20140424, 04:06  #1 
Apr 2014
1110111_{2} Posts 
Starter #  Perfect Number Minus 2
I'm new to Aliquot sequences, I noticed that a lot of the starter numbers being researched are small. Is there a record for the largest started number brought to completion?
I noticed numbers of the perfect number minus two (for true perfect and non mersenne prime formed "perfect form" numbers) seemed to complete rather easily and on rather big numbers.....are there other forms that complete easily? The largest number of this batch is 43 digits long: Other perfect numbers minus 2 (or of that form with a non mersenne prime) (2^31)*(2^3)/22 (5 cycles end in prime 3) (2^51)*(2^5)/22 (5 cycles end in prime 37) (2^71)*(2^7)/22 (11 cycles end in prime 43) (2^111)*(2^11)/22 (108 cycles end in prime 7) (2^131)*(2^13)/22 (94 cycles end in prime 43) (2^191)*(2^19)/22 (139 cycles end in prime 59) (2^231)*(2^23)/22 (498 cycles end in prime 601) (2^311)*(2^31)/22 (166 cycles end in prime 41) (2^371)*(2^37)/22 (213 cycles end in 2924/2620) (2^431)*(2^43)/22 (157 cycles end in prime 13) 43 digits (2^711)*(2^71)/22 (285 cycles end in prime 59) 
20140424, 06:02  #2 
"Curtis"
Feb 2005
Riverside, CA
1010000100000_{2} Posts 
Whatever terminated sequence that contains the largest term somewhere in the sequence would also be the record for largest starting number, since you could choose to "start" a 'new' sequence at that largest term.

20140424, 10:12  #3 
Apr 2014
7×17 Posts 
Ah that's true, thanks. I guess anything record wise worth noting may be longest known completed sequence iteration chains.

20140424, 10:29  #4  
Romulan Interpreter
"name field"
Jun 2011
Thailand
9,859 Posts 
Quote:
[edit: or if you want only nonprimes, it starts with M48+1] Last fiddled with by LaurV on 20140424 at 10:31 

20140424, 15:42  #5 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3·7·461 Posts 
Why not that pesky little sequence that starts with M48*(M48+1)/2 ?

20140424, 16:57  #6 
Apr 2014
7×17 Posts 
Yes, well spotted.
It looks like perfect numbers times 2 decompose quite nicely...was able to get as high as (2^127)*(2^1271) before the composites got out of control. (2^127)*(2^1271) total of 77 digits Decomposes in 17 iterations to this prime: 49039222283371595885559854029 
20140424, 17:54  #7 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×7×461 Posts 
No, my point was that any sequence that starts with a perfect number is terminated by definition. For your above example that would be (2^126)*(2^1271). And if I put my Capt.Obvious hat, the P48 is the largest perfect number we know at this time.
It is probably possible to backtrack a few specific cases that would land on P48 or on M48 or on M48+1 (and M48 in the next iteration). But we are too lazy. 
20140424, 18:17  #8 
Romulan Interpreter
"name field"
Jun 2011
Thailand
23203_{8} Posts 

20140424, 20:29  #9  
Apr 2014
7×17 Posts 
Yes, understood your point. I was wondering if there was any way to get anything larger (ie: 2xLargestPerfectNumber) which you eluded to by backtracking.
Interestingly, it appears that all 2xPerfectNumbers follow the same pattern in the first couple of iterations: 0  (2^p)*(2^p1) 1  2^(2*p) 2  (2^p1)*(2^p+1) or 4^p1 Iteration #3 and later on #2 leaves you with ever larger composites as the perfect number gets bigger. Quote:


20140425, 02:43  #11 
Apr 2014
7×17 Posts 
@Batalov
Actually unless I'm mistaken, I believe this sequence has a higher starting number. 0  (M48)^2 1  2^57885161 2  M48 
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