A279019 - OEIS

A279019

Least possible number of diagonals of simple convex polyhedron with n faces.

0, 0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, 240, 272, 306, 342, 380, 420, 462, 506, 552, 600, 650, 702, 756, 812, 870, 930, 992, 1056, 1122, 1190, 1260, 1332, 1406, 1482, 1560, 1640, 1722, 1806, 1892, 1980, 2070, 2162, 2256, 2352, 2450

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OFFSET

4,3

COMMENTS

Obviously, a pyramid has no diagonals. Hence minimum of diagonals of an arbitrary convex polyhedron having n faces is equal to 0.

Minimum number of diagonals among simple convex polyhedra having n faces is obtained from a polyhedron with two triangular faces, n-4 quadrangular faces and two (n-1)-sided faces. A polyhedron having 3 triangular faces, 3 pentagonal faces and 1 hexagonal face gives another example of a simple convex polyhedron with the least possible number of diagonals for n = 7. A polyhedron having 4 triangular faces and 4 hexagonal faces gives a similar example for n = 8.

Essentially the same as A103505 and A002378. - R. J. Mathar, Dec 05 2016

LINKS

Colin Barker, Table of n, a(n) for n = 4..1000

David Bremner and Victor Klee, Inner Diagonals of Convex Polytopes, Journal of Combinatorial Theory, Series A, Volume 87, Issue 1, July 1999, Pages 175-197.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

FORMULA

a(n) = n^2 - 9*n + 20 = (n-4)*(n-5).

G.f.: -2*x^6/(x-1)^3. - R. J. Mathar, Dec 05 2016

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>6. - Colin Barker, Dec 05 2016

E.g.f.: exp(x)*(20 - 8*x + x^2) - x^3/3 - 3*x^2 - 12*x - 20. - Stefano Spezia, Nov 24 2019

From Amiram Eldar, Jul 09 2023: (Start)

Sum_{n>=6} 1/a(n) = 1.

Sum_{n>=6} (-1)^n/a(n) = 2*log(2) - 1. (End)

MATHEMATICA

Table[(n-4)(n-5), {n, 4, 60}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 0, 2}, 60] (* Harvey P. Dale, Sep 23 2019 *)

PROG