login
Number of integer 4-tuples that give the lengths of the sides of a nondegenerate quadrilateral with perimeter n.
5

%I #42 Apr 04 2024 10:46:26

%S 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,

%T 167,203,210,252,259,308,316,372,380,444,453,525,534,615,625,715,725,

%U 825,836,946,957,1078,1090,1222,1234,1378,1391,1547,1560,1729,1743

%N Number of integer 4-tuples that give the lengths of the sides of a nondegenerate quadrilateral with perimeter n.

%H Felix Huber, <a href="/A057886/b057886.txt">Table of n, a(n) for n = 1..1000</a>

%H James East and Ron Niles, <a href="https://arxiv.org/abs/1710.11245">Integer polygons of given perimeter</a>, arXiv:1710.11245 [math.CO], 2017.

%H T. Jenkyns and E. Muller, <a href="http://www.jstor.org/stable/2589119">Triangular triples from ceilings to floors</a>, Amer. Math. Monthly, 107 (Aug. 2000), 634-639.

%F 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.

%F Conjectures from _Colin Barker_, Oct 27 2013: (Start)

%F a(n) = ((n-1)*((n-2)*n+18)+6*sin((Pi*n)/2)+18*cos((Pi*n)/2))/96 for n even;

%F a(n) = (n^3-7*n+6*sin((Pi*n)/2)+18*cos((Pi*n)/2))/96 for n odd.

%F G.f.: x^4*(x^3-x^2+1) / ((x-1)^4*(x+1)^3*(x^2+1)). (End)

%F 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

%e 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.

%p A057886 := proc(n) local s1, s2, s3, s4, a; a := 0; if 4 <= n then for s1 to floor(1/4*n) do for s2 from s1 to floor(1/3*n - 1/3*s1) do for s3 from max(s2, floor(1/2*n - s1 - s2) + 1) to floor(1/2*n - 1/2*s1 - 1/2*s2) do s4 := n - s1 - s2 - s3; if s1 < s2 and s2 < s3 and s3 < s4 then a := a + 3; elif s2 = s3 and (s1 = s2 or s3 = s4) then a := a + 1; else a := a + 2; end if; end do; end do; end do; end if; return a; end proc; seq(A057886(n), n = 1 .. 56); # _Felix Huber_, Mar 13 2024

%t Needs["DiscreteMath`Combinatorica`"]; Table[s=Select[Partitions[n], Length[ # ]==4 && #[[1]]<Total[Rest[ # ]] &]; cnt=0; Do[cnt=cnt+Length[ListNecklaces[4,s[[i]], Dihedral]], {i,Length[s]}]; cnt, {n,50}] (* _T. D. Noe_, Oct 24 2006 *)

%Y The Moebius transform is A057887. Cf. A005044.

%Y Cf. A062890.

%K nonn

%O 1,6

%A _John W. Layman_, Sep 19 2000

%E Corrected by _T. D. Noe_, Oct 24 2006