20120109, 04:02  #1 
Jan 2012
2^{3} 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

20120109, 04:25  #2 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
3·29·109 Posts 
That is 10^(10^(10^1.582413493301632)), go figure how much space it would take to store it, then get back to us.

20120109, 05:06  #3 
Romulan Interpreter
Jun 2011
Thailand
3^{3}·347 Posts 
What do you want to calculate about it? This is just 2^1271 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???

20120109, 06:40  #4 
Bemusing Prompter
"Danny"
Dec 2002
California
2^{3}·3^{3}·11 Posts 
I'm sure Curtis Cooper has it written down on a notepad somewhere...

20120109, 09:03  #5 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
41·229 Posts 
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.

20120110, 00:16  #6 
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**1271)1 print(MM127) 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 MiniGeek on 20120110 at 00:27 
20120110, 01:49  #7 
"William"
May 2003
New Haven
3·787 Posts 
The last 1000 digits repeat cyclically, so you calculate MM127 modulo the cycle length.
I haven't worked out the details, but I think you can track the first 1000+guard digits through successive doublings. I don't know how to calculate this one, though. 
20120110, 04:56  #8 
Jun 2003
1001100110001_{2} Posts 

20120110, 21:34  #9 
Dec 2008
you know...around...
3^{2}·71 Posts 
IICC (If I calculated correctly), here's the first 1000+ digits (thanks to Pari):
Code:
54543129001957378843931465896008210589349423685477667235404075275966293201609170424672529973992278171828573751806703589030869420288646093175468495080860272288857861747545862537756437098674222346365647119802401733149549254785201526438888502479253268057826437323564417964170617571980632548850088960037091796035202961108096679669489593720537183768359095819230433170516480756896976957583868764458099770918691408613562982373691771503874318596635831076763107059895504556733884228091178020321255026176317956600127015020809981773938409606830956146639230166636616192287793902321515403831972033102387976040468028125079678482416679257798154403144337229277566440573299804751002923089281341096461023036173103078419500018782121294711493545423257006683412552953435284142491984699944888780205024571856612842698394292710272603587159230032575847392878591995885731587973057009255468614011647081720403109593508461225776234622521599185438521239106860292082598601921729532323579501307651127212407305413121126823059540422578927614... 
20120110, 22:15  #10 
"William"
May 2003
New Haven
2361_{10} Posts 
Done with high precision logs?
Now can you teach us how to find the 314159265th digit? 
20120110, 22:42  #11 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
41·229 Posts 
The last 1000 digits appear to be
Code:
...5306813365668297450370632400602267153874336893830587374421010744080639147929797090189031583986667245151695093088884848435739003342730983774160532341947281645852376823812191537066336494080616177095839838408723503324460081378144781399003245848814312060755437197330096531689244931436290688877165655112310821396183434300597491515931447277006739103507849853624660079369090487347841908091324793171634834191372487932912287198969725285700588443513422872332657159103300342083888175598493606647054638638459487370356026995048368555266388927554035854210857534541447525830208946429733247652561005894777628061789036981761598323020941762535700923452191914591248662885115946127751862309917152642709665522155340331650424597793571384683791990076270115592503537057996418689035885639996930655711304306774936853288566384077559882136253112332940815617303382922303266378846578148511333994928346239574411949114138364447524053925671023601313393687184931477325492550655458106222489358628350506357958045615739087358437493833727 (I was also thinking that maybe someone will find a better way, like it is possible with π ?) 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
MM127  Stan  Miscellaneous Math  34  20150619 03:24 
Is MM127 a PRP?  aketilander  Operazione Doppi Mersennes  6  20121031 16:02 
MM127  Dougal  LMH > 100M  26  20091213 09:57 
Is MM127 Prime? Just a Poll  jinydu  Miscellaneous Math  57  20081108 17:48 
MM127 Checkout Page  clowns789  Lone Mersenne Hunters  44  20040930 08:06 