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2016-12-23, 16:33   #2795
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

Quote:
 Originally Posted by Drdmitry The driver has been lost again. This song will play forever. :)
too bad all it will take for it to be a driver again ( okay 2^3 is a guide technically) is gaining a 5 or a 3. this is a better link potentially if you don't want to look up the whole sequence and just see the last line.

Last fiddled with by science_man_88 on 2016-12-23 at 16:41

2016-12-23, 16:42   #2796
Drdmitry

Nov 2011

2·32·13 Posts

Quote:
 Originally Posted by science_man_88 too bad all it will take for it to be a driver again ( okay 2^3 is a guide technically) is gaining a 5 or a 3. this is a better link potentially if you don't want ot look up the whole sequence and just see the last line.
23 without a factor of 5 or 3 is not a guide. It will be stable until the power of two changes.

Last fiddled with by Drdmitry on 2016-12-23 at 16:43

2016-12-23, 16:44   #2797
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

Quote:
 Originally Posted by Drdmitry 23 without a factor of 5 or 3 is not a guide. It will be stable until the power of two changes.
http://www.rechenkraft.net/aliquot/analysis.html list it as a class 3 guide it may not class as a driver though.

Last fiddled with by science_man_88 on 2016-12-23 at 16:45

2016-12-23, 17:48   #2798
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

11100000111012 Posts

Quote:
 Originally Posted by Drdmitry 23 without a factor of 5 or 3 is not a guide. It will be stable until the power of two changes.
In SM_88's link, I followed Clifford Stern's definition:

Quote:
 A guide consists of $2^a$ ($a \gt 0$) along with a subset of the prime factors of $\sigma(2^a)$.
I interpreted it in the following way, which I believe to be at least perfectly reasonable if really just a question of meaningless semantics.

Quote:
 This motivates the following definition: a number of the form 2^b \cdot v, where v divides \sigma(2^b), is called a guide. (Note that when v=1, the guide is just the power of 2.)
Actually, technically by your statement the down"driver" wouldn't be considered a guide. I think that is certainly enough to justify the empty powers of two as guides also...

 2016-12-24, 00:42 #2799 ryanp     Jun 2012 Boulder, CO 11×17 Posts Maybe it's time to change my username to Sisyphus...
2016-12-24, 02:58   #2800
LaurV
Romulan Interpreter

Jun 2011
Thailand

22·3·739 Posts

Quote:
 Originally Posted by ryanp Maybe it's time to change my username to Sisyphus...
Told you so... Trust the Old Salt next time...
Now, don't be sad, from how the graph looks like, we may find the first cycle in the history of man, with a period 7000 or so...

 2017-01-03, 23:20 #2801 VBCurtis     "Curtis" Feb 2005 Riverside, CA 23×32×61 Posts Ugh. The 7.
 2017-01-06, 12:23 #2802 LaurV Romulan Interpreter     Jun 2011 Thailand 22×3×739 Posts good, now it is me again who owns the largest downdriver! (which is 613068, at 153 digits, not yet updated into Dubslow's table) Last fiddled with by LaurV on 2017-01-06 at 12:24 Reason: s/155/153/ tyop
 2017-01-06, 15:26 #2803 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 216468 Posts Well, don't get your hopes too low!
2017-01-07, 01:37   #2804
ryanp

Jun 2012
Boulder, CO

101110112 Posts

Quote:
 Originally Posted by Batalov Well, don't get your hopes too low!
Except that we're rapidly approaching 200 digits...

Need that downdriver ASAP.

 2017-02-22, 09:43 #2805 unconnected     May 2009 Russia, Moscow 2×11×113 Posts No progress from Jan 18, did Ryan still working on it?

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