[seqfan] Re: An oddly-behaved sequence
franktaw at netscape.net
franktaw at netscape.net
Wed Oct 7 06:59:43 CEST 2009
Good point. No, I can't prove it, but it does seem nearly certain.
a(n)^2-1 gets to be much bigger than the run of the mill sequence
elements; numbers in that range occur only sporadically and immediately
fall back to more "normal" levels. As far as actually proving it, it
may be no easier than proving that every integer does occur.
Franklin T. Adams-Watters
-----Original Message-----
From: Max Alekseyev <maxale at gmail.com>
Franklin,
I don't even see why this sequence is infinite. In other words, why it
is impossible that all divisors of a(n)^2-1 for some n represent prior
terms?
Do you have a proof for that?
Max
On Tue, Oct 6, 2009 at 6:29 PM, <franktaw at netscape.net> wrote:
> Start with 1,2,4. Thereafter, the rule is that a(n+1) is the smallest
> divisor of (a(n)^2-1) that has not yet appeared in the sequence.
>
> (The 1,2,4 start is the simplest that doesn't either give the integers
> in order, nor fail because all divisors of (a(n)^2-1) are already
> present.)
>
> The sequence starts:
>
> 1,2,4,3,8,7,6,5,12,11,10,9,16,15,14,13,21,20,19,18,17,
> 24,23,22,69,28,27,26,25,39,38,37,36,35,34,33,32,31,
> 30,29,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,
> 55,56,57,58,59,60,61,62,63,64,65,66,67,68,201,80,79,
> 78,77,76,75,74,73,72,71,70,213,106,105,104,103,102,
> 101,85,84,83,82,81,160,159,158,157,156,155,88,87,
> 86,145,96,95,94,93,92,91,90,89,99,98,97,112,111,
> 110,109,108,107,212,211,120
>
> Up through the 22, it seems like the pattern is to jump to a new
value,
> and then decrement down to the previous limit. After that, it gets
> much more complicated. The tendency is to jump up, then drop down
> after a small number of steps, and decrement for a while. But notice
> the increments from 40 to 68; this happens again at 136, 153, 204,
310,
> 265, ... (with various lengths for the runs; e.g., after 153,154, the
> next term is 255).
>
> Another curiosity: in the first 1000 terms, after 3, 198, 270, 570,
> 522, 600, 822, and 882, we have a(n+1) = a(n)^2-1.
>
> So, the obvious questions:
>
> (1) Is this a permutation of the positive integers? It seems like it
> must be, but I don't see how to prove it.
> (2) Do increment and decrement runs both occur infinitely often?
> (3) Does one of them dominate? In the first 1000 terms, there are 136
> increment steps and 706 decrements.
> (4) Does a(n+1) = a(n)^2-1 occur infinitely often?
>
> -----
> Here's the PARI program I used to generate the sequence:
>
> al(n,m=4,u=6)={local(ds,db);
> u=bitor(u,1<<m);print1(m);
> for(i=1,n,
> ds=divisors(m^2-1);
> for(k=2,#ds,m=ds[k];db=1<<m;if(!bitand(u,db),break));
> u=bitor(u,db);print1(","m))}
>
> Franklin T. Adams-Watters
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>
_______________________________________________
Seqfan Mailing list - http://list.seqfan.eu/
More information about the SeqFan
mailing list