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][/CODE](in fact i said "the easiest way" as a joke, there is no other way, and I was expected the question "what is the harder way?" from some nitpicker and I could reply "raising 31 at 4th power", hehe)

[QUOTE=BigBrother;317691]Now at line 5078, I'm already sieving the c133.[/QUOTE]
found by ECM:
prp44 = 23684003947956653714782183866686101195981013 (curve 1 stg2 B1=46000000 sigma=3481469303)

[QUOTE=RobertS;317749]found by ECM:
prp44 = 23684003947956653714782183866686101195981013 (curve 1 stg2 B1=46000000 sigma=3481469303)[/QUOTE]

Meh, I did 1000 curves @ 11M...:cry:

[QUOTE=Dubslow;317734]Or any even power. Then the sequence is effectively class 4, which means it needs to the rest (besides 2^4*31^(2n)) needs to be a product of 4 primes or less, but that's not sufficient. (There are some restrictions on what each prime has to be mod 4 or mod 8 etc. Also, [STRIKE]and[/STRIKE] an other prime raised to an even power doesn't count against that 4.)[/QUOTE]

[QUOTE=LaurV;317741]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][/CODE](in fact i said "the easiest way" as a joke, there is no other way, and I was expected the question "what is the harder way?" from some nitpicker and I could reply "raising 31 at 4th power", hehe)[/QUOTE]

As far as I can tell, every example there has 31^2, which does not contradict what I said. (None of these small multipliers has more than four odd-powered prime factors.)
[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[/code]

[QUOTE=richs;317694]How does a 2^4 * 31 driver break?[/QUOTE][QUOTE=Batalov;317969]Because of my avatar, I will reserve 804588.[/QUOTE]Most elegant is this way....... (from 804588):[code] 1947 . c114 = 2^4 * 31^2 * 9218757242216532814830455218507779571072152947622474831189587924445638638177013241155356122322458800614403093
1948 . c114 = 2 * 31 * 1217 * 4679[/code](2^4 * 31 ran from i1643:c64...)

It's good to see that there is still some life in a lot of these sequences!

[QUOTE=Dubslow;317764](None of these small multipliers has more than four odd-powered prime factors.)[/QUOTE]
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.

[QUOTE=LaurV;317985]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.[/QUOTE]

Don't think I'm any sort of genius. If you figure out why, please tell me. :razz:

I was guessing/extrapolating from [URL="http://dubslow.tk/aliquot/analysis.html"]what Clifford wrote[/URL].
[quote=Mr. Stern] Let a change in the exponent a be termed a mutation. This occurs only when the 2s count of t is equal to or less than the class of 2^a*v. In the former case, the exponent a increases and in the latter, a is reduced to the 2s count of t. The stability of a guide depends upon its class: the smaller the class, the more stable the guide. For example, a [B]class 2[/B] guide will mutate if t is the product of [B]two primes[/B] of the form 4n+1 or is a prime of the form 8n+3 or 4n+1. But a [B]class 1[/B] guide mutates only when t is [B]a prime[/B] of the form 4n+1.[/quote]

Simply extrapolate that a class n driver might be broken if it factors into n or less odd-powered primes.

(From another part of what he said, and some numerology on my part, I'm fairly sure that when v in a perfect-driver is raised to an odd power, the overall class is raised to the power of 2. That is, 2^[B]1[/B] * 3^2 is class [B]1[/B], 2^[B]2[/B] * 7^2 is class [B]2[/B], and 2^[B]4[/B] * 31^2 is class [B]4[/B], and class 4 means what it does as above. (Edit: This can easily be proven. For a perfect driver, v=2^p-1, v prime. Thus 2s_count(v) = pow_of_2(sigma(v)) = pow_of_2(2^p) = p. Thus p + (-1) = p-1 = 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=Stern, Clifford]When the class of a driver is zero or -1, a small 2s count of t is not sufficient in itself to effect a change in the exponent a because the 2s count of t is always greater than zero. Help is required from one of the components of v by having its exponent aquire an even power in order to temporarily raise the driver's class above zero. For example, when the 2^2 · 7 driver takes the form 2^2 · 7^2, its class of -1 temporarily increases by 3 (the 2s count of 7) so a mutation will occur when the 2s count of t is 2 or 1.[/quote]

[QUOTE=LaurV;317985]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 must take the pencil to understand why... And thanks for teaching me something new.[/QUOTE]

The next aliquot value is σ(N)-N.

σ(p^a*q^b*s^c)=σ(p^a)*σ(q^b)*σ(s^c)

σ(p)=p+1, so 2|σ(p)

With 5 or more distinct primes at odd powers,

2^4||N and 2^5|σ(N), so 2^4||σ(N)-N

Of course sometime 4|p+1, so you can get to 2^5 with fewer than five distinct primes

BTW: Is somebody working on the remaining C137?

not me

[QUOTE=yoyo;318307]BTW: Is somebody working on the remaining C137?[/QUOTE]

I have, there's now a c157...I've done 4500 curves @ 11M.