Recursive Sequence: Highest Prime Divisors

[Neil Fernandez]

In message <8wBlqJAGCvZ$Ew4I at borve.demon.co.uk>, Neil Fernandez
<primeness at borve.demon.co.uk> writes

>In message <E19z5oZ-0005ss-00 at hall.mail.mindspring.net>, Leroy Quet
><qqquet at mindspring.com> writes
>
>>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)),

<snip>

>I conjecture that if m and n are co-prime, all such sequences eventually
>start repeating 10, 8, 7, 9, 10, 8, 7, 9, ...

And that if they are not co-prime, the sequence either does this or
contains only a single number.

E.g. m=20, n=24 gives: 8, 5, 7, 12, 10, 8, 7, 9, ...
m=45, n=63 gives: 12, 10, 8, 7, 9, ...
m=20, n=30 gives: 10, 10, 10, ...

Further interesting sequences could be defined where starting with i=4,
a(i) := the value of m where m+n=i, m<=n, and the above recursive
sequence with initial values m and n yields the longest non-repeating
string before it goes periodic (take smallest value of m if there is a
choice); and b(i) := the length of that non-repeating string.

These sequences begin:

i         a(i)       b(i)         m,n
4         2           0           2,2
5         2           1           2,3
6         2           0           2,4
7         3           3           3,4
8         2           1           2,6
9         4           2           4,5
10        3           6           3,7
11        5           4           5,6
12        2           2           2,10
13        2           6           2,11
14        5           4           5,9
15        2           6           2,13
16        5           8           5,11
17        3           6           3,14
18        7           8           7,11
19        8           6           8,11
20        3           10          3,17

with b(i) reaching record values when i = 4, 5, 7, 10, 16, 20, ...

Neil

-- 
Neil Fernandez