SearcharxivSearch

arXiv subjects

Srikanth Cherukupally

Publications and source records attributed to Srikanth Cherukupally.

3 recordsLinked to original sources

A problem on the largest divisor $d$ of $N$ with $d\leq \sqrt{N}$

For a given number $N$, we consider the problem of computing two integers $1\leq r,f < N$ such that the set $$\mathcal{X}(N,r,f) = \{(a+b)-(f+\frac{Nr+1}{f}): ab=Nr\}$$ consists only of positive integers. Computing a solution to the problem is equivalent to finding a pair $(r,f)$ satisfying $l(Nr) < f \leq l(Nr+1)$, where $l(x)$ is the largest divisor of $x$ bounded by $\sqrt{x}$. This requires factoring both $Nr$ and $Nr+1$. We present a simple randomized algorithm that - avoiding factoring - computes pairs $(r,f)$. We give an exact formula for the total number of possible pairs $(r,f)$, and with the aid of empirical data we estimate that the ratio $$\frac{ϕ(N)-2}{|\mathfrak{F}(N)|}$$ is approximately about $c*\log \log N$. Here, $\mathfrak{F}(N)$ is the set of unique $r$ appearing among all possible pairs $(r,f)$, $ϕ(.)$ is the Euler's Totient function, and $c$ is a constant equal to 2 for prime $N$ and oscillates much for composite $N$. As a separate and independent case, we study the same problem of computing $(r,f)$ with $r>N$. We present a procedure to find such an $r$, which requires finding the least prime in an arithmetic progression.

math.NT

On the size of $\{a: 1\leq a<n, n|a^2-1, a|n^2-1\}$ for number $n$

For number $n>1$, let $\mathcal{A}(n) = \{1\leq a<n: n|a^2-1, a|n^2-1 \}$. We show that the size of $\mathcal{A}(n)$ is connected to a property concerning integer evaluations of Fibonacci-like polynomials. In the process, we prove that $|\mathcal{A}(n)|< \log_2 n$, and establish the average value of $|\mathcal{A}(n)|$ to be a little above $2$, asymptotically. But the empirical data up to $n<10^7$ indicate that $|\mathcal{A}(n)|\leq 3$, proving which is left as an open issue.

math.NT