Second-order linear recurrences having arbitrarily large defect modulo p

Kód výsledku v IS VaVaI

<a href="https://www.isvavai.cz/riv?ss=detail&h=RIV%2F67985840%3A_____%2F21%3A00542416" target="_blank" >RIV/67985840:_____/21:00542416 - isvavai.cz</a>

Výsledek na webu

<a href="http://hdl.handle.net/11104/0319826" target="_blank" >http://hdl.handle.net/11104/0319826</a>

DOI - Digital Object Identifier

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Jazyk výsledku

angličtina

Název v původním jazyce

Second-order linear recurrences having arbitrarily large defect modulo p

Popis výsledku v původním jazyce

Let (w) = w(a, b) denote the second-order linear recurrence satisfying wn+2 =awn+1 +bwn, where w0, w1, and a are integers, b = 1, and D = a2 +4b is the discriminant. We distinguish the Lucas sequences u(a, b) and v(a, b) with initial terms u0 = 0, u1 = 1, and v0 = 2, v1 = a, respectively. Let p be a prime. Given the recurrence w(a, b), let w(p), called the defect of w(a, b) modulo p, denote the number of residues not appearing in (w) modulo p. It is known that for the recurrence w(a, 1), w(p) 1 if p > 7 and p - D. Given the xed recurrence w(a, 1), where w(a, 1) = u(a, 1) or v(a, 1), we will show that limp!1 w(p) = 1. Further, given the arbitrary recurrence w(a,????1), we will demonstrate that limp!1 w(p) = 1 and limp!1 w(p)=p 1 2. We will also prove that for the arbitrary recurrence w(a, 1), we have that lim supp!1 w(p)=p = 1.

Název v anglickém jazyce

Second-order linear recurrences having arbitrarily large defect modulo p

Popis výsledku anglicky

Let (w) = w(a, b) denote the second-order linear recurrence satisfying wn+2 =awn+1 +bwn, where w0, w1, and a are integers, b = 1, and D = a2 +4b is the discriminant. We distinguish the Lucas sequences u(a, b) and v(a, b) with initial terms u0 = 0, u1 = 1, and v0 = 2, v1 = a, respectively. Let p be a prime. Given the recurrence w(a, b), let w(p), called the defect of w(a, b) modulo p, denote the number of residues not appearing in (w) modulo p. It is known that for the recurrence w(a, 1), w(p) 1 if p > 7 and p - D. Given the xed recurrence w(a, 1), where w(a, 1) = u(a, 1) or v(a, 1), we will show that limp!1 w(p) = 1. Further, given the arbitrary recurrence w(a,????1), we will demonstrate that limp!1 w(p) = 1 and limp!1 w(p)=p 1 2. We will also prove that for the arbitrary recurrence w(a, 1), we have that lim supp!1 w(p)=p = 1.

Druh

J_imp - Článek v periodiku v databázi Web of Science

CEP obor

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OECD FORD obor

10101 - Pure mathematics

Projekt

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Návaznosti

I - Institucionalni podpora na dlouhodoby koncepcni rozvoj vyzkumne organizace

Rok uplatnění

2021

Kód důvěrnosti údajů

S - Úplné a pravdivé údaje o projektu nepodléhají ochraně podle zvláštních právních předpisů

Název periodika

Fibonacci Quarterly

ISSN

0015-0517

e-ISSN

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Svazek periodika

59

Číslo periodika v rámci svazku

2

Stát vydavatele periodika

US - Spojené státy americké

Počet stran výsledku

24

Strana od-do

108-131

Kód UT WoS článku

000652052900002

EID výsledku v databázi Scopus

2-s2.0-85114135406