scholarly journals Generalized Lucas numbers of the form 3 × 2^m

2021 ◽  
Vol 27 (2) ◽  
pp. 129-136
Author(s):  
Salah Eddine Rihane ◽  
◽  
Chefiath Awero Adegbindin ◽  
Alain Togbé ◽  
◽  
...  
Keyword(s):  
Diophantine Equation ◽  
Lucas Numbers ◽  
Lucas Sequence ◽  
Generalized Lucas Numbers ◽  
Positive Integers

For an integer $k\geq 2$, let $(L_n^{(k)})_n$ be the k-generalized Lucas sequence which starts with $0,\ldots,0,2,1$ (k terms) and each term afterwards is the sum of the k preceding terms. In this paper, we look the k-generalized Lucas numbers of the form $3\times 2^m$ i.e. we study the Diophantine equation $L^{(k)}_n = 3\times 2^m$ in positive integers n, k, m with $k \geq 2$.

scholarly journals On the Sum of Powers of Two k-Fibonacci Numbers which Belongs to the Sequence of k-Lucas Numbers

2016 ◽  
Vol 67 (1) ◽  
pp. 41-46
Author(s):  
Pavel Trojovský
Keyword(s):  
Recurrence Relation ◽  
Initial Conditions ◽  
Fibonacci Numbers ◽  
Fibonacci Sequence ◽  
Lucas Numbers ◽  
Initial Values ◽  
Lucas Sequence ◽  
Positive Integers

Abstract Let k ≥ 1 and denote (Fk,n)n≥0, the k-Fibonacci sequence whose terms satisfy the recurrence relation Fk,n = kFk,n−1 +Fk,n−2, with initial conditions Fk,0 = 0 and Fk,1 = 1. In the same way, the k-Lucas sequence (Lk,n)n≥0 is defined by satisfying the same recurrence relation with initial values Lk,0 = 2 and Lk,1 = k. These sequences were introduced by Falcon and Plaza, who showed many of their properties, too. In particular, they proved that Fk,n+1 + Fk,n−1 = Lk,n, for all k ≥ 1 and n ≥ 0. In this paper, we shall prove that if k ≥ 1 and $F_{k,n + 1}^s + F_{k,n - 1}^s \in \left( {L_{k,m} } \right)_{m \ge 1} $ for infinitely many positive integers n, then s =1.

scholarly journals Powers in products of terms of Pell's and Pell–Lucas Sequences

2015 ◽  
Vol 11 (04) ◽  
pp. 1259-1274 ◽  
Author(s):  
Jhon J. Bravo ◽  
Pranabesh Das ◽  
Sergio Guzmán ◽  
Shanta Laishram
Keyword(s):  
General Result ◽  
Diophantine Equation ◽  
Initial Condition ◽  
Lucas Sequences ◽  
Lucas Sequence ◽  
Positive Integers

In this paper, we consider the usual Pell and Pell–Lucas sequences. The Pell sequence [Formula: see text] is given by the recurrence un = 2un-1 + un-2 with initial condition u0 = 0, u1 = 1 and its associated Pell–Lucas sequence [Formula: see text] is given by the recurrence vn = 2vn-1 + vn-2 with initial condition v0 = 2, v1 = 2. Let n, d, k, y, m be positive integers with m ≥ 2, y ≥ 2 and gcd (n, d) = 1. We prove that the only solutions of the Diophantine equation unun+d⋯un+(k-1)d = ym are given by u7 = 132 and u1u7 = 132 and the equation vnvn+d⋯vn+(k-1)d = ym has no solution. In fact, we prove a more general result.

scholarly journals On the Ternary Exponential Diophantine Equation Equating a Perfect Power and Sum of Products of Consecutive Integers

Mathematics ◽  
2021 ◽  
Vol 9 (15) ◽  
pp. 1813
Author(s):  
S. Subburam ◽  
Lewis Nkenyereye ◽  
N. Anbazhagan ◽  
S. Amutha ◽  
M. Kameswari ◽  
...  
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Upper Bound ◽  
Exponential Diophantine Equation ◽  
Research Article ◽  
All Solutions ◽  
Sum Of Products ◽  
Consecutive Integers ◽  
Sums Of Products ◽  
Positive Integers

