OFFSET
3,8
COMMENTS
Rotations and reversals are counted only once. For a k-gon to be nondegenerate, the longest side must be shorter than the sum of the remaining sides. Column k=3 is A005044, column k=4 is A057886, column k=5 is A124285 and column k=6 is A124286. Note that A124278 counts polygons whose sides are nondecreasing.
LINKS
Andrew Howroyd, Table of n, a(n) for n = 3..1277 (terms 3..212 from T. D. Noe)
James East, Ron Niles, Integer polygons of given perimeter, arXiv:1710.11245 [math.CO], 2017.
FORMULA
A formula is given in Theorem 1.5 of the East and Niles article.
EXAMPLE
For polygons having perimeter 7: 2 triangles, 3 quadrilaterals, 3 pentagons, 1 hexagon and 1 heptagon. The triangle begins
1
0 1
1 1 1
1 2 1 1
2 3 3 1 1
1 5 4 4 1 1
MATHEMATICA
Needs["DiscreteMath`Combinatorica`"]; Table[p=Partitions[n]; Table[s=Select[p, Length[ # ]==k && #[[1]]<Total[Rest[ # ]] &]; cnt=0; Do[cnt=cnt+Length[ListNecklaces[k, s[[i]], Dihedral]], {i, Length[s]}]; cnt, {k, 3, n}], {n, 3, 20}]
(* Second program: *)
T[n_, k_] := (DivisorSum[GCD[n, k], EulerPhi[#]*Binomial[n/#, k/#]&]/n + Binomial[Quotient[k, 2] + Quotient[n - k, 2], Quotient[k, 2]] - Binomial[ Quotient[n, 2], k - 1] - Binomial[Quotient[n, 4], Quotient[k, 2]] - If[ OddQ[k], 0, Binomial[Quotient[n + 2, 4], Quotient[k, 2]]])/2;
Table[T[n, k], {n, 3, 20}, {k, 3, n}] // Flatten (* Jean-François Alcover, Jun 14 2018, after Andrew Howroyd *)
PROG
(PARI)
T(n, k)={(sumdiv(gcd(n, k), d, eulerphi(d)*binomial(n/d, k/d))/n + binomial(k\2 + (n-k)\2, k\2) - binomial(n\2, k-1) - binomial(n\4, k\2) - if(k%2, 0, binomial((n+2)\4, k\2)))/2; }
for(n=3, 10, for(k=3, n, print1(T(n, k), ", ")); print); \\ Andrew Howroyd, Nov 21 2017
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
T. D. Noe, Oct 24 2006
STATUS
approved