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A124287
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Triangle of the number of integer-sided k-gons having perimeter n, for k=3..n.
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7
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1, 0, 1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 3, 1, 1, 1, 5, 4, 4, 1, 1, 3, 7, 9, 7, 4, 1, 1, 2, 9, 13, 15, 8, 5, 1, 1, 4, 13, 23, 25, 20, 10, 5, 1, 1, 3, 16, 29, 46, 37, 29, 12, 6, 1, 1, 5, 22, 48, 72, 75, 57, 35, 14, 6, 1, 1, 4, 25, 60, 113, 129, 125, 79, 47, 16, 7, 1, 1, 7, 34, 92, 172, 228, 231, 185
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OFFSET
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3,8
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COMMENTS
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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.
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LINKS
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FORMULA
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A formula is given in Theorem 1.5 of the East and Niles article.
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EXAMPLE
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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
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MATHEMATICA
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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;
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PROG
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(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
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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