For any positive integer $k\ge 2$, let ${\left(P_n^{\left(k\right)}\right)}_{n\ge 2-k}$ be the $k$-generalized Pell sequence which starts with $0,\cdots ,0,1$ ($k$ terms) with the linear recurrence $$P_n^{\left(k\right)}=2P_{n-1}^{\left(k\right)}+P_{n-2}^{\left(k\right)}+\cdots +P_{n-k}^{\left(k\right)}\text{for}n\ge 2.$$ Let ${\left(N_n\right)}_{n\ge 0}$ be Narayana’s sequence given by $$N_0=N_1=N_2=1\text{and}{N}_{n+3}=N_{n+2}+N_n.$$ The purpose of this paper is to determine all $k$-Pell numbers which are sums of two Narayana’s numbers. More precisely, we study the Diophantine equation $$P_p^{\left(k\right)}=N_n+N_m$$ in nonnegative integers $k$, $p$, $n$ and $m$.

Let $\left\{U_n\right\}=\left\{U_n(P,Q)\right\}$ and $\left\{V_n\right\}=\left\{V_n(P,Q)\right\}$ be the Lucas sequences of the first and second kind respectively at the parameters $P\ge 1$ and $Q\in \{-1,1\}$. In this paper, we provide a technique for characterizing the solutions of the so-called Bartz-Marlewski equation $$x^2-3xy+y^2+x=0,$$ where $(x,y)=(U_i,U_j)$ or $(V_i,V_j)$ with $i$, $j\ge 1$. Then, the procedure of this technique is applied to completely resolve this equation with certain values of such parameters.

Let ${\left(L_n\right)}_{n\ge 0}$ be the Lucas sequence. We show that the Diophantine equation $L_n-L_m=M_k$ has only the nonnegative integer solutions $(n,m,k)=(2,0,1)$, $(3,1,2)$, $(3,2,1)$, $(4,3,2)$, $(5,3,3)$, $(6,2,4)$, $(6,5,3)$ where $M_k=2^k-1$ is the $k$th Mersenne number and $n>m$.

Let $k\ge 2$ and let ${\left(P_n^{\left(k\right)}\right)}_{n\ge 2-k}$ be the $k$-generalized Pell sequence defined by $$P_n^{\left(k\right)}=2P_{n-1}^{\left(k\right)}+P_{n-2}^{\left(k\right)}+\cdots +P_{n-k}^{\left(k\right)}$$ for $n\ge 2$ with initial conditions $$P_{-(k-2)}^{\left(k\right)}=P_{-(k-3)}^{\left(k\right)}=\cdots =P_{-1}^{\left(k\right)}=P_0^{\left(k\right)}=0,P_1^{\left(k\right)}=1.$$ In this study, we handle the equation $P_n^{\left(k\right)}=y^m$ in positive integers $n$, $m$, $y$, $k$ such that $k,y\ge 2,$ and give an upper bound on $n.$ Also, we will show that the equation $P_n^{\left(k\right)}=y^m$ with $2\le y\le 1000$ has only one solution given by $P_7^{\left(2\right)}={13}^2.$

We consider $N$, the number of solutions $(x,y,u,v)$ to the equation ${(-1)}^{u}r{a}^x+{(-1)}^{v}s{b}^y=c$ in nonnegative integers $x,y$ and integers $u,v\in \{0,1\}$, for given integers $a\>1$, $b\>1$, $c\>0$, $r\>0$ and $s\>0$. When $gcd(ra,sb)=1$, we show that $N\le 3$ except for a finite number of cases all of which satisfy $max(a,b,r,s,x,y)\<2·{10}^{15}$ for each solution; when $gcd(a,b)\>1$, we show that $N\le 3$ except for three infinite families of exceptional cases. We find several different ways to generate an infinite number of cases giving $N=3$ solutions.

Let $F_n$ denote the $n^{th}$ term of the Fibonacci sequence. In this paper, we investigate the Diophantine equation $F_1^p+2F_2^p+\cdots +k{F}_k^p=F_n^q$ in the positive integers $k$ and $n$, where $p$ and $q$ are given positive integers. A complete solution is given if the exponents are included in the set $\{1,2\}$. Based on the specific cases we could solve, and a computer search with $p,q,k\le 100$ we conjecture that beside the trivial solutions only $F_8=F_1+2F_2+3F_3+4F_4$, $F_4^2=F_1+2F_2+3F_3$, and $F_4^3=F_1^3+2F_2^3+3F_3^3$ satisfy the title equation.

For an integer $k\ge 2$, let ${\left(n\right)}_n$ be the $k-$generalized Pell sequence which starts with $0,...,0,1$ ($k$ terms) and each term afterwards is given by the linear recurrence $n=2n-1+n-2+\cdots +n-k$. In this paper, we find all $k$-generalized Pell numbers with only one distinct digit (the so-called repdigits). Some interesting estimations involving generalized Pell numbers, that we believe are of independent interest, are also deduced. This paper continues a previous work that searched for repdigits in the usual Pell sequence ${\left(P_n^{\left(2\right)}\right)}_n$. ...

Let ${\left(G_n\right)}_{n\ge 1}$ be a binary linear recurrence sequence that is represented by the Lucas sequences of the first and second kind, which are $\left\{U_n\right\}$ and $\left\{V_n\right\}$, respectively. We show that the Diophantine equation $G_n=B·(g^{lm}-1)/(g^l-1)$ has only finitely many solutions in $n,m\in {\mathbb{Z}}^{+}$, where $g\ge 2$, $l$ is even and $1\le B\le g^l-1$. Furthermore, these solutions can be effectively determined by reducing such equation to biquadratic elliptic curves. Then, by a result of Baker (and its best improvement due to Hajdu and Herendi) related to the bounds of the integral...

