Aliquot sequences that start on the integer powers n^i

[RichD]

Oh no! I've got i and n reversed in my above posts. I'll try to keep them straight going forward.

Table n=19 has been initialized to i=96 but there is still work to do. Many small remaining composites which I compete with the elves. Don't want to duplicate work. I'll let you know when I get it to a somewhat stable state. BTW, all the i=odd sequences terminate or mostly will terminate. But you already knew that.

In the meantime I can start building Table n=23 next. Well, I mean initialize in FDB...

[garambois]

OK, thanks to all for your big, big help !!!
I'm going to do a big update tomorrow to get a clearer view.

Originally Posted by yoyo

Currently running up to 3^250 and 5^250

Are you sure of your exponent "250", because 5^250 has 175 digits ?

[warachwe]

Originally Posted by EdH

Done.

@warachwe: May I edit garambois' quote in your post 450 to match the current version in post #447?

Sure

Originally Posted by garambois

Your explanations, as well as those of post #452 show that the conjectures noted with two stars are certainly not true, because if we continue the calculations long enough, sooner or later, for a rather large k, for example in the case of 3^(78*k), we will find the factor 157^2 in s(s(n)).

Those conjectures may still be true, because we can still get a factor of 79 from other primes than 157.
For example, from conjecture 2), normally 5 is what provide a factor of 3 for s(n). But when k=5 (3^20) and we get a factor 5^2, we also get factor of 11 and 41, both of which also provide a factor of 3 for s(n).
In this case I think conjecture 2) is true, but I can't prove it.

P.S. Using LTE we find that 157^2 will divide s(3^(78*k)) if and only if 157 divide k (though we maybe have more than 157^2).

[warachwe]

From post #364

Originally Posted by EdH

Does this apparent observation fit in with a similar theorem?
For all aⁱ (a, i positive integers ≥ 1)
s(aⁱ) is a factor of s(a^(i*n)) (for all positive n)

for p prime, s(pⁱ) = (pⁱ-1)/(p-1) and s(p^(i*n)) = (p^(i*n)-1)/(p-1).

Since (pⁱ-1) is a factor of (p^(i*n)-1), s(pⁱ) is a factor of s(p^(i*n)) for all positive integer n.

for odd a in general, this is not true. For example

s(15(3)) = 3 · 5 · 191
s(15(3*2)) = 2 · 7 · 751 · 947
s(15(3*3)) = 3 · 76091 · 147353

[RichD]

Table n=23 has been initialized to i=88. Tables 19 & 23 can be instead in the web page with the understanding there is going to be many moving targets. FDB workers will factor small composites and I will slowly tackle the larger ones (up to C120).

In summary:
Table n=18 - all i <=77 terminate. Eventually all cells will most likely be green.
Table n=19 - It appears all i=odd sequences has or will terminate.
Table n=23 - It appears all i=odd sequences has or will terminate.

[yoyo]

Originally Posted by garambois

Are you sure of your exponent "250", because 5^250 has 175 digits ?

Yes, my system skipped it fast, because the composite is larger than C139.

I now go up to 3^335.
Nothing in base 6 was reserved. So I took all ub to 6^206.

[yoyo]

I take also up to 7^189.

[garambois]

OK, page updated.
Thank you all very much for your work !
Thank you for pointing out possible handling errors.
Base 18 added.
Base 19 added.
Base 23 added.
Bases 3, 5, 6, 7, 10, 11, 12, 13, 14, 15 and 17 have been extended.


@RichD : Can you please confirm that the attributions are good for green cells ?
All cells in base 18 are green, because if n=18^i, s(n) becomes odd for all i. Indeed, n here is always even and we know that there is a change of parity if n is a perfect square or twice a perfect square.
If i is even, then n is a perfect square, this is obvious.
If i is odd, we pose i=2*k+1 and thus 18^(2*k+1)=(2*3^2)^(2*k+1)=2^(2*k+1)*3^(2*(2*k+1))=2*2^(2*k)*3^(2*(2*k+1)) which is thus a double of a perfect square. Thus, for the bases 18=2*3^2, 50=2*5^2, 72=2*6^6, 96=2*7^2... all squares will be green !

[garambois]

Originally Posted by warachwe

Those conjectures may still be true, because we can still get a factor of 79 from other primes than 157.
For example, from conjecture 2), normally 5 is what provide a factor of 3 for s(n). But when k=5 (3^20) and we get a factor 5^2, we also get factor of 11 and 41, both of which also provide a factor of 3 for s(n).
In this case I think conjecture 2) is true, but I can't prove it.

P.S. Using LTE we find that 157^2 will divide s(3^(78*k)) if and only if 157 divide k (though we maybe have more than 157^2).

I'm beginning to understand the mechanism, thanks again to you for your explanations !

Your reasoning seems to be able to establish what we can guess with our calculations.
Seeing this, I tell myself that it is no longer worth continuing to formulate such conjectures, like those of post #447.
Therefore, we can focus on other phenomena.
But, do you also think that it is no longer worthwhile to formulate conjectures like those of post #447, or do you think that this kind of conjecture can still be of interest ?

On the other hand, don't hesitate to let us know if you have an idea to test based on our data that you might have had !

[garambois]

Originally Posted by yoyo

If I need more for my hungry volunteers I would take base 6 and 7.

The yafu@home stats have been skyrocketing for the last few days !

[RichD]

Originally Posted by garambois

@RichD : Can you please confirm that the attributions are good for green cells ?

I would say if a sequence has less than 25 terms, it was handled by FDB workers. I simply hit the refresh button and watch the sequence drop. No calculations by me.

On the other hand, if a sequence starts say above C75, then I need to run the calculations to get it down to where the workers can finish it off.

That being said, for Table n=18, "A" should be added to "i=" 6-10, 12,14, 16, 18-22, 24-26, 30-31, 33, 36, 38, 40, 42-43, 49, 57.

Likewise for n=19 and n=23.