Aliquot sequences that start on the integer powers n^i

[warachwe]

Originally Posted by warachwe

In general, if p and q (both prime) divide n, and p divide q+1, then p divide s(n).

I forgot to mention that q^2 also must not divide n, or else p may not divide s(n).

Originally Posted by warachwe

In case of n=3^(78*k), s(n) is (3^(78*k)-1/)2, which is divided by (3^78-1)/2 = 2^2 · 7 · 13^2 · 79 · 157 · 313 · 2887 · 6553 · 7333 · 10141 · 398581 · 797161
Notice that 79 divide 157+1, and 157 divide 313+1, so 79 and 157 also divide s(s(n)), so 79 also divide s(s(s(n))).

So in this case, for s(s(s(n))) may not be divided by 79 if, for example, s(s(n)) got a factor of 157^2, but that's unlikely.

[garambois]

Originally Posted by yoyo

What does this mean?

That means the sequence's coming down and we might be lucky if it ends !

Originally Posted by yoyo

My reserved sequences are now cycling through the system, between BOINC server and volunteers.

OK, fantastic !

[garambois]

Originally Posted by EdH

If I'm reading this question correctly, you can use grep. Here is an example:

$ cat base_2_matcons | grep "prime 5 " | grep "x 1 2 3 4 5"

I didn't know the instruction "grep", I'm ashamed !
Thank you for this new tool for me !
The "grep" instruction doesn't give exactly the result I expect but I have no doubt that it will be very useful to me...

[garambois]

@EdH :

Please, is it possible to modify question 3 of post #447 like this:

Question 3 :
How to prove (or invalidate) otherwise than by calculation, this kind of result (conjectures often noted **):
If n=3^(78*k), then s(n), s(n) and s(n)) are all three divisible by 79?
This kind of proof must be much, much more difficult than the one we are talking about in question 2!

should be :

Question 3 :
How to prove (or invalidate) otherwise than by calculation, this kind of result (conjectures often noted **):
If n=3^(78*k), then s(n), s(s(n)) and s(s(s(n))) are all three divisible by 79?
This kind of proof must be much, much more difficult than the one we are talking about in question 2!

And please, is it possible to modify also in the same way the statement of question 3 in the second "quote" of post #450.
Machine translation played tricks on me !
But warachwe magnificently anticipated and corrected himself !!!

[garambois]

Originally Posted by warachwe

For p prime, s(p^k) is (p^k-1)/(p-1). So S(3^(6+12*k)) is (3^(6+12*k)-1)/2, which is (3^6-1)/2 * (3^(12*k)+3^(12k-6)+3^(12k-12)...+1)

So s(n) divided by (3^6-1)/2 = 364 = 2^2 *7 * 13.

As 3^6 is 1 (mod 364), each term of (3^(12*k)+3^(12k-6)+3^(12k-12)...+1) is (1 mod 364).
Therefore 3^(12*k)+3^(12k-6)+3^(12k-12)...+1 is 2k+1 (mod 364)
So if 2k+1 is not divided by either 7 or 13, we get a factor of 2^2*7*13 in s(n)
If 2k+1 is divided by 7, such as in k=3, we get factor of 2^2*7^2*13 or more.
Similarly, If 2k+1 is divided by 13, such as in k=6, we get factor of 2^2*7*13^2 or more.

Okay, thank you so much for your help!
I suspected this kind of result could be demonstrated.
This implies that a large part of my conjectures (almost all those with a star) are in fact proven.

Originally Posted by warachwe

In general, if p and q (both prime) divide n, and p divide q+1, then p divide s(n).
In case of n=3^(78*k), s(n) is (3^(78*k)-1/)2, which is divided by (3^78-1)/2 = 2^2 · 7 · 13^2 · 79 · 157 · 313 · 2887 · 6553 · 7333 · 10141 · 398581 · 797161
Notice that 79 divide 157+1, and 157 divide 313+1, so 79 and 157 also divide s(s(n)), so 79 also divide s(s(s(n))).

