[QUOTE=science_man_88;229564]with my extension method i got Aliquot(276) to 12601 in just over 11 seconds.[/QUOTE]

I don't think I believe that. If 276 is the first member, 396 is the second member, and 21374326697892540932 is the 100th member, what do you get for members 200, 300, 400, ..., 2000?

[QUOTE=science_man_88;229569]yep 100 * 127 -99 (lost to replacement) = (#v=12601) in 11 seconds.[/QUOTE]

Which value represent 12601?
The index of the sequence Aliquot(276), so 12601 iterations?

The first 100 iterations are [url=http://factordb.com/sequences.php?se=1&eff=2&aq=276&action=range&fr=0&to=100]here[/url].

[CODE] Aliquot(n)= v=vector(1,x,n);for(b=1,100,for(a=2,127,d=Alistep(v[a-1]);v=concat(v,d);if(d==n || d==1,break()));v[1]==v[#v]);print(v)[/CODE]

once you do the alistep put in try this for yourself i got:

*** last result computed in 11,437 ms.

and no I didn't use debug though I've played around and found it has 20 levels.

[code]Alistep(n)=sigma(n)-n
Aliquot(n)=my(v=vector(1,x,n),d);for(b=1,100,for(a=2,127,d=Alistep(v[a-1]);v=concat(v,d);if(d==n || d==1,break()));v[1]==v[#v]);v

>Aliquot(10)
%3 = [10, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1][/code]

So you're calculating something, but not the right thing.

[QUOTE=CRGreathouse;229573][code]Alistep(n)=sigma(n)-n
Aliquot(n)=my(v=vector(1,x,n),d);for(b=1,100,for(a=2,127,d=Alistep(v[a-1]);v=concat(v,d);if(d==n || d==1,break()));v[1]==v[#v]);v

>Aliquot(10)
%3 = [10, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1, 8, 7, 1][/code]

So you're calculating something, but not the right thing.[/QUOTE]

it's because i forgot it's break(2) lol not break() that should be there.

never mind I see your point dang. but I think this still goes to 127 by the looks of it before repeating higher than the 100 list kar_bon gave me a link to.

if it did the math correct these are the next ones:

[CODE]23780027782003162628, 27022045182026917372, 29800418252165873732, 30868734601205564812, 30868743536688997492, 31506186871079789708, 33685079263470335092, 37869561159553616108, 44810700265393440532, 52967010256293285740, 77368308005658962068, 77526798761623418732, 81730869595023949972, 81812211224085690348, 174495211883566827540, 383889841278567188460, 851420485964196743700, 2133148752623068133100, 4920463122717210500628, 9659030418601945179372, 19140499834691254267668, 31900833057818757113004, 60928733769254068230996, 101554467127566656249004, 169306878754562576009556, 282178131257604293349484[/CODE]

never mind I didn't use my brain lol. how far would I have to go to find the 1 ot 276 ?

[QUOTE=science_man_88;229586]never mind I didn't use my brain lol. how far would I have to go to find the 1 ot 276 ?[/QUOTE]

What do you want to find for each? The vector, the number of terms, the final prime, ...?

I want to figure the whole thing out Aliquot sequence n. but if I have to settle for it maybe when it repeats or is terminated.

[QUOTE=science_man_88;229588]I want to figure the whole thing out Aliquot sequence n. but if I have to settle for it maybe when it repeats or is terminated.[/QUOTE]

You could just do
[code]vector(274,n,Aliquot(n+1))[/code]
but you'll need to modify your program to look for loops, and you'll want some way to handle sequences that don't seem to terminate.

[QUOTE=CRGreathouse;229589]You could just do
[code]vector(274,n,Aliquot(n+1))[/code]
but you'll need to modify your program to look for loops, and you'll want some way to handle sequences that don't seem to terminate.[/QUOTE]

that's what:

[CODE]if(d==n || d==1,break(2))[/CODE]
is supposed to do. check for repeats of n or 1. but I guess we'll have to check for every integer in v to find out for sure. so the main part missing is a way to scan through v and check for repeating of a single item that isn't n.