Consider the Diophantine equation yn=x+x(x+1)+⋯+x(x+1)⋯(x+k), where x, y, n, and k are integers. In 2016, a research article, entitled – ’power values of sums of products of consecutive integers’, primarily proved the inequality n= 19,736 to obtain all solutions (x,y,n) of the equation for the fixed positive integers k≤10. In this paper, we improve the bound as n≤ 10,000 for the same case k≤10, and for any fixed general positive integer k, we give an upper bound depending only on k for n.

scholarly journals A NOTE ON THE DIOPHANTINE EQUATION

2013 ◽  
Vol 89 (2) ◽  
pp. 316-321 ◽  
Author(s):  
MOU JIE DENG
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Pythagorean Triple ◽  
Positive Integers

AbstractLet $(a, b, c)$ be a primitive Pythagorean triple satisfying ${a}^{2} + {b}^{2} = {c}^{2} . $ In 1956, Jeśmanowicz conjectured that for any given positive integer $n$ the only solution of $\mathop{(an)}\nolimits ^{x} + \mathop{(bn)}\nolimits ^{y} = \mathop{(cn)}\nolimits ^{z} $ in positive integers is $x= y= z= 2. $ In this paper, for the primitive Pythagorean triple $(a, b, c)= (4{k}^{2} - 1, 4k, 4{k}^{2} + 1)$ with $k= {2}^{s} $ for some positive integer $s\geq 0$, we prove the conjecture when $n\gt 1$ and certain divisibility conditions are satisfied.

SOME NOTES ON A GENERALIZED VERSION OF PYTHAGOREAN TRIPLES

2020 ◽  
Vol 4 (2) ◽  
pp. 103
Author(s):  
Leomarich F Casinillo ◽  
Emily L Casinillo
Keyword(s):  
Diophantine Equation ◽  
Pythagorean Triples ◽  
Pythagorean Triple ◽  
Positive Integers

A Pythagorean triple is a set of three positive integers a, b and c that satisfy the Diophantine equation a^2+b^2=c^2. The triple is said to be primitive if gcd(a, b, c)=1 and each pair of integers and  are relatively prime, otherwise known as non-primitive. In this paper, the generalized version of the formula that generates primitive and non-primitive Pythagorean triples that depends on two positive integers  k and n, that is, P_T=(a(k, n), b(k, n), c(k, n)) were constructed. Further, we determined the values of  k and n that generates primitive Pythagorean triples and give some important results.

A Note on the Diophantine Equation x2 + y6 = ze, e ≥ 4

2012 ◽  
Vol 55 (2) ◽  
pp. 435-440
Author(s):  
Konstantine Zelator
Keyword(s):  
Diophantine Equation ◽  
Positive Integers

AbstractWe consider the diophantine equation x2 + y6 = ze, e ≥ 4. We show that, when e is a multiple of 4 or 6, this equation has no solutions in positive integers with x and y relatively prime. As a corollary, we show that there exists no primitive Pythagorean triangle one of whose leglengths is a perfect cube, while the hypotenuse length is an integer square.

Reductions of Hilbert's tenth problem

1958 ◽  
Vol 23 (2) ◽  
pp. 183-187 ◽  
Author(s):  
Martin Davis ◽  
Hilary Putnam
Keyword(s):  
Diophantine Equation ◽  
Hilbert's Tenth Problem ◽  
Recursively Enumerable ◽  
Hilbert’S Tenth Problem ◽  
Essential Change ◽  
Recursive Predicate ◽  
Image Position ◽  
Positive Integers

Hilbert's tenth problem is to find an algorithm for determining whether or not a diophantine equation possesses solutions. A diophantine predicate (of positive integers) is defined to be one of the formwhere P is a polynomial with integral coefficients (positive, negative, or zero). Previous work has considered the variables as ranging over nonnegative integers; but we shall find it more useful here to restrict the range to positive integers, no essential change being thereby introduced. It is clear that the recursive unsolvability of Hilbert's tenth problem would follow if one could show that some non-recursive predicate were diophantine. In particular, it would suffice to show that every recursively enumerable predicate is diophantine. Actually, it would suffice to prove far less.

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