20171221, 00:19  #1 
"Sam"
Nov 2016
2^{2}×3^{4} Posts 
Divisibility sequences based on recurrence relations
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,...an] 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,...an] 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(n1)+a(n2) 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(n1)+a(n2) for n > 2? Consider this problem for all fixed recurrence relations of any order. The same idea goes with a(n)=a(n1)+a(n2)+a(n3), for instance, except given the first three starting values a(1), a(2) and a(3). j Any help, comments, suggestions appreciated. Thank you. 
20171221, 00:42  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
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