Reserved for MF - Sequence 4788

[science_man_88]

Originally Posted by Raman

Is 2 the only way of going down by means of a downdriver. Are there no other ways? What about going down by stable guides?
There are many different drivers (perfect numbers - all only known to be even). But there is only one downdriver!
Are there any drivers for odd numbers? Is that statement equivalent to asking for existence of odd perfect numbers?
Any way that an aliquot sequence parity change will occur only for squares or twice squares. They are the only way to terminate an aliquot sequence. Isn't it?

What is the difference between 2³.3 driver and 2³.3.5 driver? Or is even 2³.5 being a driver?

It would be interesting if people could think about that Euler's Totient / Euler's Phi sequence, but not only, let not alone Sum of Factors / Restricted Divisor Function / Aliquot Sequence, namely - not - not!

you can explain odd squares changing simply be starting with odd+odd=even then noticing that odd*odd=odd then following the fact that for the prime factorization of a number N = p^i * p^j *p^k will result in (i+1)*(j+1)*(k+1) divisors and noting that if each exponent is even all the values will be odd in the product leading to an odd number of divisors that taking N away ( one of the divisors as calculated) you get an even number of odd proper divisors leading to an even sum. the reason both 2^3*3 and 2^3*5 work as drivers is that to drive a sequence the numbers d ( 3 or 5 in this case) must divide sigma(2^n) for some value n ( n=3 in this case) as sigma(2^3) =1+2+4+8 = 15 =3*5 d can be either 3 or 5 or 15.

[firejuggler]

i10305 : 2^2*3*31

[ryanp]

And the 3 is gone again, at i10317...

[Dubslow]

i10343 was a close call with 2^3 * P, but alas P was 15 mod 32 (i.e. \(\tau(P) = 4\), not 1).

[science_man_88]

2^3*3^2* values at i10412

[LaurV]

Originally Posted by science_man_88

2^3*3^2* values at i10412

That doesn't matter much, it still needs to get all the other factors to be 1 (mod 4), or get down to a single prime. The 3s come and go, but never disappear completely, unless the power of 2 changes, see my former posts in this thread.

As I know you are a "pari guy", try to compare in pari/gp:

gp > d=10; cnt=0; for(i=1,10^4,n=2^3*3^(random(5)+1)*(6*random(10^d)+(2*random(2)-1)); s=(sigma(n)-n); if(s%2^3!=0 || s%2^5==0,cnt++; print(n"\t"factorint(n)"\t\t"factorint(s))); printf("...%d / %d (%7.4f%%)... %c",cnt,i,100.0*cnt/i,13))

with

gp > d=10; cnt=0;  for(i=1,10^4,n=2^3*3^(random(5)+1)*(6*random(10^d)+(2*random(2)-1));  s=(sigma(n)-n); if(s%3!=0,cnt++;  print(n"\t"factorint(n)"\t\t"factorint(s))); printf("...%d / %d  (%7.4f%%)... %c",cnt,i,100.0*cnt/i,13))

(edit: differences marked in red; make your command prompt box 180 characters wide, or so; right click on its header, properties).

Play with it. Change d (make it 20, 30, 40). Change the powers of 2 and 3 to see how other drivers behave. These things can all be proved with modular calculus too. I did that. But I know you like pari.

edit: now after running the pari part myself, I see that some "rare" cases are possible too, where the power of 3 being even might help. I didn't consider those in my calculus
But those cases are rare, and it still needs to get the F's square-free part to a single prime, but in this case it can be any prime, not necessarily 1 mod 4.

2594944565640   [2, 3; 3, 2; 5, 1; 7, 2; 147105701, 1]          [2, 2; 3, 2; 5, 1; 487, 1; 1741, 1; 47279, 1]
3069904117464   [2, 3; 3, 2; 42637557187, 1]            [2, 2; 3, 1; 112213, 1; 3894691, 1]
3759740514936   [2, 3; 3, 2; 7, 2; 1065686087, 1]               [2, 5; 3, 2; 53, 1; 11497, 1; 46073, 1]
14550012257112  [2, 3; 3, 4; 22453722619, 1]            [2, 2; 3, 1; 19, 1; 53, 1; 23557, 1; 92051, 1]

[LaurV]

wow... i10498 is 2^3*3^4*[COLOR=Red]13[/COLOR]*...(p7)[COLOR=Red]53[/COLOR]*...(c155)[COLOR=Red]29[/COLOR], all are 1 (mod 4), so if the "...29" composite splits in two (or more) primes which are all 1 (mod 4), we may lose the 2^3 and go down again!
Scratch that! I am stupid. Wrong counting. I forgot that if you split a composite in two, you get TWO primes, haha.

so, 2^3*3^4*(3 primes, all 1 (mod 4)) - loses the 2^3
but 2^3*3^4*(4 primes, all 1 (mod 4)) - can not lose the 2^3.

Grrr...

Next time...

[Batalov]

Originally Posted by LaurV

wow... i10498 is 2^3*3^4*13*...(p7)53*...(c155)29, all are 1 (mod 4), so if the "...29" composite splits in two (or more) primes which are all 1 (mod 4), we may lose the 2^3 and go down again!

ORLY?
four factors rake in (at least) 2^4, so 2^3 will stay

[Dubslow]

Originally Posted by LaurV

wow... i10498 is 2^3*3^4*13*...(p7)53*...(c155)29, all are 1 (mod 4), so if the "...29" composite splits in two (or more) primes which are all 1 (mod 4), we may lose the 2^3 and go down again!

Hate to burst your bubble, but that aint true. With the power of two being 3, there can be at most 3 powers of two contributed by the sigma of the other non-square primes. Since there are at least 4 other primes, that's at least 4 powers of two, so no mutation.

And even if was 2 or 3 powers of 2, it would mutate to 2^2 (first case) or something higher than 2^3 (second case). And even if it was 2^3 * 3^(2k) * p for some k and prime p == 1 mod 4, it would evolve directly to 2*3 -- which would be even worse.

We have to hope for a mutation to something other than D1 before we can lose the 3 and hope for the downdriver. In the meantime, the size keeps increasing as usual...

[LaurV]

You guys won't spare me at all.. hehe.
Fortunately I realized before reading your replies, and I edited the post.
I don't know why I was silly, I had my coffee today, so I should not be silly... Or... is that from the coffee? Maybe I need to change the brand...

edit: @Dubslow: 2^3*3^4*(3 primes, all 1 mod 4) can mutate in everything else with a higher power of 2 only, which is good for us. It can not mutate into a lower power of 2.

[Dubslow]

Originally Posted by LaurV

edit: @Dubslow: 2^3*3^4*(3 primes, all 1 mod 4) can mutate in everything else with a higher power of 2 only, which is good for us. It can not mutate into a lower power of 2.

Yes, that's the second case of my second paragraph. The first case is two or fewer primes. Two primes is fine, one prime leads to D1 -- not good (assuming all are 1 mod 4 of course).