OFFSET
4,3
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.
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).
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)
EXAMPLE
a(6)=4 because 6 is the greatest possible number of diagonals of a hexahedron.
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Letsko, Dec 03 2016
STATUS
approved