View Single Post
Old 2017-12-21, 00:42   #2
science_man_88's Avatar
"Forget I exist"
Jul 2009

23·1,049 Posts

Originally Posted by carpetpool View Post
Recall the two types of divisibility sequences which are primarily involved in Lucas Sequences:

For U(n), if r divides s, then U(r) divides U(s).

For V(n), if r divides s, and r/s is odd, then V(r) divides V(s).

Most divisibility sequences terms U(n) and V(n), if not all, are generated by a recurrence relation of order n, meaning the first n terms (denoted [a1,a2,] of the sequence are given, and the rest of the terms depend on the past n previous terms.

The question is, for any recurrence relation of order n, are there finitely or infinitely many starting values or initial terms [a1,a2,] such that a(n) has the divisibility properties as U(n)? The same question can be asked about V(n).

For example, consider the recurrence relation

a(n)=a(n-1)+a(n-2) with starting values a(1) and a(2). Are there infinitely many pairs of numbers {a(1), a(2)} such that a(n) is a U(n) divisibility sequence, and there is also a corresponding V(n) sequence?

The only values I am aware which make this true is {1, 1} and {1, 3}

If a(1) = 1, and a(2) = 1, then a(n) is a divisibility sequence such that if r divides s, then a(r) divides a(s) (The U(n) sequence). These are the Fibonacci Numbers. The corresponding V(n) sequence has starting values a(1) = 1, and a(2) = 3.

Are there any more (two) sets values a(1) and a(2) which form divisibility sequences of the first kind, U(n), and the second kind V(n), with a(n)=a(n-1)+a(n-2) for n > 2? Consider this problem for all fixed recurrence relations of any order.

The same idea goes with a(n)=a(n-1)+a(n-2)+a(n-3), for instance, except given the first three starting values a(1), a(2) and a(3).
Any help, comments, suggestions appreciated. Thank you.
The parity of different parts of the recurrence will affect things, for linear and polynomial recurrences. This is because, the parity of coefficients decide how often the terms divide by 2 etc.
science_man_88 is online now   Reply With Quote