scholarly journals A remark about the positivity problem of fourth order linear recurrence sequences

Author(s):  
Vichian Laohakosol ◽  
Pinthira Tangsupphathawat
2012 ◽  
Vol 128 (1) ◽  
pp. 133-142 ◽  
Author(s):  
Pinthira Tangsupphathawat ◽  
Narong Punnim ◽  
Vichian Laohakosol

2018 ◽  
Vol 159 (3-4) ◽  
pp. 321-346 ◽  
Author(s):  
Clemens Fuchs ◽  
Christina Karolus ◽  
Dijana Kreso

Symmetry ◽  
2019 ◽  
Vol 11 (12) ◽  
pp. 1476 ◽  
Author(s):  
Lan Qi ◽  
Zhuoyu Chen

In this paper, we introduce the fourth-order linear recurrence sequence and its generating function and obtain the exact coefficient expression of the power series expansion using elementary methods and symmetric properties of the summation processes. At the same time, we establish some relations involving Tetranacci numbers and give some interesting identities.


Axioms ◽  
2019 ◽  
Vol 8 (4) ◽  
pp. 132 ◽  
Author(s):  
Paolo Emilio Ricci ◽  
Pierpaolo Natalini

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.


2014 ◽  
Vol 30 (1) ◽  
pp. 79-86
Author(s):  
ARTURAS DUBICKAS ◽  

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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