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

To add:

1. $\psi (n)$ in Cheb basis
2. $\psi (p)$ 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):

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

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

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

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

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 $P(n,k)$ is equal to the following:

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

where

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

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