# Recursive Sequence: Highest Prime Divisors

Leroy Quet qqquet at mindspring.com
Tue Sep 16 23:53:45 CEST 2003

```As John Conway and Jens have noted, the sequence, if non-constant,
eventually approaches the repeating sequence:

{...8,7,9,10,8,7,9,10... }

with the highest-prime-divisor sequence being

{...2,7,3,5,2,7,3,5,... }, of course.

I have unrigorously "proved" this myself this morning. But my proof is
probably flawed someplace.
I am glad to see the result confirmed.

But I apparently deduced that, if m and n have both the same
highest-prime-divisor, then the sequence has a period of 1, at least from
the 3rd term on.

Otherwise, all other {m,n} lead eventually to the period-4 sequence of
{8,7,9,10} repeating.

>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.
>
>....

Thanks,
Leroy Quet

```