OFFSET
3,3
COMMENTS
Rotations are counted only once, but reflections are considered different. For a polygon to be nondegenerate, the longest side must be shorter than the sum of the remaining sides (equivalently, shorter than n/2). These are row sums of A293819.
A formula is given in Section 6 of the East and Niles article.
The same article shows that a(n) is asymptotic to 2^n / n.
LINKS
Andrew Howroyd, Table of n, a(n) for n = 3..200
James East, Ron Niles, Integer polygons of given perimeter, arXiv:1710.11245 [math.CO], 2017.
FORMULA
a(n) = (Sum_{d|n} phi(n/d)*2^d)/n - 1 - 2^floor(n/2). - Andrew Howroyd, Nov 21 2017
EXAMPLE
There are 11 polygons having perimeter 7: 2 triangles (331, 322), 4 quadrilaterals (3211, 3121, 3112, 2221), 3 pentagons (31111, 22111, 21211), 1 hexagon (211111) and 1 heptagon (1111111).
MATHEMATICA
T[n_, k_] := DivisorSum[GCD[n, k], EulerPhi[#]*Binomial[n/#, k/#] &]/n - Binomial[Floor[n/2], k - 1];
a[n_] := Sum[T[n, k], {k, 3, n}]
Table[a[n], {n, 3, 40}] (* Jean-François Alcover, Jun 14 2018, after Andrew Howroyd *)
PROG
(PARI) a(n) = sumdiv(n, d, eulerphi(n/d)*2^d)/n - 1 - 2^floor(n/2); \\ Andrew Howroyd, Nov 21 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
James East, Oct 16 2017
STATUS
approved