A279015 Greatest possible number of diagonals of a polyhedron having n faces.

0, 0, 4, 10, 20, 34, 52, 73, 100, 128, 162, 199, 240, 285, 334, 387, 444, 505, 570, 639, 712, 789, 870, 955, 1044, 1137, 1234, 1335, 1440, 1549, 1662, 1779, 1900, 2025, 2154, 2287, 2424, 2565, 2710, 2859, 3012, 3169, 3330, 3495, 3664, 3837, 4014, 4195, 4380, 4569

Offset: 4

Author

Vladimir Letsko, Dec 03 2016

Keywords

• nonn
• easy

Comments

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.

Examples

			a(6)=4 because 6 is the greatest possible number of diagonals of a hexahedron.
		

Links

• Colin Barker, Table of n, a(n) for n = 4..1000
• D. Bremner, V. Klee, Inner Diagonals of Convex Polytopes, Journal of Combinatorial Theory, Series A, Volume 87, Issue 1, July 1999, Pages 175-197.
• Vladimir Letsko, Mathematical Marathon, Problem 219 (in Russian)
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A000944, A279019, A279022.

Programs

Maple

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);

Mathematica

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 *)

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

Formula

a(n) = 2*n^2 - 21*n + 64 for n=12 or n>=14.

From Colin Barker, Dec 05 2016: (Start)

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>8.

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.

(End)