# [seqfan] Re: Comtets triangles in convex polygons

Neil Sloane njasloane at gmail.com
Thu Nov 29 15:30:31 CET 2018

```In my copy of Comtet's 1974 book I have a pencilled note in the
margin saying "not integral"

All the other sequences have A-numbers written next to them,
but because these were not integers I did not add them to the OEIS.

I also have the original 2-volume French edition,  Analyse combinatoire,
which has the same Problem 8,
but it does not have the sub-problem 8(5).

Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Email: njasloane at gmail.com

On Thu, Nov 29, 2018 at 5:14 AM Richard J. Mathar <mathar at mpia-hd.mpg.de>
wrote:

> Comtet's book of 1974 has a dubious expression on page 75,
> exercise 8(5), as spelled out in
> This deals with triangles inside convex polygons constructed by diagonals.
>
> What is the correct result? Do we just have to wrap the entire
> polynomial in a round() to get correct numbers, which would
> have been written as ||n_3(n^3+18*n^2+43*n+60)/6!|| with two double
> bars in the book, to get 1, 7, 31, 97, 247,... for n>=1? This is not
> correct, because for n<=3 there are no diagonals in the polygons, so there
> are
> none of these triangles.
> That sort of error could be guessed from a similar observation in A321988.
> Or are there other typos, as in A321986?
> I'd expect to be 4 of these triangles in the 4-gon, defined by the
> two ways of defining a diagonal, and each of these has such a triangle
> on either side. This naively produces n*(n-3)/2 = A000096(n-3) for n>4,
> since there are n points for the start of a diagonal, n-3 candidates for
> the other end of the diagonal, a factor of 2 for the two sides of
> the diagonal if n=4, and a division through 2 for the double counting
> of the two directions of the diagonal. This is obviously not what