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# User:Ethan Beihl

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

## Interests

Some of my mathematical interests:

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

## Memos to self

1. ${\displaystyle \psi (n)}$ in Cheb basis
2. ${\displaystyle \psi (p)}$ in Cheb basis
3. Constant coefficients of each
4. No. nonzero terms of each

## Portolan numbers and Chebyshev Polynomials

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

${\displaystyle {\frac {\sin \left({\frac {(n-i)\pi }{3n}}\right)\sin \left({\frac {(n-j)\pi }{3n}}\right)}{\sin \left({\frac {i\pi }{3n}}\right)\sin \left({\frac {j\pi }{3n}}\right)}}={\frac {\sin \left({\frac {(n-k)\pi }{3n}}\right)}{\sin \left({\frac {k\pi }{3n}}\right)}}}$,

where ${\displaystyle 1\leq i,j,k<\left\lfloor {\frac {n}{2}}\right\rfloor -1}$ are integers.

When ${\displaystyle 10|n}$, we get this pretty thing, which I think is new to the literature: ${\displaystyle {\frac {\sin \left({\frac {7\pi }{30}}\right)\sin \left({\frac {8\pi }{30}}\right)}{\sin \left({\frac {3\pi }{30}}\right)\sin \left({\frac {2\pi }{30}}\right)}}={\frac {\sin \left({\frac {9\pi }{30}}\right)}{\sin \left({\frac {\pi }{30}}\right)}}}$.

I conjecture that ${\displaystyle 10|n}$ are the only solutions, but I don't have proof yet. Solutions occur when ${\displaystyle \psi _{12n}(x)|\left(T_{2i+n}(x)T_{2j+n}(x)T_{2k-3n}(x)-T_{2i-3n}(x)T_{2j-3n}(x)T_{2k+n}(x)\right)}$, Where ${\displaystyle T_{m}(x)}$ is a Chebyshev polynomial of the first kind and ${\displaystyle \psi _{m}(x)}$ is the minimal polynomial (over the integers) of ${\displaystyle \cos(2\pi /m)}$.

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 ${\displaystyle P(n,k)}$ is equal to the following:

${\displaystyle 1+\sum _{i=2}^{n}{ia_{i}}+\sum _{i=1}^{n-2}{(i+2)(b_{i}-c_{i})}-nk-I(n,k)-C(n,k)}$

where

• ${\displaystyle I(n,k)}$ is the number of intersection points within the polygon,
• ${\displaystyle C(n,k)}$ is the number of lines that end at opposite corners instead of on sides (I call these "vanishing endpoints"),
• ${\displaystyle a_{i}}$ for fixed (n,k) is the number of points on the interior where exactly i lines concur (${\displaystyle \sum _{i=2}^{n}{a_{i}}=I(n,k)}$),
• ${\displaystyle b_{i}}$ 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)
• ${\displaystyle c_{i}}$ is the number of vanishing endpoints where, in addition, i lines concur. (${\displaystyle \sum _{i=1}^{n}{c_{i}}=C(n,k)}$).

For the case ${\displaystyle n=4}$, ${\displaystyle b_{i},c_{i}=0}$ for ${\displaystyle i>1}$. It would be nice to know for which other n this is the case, as it simplifies the calculation immensely.