Recursive Sequence: Highest(etc) Prime Divisors

[Leroy Quet]

I wrote:

>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,...)
>

There are many variations on this theme:

a(k) = 
(highest prime dividing a(k-1)) + 
(lowest prime dividing a(k-2)),

or 

a(k) = (sum of distinct primes dividing a(k-1)) + 
(number distinct primes dividing a(k-2)),

as two arbitrary examples.
(And I have no idea if these particular examples are interesting.)


But I had an idea for a game based upon these sequences.

***
Both players start with any integer-pair {m,n}, where m and n are >= 2.
(I suggest that m and n be 2-digit or 3-digit integers.)
(Clarification: Both players have same {m,n}.)

Each player generates his/her own sequence (originating from this pair) 
by:

a(1) = m;
a(2) = n;

a(k+2) = (ANY prime divisor of a(k+1)) +
         (ANY prime divisor of a(k)).


Before play, players decide on a fixed (relatively high) integer j.

The player with the highest a(j) wins.

(And, of course, it is not necessarily advantageous to always pick the 
highest prime-divisor, for such a sequence would be bounded.)

Also, it might be fun to, instead of having a fixed j, to instead have 
the number of terms in each player's sequence, before the game is 
completed, be based upon something determined by play itself.
(This might make the game more than just having each player calculate j 
terms of their sequence without regards to what the other player does at 
all.)
For example, play ends when both players' a(j)'s sum to some value or 
above.

Or perhaps...the game can be such that players can incorporate the other 
player's a(k) into calculating their own.
(For example, perhaps any prime dividing the OPPONENT'S a(k-1) can also 
be added to get a player's a(k).)

Needs work....

--

But a serious question: 
If we can choose which prime divisors of a(k-2) and a(k-1) to add to get 
a(k), is it 
possible to get (given some particular {m,n}'s) an unbounded sequence?


Thanks,
Leroy Quet