OFFSET
3,8
COMMENTS
Rotations are counted only once, but reflections are considered different. For a k-gon to be nondegenerate, the longest side must be shorter than the sum of the remaining sides (equivalently, shorter than n/2). Column k=3 is A008742, column k=4 is A293821, column k=5 is A293822 and column k=6 is A293823.
A formula is given in Section 6 of the East and Niles article.
LINKS
Andrew Howroyd, Rows n=3..52 of triangle, flattened
James East, Ron Niles, Integer polygons of given perimeter, arXiv:1710.11245 [math.CO], 2017.
FORMULA
T(n,k) = (Sum_{d|gcd(n,k)} phi(d)*binomial(n/d, k/d))/n - binomial(floor(n/2), k-1). - Andrew Howroyd, Nov 21 2017
EXAMPLE
For polygons having perimeter 7, there are: 2 triangles (331, 322), 4 quadrilaterals (3211, 3121, 3112, 2221), 3 pentagons (31111, 22111, 21211), 1 hexagon (211111) and 1 heptagon (1111111). Note that the quadrilaterals 3211 and 3112 are reflections of each other, but these are not rotationally equivalent.
The triangle begins:
n=3: 1;
n=4: 0, 1;
n=5: 1, 1, 1;
n=6: 1, 2, 1, 1;
n=7: 2, 4, 3, 1, 1;
n=8: 1, 6, 6, 4, 1, 1;
n=9: 4, 10, 13, 10, 4, 1, 1;
...
MATHEMATICA
T[n_, k_] := DivisorSum[GCD[n, k], EulerPhi[#]*Binomial[n/#, k/#]&]/n - Binomial[Floor[n/2], k - 1];
Table[T[n, k], {n, 3, 16}, {k, 3, n}] // Flatten (* Jean-François Alcover, Jun 14 2018, translated from PARI *)
PROG
(PARI)
T(n, k)={sumdiv(gcd(n, k), d, eulerphi(d)*binomial(n/d, k/d))/n - binomial(floor(n/2), k-1)}
for(n=3, 10, for(k=3, n, print1(T(n, k), ", ")); print); \\ Andrew Howroyd, Nov 21 2017
KEYWORD
nonn,tabl
AUTHOR
James East, Oct 16 2017
STATUS
approved