precision logs? Now can you teach us how to find the 314159265th digit?
 2012-01-10, 22:42 #11 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 41·229 Posts The last 1000 digits appear to be Code: ...5306813365668297450370632400602267153874336893830587374421010744080639147929797090189031583986667245151695093088884848435739003342730983774160532341947281645852376823812191537066336494080616177095839838408723503324460081378144781399003245848814312060755437197330096531689244931436290688877165655112310821396183434300597491515931447277006739103507849853624660079369090487347841908091324793171634834191372487932912287198969725285700588443513422872332657159103300342083888175598493606647054638638459487370356026995048368555266388927554035854210857534541447525830208946429733247652561005894777628061789036981761598323020941762535700923452191914591248662885115946127751862309917152642709665522155340331650424597793571384683791990076270115592503537057996418689035885639996930655711304306774936853288566384077559882136253112332940815617303382922303266378846578148511333994928346239574411949114138364447524053925671023601313393687184931477325492550655458106222489358628350506357958045615739087358437493833727 I was thinking about the 314159265th digit from the right, actually! ...so - just like the last 1000 digits (and should fit in memory, I think). And if from left, then with even more high precision log? (I was also thinking that maybe someone will find a better way, like it is possible with π ?)

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