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# User:Ethan Beihl

Mathematics student and teacher, sequence enthusiast, and Yellow-Pig-seeker.

## Contents

## Interests

Some of my mathematical interests:

- Finite group theory
- Linear algebra
- ALife
- Recreational math
- Programming in Mathematica

## Memos to self

To add:

- in Cheb basis
- in Cheb basis
- Constant coefficients of each
- No. nonzero terms of each

## Portolan numbers and Chebyshev Polynomials

Currently seeking solutions to the following equation (see A277402):

,

where are integers.

When , we get this pretty thing, which I think is new to the literature: .

I conjecture that are the only solutions, but I don't have proof yet. Solutions occur when , Where is a Chebyshev polynomial of the first kind and is the minimal polynomial (over the integers) of .

**Update 26 Nov.**
I've now proven the the conjecture, but using the Conway-Jones theory of trigonometric diophantine equations with no reference to Chebyshev polynomials. I must say, the eventual proof was kind of a let-down; I think the Chebyshev angle would show something deeper.

### n-Portolan numbers

The portolan number is equal to the following:

where

- is the number of intersection points within the polygon,
- is the number of lines that end at opposite corners instead of on sides (I call these "vanishing endpoints"),
- for fixed (n,k) is the number of points on the interior where exactly i lines concur (),
- is the number of points on *border* of the polygon where exactly i lines concur. (i=1 means one line intersecting with an edge or corner)
- is the number of vanishing endpoints where, in addition, i lines concur. ().

For the case , for . It would be nice to know for which other n this is the case, as it simplifies the calculation immensely.