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-   -   Aliquot sequences that start on the integer powers n^i (https://www.mersenneforum.org/showthread.php?t=23612)

 kruoli 2023-04-24 11:55

[QUOTE=garambois;629173]Sorry, but I'm not sure I understand the question.
You are asking about the average time to terminate a sequence, is that correct ?
Because it varies very much depending on the sequence, but I think you already know that.
That makes me assume that I'm misunderstanding the question...[/QUOTE]

Thank you for your answer! Going from the time when I wrote that, I think I was already tired. You are correct, I [I]should[/I] know that but had likely a brain fart.

Where I wanted to go with this: Maybe newcomers to the project might be afraid a work unit takes too long. But since I do not even know how many potential candidates are there that would like to participate, this worry might be taking it too far.

 garambois 2023-04-24 14:29

To make it very simple :

Sequences of the same parity fast enough if there is no prime 3 in the terms decompositions.
Of course, if we start with 130 digits numbers, it is not the same thing as if we start with 170 digits numbers !
It is up to each one to know if he can decompose numbers of 130 or 170 digits and in how much time.

Opposite parity sequences : Same remark as before.
And it's not the same if the sequence keeps yo-yoing, or if it grows very fast if its terms are very "brittle" (I don't know if "brittle" is the right word in English to designate a term which has lots of small prime numbers in the decomposition of its terms ? Perhaps "friable" or "frangible" would be more appropriate ?)
And it's not at all the same thing to compute a sequence up to 120 digits or up to 170 digits !

In short : you can compute a sequence in minutes or hours, or in days, weeks or even months !

 birtwistlecaleb 2023-04-24 20:26

[QUOTE=garambois;629259]To make it very simple :

Sequences of the same parity fast enough if there is no prime 3 in the terms decompositions.
Of course, if we start with 130 digits numbers, it is not the same thing as if we start with 170 digits numbers !
It is up to each one to know if he can decompose numbers of 130 or 170 digits and in how much time.

Opposite parity sequences : Same remark as before.
And it's not the same if the sequence keeps yo-yoing, [B]or if it grows very fast if its terms are very "brittle" (I don't know if "brittle" is the right word in English to designate a term which has lots of small prime numbers in the decomposition of its terms ?[/B] Perhaps "friable" or "frangible" would be more appropriate ?)
And it's not at all the same thing to compute a sequence up to 120 digits or up to 170 digits !

In short : you can compute a sequence in minutes or hours, or in days, weeks or even months ![/QUOTE]

You can just say it has many small factors, but if you want to describe it like that I think abundant might be a good choice.

 Happy5214 2023-04-25 12:17

[QUOTE=garambois;629259]Opposite parity sequences : Same remark as before.
And it's not the same if the sequence keeps yo-yoing, or if it grows very fast if its terms are very "brittle" (I don't know if "brittle" is the right word in English to designate a term which has lots of small prime numbers in the decomposition of its terms ? Perhaps "friable" or "frangible" would be more appropriate ?)
And it's not at all the same thing to compute a sequence up to 120 digits or up to 170 digits ![/QUOTE]

[QUOTE=birtwistlecaleb;629294]You can just say it has many small factors, but if you want to describe it like that I think abundant might be a good choice.[/QUOTE]

"Abundant" already has a definition ([i]n[/i] is "abundant" if [i]n[/i]'s aliquot sum is greater than [i]n[/i]), so don't use it like that. [i]n[/i] can be called "[I]m[/I]-smooth" if [I]all[/I] of its factors are less than [I]m[/I], but I don't think there's a particular term for an [I]n[/I] where just [I]some[/I] factors are less than [I]m[/I].

 EdH 2023-04-25 13:46

Although still not quite right, perhaps "complex" would describe a term with many factors rather than few.

 Happy5214 2023-04-25 14:09

"Abundant" would actually be a good choice to describe the overall concept you're trying to talk about (not a number with a lot of small factors, but a number with a large aliquot sum relative to the number itself, causing the sequence to accelerate upward). I think that's the more important point, rather than the exact factor form.

 garambois 2023-04-25 15:05

Yes, I can say "abundant", that will make the concept and the message clear.
But in French we have the word "friable" which brings a nuance to the word "abundant".
The most friable numbers possible are powers of 2.

 Happy5214 2023-04-25 16:18

[QUOTE=garambois;629350]Yes, I can say "abundant", that will make the concept and the message clear.
But in French we have the word "friable" which brings a nuance to the word "abundant".
The most friable numbers possible are powers of 2.[/QUOTE]

Wikipedia describes "friable" as synonymous with "smooth", as I defined that term a few posts ago. Powers of 2 are 2-smooth (or 2-friable), but a full term in an aliquot sequence isn't smooth or friable generally, as it usually has at least one large factor to go along with the small ones, and [I]m[/I]-smooth/[I]m[/I]-friable numbers need [I]all[/I] of their prime factors to be less than or equal to [I]m[/I].

 garambois 2023-04-25 16:25

Yes Happy, sorry, you are right.
I was using the word "friable" wrong even in French, I just checked.
"Abundant" is more appropriate !

 garambois 2023-04-26 08:59

Page updated
Many thanks to all for your work !

[URL="https://www.mersenneforum.org/showpost.php?p=629413&postcount=71"]See here[/URL].

 birtwistlecaleb 2023-04-29 22:39

I think I should've said this earlier, but you didn't add my name code for bases 118 and 129, can you add these soon?

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