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Old 2014-01-03, 07:16   #1
MichelMarcus
 
Nov 2011
Saint Maur, France

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Default Help needed for a sequence

I would like to create the sequence of numbers whose aliquot sequence is always increasing until it reaches a prime number.
Among the terms one would find for instance 324 (324->523) and 392 (392->463).
But also 9112 (9112->9248->10093) or 10712 (10712->11128->11552->12451).
Another term would be 14952.
What would be the best way to proceed?

Many thanks.
Michel
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Old 2014-01-03, 10:09   #2
fivemack
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It's a nice question, and leads to some pretty tree-graphs. But I'm not quite sure what you mean by 'the best way to proceed' ... it would be nice to back-track but I don't know a clever way to do that.

Code:
for(a=5,100000000, if(!(isprime(a)),  \
 ct=0; aa=a; happy=1; while(happy==1, \
  ct=1+ct; b=sigma(aa)-aa;  if(b<=aa,happy=0,if(isprime(b),happy=2)); if(ct>50,happy=-1); aa=b); \
   if(happy==2,print([a,aa,happy,ct]))))
in pari/gp is a stupidly simplistic approach; there are aliquot sequences which keep growing as far as anyone has followed them, so you have to stick in the arbitrary limit somewhere, and if you start finding completions at ct=50 (I've not seen ct>23) then increase the value of 50 and try again.
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Old 2014-01-03, 10:30   #3
garambois
 
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Quote:
Originally Posted by MichelMarcus View Post
I would like to create the sequence of numbers whose aliquot sequence is always increasing until it reaches a prime number.
Among the terms one would find for instance 324 (324->523) and 392 (392->463).
But also 9112 (9112->9248->10093) or 10712 (10712->11128->11552->12451).
Another term would be 14952.
What would be the best way to proceed?

Many thanks.
Michel
Hello Michel,

I think your idea is very interesting !
I don't know what's the best way to proceed !
But if this can help you :
All the terms of your sequences will be even numbers (because increasing) except the last.
The last term is prime, so it is odd.
I'm sur that the last but one term of your sequences is of the form m^2 or 2*m^2.
See the proof here (sorry in french) : http://www.aliquotes.com/changement_parite.pdf

Jean-Luc
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Old 2014-01-03, 12:01   #4
MichelMarcus
 
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Yes, I ran into these kind of problems.
With an arbitrary limit, the first hurdles I got were 966 and 1134.
With factordb, I saw that 966 apparently grows monotonously until term 696 that is deficient.
Also, the graph of 1134 shows that it is going up and down.
But the process is quite lengthy and not very easy to do. Would it be possible to automate it?
You wrote that there are aliquot sequences which keep growing as far as anyone has followed them.
Do you know the smallest starting term of these sequences?
Thanks.
Michel
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Old 2014-01-03, 12:34   #5
garambois
 
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Quote:
Originally Posted by MichelMarcus View Post
Yes, I ran into these kind of problems.
With an arbitrary limit, the first hurdles I got were 966 and 1134.
With factordb, I saw that 966 apparently grows monotonously until term 696 that is deficient.
Also, the graph of 1134 shows that it is going up and down.
But the process is quite lengthy and not very easy to do. Would it be possible to automate it?
You wrote that there are aliquot sequences which keep growing as far as anyone has followed them.
Do you know the smallest starting term of these sequences?
Thanks.
Michel
I know several aliquot sequences which keep growing as far as anyone has followed them, for exemple : 5352 or 9336. I calculate those two sequences which are reserved for me on this forum on "Aliquot sequence reservations".
Or another exemple with 19560 (which is reserved for Paul Zimmermann) and which is growing at each iteration with a factor at least 2.
Jean-Luc
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Old 2014-01-03, 13:22   #6
MichelMarcus
 
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These 3 sequences, that you're telling me about, they never encounter a deficient number? Is that so ?
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Old 2014-01-03, 13:48   #7
garambois
 
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Quote:
Originally Posted by MichelMarcus View Post
These 3 sequences, that you're telling me about, they never encounter a deficient number? Is that so ?
Yes that's right and I know a lot of others aliquot sequences like that !
But they are all "OE-aliquot sequences" and perhaps if we continue, they will go down ! Nobody can know before calculs will be done !
Jean-Luc
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Old 2014-01-03, 14:03   #8
garambois
 
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Quote:
Originally Posted by garambois View Post
Hello Michel,

I think your idea is very interesting !
I don't know what's the best way to proceed !
But if this can help you :
All the terms of your sequences will be even numbers (because increasing) except the last.
The last term is prime, so it is odd.
I'm sur that the last but one term of your sequences is of the form m^2 or 2*m^2.
See the proof here (sorry in french) : http://www.aliquotes.com/changement_parite.pdf

Jean-Luc
Sorry, more complicated I knew !
I can just tell you that in a aliquot sequence, if you want the terms of the sequence change their parity, you have to pass by a term which is of the form m^2 or 2*m^2 (m an integer).
So if you begin an aliquot sequence on an even number, it's possible you have a term which is of the forme m^2 or 2*m^2, an next, the terms will be odd. And you can have several odd terms before have a prime to finish the sequence !
If you begin your sequence on a odd term, you can finish your sequence by a prime without have a term of the forme m^2 or 2*m^2 because not necessary to change the parity of the terms.

I hope I am clear because my bad english !

Jean-Luc
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Old 2014-01-03, 14:54   #9
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I made a PARI code to go backwards once but garambois is a lot better at these things. if you want I can tell you how I came to my code and you could make your own.

Last fiddled with by science_man_88 on 2014-01-03 at 14:54
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Old 2014-01-03, 15:11   #10
MichelMarcus
 
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I wrote something like:
prev(n) = {for (i=1, n, if ((sigma(i) - i) == n, return (i));); return (0);}

Going back to factordb, when I look at say 966, there is a button to download the corresponding elf file. Is there a place where these elf files can be downloaded from ?
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Old 2014-01-03, 15:11   #11
fivemack
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Quote:
Originally Posted by MichelMarcus View Post
These 3 sequences, that you're telling me about, they never encounter a deficient number? Is that so ?
That's unknown - factordb contains, as far as I'm aware, all known knowledge about those sequences - and I suspect it's false.

I am running 30-terms on [1,1e10] to see if I get any more interesting cluster points; it should be done by Monday. 166681535623 looks interesting (26th iterate of 119785014)
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