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Incomplete Tribonacci–Lucas Numbers and Polynomials [PDF]

open access: yesAdvances in Applied Clifford Algebras, 2014
In this paper, we define Tribonacci-Lucas polynomials and present Tribonacci-Lucas numbers and polynomials as a binomial sum. Then, we introduce incomplete Tribonacci-Lucas numbers and polynomials. In addition we derive recurrence relations, some properties and generating functions of these numbers and polynomials. Also, we find the generating function
Yilmaz, Nazmiye, Taskara, Necati
openaire   +2 more sources

On square Tribonacci Lucas numbers

open access: yesHacettepe Journal of Mathematics and Statistics, 2021
The Tribonacci-Lucas sequence {Sn}{Sn} is defined by the recurrence relation Sn+3=Sn+2+Sn+1+SnSn+3=Sn+2+Sn+1+Sn with S0=3, S1=1, S2=3.S0=3, S1=1, S2=3. In this note, we show that 11 is the only perfect square in Tribonacci-Lucas sequence for n≢1(mod32)n≢1(mod32) and n≢17(mod96).n≢17(mod96).
openaire   +3 more sources

Formulae of the Frobenius number in relatively prime three Lucas numbers [PDF]

open access: yesSongklanakarin Journal of Science and Technology (SJST), 2020
In this paper, we find the explicit formulae of the Frobenius number for numerical semigroups generated by relatively prime three Lucas numbers 2 , L L i i and Lil  for given integers i ≥ 3, l ≥ 4 .
Ratchanok Bokaew   +2 more
doaj   +1 more source

Solution to a pair of linear, two-variable, Diophantine equations with coprime coefficients from balancing and Lucas-balancing numbers [PDF]

open access: yesNotes on Number Theory and Discrete Mathematics, 2023
Let Bₙ and Cₙ be the n-th balancing and Lucas-balancing numbers, respectively. We consider the Diophantine equations ax + by = (1/2)(a - 1)(b - 1) and 1 + ax + by = (1/2)(a - 1)(b - 1) for (a,b) ∈ {(Bₙ,Bₙ₊₁), (B₂ₙ₋₁,B₂ₙ₊₁), (Bₙ,Cₙ), (Cₙ,Cₙ₊₁)} and ...
R. K. Davala
doaj   +1 more source

Mersenne-Lucas hybrid numbers

open access: yesMathematica Montisnigri, 2021
We introduce Mersenne-Lucas hybrid numbers. We give the Binet formula, the generating function, the sum, the character, the norm and the vector representation of these numbers. We find some relations among Mersenne-Lucas hybrid numbers, Jacopsthal hybrid numbers, Jacopsthal-Lucas hybrid numbers and Mersenne hybrid numbers.
Engin Özkan, Mine Uysal
openaire   +1 more source

A Study on Fibonacci and Lucas Bihypernomials

open access: yesDiscussiones Mathematicae - General Algebra and Applications, 2022
The bihyperbolic numbers are extension of hyperbolic numbers to four dimensions. In this paper we introduce and study the Fibonacci and Lucas bihypernomials, i.e., polynomials, which are a generalization of the bihyperbolic Fibonacci numbers and the ...
Szynal-Liana Anetta, Włoch Iwona
doaj   +1 more source

On Bicomplex Pell and Pell-Lucas Numbers

open access: yesCommunications in Advanced Mathematical Sciences, 2018
In this paper, bicomplex Pell and bicomplex Pell-Lucas numbers are defined. Also, negabicomplex Pell and negabicomplex Pell-Lucas numbers are given. Some algebraic properties of bicomplex Pell and bicomplex Pell-Lucas numbers which are connected between ...
Fügen Torunbalcı Aydın
doaj   +1 more source

Identities for Fibonacci and Lucas numbers [PDF]

open access: yesNotes on Number Theory and Discrete Mathematics, 2023
In this paper several new identities are given for the Fibonacci and Lucas numbers. This is accomplished by by solving a class of non-homogeneous, linear recurrence relations.
George Grossman   +2 more
doaj   +1 more source

Incomplete Bivariate Fibonacci and Lucas 𝑝-Polynomials

open access: yesDiscrete Dynamics in Nature and Society, 2012
We define the incomplete bivariate Fibonacci and Lucas 𝑝-polynomials. In the case 𝑥=1, 𝑦=1, we obtain the incomplete Fibonacci and Lucas 𝑝-numbers. If 𝑥=2, 𝑦=1, we have the incomplete Pell and Pell-Lucas 𝑝-numbers.
Dursun Tasci   +2 more
doaj   +1 more source

Binomial coefficients, Catalan numbers and Lucas quotients

open access: yes, 2010
Let $p$ be an odd prime and let $a,m$ be integers with $a>0$ and $m \not\equiv0\pmod p$. In this paper we determine $\sum_{k=0}^{p^a-1}\binom{2k}{k+d}/m^k$ mod $p^2$ for $d=0,1$; for example, $$\sum_{k=0}^{p^a-1}\frac{\binom{2k}k}{m^k}\equiv\left(\frac{m^
C. J. Smyth   +13 more
core   +3 more sources

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