A293819 Triangle read by rows of the number of integer-sided k-gons having perimeter n, modulo rotations but not reflections, for k=3..n.
1, 0, 1, 1, 1, 1, 1, 2, 1, 1, 2, 4, 3, 1, 1, 1, 6, 6, 4, 1, 1, 4, 10, 13, 10, 4, 1, 1, 2, 12, 21, 21, 12, 5, 1, 1, 5, 20, 37, 41, 30, 15, 5, 1, 1, 4, 23, 51, 74, 65, 43, 19, 6, 1, 1, 7, 35, 84, 126, 131, 99, 55, 22, 6, 1, 1, 5, 38, 108, 196, 239, 216, 143, 73, 26, 7, 1, 1, 10, 56, 166, 314, 422, 428
Offset: 3
Examples
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; ...
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.
Crossrefs
Programs
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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 *)
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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
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
Comments