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%I #23 Jan 20 2023 21:57:51
%S 0,0,4,10,20,34,52,73,100,128,162,199,240,285,334,387,444,505,570,639,
%T 712,789,870,955,1044,1137,1234,1335,1440,1549,1662,1779,1900,2025,
%U 2154,2287,2424,2565,2710,2859,3012,3169,3330,3495,3664,3837,4014,4195,4380,4569
%N Greatest possible number of diagonals of a polyhedron having n faces.
%C Also the greatest possible number of diagonals of a simple polyhedron with n faces. In other words, a polyhedron with n faces having the greatest possible number of diagonals must be a simple one.
%H Colin Barker, <a href="/A279015/b279015.txt">Table of n, a(n) for n = 4..1000</a>
%H D. Bremner, V. Klee, <a href="http://dx.doi.org/10.1006/jcta.1998.2953">Inner Diagonals of Convex Polytopes</a>, Journal of Combinatorial Theory, Series A, Volume 87, Issue 1, July 1999, Pages 175-197.
%H Vladimir Letsko, <a href="http://www-old.fizmat.vspu.ru/doku.php?id=marathon:problem_219">Mathematical Marathon, Problem 219</a> (in Russian)
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = 2*n^2 - 21*n + 64 for n=12 or n>=14.
%F From _Colin Barker_, Dec 05 2016: (Start)
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>8.
%F G.f.: x^6*(4 - 2*x + 2*x^2 - x^5 + 3*x^6 - 5*x^7 + 5*x^8 - 3*x^9 + x^10) / (1 - x)^3.
%F (End)
%e a(6)=4 because 6 is the greatest possible number of diagonals of a hexahedron.
%p F:=n->piecewise(4<=n and n<=5,0,6<=n and n<=10,2*n^2-20*n+52,n=11,73,n=13,128,n=12 or n>=14,2*n^2-21*n+64);
%t Drop[#, 4] &@ CoefficientList[Series[x^6*(4 - 2 x + 2 x^2 - x^5 + 3 x^6 - 5 x^7 + 5 x^8 - 3 x^9 + x^10)/(1 - x)^3, {x, 0, 53}], x] (* _Michael De Vlieger_, Dec 05 2016 *)
%o (PARI) concat(vector(2), Vec(x^6*(4 - 2*x + 2*x^2 - x^5 + 3*x^6 - 5*x^7 + 5*x^8 - 3*x^9 + x^10) / (1 - x)^3 + O(x^30))) \\ _Colin Barker_, Dec 05 2016
%Y Cf. A000944, A279019, A279022.
%K nonn,easy
%O 4,3
%A _Vladimir Letsko_, Dec 03 2016
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