20140103, 07:16  #1 
Nov 2011
Saint Maur, France
2×5^{2} Posts 
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 
20140103, 10:09  #2 
(loop (#_fork))
Feb 2006
Cambridge, England
6,361 Posts 
It's a nice question, and leads to some pretty treegraphs. But I'm not quite sure what you mean by 'the best way to proceed' ... it would be nice to backtrack 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])))) 
20140103, 10:30  #3  
Oct 2011
2^{2}×5×17 Posts 
Quote:
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 JeanLuc 

20140103, 12:01  #4 
Nov 2011
Saint Maur, France
110010_{2} Posts 
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 
20140103, 12:34  #5  
Oct 2011
340_{10} Posts 
Quote:
Or another exemple with 19560 (which is reserved for Paul Zimmermann) and which is growing at each iteration with a factor at least 2. JeanLuc 

20140103, 13:22  #6 
Nov 2011
Saint Maur, France
110010_{2} Posts 
These 3 sequences, that you're telling me about, they never encounter a deficient number? Is that so ?

20140103, 13:48  #7  
Oct 2011
524_{8} Posts 
Quote:
But they are all "OEaliquot sequences" and perhaps if we continue, they will go down ! Nobody can know before calculs will be done ! JeanLuc 

20140103, 14:03  #8  
Oct 2011
524_{8} Posts 
Quote:
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 ! JeanLuc 

20140103, 14:54  #9 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
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 20140103 at 14:54 
20140103, 15:11  #10 
Nov 2011
Saint Maur, France
62_{8} Posts 
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 ? 
20140103, 15:11  #11  
(loop (#_fork))
Feb 2006
Cambridge, England
14331_{8} Posts 
Quote:
I am running 30terms 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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