A293821 Number of integer-sided quadrilaterals having perimeter n, modulo rotations but not reflections.

1, 1, 2, 4, 6, 10, 12, 20, 23, 35, 38, 56, 60, 84, 88, 120, 125, 165, 170, 220, 226, 286, 292, 364, 371, 455, 462, 560, 568, 680, 688, 816, 825, 969, 978, 1140, 1150, 1330, 1340, 1540, 1551, 1771, 1782, 2024, 2036, 2300, 2312, 2600, 2613, 2925, 2938, 3276, 3290, 3654, 3668, 4060

Offset: 4

Author

James East, Oct 16 2017

Keywords

• nonn

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).

A formula is given in Section 6 of the East and Niles article.

Examples

			For example, there are 4 rotation-classes of perimeter-7 quadrilaterals: 3211, 3121, 3112, 2221. Note that 3211 and 3112 are reflections of each other, but these are not rotationally equivalent.
		

Links

• James East, Ron Niles, Integer polygons of given perimeter, arXiv:1710.11245 [math.CO], 2017.

Crossrefs

Column k=4 of A293819.

Cf. A008742 (triangles), A293820 (polygons), A293822 (pentagons).

Programs

Mathematica

T[n_, k_] := DivisorSum[GCD[n, k], EulerPhi[#]*Binomial[n/#, k/#] &]/n - Binomial[Floor[n/2], k - 1];
a[n_] := T[n, 4];
Table[a[n], {n, 4, 59}] (* Jean-François Alcover, Jan 29 2019, after Andrew Howroyd in A293819 *)

Formula

Conjectures from Colin Barker, Nov 01 2017: (Start)

G.f.: x^3*(1 - x^2 + 2*x^3) / ((1 - x)^4*(1 + x)^3*(1 + x^2)).

a(n) = (1/96)*(-3*(-1 + (-1)^n + 4*i*(-i)^n - 4*i*i^n) + (7 - 15*(-1)^n)*n + 3*(-1 + (-1)^n)*n^2 + 2*n^3) where i=sqrt(-1).

(End)