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 A057886 Number of integer 4-tuples that give the lengths of the sides of a nondegenerate quadrilateral with perimeter n. 4
 0, 0, 0, 1, 1, 2, 3, 5, 7, 9, 13, 16, 22, 25, 34, 38, 50, 54, 70, 75, 95, 100, 125, 131, 161, 167, 203, 210, 252, 259, 308, 316, 372, 380, 444, 453, 525, 534, 615, 625, 715, 725, 825, 836, 946, 957, 1078, 1090, 1222, 1234, 1378, 1391, 1547, 1560, 1729, 1743 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,6 LINKS James East, Ron Niles, Integer polygons of given perimeter, arXiv:1710.11245 [math.CO], 2017. T. Jenkyns and E. Muller, Triangular triples from ceilings to floors, Amer. Math. Monthly, 107 (Aug. 2000), 634-639. FORMULA Conjecture: a(1)=0 and, for n>1, a(n)=a(n-1)+d(n-1), where d(n)=floor(n/4)*floor((n-2)/4) if n is even and d(n)=floor((n+1)/4) if n is odd. Conjectures from Colin Barker, Oct 27 2013: (Start)   a(n) = ((n-1)*((n-2)*n+18)+6*sin((Pi*n)/2)+18*cos((Pi*n)/2))/96 for n even;   a(n) = (n^3-7*n+6*sin((Pi*n)/2)+18*cos((Pi*n)/2))/96 for n odd.   G.f.: x^4*(x^3-x^2+1) / ((x-1)^4*(x+1)^3*(x^2+1)). (End) Conjecture: a(n) = ( 2*n^3-3*n^2+13*n-18 - 3*(n^2-9*n+6)*(-1)^n + 12*(2+(-1)^n)*(-1)^((2*n+(-1)^n-1)/4) )/192. - Luce ETIENNE, Nov 06 2014 EXAMPLE There are five quadrilaterals with perimeter 8, with sides (1,1,3,3), (1,2,2,3), (1,2,3,2), (1,3,1,3) and (2,2,2,2), so a(8)=5. MATHEMATICA Needs["DiscreteMath`Combinatorica`"]; Table[s=Select[Partitions[n], Length[ # ]==4 && #[[1]]

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Last modified April 25 13:31 EDT 2019. Contains 322461 sequences. (Running on oeis4.)