Recursive Sequence: Highest Prime Divisors
Neil Fernandez
primeness at borve.demon.co.uk
Tue Sep 16 13:56:59 CEST 2003
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
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