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Pranabesh Das

Publications and source records attributed to Pranabesh Das.

8 recordsLinked to original sources
arXiv2025

Products of Tribonacci Numbers that are the Products of Factorials

In 2014 Marques and Lengyel gave all of the solutions to the equation $T_n=m!$, where $T_n$ is the $n$th term of the Tribonacci sequence $0,1,1,2,4,7,13,24,\ldots$. In 2023 Alahmadi and Luca generalized their result to the equation $T_n=m_1!m_2!\cdots m_k!$ for every $k\in\mathbb{N}$, where $m_1\leq m_2\leq\ldots\leq m_k$ listing all the solutions to this equation. Here we generalize these results further and give all the solutions to $T_nT_{n+1}T_{n+2}\cdots T_{n+r}=m_1!m_2!\cdots m_k!$ and $ |T_{-n}T_{-n-1}T_{-n-2}\cdots T_{-n-r}|=m_1!m_2!\cdots m_k!$ for every $n,r\in\mathbb{N}$, where $m_1\leq m_2\leq\ldots\leq m_k$.

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arXiv2025

Repdigits as Product of Consecutive Shifted Tribonacci Numbers

A repdigit is a positive integer that has only one distinct digit in its decimal expansion, i.e., a number has the form $d(10^m-1)/9$ for some $m\geq 1$ and $1 \leq d \leq 9$. Let $\left(T_n\right)_{n\ge0}$ be the Tribonacci sequence. This paper deals with the presence of repdigits in the products of consecutive shifted Tribonacci numbers.

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arXiv2025

On Pell numbers representable as product of two generalized Fibonacci numbers

A generalization of the well-known Fibonacci sequence is the $k$-Fibonacci sequence with some fixed integer $k\ge 2$. The first $k$ terms of this sequence are $0,0, \ldots, 1$, and each term afterwards is the sum of the preceding $k$ terms. In this paper, we find all Pell numbers that can be written as a product of two $k$-Fibonacci numbers. The proof of our main theorem uses lower bounds for linear forms in logarithms, properties of continued fractions, and a variation of a result of Dujella and Peth\H{o} in Diophantine approximation. This work generalizes a prior result of Alekseyev which dealt with determining the intersection of the Fibonacci and Pell sequences, a work of Ddamulira, Luca and Rakotomalala who searched for Pell numbers which are products of two Fibonacci numbers, and a result of Bravo, G\'omez, and Herrera, who found all Pell numbers appearing in the $k$-Fibonacci sequence.

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arXiv2023

Rational solutions to the Variants of Erdős- Selfridge superelliptic curves

For the superelliptic curves of the form $$ (x+1) \cdots(x+i-1)(x+i+1)\cdots (x+k)=y^\ell$$ with $x,y \in \mathbb{Q}$, $y\neq 0$, $k \geq 3$, $1\leq i\leq k$, $\ell \geq 2,$ a prime, Das, Laishram, Saradha, and Edis showed that the superelliptic curve has no rational points for $\ell\geq e^{3^k}$. In fact, the double exponential bound, obtained in these papers is far from reality. In this paper, we study the superelliptic curves for small values of $k$. In particular, we explicitly solve the above equation for $4 \leq k \leq 8.$

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arXiv2020

Repdigits in Narayana's Cows Sequence and their Consequences

Narayana's cows sequence satisfies the third-order linear recurrence relation $N_n=N_{n-1}+N_{n-3}$ for $n \geq 3$ with initial conditions $N_0=0$ and $N_1=N_2=1$. In this paper, we study $b$-repdigits which are sums of two Narayana numbers. We explicitly determine these numbers for the bases $2\le b\leq100$ as an illustration. We also obtain results on the existence of Mersenne prime numbers, 10-repdigits, and numbers with distinct blocks of digits in the Narayana sequence. The proof of our main theorem uses lower bounds for linear forms in logarithms and a version of the Baker-Davenport reduction method in Diophantine approximation.

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arXiv2020

Perfect powers in sum of three fifth powers

In this paper we determine the perfect powers that are sums of three fifth powers in an arithmetic progression. More precisely, we completely solve the Diophantine equation $$ (x-d)^5 + x^5 + (x + d)^5 = z^n,~n\geq 2, $$ where $d,x,z \in \mathbb{Z}$ and $d = 2^a5^b$ with $a,b\geq 0$.

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arXiv2017

Perfect powers in alternating sum of consecutive cubes

In this paper, we consider the problem about finding out perfect powers in an alternating sum of consecutive cubes. More precisely, we completely solve the Diophantine equation $(x+1)^3 - (x+2)^3 + \cdots - (x + 2d)^3 + (x + 2d + 1)^3 = z^p$, where $p$ is prime and $x,d,z$ are integers with $1 \leq d \leq 50$.

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