Let $k\ge 5$ be an odd integer and $\eta$ be any given real number. We prove that if ${\lambda }_1$, ${\lambda }_2$, ${\lambda }_3$, ${\lambda }_4$, $\mu$ are nonzero real numbers, not all of the same sign, and ${\lambda }_1/{\lambda }_2$ is irrational, then for any real number $\sigma$ with $0<\sigma <1/\left(8\vartheta \right(k\left)\right)$, the inequality $$|{\lambda }_1{p}_1^2+{\lambda }_2{p}_2^2+{\lambda }_3{p}_3^2+{\lambda }_4{p}_4^2+\mu p_5^k+\eta |<{\left(\underset{1\le j\le 5}{max}{p}_j\right)}^{-\sigma }$$ has infinitely many solutions in prime variables $p_1,p_2,\cdots ,p_5$, where $\vartheta \left(k\right)=3\times 2^{(k-5)/2}$ for $k=5,7,9$ and $\vartheta \left(k\right)=[(k^2+2k+5)/8]$ for odd integer $k$ with $k\ge 11$. This improves a recent result in W. Ge, T. Wang (2018).

Suppose that ${\lambda }_1,{\lambda }_2,{\lambda }_3,{\lambda }_4$ are nonzero real numbers, not all negative, $\delta >0$, $𝒱$ is a well-spaced set, and the ratio ${\lambda }_1/{\lambda }_2$ is algebraic and irrational. Denote by $E(𝒱,N,\delta )$ the number of $v\in 𝒱$ with $v\le N$ such that the inequality $$|{\lambda }_1{p}_1^2+{\lambda }_2{p}_2^3+{\lambda }_3{p}_3^4+{\lambda }_4{p}_4^5-v|<v^{-\delta }$$ has no solution in primes $p_1$, $p_2$, $p_3$, $p_4$. We show that $$E(𝒱,N,\delta )\ll N^{1+2\delta -1/72+\epsilon }$$ for any $\epsilon >0$.

Let $W(m,n;\underline{\psi })$ denote the set of ${\psi }_1,...,{\psi }_n$–approximable points in ${ℝ}^{mn}$. The classical Khintchine–Groshev theorem assumes a monotonicity condition on the approximating functions $\underline{\psi }$. Removing monotonicity from the Khintchine–Groshev theorem is attributed to different authors for different cases of $m$ and $n$. It can not be removed for $m=n=1$ as Duffin–Schaeffer provided the counter example. We deal with the only remaining case $m=2$ and thereby remove all unnecessary conditions from the Khintchine–Groshev theorem. ...

For any positive integer $D$ which is not a square, let $(u_1,v_1)$ be the least positive integer solution of the Pell equation $u^2-D{v}^2=1,$ and let $h\left(4D\right)$ denote the class number of binary quadratic primitive forms of discriminant $4D$. If $D$ satisfies $2\nmid D$ and $v_{1}h\left(4D\right)\equiv 0(modD)$, then $D$ is called a singular number. In this paper, we prove that if $(x,y,z)$ is a positive integer solution of the equation $x^y+y^x=z^z$ with $2\mid z$, then maximum $max\{x,y,z\}<480000$ and both $x$, $y$ are singular numbers. Thus, one can possibly prove that the equation has no positive integer solutions...

Let ${\Phi }_1,...,{\Phi }_{k+1}$ (with $k\ge 1$) be vector fields of class $C^k$ in an open set $U{\subset }^{N+m}$, let $𝕄$ be a $N$-dimensional $C^k$ submanifold of $U$ and define $$𝕋:=\{z\in 𝕄:{\Phi }_1\left(z\right),...,{\Phi }_{k+1}\left(z\right)\in T_z𝕄\}$$ where $T_z𝕄$ is the tangent space to $𝕄$ at $z$. Then we expect the following property, which is obvious in the special case when $z_0$ is an interior point (relative to $𝕄$) of $𝕋$: If $z_0\in 𝕄$ is a $(N+k)$-density point (relative to $𝕄$) of $𝕋$ then all the iterated Lie brackets of order less or equal to $k$ $${\Phi }_{i_1}\left(z_0\right),[{\Phi }_{i_1},{\Phi }_{i_2}]\left(z_0\right),[[{\Phi }_{i_1},{\Phi }_{i_2}],{\Phi }_{i_3}]\left(z_0\right),...(h,i_h\le k+1)$$ belong to $T_{z_0}𝕄$. Such a property has been proved in [9] for $k=1$ and its proof in the...

The sequence of balancing numbers $\left(B_n\right)$ is defined by the recurrence relation $B_n=6B_{n-1}-B_{n-2}$ for $n\ge 2$ with initial conditions $B_0=0$ and $B_1=1.$ $B_n$ is called the $n$th balancing number. In this paper, we find all repdigits in the base $b,$ which are sums of four balancing numbers. As a result of our theorem, we state that if $B_n$ is repdigit in the base $b$ and has at least two digits, then $(n,b)=(2,5),(3,6)$. Namely, $B_2=6={\left(11\right)}_5$ and $B_3=35={\left(55\right)}_6.$