scholarly journals Positivity of third order linear recurrence sequences

2009 ◽  
Vol 157 (15) ◽  
pp. 3239-3248 ◽  
Author(s):  
Vichian Laohakosol ◽  
Pinthira Tangsupphathawat
Keyword(s):  
Linear Recurrence ◽  
Third Order ◽  
Order Linear ◽  
Recurrence Sequences

scholarly journals Decomposable polynomials in second order linear recurrence sequences

2018 ◽  
Vol 159 (3-4) ◽  
pp. 321-346 ◽  
Author(s):  
Clemens Fuchs ◽  
Christina Karolus ◽  
Dijana Kreso
Keyword(s):  
Second Order ◽  
Linear Recurrence ◽  
Order Linear ◽  
Recurrence Sequences

scholarly journals General Linear Recurrence Sequences and Their Convolution Formulas

Axioms ◽  
2019 ◽  
Vol 8 (4) ◽  
pp. 132 ◽  
Author(s):  
Paolo Emilio Ricci ◽  
Pierpaolo Natalini
Keyword(s):  
Generating Functions ◽  
Recurrence Relations ◽  
Second Order ◽  
General Linear ◽  
Linear Recurrence ◽  
Order Linear ◽  
Matrix Polynomials ◽  
Recurrence Sequences ◽  
Straightforward Extension

We extend a technique recently introduced by Chen Zhuoyu and Qi Lan in order to find convolution formulas for second order linear recurrence polynomials generated by 1 1 + a t + b t 2 x . The case of generating functions containing parameters, even in the numerator is considered. Convolution formulas and general recurrence relations are derived. Many illustrative examples and a straightforward extension to the case of matrix polynomials are shown.

scholarly journals Distribution of some quadratic linear recurrence sequences modulo 1

2014 ◽  
Vol 30 (1) ◽  
pp. 79-86
Author(s):  
ARTURAS DUBICKAS ◽  
Keyword(s):  
Limit Point ◽  
Fibonacci Sequence ◽  
Second Order ◽  
Linear Recurrence ◽  
Order Linear ◽  
Recurrence Sequences

We show that if a is an even integer then for every ξ ∈ R the smallest limit point of the sequence ||ξan||∞n=1 does not exceed |a|/(2|a| + 2) and this bound is best possible in the sense that for some ξ this constant cannot be improved. Similar (best possible) bound is also obtained for the smallest limit point of the sequence ||ξxn||∞n=1, where (xn)∞n=1 satisfies the second order linear recurrence xn = axn−1 + bxn−2 with a, b ∈ N satisfying a > b. For the Fibonacci sequence (Fn)∞n=1 our result implies that supξ∈R lim infn→∞ ||ξFn|| = 1/5, and e.g., in case when a > 3 is an odd integer, b = 1 and θ := a/2 + p a 2/4 + 1 it shows that supξ∈R lim infn→∞ ||ξθn|| = (a − 1)/2a.

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