Reserved for MF - Sequence 4788

[LaurV]

what are you talking there? Almost any multiple of 31 would do it (look to the 3rd column after zero)

(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]

(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)

[RobertS]

Originally Posted by BigBrother

Now at line 5078, I'm already sieving the c133.

found by ECM:
prp44 = 23684003947956653714782183866686101195981013 (curve 1 stg2 B1=46000000 sigma=3481469303)

[BigBrother]

Originally Posted by RobertS

found by ECM:
prp44 = 23684003947956653714782183866686101195981013 (curve 1 stg2 B1=46000000 sigma=3481469303)

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

[Dubslow]

Originally Posted by Dubslow

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, and an other prime raised to an even power doesn't count against that 4.)

Originally Posted by LaurV

what are you talking there? Almost any multiple of 31 would do it (look to the 3rd column after zero)

(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]

(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)

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.)

>>> 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

[schickel]

Originally Posted by richs

How does a 2^4 * 31 driver break?

Originally Posted by Batalov

Because of my avatar, I will reserve 804588.

Most elegant is this way....... (from 804588):

 1947 .  c114 = 2^4 * 31^2 * 9218757242216532814830455218507779571072152947622474831189587924445638638177013241155356122322458800614403093
 1948 .  c114 = 2 * 31 * 1217 * 4679

(2^4 * 31 ran from i1643:c64...)

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

[LaurV]

Originally Posted by Dubslow

(None of these small multipliers has more than four odd-powered prime factors.)

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.

[Dubslow]

Originally Posted by LaurV

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.

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

I was guessing/extrapolating from what Clifford wrote.

Originally Posted by 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 class 2 guide will mutate if t is the product of two primes of the form 4n+1 or is a prime of the form 8n+3 or 4n+1. But a class 1 guide mutates only when t is a prime of the form 4n+1.

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, 2^1 * 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^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.))

Originally Posted by 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.

[wblipp]

Originally Posted by LaurV

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.

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

[yoyo]

BTW: Is somebody working on the remaining C137?

[LaurV]

not me

[BigBrother]

Originally Posted by yoyo

BTW: Is somebody working on the remaining C137?

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