There was an error in the wording of question 3 of post #447, but you anticipated the real question I wanted to ask...
Here again, thank you very much for your help !
I am really very surprised !
I thought that my conjectures noted with two stars implied mechanisms impossible to understand.

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)).

These explanations significantly change the order of priorities for the calculations for the rest of the project. Above all, these explanations exempt us from continuing to look for this kind of conjectures.

[garambois]

What are we going to do now ?

If I take into account the events of the last few days, the study of the occurrences of prime numbers in all the terms of a sequence no longer seems to be a priority. I will tackle this problem in the light of the explanations in post #452.

It therefore seems that it will be more useful to focus on the prime numbers that end sequences.
We have to try to find at last some conjectures not yet known !!!

However, the sequences that we are fairly certain that they end with a prime number are those that begin with numbers n^i when n and i have the same parity. If n and i don't have the same parity, some sequences also end but they are much rarer, but this makes them very interesting.

So I propose :

1) To try to continue to advance (or even better, to end) sequences that have orange boxes for the different bases.

2) To start new bases and especially bases that are prime numbers:
- Base 18
- Base 19
- Base 20
- Base 23
- Base 29
- Base 31
- Base 79 (seems to play a special role !)

3) To try to extend for each base the number of cells n^i with n and i having the same parity. I think to extend to the next update some bases in the following way :

for base 3 i from 252 to 335
for base 5 i from 172 to 229
for base 6 i from 155 to 206
for base 7 i from 142 to 189
for base 10 i from 121 to 160
for base 11 i from 116 to 153
for base 12 i from 113 to 148
for base 13 i from 108 to 143
for base 14 i from 105 to 140
for base 15 i from 104 to 139
for base 17 i from 98 to 131

Then,

Originally Posted by RichD

@garambois: I can release the 18 open sequences from table 770 to yoyo and initialize tables 1155 and 385 or some other table(s) of your interest. Then repeat, release and initialize. You may have some resources coming your way...

@RichD or others :
Is it possible for you (or others who wish to do so) to do the preliminary work for either of the bases presented in point 2) ?

Originally Posted by yoyo

@garambois: Which lets say 200 next sequences do you need?

@ yoyo :
Is it possible for you at first to continue working on the orange cells on base 3? (point 1) above).
And if you are willing to devote more computing resources to this project, you can add the orange cells from another base, for example base 5 (especially the cells that are stopped at 120 or 121 digits). Just let me know so that I can note the reservations on the page.



With regard to point 3), I know that this request is not popular because it is very resource-intensive.
But I think it is an important point. Let's wait until we see the update to get a good visualization of the work to be done.
Everyone will then be able to see if they want to start according to their possibilities.

[yoyo]

I ran everything until the remaining composite is greater than C139.

Currently running up to 3^250 and 5^250
Seems that
5^185
5^179
have finished so far.

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

[RichD]

Originally Posted by garambois

[2) To start new bases and especially bases that are prime numbers:
- Base 18
- Base 19
- Base 20
- Base 23
- Base 29
- Base 31
- Base 79 (seems to play a special role !)

I will start preliminary work on Base 18 & Base 19. Work will (slowly) continue on Base 770 in the background.

[EdH]

Originally Posted by garambois

@EdH :
Please, is it possible to modify question 3 of post #447 like this:
. . .

Done.

Originally Posted by garambois

. . .
And please, is it possible to modify also in the same way the statement of question 3 in the second "quote" of post #450.
. . .

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

[RichD]

Base 18 table is ready for insertion. I took n to 96. Everything n <= 64 terminates (green). n > 64 has remaining composites near C80 or above and will need more work. They are all odd so they most likely will terminate. I will add them to the work queue which includes Base 770.

[EdH]

While we wait for the next update, to get a feel for the work relating to proposition 3, I will do the preliminary work for table 79.