Papers
Topics
Authors
Recent
Detailed Answer
Quick Answer
Concise responses based on abstracts only
Detailed Answer
Well-researched responses based on abstracts and relevant paper content.
Custom Instructions Pro
Preferences or requirements that you'd like Emergent Mind to consider when generating responses
Gemini 2.5 Flash
Gemini 2.5 Flash 27 tok/s
Gemini 2.5 Pro 46 tok/s Pro
GPT-5 Medium 23 tok/s Pro
GPT-5 High 29 tok/s Pro
GPT-4o 70 tok/s Pro
Kimi K2 117 tok/s Pro
GPT OSS 120B 459 tok/s Pro
Claude Sonnet 4 34 tok/s Pro
2000 character limit reached

Strong arithmetic property of certain Stern polynomials (1909.10844v1)

Published 24 Sep 2019 in math.NT

Abstract: Let $B_{n}(t)$ be the $n$th Stern polynomial, i.e., the $n$th term of the sequence defined recursively as $B_{0}(t)=0, B_{1}(t)=1$ and $B_{2n}(t)=tB_{n}(t), B_{2n+1}(t)=B_{n}(t)+B_{n-1}(t)$ for $n\in\N$. It is well know that $i$th coefficient in the polynomial $B_{n}(t)$ counts the number of hyperbinary representations of $n-1$ containing exactly $i$ digits 1. In this note we investigate the existence of odd solutions of the congruence \begin{equation*} B_{n}(t)\equiv 1+rt\frac{t{e(n)}-1}{t-1}\pmod{m}, \end{equation*} where $m\in\N_{\geq 2}$ and $r\in{0,\ldots,m-1}$ are fixed and $e(n)=\op{deg}B_{n}(t)$. We prove that for $m=2$ and $r\in{0,1}$ and for $m=3$ and $r=0$, there are infinitely many odd numbers $n$ satisfying the above congruence. We also present results of some numerical computations.

Summary

We haven't generated a summary for this paper yet.

List To Do Tasks Checklist Streamline Icon: https://streamlinehq.com

Collections

Sign up for free to add this paper to one or more collections.

Dice Question Streamline Icon: https://streamlinehq.com

Follow-Up Questions

We haven't generated follow-up questions for this paper yet.

Authors (1)