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Number of pairs of parallel diagonals in a regular n-gon.
3

%I #32 Nov 22 2023 01:38:09

%S 0,0,0,3,7,16,27,45,66,96,130,175,225,288,357,441,532,640,756,891,

%T 1035,1200,1375,1573,1782,2016,2262,2535,2821,3136,3465,3825,4200,

%U 4608,5032,5491,5967,6480,7011,7581,8170,8800,9450,10143,10857,11616,12397,13225

%N Number of pairs of parallel diagonals in a regular n-gon.

%H Paolo Xausa, <a href="/A211379/b211379.txt">Table of n, a(n) for n = 3..10000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/RegularPolygonDivisionbyDiagonals.html">Regular Polygon Division by Diagonals</a>.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,1,-4,1,2,-1).

%F a(n) = (1/2)*n*(binomial(n/2-1,2) + binomial(n/2-2,2)) = (1/8)*n*(n-4)^2 for n even;

%F a(n) = n*binomial((n-3)/2,2) = (1/8)*n*(n-3)*(n-5) for n odd.

%F G.f.: -x^6*(x^2-x-3) / ((x-1)^4*(x+1)^2). - _Colin Barker_, Feb 14 2013

%e a(6) = 3 since by numbering the vertices from 1 to 6 there are three pairs of parallel diagonals, i.e., {[1, 3], [4, 6]}, {[1, 5], [2, 4]}, {[2, 6], [3, 5]}.

%e a(7) = 7 since there are the seven pairs {[1, 3], [4, 7]}, {[1, 4], [5, 7]}, {[1, 5], [2, 4]}, {[1, 6], [2, 5]}, {[2, 6], [3, 5]}, {[2, 7], [3, 6]}, {[3, 7], [4, 6]}.

%e a(8) = 16 since there are the sixteen pairs {[1, 3], [4, 8]}, {[1, 3], [5, 7]}, {[1, 4], [5, 8]}, {[1, 5], [2, 4]}, {[1, 5], [6, 8]}, {[1, 6], [2, 5]}, {[1, 7], [2, 6]}, {[1, 7], [3, 5]}, {[2, 4], [6, 8]}, {[2, 6], [3, 5]}, {[2, 7], [3, 6]}, {[2, 8], [3, 7]}, {[2, 8], [4, 6]}, {[3, 7], [4, 6]}, {[3, 8], [4, 7]}, {[4, 8], [5, 7]}.

%p a:=n->piecewise(n mod 2 = 0,1/8*n*(n-4)^2,n mod 2 = 1, 1/8*n*(n-3)*(n-5),0);

%t A211379[n_]:=n/8If[OddQ[n],(n-3)(n-5),(n-4)^2];Array[A211379,100,3] (* or *)

%t LinearRecurrence[{2,1,-4,1,2,-1},{0,0,0,3,7,16},100] (* _Paolo Xausa_, Nov 21 2023 *)

%o (Python)

%o def A211379(n): return n*(n*(n-8)+16-(n&1))>>3 # _Chai Wah Wu_, Nov 22 2023

%Y Cf. A000096, A367204.

%K nonn,easy

%O 3,4

%A _Martin Renner_, Feb 07 2013