20121109, 20:00  #2003 
Romulan Interpreter
Jun 2011
Thailand
5^{2}·7·53 Posts 
what are you talking there? Almost any multiple of 31 would do it (look to the 3rd column after zero)
Code:
(02:56:07) gp > forstep(i=1,1000,2, a=2^4*31*i;s=sigma(a)a;v=factorint(s); if(v[1,2]!=4, print(i", "i%31", "i\31",\t"a" , "s", "v))) 31, 0, 1, 15376, 15407, [7, 1; 31, 1; 71, 1] 93, 0, 3, 46128, 77004, [2, 2; 3, 3; 23, 1; 31, 1] 155, 0, 5, 76880, 107818, [2, 1; 31, 1; 37, 1; 47, 1] 217, 0, 7, 107632, 138632, [2, 3; 13, 1; 31, 1; 43, 1] 279, 0, 9, 138384, 261795, [3, 1; 5, 1; 31, 1; 563, 1] 341, 0, 11, 169136, 200260, [2, 2; 5, 1; 17, 1; 19, 1; 31, 1] 403, 0, 13, 199888, 231074, [2, 1; 31, 1; 3727, 1] 465, 0, 15, 230640, 508152, [2, 3; 3, 1; 31, 1; 683, 1] 527, 0, 17, 261392, 292702, [2, 1; 31, 1; 4721, 1] 589, 0, 19, 292144, 323516, [2, 2; 31, 1; 2609, 1] 713, 0, 23, 353648, 385144, [2, 3; 31, 1; 1553, 1] 775, 0, 25, 384400, 569873, [31, 2; 593, 1] 837, 0, 27, 415152, 816168, [2, 3; 3, 1; 31, 1; 1097, 1] 899, 0, 29, 445904, 477586, [2, 1; 31, 1; 7703, 1] Last fiddled with by LaurV on 20121109 at 20:07 
20121109, 20:37  #2004 
Aug 2009
somewhere
197 Posts 

20121109, 21:34  #2005 
Feb 2005
The Netherlands
2·109 Posts 

20121109, 22:17  #2006  
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 
Quote:
Quote:
Code:
>>> for i in range(1,1000,2): ... n = 2**4*31*i ... s = a.aliquot(n) ... v = a.factor(s) ... if v[2] != 4: ... print("{}, {}, {}, \t{}={}, \t{}, {}".format(i, i%31, i//31, n, a.factor(n), s, v)) ... 31, 0, 1, 15376=2^4 * 31^2, 15407, 7 * 31 * 71 93, 0, 3, 46128=2^4 * 3 * 31^2, 77004, 2^2 * 3^3 * 23 * 31 155, 0, 5, 76880=2^4 * 5 * 31^2, 107818, 2 * 31 * 37 * 47 217, 0, 7, 107632=2^4 * 7 * 31^2, 138632, 2^3 * 13 * 31 * 43 279, 0, 9, 138384=2^4 * 3^2 * 31^2, 261795, 3 * 5 * 31 * 563 341, 0, 11, 169136=2^4 * 11 * 31^2, 200260, 2^2 * 5 * 17 * 19 * 31 403, 0, 13, 199888=2^4 * 13 * 31^2, 231074, 2 * 31 * 3727 465, 0, 15, 230640=2^4 * 3 * 5 * 31^2, 508152, 2^3 * 3 * 31 * 683 527, 0, 17, 261392=2^4 * 17 * 31^2, 292702, 2 * 31 * 4721 589, 0, 19, 292144=2^4 * 19 * 31^2, 323516, 2^2 * 31 * 2609 713, 0, 23, 353648=2^4 * 23 * 31^2, 385144, 2^3 * 31 * 1553 775, 0, 25, 384400=2^4 * 5^2 * 31^2, 569873, 31^2 * 593 837, 0, 27, 415152=2^4 * 3^3 * 31^2, 816168, 2^3 * 3 * 31 * 1097 899, 0, 29, 445904=2^4 * 29 * 31^2, 477586, 2 * 31 * 7703 

20121112, 04:54  #2007 
"Frank <^>"
Dec 2004
CDP Janesville
4112_{8} Posts 
Most elegant is this way....... (from 804588):
Code:
1947 . c114 = 2^4 * 31^2 * 9218757242216532814830455218507779571072152947622474831189587924445638638177013241155356122322458800614403093 1948 . c114 = 2 * 31 * 1217 * 4679 It's good to see that there is still some life in a lot of these sequences! 
20121112, 05:51  #2008 
Romulan Interpreter
Jun 2011
Thailand
10010000111011_{2} Posts 
Hmmm... That is what I missed in your first post, or at least, sounded very complicate as you said it... So, you say that having 5 or more primes different of 2 or 31 in the list, all at odd powers, will make impossible to kill a 2, or breed a new 2. Now, I did not know this, and after a couple of unsuccessful attempts to get a counterexample with pari/gp, I swore I would take the pencil (but not yet). I took gp and did "select 5 random primes, do their product, times 2^4, times an even random power of 31, factor its sigma minus itself, and if the power of 2 is not 4, then print it; repeat forever". When I use 4 primes, it murders few drivers every second (i.e. printing lines), but with 5 or more primes, it prints nothing after 15 minutes. Now I must take the pencil to understand why... And thanks for teaching me something new.
Last fiddled with by LaurV on 20121112 at 05:53 Reason: forgot the "repeat forever" part 
20121112, 07:21  #2009  
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 
Quote:
I was guessing/extrapolating from what Clifford wrote. Quote:
(From another part of what he said, and some numerology on my part, I'm fairly sure that when v in a perfectdriver is raised to an odd power, the overall class is raised to the power of 2. That is, 2^1 * 3^2 is class 1, 2^2 * 7^2 is class 2, and 2^4 * 31^2 is class 4, and class 4 means what it does as above. (Edit: This can easily be proven. For a perfect driver, v=2^p1, v prime. Thus 2s_count(v) = pow_of_2(sigma(v)) = pow_of_2(2^p) = p. Thus p + (1) = p1 = power of two in the driver, as in the example below. (The only part I don't understand is how you can just add the two separate 2s counts to get the class.)) Quote:
Last fiddled with by Dubslow on 20121112 at 07:27 Reason: forgot link 

20121112, 20:42  #2010  
"William"
May 2003
New Haven
3×787 Posts 
Quote:
σ(p^a*q^b*s^c)=σ(p^a)*σ(q^b)*σ(s^c) σ(p)=p+1, so 2σ(p) With 5 or more distinct primes at odd powers, 2^4N and 2^5σ(N), so 2^4σ(N)N Of course sometime 4p+1, so you can get to 2^5 with fewer than five distinct primes Last fiddled with by wblipp on 20121112 at 20:47 Reason: Fix "five or six" errors 

20121114, 11:52  #2011 
Oct 2006
Berlin, Germany
607 Posts 
BTW: Is somebody working on the remaining C137?

20121114, 12:00  #2012 
Romulan Interpreter
Jun 2011
Thailand
243B_{16} Posts 
not me

20121115, 17:27  #2013 
Feb 2005
The Netherlands
11011010_{2} Posts 

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