Recursive Sequence: Highest Prime Divisors
Kennedy
kennedy at oldnews.org
Tue Sep 16 05:18:33 CEST 2003
Leroy,
After fiddling around with your sequence for a few minutes,
I got the sense that it is akin to the as yet intractable 3n+1 problem,
about which Paul Erdös supposedly commented:
"Mathematics is not yet ready for such problems."
Regards,
Kennedy
----- Original Message -----
From: "Leroy Quet" <qqquet at mindspring.com>
To: "Leroy Quet" <qqquet at mindspring.com>; <seqfan at ext.jussieu.fr>
Sent: Monday, September 15, 2003 7:41 PM
Subject: Recursive Sequence: Highest Prime Divisors
> If we start with two positive integers {m,n}, and let
> a(1) = m,
> a(2) = n,
>
> and we let, for k >= 3,
>
> a(k) =
> (highest prime dividing a(k-1)) +
> (highest prime dividing a(k-2)),
>
>
> then we get a (possibly eventually-periodic) sequence.
>
> (For example: m = 2, n = 3, leads to ->
>
> 2, 3, 5, 8, 7, 9, 10, 8, 7, 9,...)
>
> I am not at all certain, but I wonder if every {m,n} leads to a sequence
> that is eventually periodic.
>
> Is this so??
>
> Obviously, if the highest primes dividing a(j) and a(j-1) are those which
> are the highest primes which also divide a(k) and a(k-1), respectively
> (for some k not = j), then the sequence is periodic.
>
> There may be something in the EIS, but the fact that {m,n} can vary makes
> a search not too easy, since I do not know which {m,n}-sequence is
> fundamental enough to be in the EIS, yet such a sequence is not trivial.
>
> Thanks,
> Leroy Quet
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